Mathematics-1, GTU–2018 9789353162818, 9353162815
This book has been designed as per the Mathematics-1 course offered in the first year to the undergraduate engineering s
712 101 100MB
English Pages 1050 Year 2018
Polecaj historie
Table of contents :
Title
Contents
Unit 1
1 Indeterminate Forms
2 Improper Integrals
3 Gamma and Beta Functions
4 Applications of Definite Integrals
Unit 2
5 Sequencesand Series
6 Taylor’s and Maclaurin’s Series
Unit 3
7 Fourier Series
Unit 5
9 Multiple Integrals
Unit 6
10 Matrices
Index
Citation preview
Mathematics-1 Gujarat Technological University 2018
About the Authors Ravish R Singh is presently Director at Thakur Ramnarayan College of Arts and Commerce, Mumbai. He obtained a BE degree from University of Mumbai in 1991, an MTech degree from IIT Bombay in 2001, and a PhD degree from Faculty of Technology, University of Mumbai, in 2013. He has published several books with McGraw Hill Education (India) on varied subjects like Engineering Mathematics, Applied Mathematics, Electrical Networks, Network Analysis and Synthesis, Electrical Engineering, Basic Electrical and Electronics Engineering, etc., for all-India curricula as well as regional curricula of some universities like Gujarat Technological University, Mumbai University, Pune University, Jawaharlal Nehru Technological University, Anna University, Uttarakhand Technical University, and Dr A P J Abdul Kalam Technical University. Dr Singh is a member of IEEE, ISTE, and IETE, and has published research papers in national and international journals. His fields of interest include Circuits, Signals and Systems, and Engineering Mathematics. Mukul Bhatt is presently Assistant Professor at Department of Mathematics and Statistics, Thakur Ramnarayan College of Arts and Commerce, Mumbai. She obtained her MSc (Mathematics) degree from H N B Garhwal University in 1992, and a PhD degree from Faculty of Science, PAHER University, Udaipur, Rajasthan in 2017. She has published several books with McGraw Hill Education (India) Private Limited on Engineering Mathematics and Applied Mathematics for all-India curricula as well as regional curricula of some universities like Gujarat Technological University, Mumbai University, Pune University, Jawaharlal Nehru Technological University, Anna University, Uttarakhand Technical University, and Dr A P J Abdul Kalam Technical University. Dr Bhatt has twenty six years of teaching experience at various levels in engineering colleges and her fields of interest include Integral Calculus, Complex Analysis, and Operation Research. She is a member of ISTE.
Mathematics-1 Gujarat Technological University 2018
Ravish R Singh Director Thakur Ramnarayan College of Arts & Commerce Mumbai, Maharashtra Mukul Bhatt Assistant Professor Thakur Ramnarayan College of Arts & Commerce Mumbai, Maharashtra
McGraw Hill Education (India) Private Limited Chennai McGraw Hill Education Offices Chennai new York St Louis San Francisco auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto
McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Mathematics-1, GTU–2018 Copyright © 2019 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listing (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. 1 2 3 4 5 6 7 8 9
D103074
22 21 20 19 18
Printed and bound in India. Print-Book ISBN (13): 978-93-5316-280-1 ISBN (10): 93-5316-280-7 E-Book ISBN (13): 978-93-5316-281-8 ISBN (10): 93-5316-281-5 Director—Science & Engineering Portfolio: Vibha Mahajan Senior Portfolio Manager—Science & Engineering: Hemant K Jha Associate Portfolio Manager—Science & Engineering: Tushar Mishra Production Head: Satinder S Baveja Copy Editor: Taranpreet Kaur Assistant Manager—Production: Anuj K Shriwastava General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.
Typeset at APS Compugraphics, 4G, PKT 2, Mayur Vihar Phase-III, Delhi 96, and printed at Cover Designer: APS Compugraphics Cover Image Source: Shutterstock Cover Printer:
Visit us at: www.mheducation.co.in Write to us at: [email protected] CIN: U22200TN1970PTC111531 Toll Free Number: 1800 103 5875
Dedicated to Aman and Aditri Ravish R Singh Soumya and Siddharth Mukul Bhatt
Contents Preface Roadmap to the Syllabus
xi xv
UNIT-1 1. Indeterminate Forms
1.1–1.63
1.1 1.2
Introduction 1.1 L’Hospital’s Rule 1.1 0 1.3 Type 1: Form 1.2 0 • 1.4 Type 2: form 1.16 • 1.5 Type 3: 0 × • Form 1.23 1.6 Type 4: • − • Form 1.30 1.7 Type 5: 1•, • 0, 0 0 Forms 1.38 Points to Remember 1.60 Multiple Choice Questions 1.60 2. Improper Integrals
2.1–2.25
2.1 Introduction 2.1 2.2 Improper Integrals 2.1 2.3 Improper Integrals of the First Kind 2.2 2.4 Improper Integrals of the Second Kind 2.9 2.5 Improper Integral of the Third Kind 2.16 2.6 Convergence and Divergence of Improper Integrals 2.17 Points to Remember 2.22 Multiple Choice Questions 2.23 3. Gamma and Beta Functions 3.1 3.2 3.3 3.4 3.5 3.6
Introduction 3.1 Gamma Function 3.1 Properties of Gamma Function 3.2 Beta Function 3.11 Properties of Beta Functions 3.12 Beta Function as Improper Integral 3.28
3.1–3.39
viii
Contents
Points to Remember 3.36 Multiple Choice Questions 3.37 4. Applications of Definite Integrals
4.1–4.67
4.1 Introduction 4.1 4.2 Volume Using Cross-sections 4.1 4.3 Length of Plane Curves 4.6 4.4 Area of Surface of Solid of Revolution 4.46 Points to Remember 4.65 Multiple Choice Questions 4.66
UNIT-2 5. Sequences and Series
5.1–5.121
5.1 Introduction 5.1 5.2 Sequence 5.2 5.3 Infinite Series 5.8 5.4 The nth Term Test for Divergence 5.9 5.5 Geometric Series 5.10 5.6 Telescoping Series 5.15 5.7 Combining Series 5.18 5.8 Harmonic Series 5.19 5.9 p-Series 5.20 5.10 Comparison Test 5.20 5.11 D’Alembert’s Ratio Test 5.40 5.12 Raabe’s Test 5.67 5.13 Cauchy’s Root Test 5.73 5.14 Cauchy’s Integral Test 5.82 5.15 Alternating Series 5.87 5.16 Absolute and Conditional Convergent of a Series 5.94 5.17 Power Series 5.101 Points to Remember 5.115 Multiple Choice Questions 5.117 6. Taylor’s and Maclaurin’s Series 6.1 Introduction 6.1 6.2 Taylor’s Series 6.1 6.3 Maclaurin’s Series 6.27 Points to Remember 6.67 Multiple Choice Questions 6.68
6.1–6.70
Contents
ix
UNIT-3 7. Fourier Series
7.1–7.126
7.1 Introduction 7.1 7.2 Periodic Functions 7.1 7.3 Orthogonality of Trigonometric System 7.2 7.4 Dirichlet’s Conditions for Representation by a Fourier Series 7.5 7.5 Trigonometric Fourier Series 7.6 7.6 Fourier Series of Functions of Period 2l 7.7 7.7 Fourier Series of Even and Odd Functions 7.66 7.8 Half-Range Fourier Series 7.93 Points to Remember 7.120 Multiple Choice Questions 7.122
UNIT-4 8. Partial Derivatives
8.1–8.184
8.1 Introduction 8.1 8.2 Functions of Two or More Variables 8.2 8.3 Limit and Continuity of Functions of Several Variables 8.2 8.4 Partial Derivatives 8.10 8.5 Higher-Order Partial Derivatives 8.11 8.6 Total Derivatives 8.59 8.7 Implicit Differentiation 8.94 8.8 Gradient and Directional Derivative 8.103 8.9 Tangent Plane and Normal Line 8.107 8.10 Local Extreme Values (Maximum and Minimum Values) 8.116 8.11 Extreme Values with Constrained Variables 8.134 8.12 Method of Lagrange Multipliers 8.145 Points to Remember 8.177 Multiple Choice Questions 8.179
UNIT-5 9. Multiple Integrals 9.1 9.2 9.3 9.4 9.5 9.6
Introduction 9.1 Double Integrals 9.1 Change of Order of Integration 9.31 Double Integrals in Polar Coordinates 9.66 Multiple Integrals by Substitution 9.77 Triple Integrals 9.109
9.1–9.175
x
Contents
9.7 Area by Double Integrals 9.141 Points to Remember 8.169 Multiple Choice Questions 8.171
UNIT-6 10. Matrices
10.1–10.141
10.1 Introduction 10.1 10.2 Matrix 10.2 10.3 Some Definitions Associated with Matrices 10.2 10.4 Elementary Row Operations in Matrix 10.6 10.5 Row Echelon and Reduced Row Echelon Forms of a Matrix 10.7 10.6 Rank of a Matrix 10.13 10.7 Inverse of a Matrix by Gauss–Jordan Method 10.18 10.8 System of Non-Homogeneous Linear Equations 10.22 10.9 System of Homogeneous Linear Equations 10.48 10.10 Eigenvalues and Eigenvectors 10.64 10.11 Properties of Eigenvalues 10.65 10.12 Linear Dependence and Independence of Eigenvectors 10.76 10.13 Properties of Eigenvectors 10.76 10.14 Cayley-Hamilton Theorem 10.108 10.15 Similarity Transformation 10.119 10.16 Diagonalization of a Matrix 10.119 Multiple-Choice Questions 10.137 Index
I.1–I.3
Preface Mathematics is a key area of study in any engineering course. A sound knowledge of this subject will help engineering students develop analytical skills, and thus enable them to solve numerical problems encountered in real life, as well as apply mathematical principles to physical problems, particularly in the field of engineering.
Users This book is designed for the first year GTU engineering students pursuing the course Mathematics-1, SUBJECT CODE: 3110014 in their first year Ist Semester. It covers the complete GTU syllabus for the course on Mathematics-1, which is common to all the engineering branches.
Objective The crisp and complete explanation of topics will help students easily understand the basic concepts. The tutorial approach (i.e., teach by example) followed in the text will enable students develop a logical perspective to solving problems.
Features Each topic has been explained from the examination point of view, wherein the theory is presented in an easy-to-understand student-friendly style. Full coverage of concepts is supported by numerous solved examples with varied complexity levels, which is aligned to the latest GTU syllabus. Fundamental and sequential explanation of topics are well aided by examples and exercises. The solutions of examples are set following a ‘tutorial’ approach, which will make it easy for students from any background to easily grasp the concepts. Exercises with answers immediately follow the solved examples enforcing a practice-based approach. We hope that the students will gain logical understanding from solved problems and then reiterate it through solving similar exercise problems themselves. The unique blend of theory and application caters to the requirements of both the students and the faculty. Solutions of GTU examination questions are incorporated within the text appropriately.
xii
Preface
Highlights ∑ Crisp content strictly as per the latest GTU syllabus of Mathematics-1 (Regulation 2018) ∑ Comprehensive coverage with lucid presentation style ∑ Each section concludes with an exercise to test understanding of topics ∑ Solutions of GTU examination questions included appropriately within the chapters ∑ Rich exam-oriented pedagogy: Solved examples within chapters: 850+ Unsolved exercises: 500+ MCQs at the end of chapters: 300+
Chapter Organization The content spans the following 10 chapters which wholly and sequentially cover each module of the syllabus. Chapter 1 introduces Indeterminate Forms. Chapter 2 discusses Improper Integrals. Chapter 3 presents Gamma and Beta Functions. Chapter 4 covers Applications of Definite Integrals. Chapter 5 deals with Sequences and Series. Chapter 6 presents Taylor’s and Maclaurin’s Series. Chapter 7 discusses Fourier Series. Chapter 8 presents Partial Derivatives. Chapter 9 covers Multiple Integrals. Chapter 10 deals with Matrices.
Acknowledgements We are grateful to the following reviewers who reviewed sample chapters of the book and generously shared their valuable comments: Prof. Bhavini Pandya Prof. Som Sahani
SVIT, Vasad Babaria Institute of Technology, Baroda
We would also like to thank all the staff at McGraw Hill Education (India), especially Vibha Mahajan, Shalini Jha, Hemant K Jha, Tushar Mishra, Satinder Singh Baveja, Taranpreet Kaur and Anuj Shriwastava for coordinating with us during the editorial, copyediting, and production stages of this book.
Preface
xiii
Our acknowledgements would be incomplete without a mention of the contribution of all our family members. We extend a heartfelt thanks to them for always motivating and supporting us throughout the project. Constructive suggestions for the improvement of the book will always be welcome. Ravish R Singh Mukul Bhatt
Publisher’s Note Remember to write to us. We look forward to receiving your feedback, comments and ideas to enhance the quality of this book. You can reach us at [email protected]. Please mention the title and authors’ name as the subject. In case you spot piracy of this book, please do let us know.
Roadmap to the Syllabus Mathematics-1 Subject Code: 3110014
Unit-I Indeterminate Forms and L’Hôspital’s Rule. Improper Integrals, Convergence and divergence of the integrals, Beta and Gamma functions and their properties. Applications of definite integral, Volume using cross-sections, Length of plane curves, Areas of Surfaces of Revolution. CHAPTER 1: Introduction to Some Special Functions Go To
CHAPTER 2: Improper Integrals CHAPTER 3: Gamma and Beta Functions CHAPTER 4: Applications of Definite Integrals
Unit-2 Convergence and divergence of sequences, The Sandwich Theorem for Sequences, The Continuous Function Theorem for Sequences, Bounded Monotonic Sequences, Convergence and divergence of an infinite series, geometric series, telescoping series, nth term test for divergent series, Combining series, Harmonic Series, Integral test, The p-series, The Comparison test, The Limit Comparison test, Ratio test, Raabe’s Test, Root test, Alternating series test, Absolute and Conditional convergence, Power series, Radius of convergence of a power series, Taylor and Maclaurin series. CHAPTER 5: Sequences and Series Go To
CHAPTER 6: Taylor’s and Maclaurin’s Series
xvi
Roadmap to the Syllabus
Unit-3 Fourier Series of 2l periodic functions, Dirichlet’s conditions for representation by a Fourier series, Orthogonality of the trigonometric system, Fourier Series of a function of period 2l, Fourier Series of even and odd functions, Half range expansions. Go To
CHAPTER 7: Fourier Series
Unit-4 Functions of several variables, Limits and continuity, Test for non existence of a limit, Partial differentiation, Mixed derivative theorem, differentiability, Chain rule, Implicit differentiation, Gradient, Directional derivative, tangent plane and normal line, total differentiation, Local extreme values, Method of Lagrange Multipliers. Go To
CHAPTER 8: Partial Derivatives
Unit-5 Multiple integral, Double integral over Rectangles and general regions, double integrals as volumes, Change of order of integration, double integration in polar coordinates, Area by double integration, Triple integrals in rectangular, cylindrical and spherical coordinates, Jacobian, multiple integral by substitution. Go To
CHAPTER 9: Multiple Integrals
Unit-5 Elementary row operations in Matrix, Row echelon and Reduced row echelon forms, Rank by echelon forms, Inverse by Gauss-Jordan method, Solution of system of linear equations by Gauss elimination and Gauss-Jordan methods. Eigen values and eigen vectors, Cayley-Hamilton theorem, Diagonalization of a matrix. Go To
CHAPTER 10: Matrices
UNIT-1 Chapter Chapter Chapter Chapter
1. 2. 3. 4.
Indeterminate Forms Improper Integrals Gamma and Beta Functions Applications of Definite Integrals
1 Indeterminate CHAPTER
Forms chapter outline 1.1 1.2 1.3 1.4 1.5 1.6 1.7
1.1
Introduction L’Hospital’s Rule Type 1 : 0 Form 0 • Type 2 : Form • Type 3 : 0 × • Form Type 4 : • − • Form Type 5 : 1•, • 0, 0 0 Forms
IntroductIon
We have studied certain rules to evaluate the limits. But some limits cannot be evaluated by using these rules. These limits are known as indeterminate forms. There are seven types of indeterminate forms: ∞ 0 (i) (ii) ∞ (iii) 0 × ∞ (iv) ∞ − ∞ 0 (v) 1∞ (vi) 0° (vii) ∞° These limits can be evaluated by using L’Hospital’s rule.
1.2
L’HospItaL’s ruLe
Statement If f (x) and g (x) are two functions of x which can be expanded by Taylor’s series in the neighbourhood of x = a and if lim f ( x) = f (a ) = 0, lim g ( x) = g (a ) = 0, x→a x→a then
1.2
Chapter 1
lim x→a
Indeterminate Forms
f ( x) f ′ ( x) = lim x → a g ( x) g ′ ( x)
Proof Let
x=a+h f ( x) f ( a + h) lim = lim x → a g ( x) h → 0 g ( a + h) h2 f ″ (a ) + ....... 2! = lim h→ 0 h2 g (a ) + hg ′ (a ) + g ″ (a ) + ....... 2! h2 hf ′ (a ) + f ″ (a ) + ....... 2! = lim 2 h→ 0 h hg ′ (a ) + g ″ (a ) + ....... 2! h f ′ (a ) + f ″ (a ) + ....... 2! = lim h→ 0 h g ′ (a ) + g ″ (a ) + ....... 2! f (a ) + hf ′ (a ) +
[By Taylor’s theorem]
[Q f (a ) = 0, g (a ) = 0]
f ′(a) g ′(a) f ′ ( x) , provided g ′ (a ) ≠ 0. = lim x → a g ′ ( x) =
Note The following standard limits can be used to solve the problems: tan x =1 x→0 x
sin x =1 x→0 x
(ii) lim
ex − 1 =1 x→0 x
(v) lim(1 + x) x = e
(i) lim
(iv) lim
sin −1 x =1 x→0 x
(vii) lim
1 x→0
(vii) lim x→0
sin h x =1 x
ax −1 = log a x→0 x
(iii) lim
1 (vi) lim 1 + = e x →∞ x x
tan −1 x =1 x→0 x
(ix) lim
0 Form 0 Problems under this type are solved by using L’Hospital’s rule considering the fact that
1.3 type 1 :
1.3
lim x→ a
f ( x) f ′ ( x) = lim x → a g ( x) g ′ ( x)
Type 1:
0 0
Form
1.3
if lim f ( x) = 0 and lim g ( x) = 0. x→ a
x→ a
example 1 (1 + x) n − 1 = n. Prove that lim x→0 x
Solution (1 + x) n − 1 x→0 x
0 0 form
l = lim
Let
n(1 + x) n −1 x→0 1 =n
[Applying L’Hospital’s rule]
= lim
example 2 xe x - log (1 + x )
Evaluate lim
x2
x Æ0
.
Solution xe x − log (1 + x) x→0 x2 1 e x + xe x − 1+ x = lim x→0 2x
l = lim
Let
e x + e x + xe x + = lim x→0
=
3 2
example 3 Evaluate lim x →1
x log x − ( x − 1) . ( x − 1) log x
2
0 0 form
0 0 form [Applying L’Hospital’s rule]
1 (1 + x) 2 [Applying L’Hospital’s rule]
1.4
Chapter 1
Indeterminate Forms
Solution x log x - ( x - 1) x Æ1 ( x - 1) log x
È0 ˘ Í 0 form ˙ Î ˚
l = lim
Let
1 + log x - 1 x = lim 1 x Æ1 log x + ( x - 1) ◊ x x◊
= lim
x Æ1
= lim x Æ1
=
[ Applying L’Hospital’s rule] È0 ˘ Í 0 form ˙ Î ˚
log x 1 log x + 1 x
1 x
[ Applying L’Hospital’s rule]
1 1 + x x2
1 2
example 4 e x - e - x - 2 log(1 + x ) . x Æ0 x sin x
Evaluate lim Solution
e x - e - x - 2 log(1 + x ) x Æ0 x sin x
Let l = lim
1 1+ x sin x + x cos x
e x + e- x - 2 ◊ = lim
x Æ0
2 (1 + x )2 = lim x Æ 0 cos x + cos x - x sin x
È0 ˘ Í 0 form ˙ Î ˚ È0 ˘ Í 0 form ˙ Î ˚
[ Applying L’Hospital’s rule]
e x - e- x +
2 2 =1 =
[ Applying L’Hospital’s rule]
1.3
Type 1:
0 0
Form
1.5
example 5 Evaluate lim x Æ0
e x + e- x - x2 - 2 sin 2 x - x 2
.
[Winter 2016; Summer 2014]
Solution Let
l = lim
e x + e- x - x2 - 2 2
sin x - x
x Æ0
2
e - e- x - 2 x x Æ 0 2 sin x cos x - 2 x
È0 ˘ Í 0 form ˙ Î ˚
x
= lim
= lim x Æ0
= lim x Æ0
e x + e- x - 2 2( - sin x )sin x + 2 cos2 x - 2
[ Applying L’Hospital’s rule] [ Applying L’Hospital’s rule]
e x + e- x - 2 -2 sin 2 x + 2 cos2 x - 2
e x + e- x - 2 x Æ 0 2 cos 2 x - 2
= lim
e x - e- x x Æ 0 ( -2)2 sin 2 x
[ Applying L’Hospital’s rule]
= lim
e x - e- x x Æ 0 - 4 sin 2 x
= lim
e x + e- x x Æ 0 -8 cos 2 x 2 = -8 1 =4
[Applying L’Hospital’ss rule]
= lim
example 6 Evaluate lim
2x - 1
x Æ0
(1 +
1 x) 2
.
-1
Solution Let
l = lim
x Æ0
2x - 1 (1 +
1 2 x)
-1
È0 ˘ Í 0 form ˙ Î ˚
1.6
Chapter 1
Indeterminate Forms
2 x log 2
= lim
1
x Æ0
1 (1 + x ) 2 2 = 2 log 2
[ Applying L’Hospital’s rule]
example 7 Evaluate lim
(
log 1 + kx 2 1 − cos x
x→0
).
Solution l = lim
Let
(
log 1 + kx 2
)
1 − cos x
x→0
1 ⋅ 2kx kx 2 + 1 = lim x→0 sin x 1 = 2k lim x→0 sin x 1 + kx 2 ⋅ x sin x = 2k = 1 Q lim x→0 x
(
0 0 form [Applying L’Hospital’s rule]
)
example 8 Evaluate lim x Æ0
log (1 + x 3 ) sin3 x
.
Solution Let
l = lim
x Æ0
log (1 + x 3 ) 3
sin x
1 ⋅ 3x 2 3 1 x + = lim x → 0 3 sin 2 x cos x
È0 ˘ Í 0 form ˙ Î ˚ [Applying L’Hospital’s rule]
2
1 Ê x ˆ = lim Á ˜ x Æ 0 Ë sin x ¯ (1 + x 3 ) cos x =1
x È ˘ = 1˙ Í xlim Î Æ0 sin x ˚
1.3
Type 1:
0 0
Form
1.7
example 9 x − xx . Evaluate lim x →1 1 + log x − x Solution Let
x − xx x →1 1 + log x − x
l = lim
0 0 form
x − e x log x x →1 1 + log x − x
= lim
1 − e x log x (1 + log x) x →1 1 −1 x
= lim
1 −e x log x (1 + log x) 2 − e x log x x = lim x →1 1 − 2 x =2
g L’Hospital’s rule] [Applying
[Applying L’Hospital’s rule]
example 10 Evaluate lim
xÆ y
xy - yx xx - yy
.
Solution Let
xy − yx x→ y x x − y y
l = lim
= lim x→ y
yx y −1 − y x log y x x (1 + log x) − 0
y y − y y log y y y (1 + log y ) 1 − log y = 1 + log y =
example 11 Evaluate lim p xÆ 2
sin ( x cos x ) . cos( x sin x )
0 0 form [Applying L’Hospital’s rule]
1.8
Chapter 1
Indeterminate Forms
Solution l = lim
Let
x→
p 2
= limp x→ 2
=
sin ( x cos x) cos ( x sin x)
0 0 form
cos ( x cos x)(cos x − x sin x) − sin ( x sin x)(sin x + x cos x)
[Applying L’Hospital’s rule]
p 2
example 12 cos 2 π x Evaluate lim 2 x . 1 e − 2 xe x→
[Winter 2015]
2
Solution Let
cos 2 p x e 2 x − 2 xe
0 0 form
= lim
2 cos p x ( −p sin p x) 2e 2 x − 2 e
[Applying L’Hospital’s rule]
= lim1
−p sin 2p x 2( e 2 x − e )
0 0 form
−2p 2 cos 2p x 2 ⋅ 2e 2 x
[Applying L’Hospital’s rule]
l = lim1 x→
2
1 x→ 2
x→
2
= lim x→
1 2
p2 2e
=
example 13 Prove that θlim →α
1 − cos (θ − α ) 1 2 = sec α . (sin θ − sin α ) 2 2
Solution Let
l = lim
q Æa
1 - cos(q - a ) (sin q - sin a )
2
È0 ˘ Í 0 form ˙ Î ˚
1.3
Type 1:
0 0
Form
1.9
= lim
sin (q − a ) 2 (sin q − sin a ) cos q
[Applying L’Hospital’s rule]
= lim
sin (q − a ) (sin 2q − 2 sin a cos q )
0 0 form
q →a
q →a
cos (q − a ) q →a 2 cos 2q + 2 sin a sin q cos 0 = 2 cos 2a + 2 sin a sin a = lim
[Applying L’Hospital’s rule]
1 2(1 − 2 sin 2 a ) + 2 sin 2 a 1 = 2 − 2 sin 2 a 1 = 2 cos 2 a 1 = sec 2 a 2 =
example 14 5 sin x − 7 sin 2 x + 3 sin 3 x . x→0 tan x − x
Evaluate lim Solution Let l = lim x→0
5 sin x − 7 sin 2 x + 3 sin 3 x tan x − x
5 cos x − 14 cos 2 x + 9 cos 3 x sec 2 x − 1 −5 sin x + 28 sin 2 x − 27 sin 3 x = lim x→0 2 sec 2 x tan x sin x sin 2 x sin 3 x −5 + 56 ⋅ − 81 x x 2 3x = lim x→0 tan x 2 sec 2 x ⋅ x = lim x→0
−5 + 56 − 81 2 = −15
=
0 0 form
[Applying L’Hospital’s rule] 0 0 form [Applying L’Hospital’s rule]
sin nx tan x = 1 and lim = 1 Q lim x→0 x → 0 nx x
1.10
Chapter 1
Indeterminate Forms
example 15 3
2
2 x 2 − 2e x + 2 cos ( x 2 ) + sin 3 x Evaluate lim . x→0 x2
Solution 3 2
2 x 2 − 2e x + 2 cos x 2 + sin 3 x Let l = lim x→0 x2 3 3 1 2 4 x − 2e x (2 x) − 2 sin x 2 x 2 + 3 sin 2 x cos x 2 = lim x→0 2x
0 0 form
0 0 form [Applying L’Hospital’s rule]
3 3 2 2 3 1 1 sin x 2 + 6 sin x cos 2 x − 3 sin 3 x 4 − 4 (e x + xe x ⋅ 2 x) − 3 x cos x 2 ⋅ x 2 + 2 2 x = lim x→0 2
[Applying L’Hospital’s rule] 3 2
4 − 4 − lim x→0
=
sin x x ⋅ 2 x x
2 3
1 1 sin x 2 = − ⋅ lim ⋅x 3 2 2 x→0 2 x =0
3 2 Qlim sin x = 1 3 x→0 x2
example 16 Evaluate lim x→0
A x2
sin kx k sin lx − l .
Solution Let
l = lim x→0
A sin kx k − x 2 sin lx l
l sin kx − k sin lx lx 2 sin lx A l sin kx − k sin lx = lim sin lx l x→0 x2 ⋅ lx lx
= A lim x→0
0 0 form
1.3
=
A l sin kx − k sin lx lim 2 x→0 l x3
=
A lk cos kx − kl cos lx lim l 2 x→0 3x 2
0 0 form 0 0 form
A −lk 2 sin kx + kl 2 sin lx lim 6x l 2 x→0 A sin kx sin lx = 2 lim −lk 2 ⋅ ⋅ k + kl 2 ⋅ ⋅l kx lx 6l x → 0 =
(
A −lk 3 + kl 3 6l 2 Ak 2 = l − k2 6l
=
(
Type 1:
0 0
Form
1.11
sin lx = 1 Q lim x→0 lx [Appling L’Hospital’s rule] [Applying L’Hospital’s rule]
sin kx sin lx = 1 = lim Q lim 0 x→0 x → kx lx
)
)
example 17 1 2
x tan x
Evaluate lim x→0
x
(e − 1)
.
3 2
Solution l = lim
Let
x Æ0
3 (e x - 1) 2
x x
= lim x→0
(e x − 1)
3 2
x x
= lim x→0
(e x − 1)
3 2
⋅
3
x 1 = lim x x → 0 e −1 e =1
Now,
lim
\
x 2 lim x = (1) 2 = 1 x → 0 e − 1
Hence,
l=1
tan x = 1 Q lim x→0 x [Applying L’Hospital’s rule]
x
3
tan x x
tan x ⋅ lim x→0 x
x 2 = lim x x → 0 e − 1
x→0
È0 ˘ Í 0 form ˙ Î ˚
x tan x
3
1.12
Chapter 1
Indeterminate Forms
example 18 x 2. Evaluate lim x →0 log x cos x log sec x cos sec
2
Solution Let
x 2 l = lim x → 0 log cos x x log sec x cos sec
2
x x log sec 2 2 = lim ⋅ x → 0 log sec x log cos x log cos
x − log cos x 2 2 ⋅ = lim x → 0 ( − log cos x ) log cos x log cos
x log cos 2 = lim x→0 log cos x
Now,
2
0 0 form
x 1 1 ⋅ − sin x x 2 2 log cos cos 2 = lim 2 lim x → 0 log cos x x→0 1 ( − sin x) cos x
x 2 = lim x → 0 2 tan x x tan 1 2 = lim x→0 4 x 2 tan
=
1 4
x ⋅ tan x
[Applying L’Hospital’s rule]
1.3
Type 1:
0 0
Form
1.13
2
∴
x 2 log cos 1 1 2 lim = = x→0 log cos x 4 16 l=
Hence,
tan x = 1 Q lim x→0 x
1 16
example 19 1
(1 + x) x − e e =− . Prove that lim x→0 x 2 Solution 1
Let
(1 + x) x − e l = lim x→0 x 1
= lim
log(1+ x )
ex
x→0
−e
[Applying L’Hospital’s rule]
x 1
log(1+ x )
ex = lim x→0
1 1 − x 2 log (1 + x) + x (1 + x) 1
1
= lim (1 + x) x lim x→0
0 0 form
[ − log (1 + x) ⋅ (1 + x) + x ]
x→0
x (1 + x) 2
− log (1 + x) − 1 + 1 = e lim x→0 2 x + 3x 2
1 − 1 + x = e lim x→0 2 + 6x e =− 2
0 0 form [Applying L’Hospital’s rule] 0 0 form
[Applying L’Hospital’s rule] 0 0 form
1.14
Chapter 1
Indeterminate Forms
example 20
( Prove that lim x→0
)
2n
1− x −1
(1 − cos x )
= 2− n .
n
Solution Let
( l = lim x→0
2n
) ⋅( (1 − cos x ) (
) 1 − x + 1)
1− x −1
2n
1− x +1
n
2n
(1 − x − 1) 2 n
= lim
2 2 sin
x→0
x 2
n
(
)
1− x +1
2n
( − x) 2 n
= lim
x 2n sin 2
x→0
x = lim 2 x→0 x sin 2
2n
(
)
1− x +1
⋅ 2n
2n 2n
2n
2n
(
)
1− x +1
2n
{
Q ( − x) 2 n = ( − x) 2
}
n
= x2n
1 2n
=
example 21 Evaluate lim x→0
x 2 + 2 cos x − 2 . x sin x
Solution Let
l = lim x→0
x 2 + 2 cos x − 2 x sin x
2 x − 2 sin x sin x + x cos x 2 − 2 cos x = lim x → 0 cos x + cos x − x sin x =0
= lim x→0
0 0 form
0 0 form [Applying L’Hospital’s rule] [Applying L’Hospital’s rule]
1.3
Type 1:
0 0
Form
1.15
example 22 Evaluate lim x→0
e x sin x − x − x 2 . x 2 + x log(1 − x)
Solution e x sin x − x − x 2 x → 0 x 2 + x log(1 − x )
0 0 form
Let l = lim
e x sin x + e x cos x − 1 − 2 x 0 0 form x→0 x 2 x + log(1 − x) − 1− x x e (sin x + cos x) + e x (cos x − sin x) − 2 = lim x→0 1 1 x 2− − − 1 − x 1 − x (1 − x) 2 = lim
2(e x cos x − 1) x→0 1 x 2 1 − − 1 − x (1 − x) 2 2(e x cos x − e x sin x) x→0 1 1 2x − 2 − 2− 2 (1 − x)3 (1 − x) (1 − x)
= lim
2 3
exercIse 1.1 x 2 log a − a 2 log x 1 = log a − . 2 2 x →a 2 x −a
1. Prove that lim
log(1 − x 2 ) = 2. x → 0 log cos x
2. Prove that lim
3. Prove that limπ x→
4
1 − tan x
= 2.
1 − 2 sin x
e x − 1 + 2x = 1. x → 0 log(1 + x 2 )
4. Prove that lim
[Applying L’Hospital’s rule]
0 0 form
= lim
=−
[Applyying L’Hospital’s rule]
[Applying L’Hospital’s rule]
1.16
Chapter 1
Indeterminate Forms
a + x tan−1 a 2 − x 2
5. Evaluate lim x →a
.
a−x tan−1 a 2 − x 2 as x → a, a − x → 0 Hint : lim(x + a) x →a a2 − x 2
[ans.: 2a] 1− x e x + loge e . 6. Evaluate lim x →0 tan x − x
7. Evaluate lim x →1
1 − x + log x 1 − 2x − x 2
1 Ans.: − 2
. [ans.: –1]
x
8. Prove that lim x →0
xe − log(1 + x) 3 = . 2 x2
e x − 1 + 2x = 1. x → 0 log(1 + x 2 )
9. Prove that lim
10. Prove that lim π x→ 4
11. Prove that lim x →3
1 − tan x
= 2.
1 − 2 sin x 3x − 12 − x
=
2 x − 3 19 − 5x
8 . 69
alog x − x a = log . x →1 e log x
12. Prove that lim
13. Prove that lim x →a
= 2
2
x −a tan x − x 1 14. Prove that lim = . x →0 3 x3
1.4 type 2:
1
x − a + x −a
.
2a
Form
Problems under this type are also solved by using L’Hospital’s rule considering the fact that f ( x) f ′ ( x) lim = lim if lim f ( x) = ∞ and lim g ( x) = ∞. x →a g ( x) x →a g ′ ( x) x →a x →a
1.4
Type 2:
• •
Form
1.17
example 1 log x = 0, (n > 0). x →∞ x n
Prove that lim Solution
∞ ∞ form
log x xn 1 = lim xn −1 x →∞ nx 1 1 = lim n n x →∞ x =0
l = lim
Let
x →∞
[Applying L’Hospital’s rule]
example 2 Evaluate lim x →0
log sin x . cot x
Solution log sin x cot x 1 ⋅ cos x x sin = lim x → 0 − cosec 2 x = lim − ( cos x sin x )
l = lim
Let
x→0
∞ ∞ form [Applying L’Hospital’s rule]
x→0
=0
example 3
π log x − 2 Prove that lim = 0. π tan x x→ 2
Solution Let
l = lim xÆ
p
pˆ Ê log Á x - ˜ Ë 2¯ tan x
2
= lim xÆ
p 2
È• ˘ Í • form ˙ Î ˚
Ê 1 ˆ Á p˜ ÁË x - ˜¯ 2 sec 2 x
[Applying L’Hospital’s rule]
1.18
Chapter 1
Indeterminate Forms
= lim p xÆ 2
= lim p xÆ 2
cos2 x p x2
È0 ˘ Í 0 form ˙ Î ˚
2 cos x ( - sin x ) 1
[Applying L’Hospital’s rule]
=0
example 4 log ( x – a ) = 1. x → a log ( a x – a a )
Prove that lim Solution Let l = lim
log ( x - a ) x
a
log (a - a ) 1 ( x - a) = lim 1 . x xÆa a log a a x - aa a x 1 . lim a − a = lim x x → a a log a x→a x − a xÆa
x 1 . lim a log a 1 a a log a x → a 1 = a ⋅ a a log a a log a =1
=
È• ˘ Í • form ˙ Î ˚
[Applying L’Hospital’s rule] 0 0 form [Applying L’Hospital’s rule to second term]
example 5 Prove that lim log x tan x = 1. x→0
Solution Let
l = lim log x tan x x→0
= lim x→0
log tan x log x
∞ ∞ form
[Change of base property]
1.4
1 . 2 sec x = lim tan x x→0 1 x x . = lim limsec 2 x x → 0 tan x x → 0
Type 2:
• •
Form
1.19
[Applying L’Hospital’s rule]
tan x = 1 Q lim x→0 x
=1
example 6 Evaluate lim log tan x tan 2 x . x →0
Solution Let
l = lim log tan x tan 2 x x→0
∞ ∞ form
log tan 2 x log tan x 1 ⋅ 2 sec 2 2 x 2 x tan = lim x→0 1 ⋅ sec 2 x tan x = lim x→0
[Applying L’Hospital’s rule]
tan x ⋅ sec 2 2 x x = lim x → 0 tan 2 x ⋅ sec 2 x 2x tan x = 1 Q lim x→0 x
=1
example 7 sin h−1 x = 1. x →∞ cos h −1 x
Prove that lim Solution Let
sin h −1 x x →∞ cos h −1 x
l = lim
= lim
x →∞
( log ( x +
) − 1)
log x + x 2 + 1 x
2
∞ ∞ form
1.20
Chapter 1
Indeterminate Forms
= lim
(
1 ⋅ 1 + ⋅ 2 x 2 x + x +1 2 x +1 1
)
2
x →∞
(x +
1 x2 − 1
)
⋅ 1 +
1 2 x2 − 1
⋅ 2 x
[Applying L’Hospital’s rule]
x2 + 1 + x
(x + = lim
x2 + 1
(x +
x2 − 1
x →∞
x2 + 1
x2 − 1 + x
x →∞
= lim
) )
x2 − 1
x2 − 1 x2 + 1
1 x2 = lim x →∞ 1 1+ 2 x =1 1−
example 8 1
2
3
x
e x + e x + e x + ... + e x Prove that lim = e − 1. x →∞ x Solution 1
Let
2
3
x
e x + e x + e x + ... + e x l = lim x →∞ x x 1 1 e x 1 − e x 1 = lim ⋅ 1 x →∞ x x 1− e
[Sum of GP]
1
e x (e − 1) 1 ⋅ = lim 1 x →∞ x x e −1
Putting
1 = y, when x → ∞, y → 0 x (e − 1)e y y l = lim y→0 ey −1
0 0 form
1.4
(e − 1) ( ye y + e y ) y →0 ey = e −1
Type 2:
• •
Form
1.21
[Applying L’Hospital’s rule]
= lim
example 9 xn = 0. x →∞ e kx
Prove that lim Solution Let
xn x →∞ e kx
∞ ∞ form
nx n −1 x →∞ ke kx
∞ ∞ form
[Applying L’Hospital’s rule]
n (n − 1) x n − 2 x →∞ k 2 e kx
∞ ∞ form
[Applying L’Hospital’s rule]
l = lim = lim = lim
Applying L’Hospital’s rule (n – 2) times in the above expression, n (n − 1) (n − 2)...2 ⋅1 k n e kx n! = lim n kx x →∞ k e =0 [Q lim e kx = ∞]
l = lim
x →∞
x →∞
example 10 12 + 22 + 32 + ... + x 2 1 = . x →∞ x3 3
Prove that lim Solution Let
12 + 22 + 32 + ... + x 2 x →∞ x3 x ( x + 1) (2 x + 1) = lim x →∞ 6 x3 2 x3 + 3x 2 + x = lim x →∞ 6 x3 3 1 2+ + 2 x x = lim x →∞ 6
l = lim
n (n + 1) (2n + 1) 2 Q∑ n = 6
1.22
Chapter 1
Indeterminate Forms
2 6 1 = 3 =
example 11 Prove that lim x →∞
1
ex 1 x 1 + x
x
= e2 .
Solution Let
l = lim
x Æ•
= lim
x Æ•
ex ÈÊ ÍÁ 1 + ÍÎË
x 1ˆ ˘ ˙ ˜ x¯ ˙ ˚
x
È• ˘ Í • form ˙ Î ˚
ex Ê ÁË 1 +
1ˆ x ˜¯
x2
Taking logarithm on both sides, x2 ˘ È Ê 1ˆ log l = lim Í log e x - log Á 1 + ˜ ˙ x Æ• Í Ë x¯ ˙ ˚ Î È 1ˆ˘ Ê = lim Í x - x 2 log Á 1 + ˜ ˙ x Æ• Î Ë x¯˚ È1 1ˆ˘ Ê = lim x 2 Í - log Á 1 + ˜ ˙ x Æ• Ë x x¯˚ Î
1 Ê 1ˆ - log Á 1 + ˜ Ë x x¯ = lim 1 x Æ• x2 1 1 Ê 1ˆ - 2 1 ÁË x 2 ˜¯ x 1+ x = lim 2 x Æ• - 3 x
È0 ˘ Í 0 form ˙ Î ˚
[ Applying L’Hospital’s rule]
1.5 Type 3 : 0 × ∞ Form 1.23
1
1−
1+ = lim
2 x 1
x →∞
=
=
1 lim 2 x →∞
1 x
1+
1 x
1 2 1
l = e2
Hence,
exercIse 1.2 log x = 0. cot x log(1 − x) 2. Prove that lim = 0. x →1 cot π x logsin x cos x = 4. 3. Prove that lim x →0 x log x cos sin 2 2 1. Prove that lim x →0
4. Prove that limlogsin x sin 2 x = 1. x →0
log(1 + e 3 x ) = 3. x →∞ x log x 2 = 0. 6. Prove that lim x → 0 cot x 2 xm 7. Prove that lim x = 0 (m > 0) . x →∞ e
5. Prove that lim
2
[Hint: Put x 2 = y ]
3
n
1 1 1 1 + + + LL + e e e e = 0. 8. Prove that lim n →∞ n 9. Prove that limlog x sin 2 x = x →0
1 . 2
1.5 type 3 : 0 × ∞ Form To solve the problems of the type
lim [ f ( x) ⋅ g ( x)], when lim f ( x) = 0, lim g ( x) = ∞ (i.e. 0 × ∞ form) x→ a
x→ a
x→ a
1.24
Chapter 1
Indeterminate Forms
We write lim [ f ( x) ⋅ g ( x) ] = lim x→a
x→a
f ( x) g ( x) or lim ⋅ x→a 1 1 g ( x) f ( x)
These new forms are of the type
0 ∞ or respectively, which can be solved using 0 ∞
L’Hospital’s rule.
example 1 Prove that lim x log x = 0. x→0
Solution Let
l = lim x log x
[0 × ∞ form]
log x 1 x 1 l = lim x x→0 1 − 2 x = lim( − x)
∞ ∞ form
x→0
= lim x→0
[Applying L’Hospital’s rule]
x→0
=0
example 2 Prove that lim sin x log x = 0 . x→0
Solution Let
l = lim sin x log x x→0
= lim x→0
log x cosec x
1 x = lim x → 0 − cosec x cot x tan x = − lim sin x ⋅ x→0 x tan x = − lim sin x ⋅ lim x→0 x→0 x =0
[0 × ∞ form] ∞ ∞ form [Applying L’Hospital’s rule]
tan x = 1 Qlim x→0 x
1.5 Type 3 : 0 × ∞ Form 1.25
example 3 a 2 x ⋅ sin x = a. Prove that lim x →∞ 2 Solution Let Putting
a l = lim 2 x ⋅ sin x x →∞ 2 1 1 2x = , t = x , t 2
When x → ∞, 2 x → ∞,
t→0
sin at t a sin at = lim t →0 at
l = lim t →0
= a ⋅1
sin x = 1 Q lim x→0 x
=a
example 4 x Evaluate lim log 2 − cot( x − a ). x→ a a Solution Let
x l = lim log 2 − cot( x − a ) x→a a xˆ Ê log Á 2 - ˜ Ë a¯ = lim x Æ a tan( x - a ) Ê 1ˆ x ˆ ÁË a ˜¯ Ê 2 ÁË a ˜¯
[0 × ∞ form]
È0 ˘ Í 0 form ˙ Î ˚
1
= lim
xÆa
=-
1 a
sec 2 ( x - a )
[ Applying L’Hospital’s rule]
1.26
Chapter 1
Indeterminate Forms
example 5 1x Prove that lim a − 1 x = log a . x →∞ Solution Let
(a − 1) ⋅ x (a − 1) = lim
l = lim
x →∞
1 x
1 x
x →∞
Putting
1 x
[Winter 2013]
[0 × ∞ form] 0 0 form
1 = t , when x → ∞, t → 0 x l = lim t →0
at − 1 t
a t log a t →0 1 = log a = lim
0 0 form [Applying L’Hospital’s rule]
example 6 Ê p xˆ Prove that lim tan 2 Á ˜ (1 + sec p x ) = - 2. Ë 2 ¯ x Æ1
Solution Let
Ê pxˆ l = lim tan 2 Á ˜ ( 1 + sec p x ) x Æ1 Ë 2 ¯ 1 + sec p x = lim x Æ1 Ê pxˆ cot 2 Á ˜ Ë 2 ¯ p sec p x tan p x pxˆ p Ê pxˆ Ê 2 cot Á ˜ Á - cosec 2 Ë 2 ¯Ë 2 ˜¯ 2 Ê ˆÊ ˆ sec p x ˜ Á tan p x ˜ Á = - Á lim ˜ Á lim px ˜ x Æ1 x Æ1 2 px ˜ Á ˜Á cosec cot Ë 2 ¯Ë 2 ¯
[• ¥ 0 form ] È0 ˘ Í 0 form ˙ Î ˚
= lim
[ Applying L’Hospital’s rule]
Ê ˆ tan p x Á sec p ˜ = -Á ˜ lim p px x Æ1 Á cosec 2 ˜ cot Ë ¯ 2 2
È0 ˘ Í 0 form ˙ Î ˚
x Æ1
1.5 Type 3 : 0 × ∞ Form 1.27
π sec 2 π x x →1 2 π x π −cosec 22
= −( −1) lim
= −2
[Applying L’Hospital’s rule]
sec 2 π π cosec 2 2
= −2
example 7 a+ x ⋅ tan −1 a 2 − x 2 . a−x
Evaluate lim x→ a
Solution Let
a+ x ◊ tan -1 a 2 - x 2 a-x
l = lim xÆa
= lim xÆa
a+ x
a+ x ◊
a-x
= lim(a + x )
tan
\
tan -1 a 2 - x 2
xÆa
a2 - x2
tan -1 a 2 - x 2 a2 - x2
xÆa
Let When
a2 - x2
a2 - x2
= lim(a + x ) ◊ lim
= 2 a lim
tan -1 a 2 - x 2
a+ x -1
xÆa
xÆa
[• ¥ 0 form ]
a2 − x2 = α x → a, α → 0 tan -1 a a Æ0 a È ˘ tan -1 a = 2a = 1˙ ÍQ lim ÍÎ a Æ0 a ˙˚
l = 2 a ◊ lim
example 8 sin −1 Evaluate lim x→ a
a−x 1 cosec a 2 − x 2 = . a+ x 2a
1.28
Chapter 1
Indeterminate Forms
Solution l = lim sin −1
Let
x→a
a−x cosec a 2 − x 2 a+ x
[0 × ∞ form]
a−x a+ x = lim 2 x→ a sin a − x 2 sin −1
Here, applying L’Hospital’s rule will make the expression complicated, so we rearrange the terms to apply the limits directly. Let When
a−x = α , a2 − x2 = β a+ x x → a, α → 0 and β → 0
1 β → 0 sin β
l = lim sin −1 α ⋅ lim
\
α →0
sin −1 α β 1 = lim ⋅ ⋅ α lim α →0 β →0 α sin β β = lim α ⋅ lim α →0
= lim x→ a
β →0
1 β
a−x 1 ⋅ a + x a2 − x2
sin −1 α = 1 Q αlim →0 α sin β = 1 and lim β →0 β
[Resubstituting a and b ]
a−x 1 ⋅ a+ x a+ x a−x 1 = lim x→a a + x 1 = 2a = lim x→a
example 9 m n Evaluate lim x (log x) , where m and n are positive integers. x →0
Solution Let
l = lim x m (log x) n x→0
(log x) n x→0 1 xm
= lim
[0 × • form] ∞ ∞ form
1.5 Type 3 : 0 × ∞ Form 1.29
n (log x) n −1 = lim
1 x
[Applying L’Hospital’s rule]
− m ( x) − m −1
x→0
( −1)1 n (log x) n −1 x→0 m ( x) − m
∞ ∞ form
= lim
( − 1)1 n (n − 1) (log x) n − 2 ⋅ = lim
1 x
m ( − m)1 ( x) − m −1
x→0
( −1) 2 n (n − 1) (log x) n − 2 x→0 m 2 ( x) − m
= lim
[Applying L’Hospital’s rule] ∞ ∞ form
Applying L’Hospital’s rule (n – 2) times in the above expression,
( −1) n n !(log x)0 x→0 m n ( x) − m
l = lim
( −1) n n ! m ⋅x x→0 mn =0
= lim
exercIse 1.3 1. Prove that lim x log x = 0. x →0
2. Prove that lim x 2 e − x = 0. x →∞
(
3. Prove that lim x 2 1 − e x →∞
−
2 gy x2
) = 2gy.
4. Prove that limtan x log x = 0. x →0
4 πx 2 5. Prove that lim(x − 1)tan = − . x →1 2 π
πx = 0. 2 πx 7. Prove that limlog(1 − x) cot = 0. x →1 2 6. Prove that lim(1 + sec π x)tan x →1
1+ x 8. Prove that limlog 1 − x cot x = 2. x →0 9. Prove that lim x →2
2+x tan−1 4 − x 2 = 4. 2−x
1.30
Chapter 1
1.6 type 4 :
Indeterminate Forms
∞ − ∞ Form
To evaluate the limits of the type lim[ f ( x) − g ( x)] , when lim f ( x) = ∞ and, x→a
x→a
lim g ( x) = ∞ [i.e., (• – •) form], we reduce the expression in the form of x→a
0 ∞ by or 0 ∞
taking LCM or by rearranging the terms and then apply L’Hospital’s rule.
example 1 Prove that lim (cosh −1 x − log x) = log 2 . x →∞
Solution l = lim (cosh −1 x − log x)
Let
x →∞
(
[∞ − ∞ form]
)
= lim log x + x 2 − 1 − log x x →∞ x + x2 − 1 = lim log x →∞ x 1 = lim log 1 + 1 − 2 x→∞ x = log 2
example 2 1 1 Evaluate lim − 2 log(1 + x) . x→0 x x Solution 1 1 Let l = lim − 2 log(1 + x) x→0 x x
x − log(1 + x) = lim x→0 x2 1 1− = lim 1 + x x→0 2x
[∞ − ∞ form]
0 0 form 0 0 form
[Applying L’Hospital’s rule]
1.6 Type 4 : ∞ − ∞ Form 1.31
1 2 = lim 1 + x x→0 2 1 = 2
[Applying L’Hospital’s rule]
example 3 Ê x 1 ˆ Evaluate lim Á . x Æ1 Ë x - 1 log x ˜¯
[Winter 2016]
Solution Ê x 1 ˆ l = lim Á x Æ1 Ë x - 1 log x ˜¯
Let
[• – • form]
È x log x - x + 1 ˘ = lim Í ˙ x Æ1 Î log x ( x - 1) ˚
È0 ˘ Í 0 form ˙ Î ˚
È ˘ Í 1 + log x - 1 ˙ = lim Í ˙ x Æ1 x - 1 Í + log x ˙ ÍÎ x ˙˚
[Applying L’Hospital’s rule]
log x x -1 + log x x 1 x = lim 1 x Æ1 1 + x2 x = lim
x Æ1
=
È0 ˘ Í 0 form ˙ Î ˚
[Applying L’Hospital’s rule]
1 2
example 4 1 1 Evaluate lim − . x→ 2 x − 2 log( x − 1) Solution Let
1 1 l = lim − x→2 x − 2 log( x − 1)
[∞ − ∞ form]
1.32
Chapter 1
Indeterminate Forms
log( x − 1) − ( x − 2) ( x − 2) log( x − 1) 1 −1 x −1 = lim x→ 2 x − 2 + log( x − 1) x −1
0 0 form
= lim x→ 2
[Applying L’Hospital’s rule]
1 − ( x − 1) ( x − 2) + ( x − 1) log( x − 1) −1 = lim x→2 1 1 + ( x − 1) ⋅ + log( x − 1) ( x − 1) 1 =− 2
= lim x→2
0 0 form [Applying L’Hospital’s rule]
example 5 x a Prove that lim − cot = 0 . x→0 x a Solution Let Putting
x a l = lim − cot x→0 x a
[∞ − ∞ form]
x = y, when x → 0, y → 0 a
1 l = lim − cot y y→0 y 1 1 = lim − y→0 y tan y
[∞ − ∞ form]
tan y − y = lim y → 0 y tan y
0 0 form
tan y − y 1 = lim ⋅ lim y→0 y 2 y → 0 tan y y
tan y − y ⋅1 y→0 y2
0 0 form
tan y = 1 Q lim y→0 y
sec 2 y − 1 y→0 2y
0 0 form
[Applying L’Hospital’s rule]
= lim
= lim
1.6 Type 4 : ∞ − ∞ Form 1.33
= lim y→0
2 sec y ⋅ sec y tan y 2
[Applying L’Hospital’s rule]
=0
example 6 Èp ˘ p p Prove that lim Í ˙= . p x x Æ0 Î 4 x 2 x(e + 1) ˚ 4 Solution Let
È 1 ˘ p 1 lim Í ˙ x p 2 x Æ0 Î 2 x x(e + 1) ˚
[• - • form ]
=
p ep x + 1 - 2 lim 2 x Æ0 2 x(ep x + 1)
È0 ˘ Í 0 form ˙ Î ˚
=
p ep x p lim 2 x Æ0 2 È(ep x + 1) + x(p ep x )˘ Î ˚
[ Applying L’Hospital’s rule]
=
p2 e0 ◊ 0 4 (e + 1)
=
p2 8
l=
example 7 1 1 1 Prove that lim x + log x + − log x = . x →∞ 2 2 2 Solution Let
1 1 l = lim x + log x + − log x x →∞ 2 2
[∞ − ∞ form ]
1 1 x+ = lim x + log 2 x →∞ 2 x 1 1 1 = lim x log 1 + + log 1 + x →∞ 2x 2x 2 1 1 = lim log 1 + x →∞ 2 2x
=
1 1 log e + log1 2 2
=
1 2
2x
1 1 + lim log 1 + x →∞ 2 2x ax 1 1 + = e Q lim x →∞ ax
1.34
Chapter 1
Indeterminate Forms
example 8 1 ˆ Ê 1 Evaluate lim Á 2 - 2 ˜ . x Æ0 Ë x sin x ¯
[Winter 2014; Summer 2015]
Solution Let
1 Ê 1 l = lim Á 2 - 2 x Æ0 Ë x sin
ˆ ˜ x¯
[• - • form ]
Ê sin 2 x - x 2 ˆ = lim Á 2 2 ˜ x Æ 0 Ë x sin x ¯
È0 ˘ Í 0 form ˙ Î ˚
Ê sin 2 x - x 2 ˆ = lim Á ˜ x Æ0 Ë x4 ¯
È ˘ sin 2 x = 1˙ ÍQ lim 2 ÍÎ x Æ0 x ˙˚
Ê 2 sin x cos x - 2 x ˆ = lim Á ˜¯ x Æ0 Ë 4 x3
[ Applying L’Hospital’s rule]
Ê sin 2 x - 2 x ˆ = lim Á ˜¯ x Æ0 Ë 4 x3
Ê 2 cos 2 x - 2 ˆ = lim Á ˜¯ x Æ0 Ë 12 x 2
[ Applying L’Hospital’s rule]
Ê - 4 sin 2 x ˆ = lim Á x Æ0 Ë 24 x ˜¯
[ Applying L’Hospital’s rule]
Ê -8 cos 2 x ˆ = lim Á x Æ0 Ë 24 ˜¯
[ Applying L’Hospital’s rule]
-8 24 1 =3 =
example 9 1 2 Evaluate lim − cot x . x→0 x 2 Solution Let
Ê 1 ˆ l = lim Á 2 - cot 2 x ˜ x Æ0 Ë x ¯
[• - • form ]
1.6 Type 4 : ∞ − ∞ Form 1.35
1 ˆ Ê 1 = lim Á 2 ˜ x Æ0 Ë x tan 2 x ¯ = lim
x Æ0
= lim
x Æ0
tan 2 x - x 2 2
˘ È0 Í 0 form ˙ ˚ Î
2
x tan x tan 2 x - x 2 x4 ◊
tan 2 x x2
Ê tan 2 x - x 2 ˆ Ê x2 = lim Á ◊ lim Á 2 ˜ 4 x Æ0 Ë x ¯ x Æ0 Ë tan
ˆ ˜ x¯
Ê tan 2 x - x 2 ˆ = lim Á ˜ ◊1 x Æ0 Ë x4 ¯ = lim
x Æ0
= lim
x Æ0
= lim
x Æ0
tan x È ˘ = 1˙ ÍQ xlim Î Æ0 x ˚
tan 2 x - x 2
È0 ˘ Í 0 form ˙ ˚ Î
x4
2 tan x ◊ sec 2 x - 2 x 4 x3
(
[ Applying L’Hospital’s rule]
)
2 tan x 1 + tan 2 x - 2 x 4x
3
=
1 tan x + tan 3 x - x lim 2 x Æ0 x3
=
1 tan 3 x Ê tan x - x ˆ 1 lim Á lim + 2 x Æ0 Ë x 3 ˜¯ 2 x Æ0 x 3
=
1 Ê tan x - x ˆ 1 lim + 2 x Æ0 ÁË x 3 ˜¯ 2
=
Ê sec 2 x - 1ˆ 1 1 + lim Á 2 2 x Æ0 Ë 3 x 2 ˜¯
1 1 2 sec 2 x ◊ tan x + lim 2 2 x Æ0 6x 1 1 tan x = + lim sec 2 x ◊ lim x Æ0 x 2 6 x Æ0 1 1 = + 2 6 2 = 3 =
tan x È ˘ = 1˙ ÍQ xlim Æ 0 x Î ˚
È0 ˘ Í 0 form ˙ Î ˚
[ Applying L’Hospital’s rule]
[ Applying L’Hospital’s rule]
1.36
Chapter 1
Indeterminate Forms
example 10 a cot x b 1 If lim + 2 = , find a and b. x→0 x x 3 Solution 1 a cot x b = lim + 2 3 x→0 x x b a = lim + 2 x → 0 x tan x x ax + b tan x = lim 2 x→0 x tan x = lim x→0
(ax + b tan x) tan x ( x 2 ⋅ x) x
= lim
ax + b tan x x ⋅ lim x → 0 tan x x3
= lim
ax + b tan x ⋅1 x3
x→0
x→0
0 0 form
a + b sec 2 x = lim x→0 3x 2
0 0 forrm
x = 1 Q lim x → 0 tan x
[Applying L’Hospital’s rule]
a + b sec 0 0 a+b = 0 =
a + b = 0, a = –b 1 −b + b sec x = lim 3 x→0 3x 2 2
Thus,
= lim x→0
b ⋅ 2 sec x sec x tan x 6x
b = lim sec 2 x→0 3 b 3 b =1 =
From Eq. (1), a = –1 Hence, a = –1, b = 1
x
... (1) 0 0 form [Applying L’Hospital’s rule]
tan x lim x→0 x tan x = 1 Q xlim →0 x
1.6 Type 4 : ∞ − ∞ Form 1.37
example 11 f ′ ( x) 1 − Evaluate lim . x→ a f ( x) − f (a) x − a Solution f ′ ( x) 1 Let l = lim − x→a f ( x) − f (a) x − a
[∞ − ∞ form]
= lim
( x − a) f ′ ( x) − [ f ( x) − f (a) ] [ f ( x) − f (a) ] ( x − a)
0 0 form
= lim
f ′ ( x) + ( x − a ) f ′′ ( x) − f ′ ( x) [ f ( x) − f (a ) ] + ( x − a) f ′ ( x)
0 0 form [Applying L’Hospital’s rule]
x→a
x→a
f ′′ ( x) + ( x − a ) f ′′′ ( x) f ′ ( x) + f ′ ( x) + ( x − a ) f ′′ ( x) f ′′ (a ) = 2 f ′ (a)
= lim x→a
exercIse 1.4 1 1. Prove that lim cot x − = 0. x →0 x
1 − cot(x − a) = 0 . 2. Prove that lim x →a x − a 3. Prove that lim (sec x − tan x) = 0. x→
π 2
2 x sec x 2 4. Prove that lim tan x − = π . π π x→ 2
1 1 1 =− . − 5. Prove that lim x →0 x − a 2 log(x + 1 − a)
1 1 1 − 6. Prove that lim =− . x →3 x − 3 2 log(x − 2)
[Applying L’Hospital’s rule]
1.38
Chapter 1
Indeterminate Forms
1 7. Evaluate lim x − x 2 log 1 + x →∞ x
1 Hint : Put x = y 1 Ans.: 2
1 1 1 8. Prove that lim − x = . x →0 x e − 1 2
1.7 type 5 : 1∞, ∞ 0, 0 0 Forms To evaluate the limits of the type lim[ f ( x)]g ( x ) , which takes any one of the form x →a
1•, •0, 00 for f (x) > 0, we proceed as follows:
l = lim [ f ( x)]g ( x )
Let
x →a
where f (x) > 0
log l = lim [ g ( x) ⋅ log f ( x)] x→a
which takes the form • × 0, i.e., type 3 form.
example 1 1
(a x + x) x = ae. Prove that lim x→0
[Summer 2014]
Solution 1
Let
l = lim (a x + x ) x
[1• form]
x Æ0
log l = lim
x Æ0
1 ◊ log (a x + x ) x
log (a x + x ) x Æ0 x
= lim
1 (a x log a + 1) a x + = lim x Æ0 1 a 0 log a + 1 = a0 + 0 loge a + loge e = 1
[• × 0 form ] ˘ È0 Í 0 form ˙ ˚ Î
x
[ Applying L’Hospital’s rule]
1.7
Type 5: 1•, •0, 00 Forms
1.39
= log ae l = ae
Hence,
example 2 1 x
3x
Prove that lim(e − 5 x) = e −2 . x→0
Solution 1
[1• form]
l = lim(e3 x − 5 x) x
Let
x→0
1 log l = lim log(e3 x − 5 x) x→0 x log(e3 x − 5 x) = lim x→0 x 1 ⋅ (3e3 x − 5) (e 3 x − 5 x ) = lim x→0 1 0 3e − 5 = e0 = −2 Hence,
0 0 form
[Applying L’Hospital’s rule]
l = e -2
example 3 Evaluate lim(log x)
1 1 − log x
.
x→e
Solution 1
Let
l = lim(log x)1− log x
[1• form]
x→e
1 ⋅ log(log x) 1 − log x 1 1 ⋅ log x x = lim x→e 1 − x = −1
log l = lim x→e
Hence,
l = e −1 =
1 e
0 0 form
[Applying L’Hospital’s rule]
1.40
Chapter 1
Indeterminate Forms
example 4 1 1 Ê a x + bx + c x ˆ 3x 9. Prove that lim Á ( ) = abc ˜ x Æ0 Ë 3 ¯
[Summer 2015]
Solution 1
Ê a x + bx + c x ˆ 3x l = lim Á ˜ x Æ0 Ë 3 ¯
Let
[1• form]
Ê ax + bx + cx ˆ 1 log Á ˜ x Æ0 3 x 3 Ë ¯
log l = lim
Ê ax + bx + cx ˆ log Á ˜ 3 ¯ Ë = lim x Æ0 3xx
È0 ˘ Í 0 form ˙ Î ˚
3 Ê ˆ (a x log a + b x log b + c x log c) ◊ ÁË x x x˜ 3 a +b +c ¯ = lim x Æ0 3 [Applying L’Hospital’s rule] 1 Ê ˆ (a 0 log a + b0 log b + c 0 log c) =Á 0 Ë a + b0 + c 0 ˜¯ 3 =
1 log abc 9
= log
Hence,
l=
1 (abc) 9
1 (abc) 9
example 5 Ê1 + 2 + lim Á x Æ 0Ë 3 x
Evaluate Solution Let
x
1 xˆx 3
[Summer 2017]
˜ ¯
1
Ê 1x + 2 x + 3 x ˆ x l = lim Á ˜ xÆ0 Ë 3 ¯ log l = lim
xÆ 0
Ê 1x + 2 x + 3 x ˆ 1 log Á ˜ x 3 Ë ¯
[1• form]
1.7
Ê 1x + 2 x + 3 x ˆ log Á ˜ 3 ¯ Ë = lim xÆ0 x = lim xÆ 0
3
◊
1x + 2 x + 3 x
Type 5: 1•, •0, 00 Forms
1.41
È0 ˘ Í 0 form ˙ Î ˚
(1x log 1 + 2 x log 2 + 3 x log 3) 3 [Applying L’Hospital’s rule]
1 = (log 2 + log 3) 3 1 = log(6) 3 1
= log(6) 3 1
Hence,
l = (6) 3
example 6 x
1 1 1 1 1x x x x 4 . a + b + c + d Prove that lim = abcd ( ) x →∞ 4 Solution
Let Putting
1 1 1 1 l = lim a x + b x + c x + d x x →∞ 4
x
1 = y, when x Æ •, y Æ 0 x 1
Ê ay + by + cy + d y ˆ y l = lim Á ˜ yÆ0 Ë 4 ¯
[1• form]
ay + by + cy + d y 1 log l = lim log y→0 y 4 ay + by + cy + d y log 4 = lim y→0 y
0 0 form
y y y y 4 a log a + b log b + c log c + d log d = lim y y→0 a + b y + c y + d y 4
[Applying L’Hospital’s rule]
1.42
Chapter 1
Indeterminate Forms
log a + log b + log c + log d 4 1 = log (abcd ) 4 =
1
= log(abcd ) 4 1
Hence,
l = (abcd ) 4
example 7 ax + 1 a Prove that lim =e . x →∞ ax − 1 2
x
Solution Let
ax + 1 l = lim x →∞ ax − 1
x
1 1+ ax = lim x →∞ 1 1− ax
x
[1∞ form]
1 1 + ax log l = lim x log x →∞ 1− 1 ax 1 1 ax = lim log 1 + − log 1 − x →∞ a ax ax ax − ax 1 1 1 = lim log 1 + + log 1 − x →∞ a ax ax ax 1 1 Q lim 1 + = e = (log e + log e) x →∞ ax a 1 = (1 + 1) a 2 = a
Hence,
l=e
2 a
1.7
Type 5: 1•, •0, 00 Forms
1.43
example 8 1 a x + Evaluate lim x→ a 2 a x Solution
1 x−a
.
1
1 a x x−a l = lim + x→ a 2 a x
Let
log l = lim
xÆa
[1∞ form]
È1 Ê a 1 x ˆ˘ log Í Á + ˙ x-a a ˜¯ ˙˚ ÍÎ 2 Ë x a+x 2 ax x-a
log = lim
xÆa
= lim x→a
log(a + x) − log 2 ax x−a
1 log(a + x) − log 2 a − log x 2 = lim x→a x−a 1 1 − + a x 2 x = lim x→a 1
0 0 form [Applying L’Hospital’s rule]
1 1 2a 2a =0 =
Hence,
l = e0 =1
example 9 3 2
Prove that lim(cos 2 x) x = e −6 . x→0
Solution Let
3 2
l = lim(cos 2 x) x x→0
[1• form]
1.44
Chapter 1
log l = lim
x Æ0
= lim
x Æ0
Indeterminate Forms
3
◊ log(cos 2 x ) x2 3 log(cos 2 x ) x
È0 ˘ Í 0 form ˙ Î ˚
2
3 ( -2 sin 2 x ) ◊ 2x cos 2 x Ê tan 2 x ˆ = lim - 6 Á x Æ0 Ë 2 x ˜¯ = lim
[ Applying L’Hospital’s rule]
x Æ0
= -6
l = e -6
Hence,
example 10 1
1
Ê tan x ˆ x 2 3 Prove that lim Á ˜ =e . x Æ0 Ë x ¯ Solution 1
Let
tan x x2 l = lim x→0 x log l = lim x→0
[1∞ form]
1 tan x log 2 x x
tan x log x = lim x→0 x2
x x sec 2 x − tan x 1 ⋅ x → 0 tan x x2 2x
= lim
= lim x→0
tan x = 1 Q lim x→0 x
x sec 2 x − tan x 2 x3
sec 2 x + x ⋅ 2 sec 2 x tan x − sec 2 x x→0 6x2 2 sec x tan x = lim ⋅ x→0 x 3 1 = 3
0 0 form
[Applying L’Hospital’s rule] x 0 = 1 0 form Q lim x → 0 tan x
= lim
1
Hence,
l = e3
[Applying L’Hospital’s rule]
1.7
Type 5: 1•, •0, 00 Forms
1.45
example 11 xˆ Ê Prove that lim Á 2 - ˜ xÆa Ë a¯ Solution Let
xˆ Ê l = lim Á 2 - ˜ xÆa Ë a¯
tan
tan
px 2a
=
2 p e
.
px 2a
[1• form]
xˆ Ê p xˆ Ê log l = lim tan Á ˜ log Á 2 - ˜ xÆa Ë 2a ¯ Ë a¯ xˆ Ê log Á 2 - ˜ Ë a¯ = lim xÆa Ê p xˆ cot Á ˜ Ë 2a ¯ = lim
xÆa
=
È0 ˘ Í 0 form ˙ Î ˚
1 Ê 1ˆ - ˜ [Applying L’Hospital’s rule] Á Ë ¯ xˆ a Ê Ê 2 p xˆ Ê p ˆ 2 cosec ÁË ÁË a ˜¯ 2 a ˜¯ ÁË 2 a ˜¯ 1
2 p 2
Hence,
l = ep
example 12 Evaluate lim(cos x )cot x .
[Winter 2013]
x Æ0
Solution Let
l = lim(cos x )cot x
[1• form]
x Æ0
log l = lim cot x ◊ log(cos x ) x Æ0
= lim
È0 ˘ Í 0 form ˙ Î ˚
1 ( - sin x ) x cos = lim x Æ0 sec 2 x =0
[ Applying L’Hospital’s rule]
log(cos x ) x Æ0 tan x
1.46
Chapter 1
Indeterminate Forms
l = e0 = 1
Hence,
example 13 2
Evaluate lim(cosec x) tan x . π x→ 2
Solution l = lim (cosec x )tan
Let
2
x
[1• form]
p xÆ 2
log l = lim tan 2 x ◊ log(cosec x ) xÆ
p 2
= lim p xÆ 2
= lim xÆ
p 2
= lim xÆ
=
Hence,
l=
p 2
È0 ˘ m˙ Í 0 form Î ˚
log(cosec x ) cot 2 x 1 ( -cosec x cot x ) cosec x 2 cot x( -cosec 2 x )
[ Applying L’Hospital’s rule]
sin 2 x 2
1 2 1 2 e
example 14 1
1
Ê sin h x ˆ x 2 6 Prove that lim Á ˜ =e . x Æ0 Ë x ¯ Solution 1
Let
sinh x x2 l = lim x→0 x log l = lim
x Æ0
1 x2
[1• form]
Ê sinh x ˆ log Á Ë x ˜¯
sinh x = 1 Q lim x→0 x
1.7
= lim
Ê sinh x ˆ log Á Ë x ˜¯
x Æ0
x
x Æ0
= lim
x cosh x - sinh x 3
2x x sinh x + cosh x - cosh x 6 x2
x Æ0
1.47
È0 ˘ Í 0 form ˙ Î ˚
2
x Ê x cosh x - sinh x ˆ 1 = lim Á ˜¯ ◊ 2 x x Æ 0 sinh x Ë x2 = lim
Type 5: 1•, •0, 00 Forms
1 sinh x = lim ◊ x Æ0 6 x 1 = 6
[ Applying L’Hospital’s rule] x ˘È È0 ˘ = 1˙ Í 0 form ˙ ÍQ xlim Æ 0 x sinh Î ˚Î ˚ [ Applying L’Hospital’s rule]
sinh x ˘ È = 1˙ ÍQ xlim Æ 0 x ˚ Î
1
Hence,
l = e6
example 15 p ˆ Ê Evaluate lim Á sin 2 ˜ x Æ0 Ë 2 - ax ¯ Solution p ˆ Ê Let l = lim Á sin 2 x Æ0 Ë 2 - ax ˜¯
log l = lim sec 2 x Æ0
sec2
sec2
p 2 - bx
p 2 - bx
.
[1• form]
p p ˆ Ê ◊ log Á sin 2 Ë 2 - bx 2 - ax ˜¯
p ˆ Ê log Á sin 2 Ë 2 - ax ˜¯ = lim p x Æ0 cos2 2 - bx ˘ p Ê p ˆ Ê p ˆ È ( - a )˙ ◊ 2 sin Á ˜¯ ◊ Í ˜¯ ◊ cos ÁË 2 p Ë ax 2 2 - ax Î (2 - ax ) ˚ sin 2 ax 2 = lim x Æ0 ˘ p Ê p ˆÈ Ê p ˆ˘È 2 cos Á - sin Á ( - b)˙ Ë 2 - bx ˜¯ ÍÎ Ë 2 - bx ˜¯ ˙˚ ÍÎ (2 - bx )2 ˚ 1
[ Applying L’Hospital’s rule]
1.48
Chapter 1
Indeterminate Forms
Ê p ˆ cot Á Ë 2 - ax ˜¯ 2 a(2 - bx )2 ◊ lim = lim x Æ0 Ê 2p ˆ x Æ0 b(2 - ax )2 - sin Á Ë 2 - bx ˜¯
=-
2a lim b x Æ0
˘ p Ê p ˆ È -cosec 2 Á ◊ Í( - a )˙ ˜ 2 Ë 2 - ax ¯ Î (2 - ax ) ˚ ˘ 2p Ê 2p ˆ È ◊ cos Á ( - b)˙ Ë 2 - bx ˜¯ ÍÎ (2 - bx )2 ˚
È0 ˘ Í 0 form ˙ Î ˚
[ Applying L’Hospital’s rule]
Ê pa ˆ 2a Á 4 ˜ =◊Á p b˜ 2 b Á ˜ Ë 4 ¯ =-
a2 b2 a2 −
Hence, l = e
b2
example 16 qˆ Ê Evaluate lim Á cos ˜ nÆ• Ë n¯ Solution Let
n2
.
qˆ Ê l = lim Á cos ˜ n Æ• Ë n¯
n2
[1• form]
qˆ Ê log l = lim n2 log Á cos ˜ n Æ• Ë n¯
Putting when
1 = x, n
n Æ •, x Æ 0
log l = lim
x Æ0
log(cos q x ) 2
x 1 ( -q sin q x ) = lim cos q x x Æ0 2x q 1 q sin q x = - lim ◊ 2 x Æ0 cos q x qx
È0 ˘ Í 0 form ˙ Î ˚ [ Applying L’Hospital’s rule]
1.7
=-
l=e
Hence,
q2 2
-
Type 5: 1•, •0, 00 Forms
1.49
sin q x È ˘ = 1˙ ÍQ xlim Î Æ0 q x ˚
q2 2
example 17 1ˆ Ê Evaluate lim Á x sin ˜ x Æ• Ë x¯
x2
.
Solution 1ˆ Ê l = lim Á x sin ˜ x Æ• Ë x¯
Let
x2
[1• form]
1 log l = lim x 2 log x sin x →∞ x 1 log x sin x = lim x →∞ 1 x2
Let when
1 = y, x x → ∞, y → 0 1 log sin y y log l = lim 2 y→0 y y cos y sin y − 2 sin y y y = lim y→0 2y
= lim
y cot y − 1 2 y2
= lim
y − tan y 2 y 2 tan y
y→0
y→0
1 − sec 2 y y → 0 4 y tan y + 2 y 2 sec 2 y
= lim
− tan 2 y y → 0 4 y tan y + 2 y 2 sec 2 y
= lim
0 0 form
[Applying L’Hospital’s rule]
0 0 form [Applying L’Hospital’s rule]
1.50
Chapter 1
Indeterminate Forms
tan 2 y 1 y2 = − lim 2 y → 0 2 tan y + sec 2 y y 1 1 =− ⋅ 2 2 +1 =−
tan y = 1 Q lim y→0 y
1 6
example 18 Evaluate lim(cot x)sin x . x →0
Solution Let
l = lim(cot x)sin x x→0
log l = lim sin x ⋅ log(cot x) x→0
= lim x→0
log(cot x) cosec x
1 ( −cosec 2 x) x cot = lim x → 0 − cosec x cot x = lim x→0
cosec x cot 2 x
1 sin 2 x ⋅ x → 0 sin x cos 2 x
= lim
sin x x → 0 cos 2 x
= lim =0 Hence,
l = e0 = 1
[∞ 0 form]
[0 × ∞ form] ∞ ∞ form
[Applying L’Hospital’s rule]
1.7
Type 5: 1•, •0, 00 Forms
1.51
example 19 tan x
Ê 1ˆ Evaluate: lim Á ˜ x Æ0 Ë x ¯
[Summer 2016]
.
Solution Ê 1ˆ l = lim Á ˜ xÆ0 Ë x¯
Let
tan x
[•∞ form ]
Ê 1ˆ log l = lim tan x log Á ˜ xÆ0 Ë x¯ = lim xÆ0
= lim
xÆ0
È• ˘ Í • form ˙ Î ˚
log x cot x Ê 1ˆ ÁË - x ˜¯
- cosec 2 x
1 x = lim 1 xÆ0 sin 2 x = lim xÆ0
= lim
sin 2 x x sin 2 x
xÆ0
x2
◊x 2
Ê sin x ˆ = lim Á x ˜ ◊ xlim xÆ0 Ë x ¯ Æ0 = 1.0 Hence,
=0 l = e0 = 1
example 20 1 Prove that lim x→0 x
1− cos x
= 1.
[Applying L’Hospital’s rule ]
1.52
Chapter 1
Indeterminate Forms
Solution 1 l = lim x→0 x
Let
1− cos x
[∞ 0 form]
1 log l = lim(1 − cos x) log x→0 x x = lim 2 sin 2 ( − log x) x→0 2 2
= lim
x x 2 sin 2 2
x→0
x 2
2
2
x 2 ( − log x) = lim x→0 2 = lim x→0
1 ( − log x) 2 1 2 x
1 − 1 = lim x 2 2 x→0 − 3 x =
( − log x)
x sin 2 = 1 Q lim x→0 x 2 ∞ ∞ form
[Applying L’Hospital’s rule]
x2 1 lim 2 x→0 2
=0 Hence,
l = e0 = 1
example 21 sinh −1 x
Prove that lim e cosh
−1
x
= e.
x →∞
Solution sinh −1 x
Let
l = lim e cosh
−1
x
x →∞
(
= lim e x →∞
1 sinh −1 x cosh −1 x
)
[∞ 0 form]
1.7
Type 5: 1•, •0, 00 Forms
1.53
sinh −1 x ⋅ log e x →∞ cosh −1 x
log l = lim = lim
x →∞
∞ ∞ form
log ( x + x 2 + 1) log ( x + x − 1)) 2
1 +
⋅ 2 x x + x +1 2 x +1 = lim x →∞ 1 1 ⋅ 2 x 1 + 2 2 x + x −1 2 x −1 1
2
= lim
1
2
[Appplying L’Hospital’s rule]
x2 − 1
x →∞
x2 + 1 1 1− 2 x = lim x →∞ 1 1+ 2 x =1 Hence,
l = e1 = e
example 22 Prove that lim x→0
ex 1 x 1 + x
x
= 1.
Solution Let
1 x l = lim 1 + x→0 x 1 = lim 1 + x→0 x
x
x2
[∞ 0 form]
1 log l = lim x 2 log 1 + x→0 x 1 log 1 + x = lim x→0 1 x2
∞ ∞ form
1.54
Chapter 1
Indeterminate Forms
1 1 − 1 x 2 1+ x = lim x→0 2 − 3 x x = lim x→0 1 2 1 + x
[Applying L’Hospital’s rule]
x2 x → 0 2 ( x + 1)
= lim =0 l = e0
\
=1 Hence,
lim x Æ0
ex x ÈÊ 1ˆ ˘ ÍÁ 1 + ˜ ˙ x¯ ˙ ÍÎË ˚
x
=
e0 =1 1
example 23 1
Ê 1ˆ x Prove that lim Á ˜ = 1 . x Æ• Ë x ¯ Solution 1
Let
1 x l = lim x →∞ x
[00 form]
1 1 log l = lim log x →∞ x x − log x x 1 = − lim x x →∞ 1 = lim
x →∞
Hence,
=0 l = e0 = 1
∞ ∞ form [Applying L’Hospital’s rule]
1.7
Type 5: 1•, •0, 00 Forms
1.55
example 24 1 2 log(1- x )
Prove that lim(1 - x )
[Winter 2013]
= e.
x Æ1
Solution
1 2 log(1- x )
[00 form]
l = lim(1 - x )
Let
x Æ1
1 log (1 - x 2 ) x Æ1 log(1 - x ) 2x (1 - x 2 ) = lim 1 x Æ1 ( -1) (1 - x )
log l = lim
È• ˘ Í • form ˙ Î ˚
[Applying L’Hospital’s rule]
2 x(1 − x) (1 − x)(1 + x) 2x = lim x →1 1 + x =1 l=e = lim x →1
Hence,
example 25 p -x
[Summer 2016]
Evaluate lim (cos x ) 2 . p xÆ 2
Solution p
Let
-x
l = lim (cos x ) 2 xÆ
[0∞ form]
p 2
Êp ˆ log l = lim Á - x ˜ log cos x p Ë 2 ¯ xÆ
[0 ¥ • form]
2
log cos x 1 Êp ˆ ÁË 2 - x ˜¯ sin x x cos = lim p 1 xÆ (-1) 2 2 Êp ˆ ÁË 2 - x ˜¯ = lim
p xÆ 2
È• ˘ Í • form ˙ Î ˚
1.56
Chapter 1
Indeterminate Forms
- tan x 1
= lim
p xÆ 2
Êp ˆ ÁË 2 - x ˜¯
2
2
Êp ˆ = lim - Á - x ˜ tan x p Ë ¯ 2 xÆ 2
Êp ˆ ÁË 2 - x ˜¯ = lim p cot x xÆ
2
2
= lim xÆ
Êp ˆ -2 Á - x ˜ (-1) Ë2 ¯
p 2
-cosec 2 x
Êp ˆ = lim 2 Á - x˜ sin 2 x p Ë 2 ¯ xÆ 2
=0 l = e0 = 1
Hence,
example 26 Evaluate lim x→0
1 − xx . x log x
Solution Let
l = lim x→0
1 − xx x log x
1 − e x log x x → 0 x log x
= lim
1 −e x log x x ⋅ + log x x = lim x→0 1 x ⋅ + log x x x log x = lim ( −e )
0 0 form [Applying L’Hospital’s rule]
x→0
= lim( − x x ) x→0
= − lim x x x→0
Let
L = lim x x x→0
00 form
1.7
Type 5: 1•, •0, 00 Forms
1.57
log L = lim log( x x ) x→0
= lim x log x x→0
log x 1 x 1 = lim x x→0 1 − 2 x = lim( − x) = lim x→0
∞ ∞ form
[Applying L’Hospital’s rule]
x→0
=0
L = e0 = 1 l = −1
Hence,
example 27 1 − x sin x = −1 . x → 0 x log x
Prove that lim Solution Let
l1 = lim x sin x
[00 form]
x→0
log l1 = lim sin x ⋅ log x x→0
log x cosec x 1 x = lim x → 0 − cosec x cot x sin 2 x = lim − x→0 x cos x = lim x→0
∞ ∞ form [Applying L’Hospital’s rule]
2
x sin x = − lim ⋅ x→0 x cos x =0
sin x Q lxi→m0 x = 1
log l1 = 0 l1 = e0 = 1 ∴ lim x sin x = 1 x→0
...(1)
1.58
Let
Chapter 1
Indeterminate Forms
l2 = lim x log x
[0 × ∞ form]
x→0
∞ ∞ form
log x 1 x 1 = lim x x→0 1 − 2 x = lim( − x) = lim x→0
[Applying L’Hospital’ss rule]
x→0
...(2)
=0
Let
1 − x sin x x → 0 x log x
l = lim
0 0 form
1 - esin x log x x Æ0 x log x
0 0 form
= lim
1 - x sin x lim = lim x Æ 0 x log x x Æ0
[Using Eqs (1) and (2)]
Ê sin x ˆ + cos x ◊ log x ˜ - esin x log x Á Ë x ¯ 1 + log x
[Applying L’Hospital’s rule]
sin x 1 − x sin x + cos x ⋅ log x x = lim x→0 1 +1 log x
...(3)
[Dividing numerator and denominator by log x] 1 1 + cos 0 −∞ =− 1 +1 −∞ = −1
sin x = 1 Using Eq. (1) and lim x x →0
exercIse 1.5 1
1 a x + a2x + ... + anx x n = a a a 1. Prove that lim 1 ( ... ) . n 1 2 x →0 n 1 1 1 x1 x x x + + + 1 2 3 4 2. Prove that lim x →∞ 4
4x
= 24 .
1.7
1
1
3. Prove that lim(1 + sin x) 2 x = e 2 . x →0
1 1 4. Prove that lim(x)1− x = . x →1 e 2
5. Prove that lim(sin x)tan
1 . = e
x
π x→ 2
1 1
− sin x x 2 = e 6. 6. Prove that lim x →0 x x
a a 7. Prove that lim 1 + = e . x →∞ x x
x + 1 2 8. Prove that lim =e. x →∞ x − 1 x
2 x + 1 9. Prove that lim = e. x →∞ 2 x − 1
10. Prove that lim(1 + sin x)cosec x = e. x →0
11. Prove that lim(1 + sin x)cot x = e. x →0
12. Prove that lim(1 + tan x)cot x = e. x →0
1
13. Prove that lim(1 − tan x) x = x →0
1 2
14. Prove that lim(cos x) x =
1 . e
1
x →0
.
e 1
sin x x 15. Prove that lim x = 0 . x →0 a2 cosec2 bx
16. Prove that lim(cos ax)
−
=e
x →0
17. Prove that lim(cos x)cot
2
x
=
x →0
1 e
18. Prove that lim(cosec x)sin x = 1. x →0
x
2 2 19. Prove that lim 1 + = e . x →∞ x
.
2 b2
.
Type 5: 1•, •0, 00 Forms
1.59
1.60
Chapter 1
Indeterminate Forms
x 20. Prove that lim 2 − x →a a
cot( x − a )
−
1
= e a.
21. Prove that lim(tan x)cos 2 x = 1 . x→
π 4
22. Prove that lim(sec x)cot x = 1 . x→
π 2
1 23. Prove that lim x →0 x
2 sin x
= 1.
24. Prove that lim(sin x)tan x = 1. x →0
1
25. Prove that lim(1 − x n )log(1− x ) = e . x →1
26. Prove that lim x
πx tan 2
x →0
= e. 1
27. Prove that lim(e 3 x − 5x) x = e −2. x →0
3 2
28. Prove that lim(cos 2 x) x x →0
29. Prove that lim(cos x)cos
2
x
= e −6. = 1.
π x→ 2
points to remember L’Hospital’s Rule If lim f (x) = 0, lim g (x) = 0, then lim xÆa
xÆa
xÆa
f ( x) f ¢( x ) = lim g( x ) x Æ a g ¢( x )
multiple choice Questions Select the most appropriate response out of the various alternatives given in each of the following questions: 1. The value of lim x Æ•
(a) •
log x
, n > 0 is xn (b) –•
(c) 1
(d) 0
Multiple Choice Questions
tan x is x Æ0 x (b) 1
1.61
2. The value of lim (a) 0
(c) p
(1 - a )e x exists and is finite then it is equal to x Æ0 x (a) 1 (b) 0 (c) –1
(d) •
3. If lim
4. The integer p for which lim
px + sin x x2
x Æ0
(a) 0
(b) –1
(d)
1 2
is finite is
(c) 1
(d) 2
(c) log 3
(d) 0
(c) e2
(d) 0
(c) –1
(d) 2
(c) –1
(d) 2
(c) –1
(d)
1 2
(c) –1
(d)
1 2
(c) 2
(d) 3
(c) 1
(d) 5
5. The value of lim(cosh -1 x - log x ) is x Æ0
(a) log 1
(b) log 2 2
6. The value of lim x e
-x
is
x Æ•
1
(a) 1
(b) e
7. The value of lim (cos x ) xÆ
cos x
is
p 2
(a) 0
(b) 1
8. The value of lim log tan x tan 2 x is x Æ0
(a) 0
(b) 1 x
9. The value of lim(1 - x ) is x Æ0
(a) 0
(b) 1
10. The value of lim
1 + 2 + 3 + ... + x x2
x Æ•
(a) 0
(b) 1
is
1
Ê 1ˆ x 11. The value of lim Á ˜ is x Æ• Ë x ¯ (a) 0
(b) 1
12. The value of lim sin x log x is x Æ0
(a) –1
(b) 0
13. The value of lim(cos x )
cot x
[Summer 2016]
is
x Æ0
(a) 0
(b) –1
14. The value of
1 lim(e3 x - 5 x ) x x Æ0
(a) 0
(b) 1
(c) 1
(d) 2
(c) e–2
(d) e2
is
1.62
Chapter 1
15. If lim
Indeterminate Forms
sin 2 x + p sin x
x Æ0
x3
(a) –2
is finite then value of p is
(b) 1
16. The value of lim x Æ0
(a) 0
(c) –1
tan x is x (b) 1
(d) 2 [Winter 2013]
(c) p
(d) •
x 17. The value of lim x Æ0
(a) 0
(c) p
(d) •
x2 - x - 2
18. The value of lim
x2 - 4
x Æ0
(a) 4
[Winter 2013]
is x (b) ±1
(b)
[Winter 2014]
is
3 4
(c) 0
3 4
(d) –
19. The value of lim x x is
[Winter 2015]
x Æ0
(a) 1
(b) –1
(c) e
(d)
1 - n2 is n Æ• Sn
20. The value of lim (a) –2
(b) 1
1 e [Summer 2014]
(c) 2
(d) 0
1
21. The value of lim x x is
[Summer 2015]
x Æ•
(a) •
(b) – •
(c) 1
(d) 0
n
Ê 1ˆ 22. The value of lim Á 1 + ˜ is n Æ• Ë n¯ (a) 1
(b) –1
23. The value of lim x Æ•
(a) 1 x Æ6
(a) 0
(c) e
sin x is x
(b) –1
24. The value of lim
[Winter 2015]
sin( x - 6) is x-6 (b) 1
(d)
1 e [Summer 2016]
(c) 0
(d) none of these [Winter 2016]
(c) –1
(d) 0.5
Multiple Choice Questions
sin 2 x is x Æ0 x
25. The value of lim (a) 2
(b) 1
26. The value of lim p xÆ 2
(a) 0
[Summer 2017] (c) –1
(d)
cos x is p x2 (b) 1
1.63
1 2 [Summer 2017]
(c) –1
(d)
p 2
answers 1. (d) 2. (b) 3. (b) 4. (a) 5. (b) 6. (d) 7. (a) 8. (b) 9. (a) 10. (d) 11. (b) 12. (b) 13. (c) 14. (c) 15. (a) 16. (b) 17. (b) 18. (b) 19. (a) 20. (a) 21. (c) 22. (c) 23. (c) 24. (b) 25. (a) 26. (c)
CHAPTER
Improper Integrals
2
chapter outline 2.1 2.2 2.3 2.4 2.5 2.6
2.1
Introduction Improper integrals Improper Integrals of the First Kind Improper Integrals of the Second Kind Improper Integral of the Third Kind Convergence and Divergence of Improper Integrals
IntroductIon b
The definition of a definite integral
Ú f ( x) dx
requires the interval [a, b] be finite. The
a
fundamental theorem of calculus requires that f (x) be continuous on [a, b] or at least bounded. In this chapter, we will study a method of evaluating integrals that fail these requirements either because their limits of integration are infinite, or because a finite number of discontinuities exist on the interval [a, b]. Integrals that fail either of these requirements are known as improper integrals. Improper integrals cannot be computed using a normal Riemann integral.
2.2
Improper Integrals b
The integral
Ú f ( x) dx a
is called an improper integral if
(i) one or both limits of integration are infinite (ii) function f (x) becomes infinite at a point within or at the end points of the interval of integration. Improper integrals are classified into three kinds.
2.2
Chapter 2
2.3
Improper Integrals
Improper Integrals of the fIrst KInd
It is a definite integral in which one or both limits of integration are infinite, for • example,
Úe
-x
dx is an improper integral of the first kind since the upper limit of
0
integration is infinite. These integrals are evaluated as follows: (i) If f (x) is continuous on [ a, •) then •
Ú
b
f ( x ) dx = lim Ú f ( x ) dx b Æ•
a
… (1)
a
(ii) If f (x) is continuous on ( -•, b] then b
b
Ú
f ( x ) dx = lim
a Æ-•
-•
Ú f ( x) dx
… (2)
a
(iii) If f (x) is continuous on ( -•, •) then •
Ú
b
Ú f ( x) dx
f ( x ) dx = lim
a Æ-• b Æ• a
-•
0
= lim
a Æ-•
Ú
b
f ( x ) dx + lim Ú f ( x ) dx b Æ•
a
… (3)
0
The improper integral is said to converge (or exist) when the limit in RHS of (1), (2) and (3) exist (or finite). Otherwise, it is said to diverge.
example 1 •
Evaluate
Ú
1
dx .
x
1
Solution •
Ú 1
1 x
b
dx = lim Ú b Æ•
1
1
= lim 2 x b Æ•
dx
x b 1
= lim (2 b - 2) b Æ•
=•
example 2 •
Evaluate
Ú 1
1 x2
dx .
2.3
Improper Integrals of the First Kind
2.3
Solution ∞
∫ 1
b
1 1 dx = lim ∫ 2 dx 2 b →∞ x x 1 b
= lim − b →∞
1 x1
1 = lim − + 1 b →∞ b =1
example 3 •
Evaluate
Ú 0
dx x2 + 1
[Winter 2014]
.
Solution •
b
dx
Ú x2 + 1
= lim Ú b Æ•
0
0
dx x2 + 1
= lim tan -1 x b Æ•
b 0
= lim ÈÎ tan (b) - tann -1 0 ˘˚ b Æ• -1
= lim tan -1 (b) b Æ•
= tan -1 (•) =
p 2
example 4 0
Evaluate
Ú
x sin x dx .
-•
Solution 0
Ú -•
0
Ú x sin x dx a Æ-•
x sin x dx = lim
a 0
= lim - x cos x + sin x a a Æ-•
= lim (a cos a - sin a ) a Æ-•
= -•
[Q sin a and cos a oscillate between ± 1]
2.4
Chapter 2
Improper Integrals
example 5 0
∫e
Evaluate
2x
dx.
−∞
Solution 0
0
Úe
2x
Úe a Æ-•
dx = lim
-•
2x
dx
a
e2 x = lim a Æ-• 2
0
a
Ê1 1 ˆ = lim Á - e2 a ˜ a Æ-• Ë 2 ¯ 2 1 -0 2 1 = 2 =
example 6 •
Evaluate
Ú
1
1
dx = lim
-• 1 +
x2
dx .
Solution •
Ú
-• 1 +
0
x
2
1
Ú a Æ-• 1 + x 2
b
dx + lim Ú b Æ•
a
1
0 1+
0
x2
= lim tan -1 x + lim tan -1 x a Æ-•
= lim (0 - tan a Æ-•
p p + 2 2 =p =
example 7 •
Evaluate
Ú -•
e x dx .
a
-1
b Æ•
dx
b 0
-1
a ) + lim (tan b - 0) b Æ•
2.3
Improper Integrals of the First Kind
Solution •
Ú
0
e x dx = lim
a Æ-•
-•
b
x x Ú e dx + lim Ú e dx b Æ•
a
= lim e x a Æ-•
0
0
+ lim e x
a
b Æ•
b 0
a
= lim (1 - e ) + lim (eb - 1) a Æ-•
b Æ• b
= (1 - 0) + lim (e - 1) b Æ•
=•
example 8 •
1
Ú
Evaluate
e + e- x
dx .
x
-•
Solution ∞
∞
1 ex d x = ∫ e x + e− x ∫ e2 x + 1 dx −∞ −∞ b
0
ex ex dx d x + lim a →− ∞ ∫ e 2 x + 1 b →∞ ∫ e 2 x + 1 a 0
= lim
Putting u = e x , du = e x dx , ex
du
Ú e2 x + 1 dx = Ú u2 + 1 = tan •
Ú
-•
1 x
e +e
-x
-1
u = tan -1 e x
dx = lim tann -1 e x a Æ-•
0 a
+ lim tan -1 e x b Æ•
b 0
pˆ Êp ˆ Ê = lim Á - tan -1 e a ˜ + lim Á tan -1 eb - ˜ a Æ-• Ë 4 ¯ bÆ• Ë 4¯ Êp ˆ Êp pˆ = Á - 0˜ + Á - ˜ Ë4 ¯ Ë 2 4¯ =
example 9 •
Evaluate
Ú
x
2 2 -• (1 + x )
dx .
p 2
2.5
2.6
Chapter 2
Improper Integrals
Solution •
0
x
Ú
2 2 -• (1 + x )
dx = lim
a Æ-•
= lim
a Æ-•
x
Ú (1 + x 2 )2
dx + lim Ú
0
2x 2x dx + lim Ú (1 + x 2 )-2 dx b Æ• 2 2 0
b Æ•
a
2 -2 Ú (1 + x ) a
= lim a Æ-•
b
x
2 2 0 (1 + x ) b
0
1
dx
2(1 + x 2 ) a
+ lim b Æ•
1
b
2(1 + x 2 ) 0
È n [ f ( x)]n +1 ˘˙ ÍQ Ú ÈÎ f ( x )˘˚ f ¢( x ) dx = n +1 ˚ Î È 1 ˘ È 1 1 1˘ = lim Í - + + ˙ ˙ + lim Í 2 2 a Æ-• Î 2 2(1 + a ) ˚ bÆ• Î 2(1 + b ) 2 ˚ 1 1 =- + 2 2 =0
example 10 •
Evaluate
Ú 0
dv -1
2
(1 + v )(1 + tan v)
[Winter 2016]
.
Solution 1 1 + v 2 dv Ú (1 + v2 )(1 + tan -1 v) = blim Ú Æ• 1 + tan -1 v 0 0
•
b
dv
= lim log (1 + tan -1 v) b Æ•
b 0
˘ È f ¢( v ) dv = log | f (v) |˙ ÍQ Ú f (v) ˚ Î
= lim È log 1 + tan -1 b - log 1˘ ˚ b Æ• Î = log(1 + tan -1 •) - 0 Ê pˆ = log Á 1 + ˜ Ë 2¯
2.3
Improper Integrals of the First Kind
2.7
example 11 •
Evaluate
Ú
x+3
2 2 ( x - 1)( x + 1)
dx .
Solution •
b
x+3
( -2 x - 1) ˘ È 2 + 2 Ú ( x - 1)( x 2 + 1) dx = blim Ú Í ˙ dx [ Using partial fraction expansion ] Æ• Î x - 1 x +1 ˚ 2 2 b b Èb 2 ˘ 2x 1 = lim Í Ú dx ˙ dx - Ú 2 dx - Ú 2 b Æ• Í x - 1 ˙˚ 2 x +1 2 x +1 Î2 b˘ È = lim Í 2 log( x - 1) - log( x 2 + 1) - tan -1 x ˙ 2˚ b Æ• Î
˘ È f ¢( x ) dx = log f ( x ) ˙ ÍQ Ú f ( x) ˚ Î b˘ È ( x - 1)2 -1 Í = lim log 2 - tan x ˙ b Æ• Í ˙ x +1 2˚ Î
È ˘ 1 (b - 1)2 = lim Í log 2 - tan -1 b - log + tan -1 2 ˙ b Æ• Í 5 b +1 Î ˚˙ 2 È ˘ Ê 1ˆ Í ˙ ÁË 1 - b ˜¯ = lim Í log - tan -1 b + log 5 + tan -1 2 ˙ ˙ 1 b Æ• Í 1+ 2 ˙ Í b ÍÎ ˚˙
= log 1 - tan -1 • + log 5 + tan -1 2 = 0=-
p + log 5 + tan -1 2 2
p + log 5 + tan -1 2 2
example 12 Prove that p-integral p £ 1.
•
Ú 1
1 xp
dx converges when p > 1 and diverges when
2.8
Chapter 2
Improper Integrals
Solution •
1
Ú xp
b
dx = lim Ú b Æ•
1
1
1 xp
dx b
x - p +1 = lim , p π1 b Æ• - p + 1 1
È b - ( p -1) - 1 ˘ = - lim Í ˙, p π 1 b Æ• Í Î p - 1 ˙˚
Case (I) When p > 1 •
Ú 1
È 1 ˘ Í b p -1 - 1 ˙ dx = - lim Í ˙ b Æ• Í p - 1 ˙ xp ÍÎ ˙˚ È1 ˘ Í • - 1˙ = -Í ˙ Í p -1 ˙ ÍÎ ˚˙ 1 = [finite] p -1 1
Case (II) When p < 1 •
Ú 1
È b1- p - 1 ˘ lim d = x Í ˙ b Æ• Í p - 1 ˙ xp Î ˚ È • - 1˘ = -Í ˙ Î p -1˚ 1
= -•
Case (III) When p = 1 •
1
Ú x dx =
•
log x 1 = •
1
•
Hence, p integral
1
Ú xp
dx converges when p > 1 and diverges when p £ 1.
1
exercIse 2.1 Evaluate the following improper integrals: ∞
1.
1
∫ x dx 1
[ans.: •]
2.4
Improper Integrals of the Second Kind
π Ans.: 2
∞
1 ∫0 1 + x 2 dx
2.
1 Ans.: − 2
−1
2 dx 5 x −∞
∫
3.
∞
1 Ans.: log 2
1
∫ x log
4.
2
x
2
∞
∫e
5.
−x
2.9
1 Ans.: 2
sin xdx
0 ∞
∫x e 2
6.
−x
dx
[ans.: 2]
0 ∞
∫ | x |e
7.
− x2
dx
[ans.: 1]
−∞ ∞
1
∫x
8.
x −1 2
1
π Ans.: 2
dx
∞
π Ans.: 2
1 dx 2 x + x +5 2 −∞
∫
9.
∞
∫
10.
e−
dx x
1
2.4
2 Ans.: e
x
Improper Integrals of the second KInd
It is a definite integral in which integrand become infinite (or unbounded or discontinuous) at one or more points within or at the end points of the interval of integration, for example, 1
(i)
1
Ú x dx
is an improper integral of the second kind as
0
2
(ii)
Ú
1 2
1 is not continuous at x = 0. x
dx is an improper integral of the second kind because
x -1 continuous at x = –1 and x = 1. -2
1 2
x -1
is not
These integrals are evaluated as follows: (i) If f (x) is unbounded at x = a then b
b
Ú f ( x)dx = clim Ú f ( x)dx Æa a
c
… (1)
2.10
Chapter 2
Improper Integrals
(ii) If f (x) is unbounded at x = b then b
Ú
c
f ( x )dx = lim Ú f ( x )dx cÆb
a
… (2)
a
(iii) If f (x) is unbounded at x = a and x = b then b
c2
0
Ú f ( x) dx = clim Ú f ( x)dx + clim Ú 1 Æa 2 Æb a
c1
f ( x )dx
… (3)
0
The improper integral is said to converge (or exist) when the limit in RHS of (1), (2) and (3) exist (or finite). Otherwise, it is said to diverge.
example 1 3
Evaluate
Ú 0
1 3- x
[Summer 2014, 2017]
dx .
Solution The integrand
1 3- x
is unbounded at x = 3. 3
Ú 0
c
1 3- x
dx = lim Ú c Æ3
0
1 3- x
dx
= lim -2 3 - x c Æ3
c 0
= lim( -2 3 - c + 2 3 ) c Æ3
=2 3
example 2 5
Check the convergence of
1
Ú x 2 dx.
[Summer 2016]
0
Solution The integrand
1 x2
is unbounded at x = 0. 5
1
5
1
Ú x 2 dx = clim Ú dx Æ0 x 2 0
c
2.4
Improper Integrals of the Second Kind
= lim cÆ0
1 x
2.11
5
c
È 1 Ê 1ˆ ˘ = lim Í- - Á - ˜ ˙ cÆ0 Î 5 Ë c ¯ ˚ =• Hence, the integrand is divergent.
example 3 1
dx
Ú0 1 - x .
Check the convergence of
If convergent, then evaluate the [Winter 2016]
same. Solution The integrand 1
1 is unbounded at x = 1. 1- x dx
c
1
dx Ú0 1 - x = clim Ú Æ1 0 1 - x = lim - log(1 - x ) c Æ1
c 0
= lim [ - log(1 - c) + log1] c Æ1
= –log (1 – 1) =• Hence, the integrand is divergent.
example 4 3
Check the convergence of
dx
Ú
[Summer 2015]
.
9- x
0
2
Solution The integrand
1 9 - x2
is unbounded at x = 3. 3
Ú 0
c
dx 9- x
2
= lim Ú c Æ3
0
dx 9 - x2
2.12
Chapter 2
Improper Integrals
Ê xˆ = lim sin -1 Á ˜ c Æ3 Ë 3¯
c
0
˘ È Ê cˆ = lim Ísin -1 Á ˜ - sin -1 0 ˙ c Æ3 Î Ë 3¯ ˚ = sin -1 (1) - 0 p = [finite] 2 Hence, the integrand is convergent.
example 5 p 2
Evaluate
Ú sec xdx . 0
Solution The integrand sec x is not continuous at x = p 2
p . 2
c
sec x dx Ú sec x dx = clim p Ú Æ 0
2
0
= lim log sec x + tan x p cÆ 2
c 0
= lim log sec c + tan c cÆ
p 2
= log sec
p p - tan 2 2
=•
example 6 3
Evaluate
dx
Ú 0
(x
2 - 1) 3
[Summer 2015]
.
Solution 1
The integrand (x
2 - 1) 3
is unbounded at x = 1.
2.4
3
Ú 0
c1
dx (x
2 - 1) 3
Ú c1 Æ1 0
= lim
c1 Æ1
3
dx
= lim
(x
(x
2 - 1) 3
1 - 1) 3
1 3
Improper Integrals of the Second Kind
Ú c2 Æ1
dx
+ lim
2
c2
( x - 1) 3
c1
+ lim
(x
3
1 - 1) 3
1 3
c2 Æ1
0
c2
1 1˘ È È 3 = lim 3 Í(c1 lim 3 Í(3 - 1) - (c2 - 1) 3 ˙ c1 Æ1 Í c2 Æ1 Í ˙ ˙˚ Î ˚ Î 1˘ È 1 ˘ È = 3 Í0 - ( -1) 3 ˙ + 3 Í 2 3 - 0 ˙ ˙˚ ÍÎ ˚˙ ÎÍ È 1 ˘ = 3 Í2 3 + 1˙ ÍÎ ˙˚ 1 - 1) 3
1˘ - ( -1) 3 ˙ +
example 7 1
Evaluate
Ú -1
1 2 x3
dx .
Solution 1 The integrand
2 x3
is unbounded at x = 0. 1
Ú
1
2 -1 3 x
c1
Ú c1 Æ 0
dx = lim
= lim
c1 Æ 0
= lim
1
2 -1 3 x
1
Ú c2 Æ 0
dx + lim
c2
c 1 1 3x 3 -1
È 1 Í3c13
c1 Æ 0 Í
+ lim
c2 Æ 0
1 2 x3
1 3x 3
dx
1
c2 1˘ È lim Í3 - 3c23 ˙ ˙˚ ˙˚ c2 Æ0 ÍÎ
1˘ - 3( -1) 3 ˙ +
Î 1
= [0 - 3( -1) 3 ] + [3 - 0] =6
2.13
2.14
Chapter 2
Improper Integrals
example 8 5
Evaluate
1
∫ ( x − 2)
2
dx .
0
Solution 1 is unbounded at x = 2. ( x − 2) 2
The integrand
5
1
Ú ( x - 2 )2
c1
dx = lim
c1 Æ 2
0
5
1
Ú ( x - 2 )2
dx + lim
c2 Æ 2
0
1
Ú ( x - 2 ) 2 dx
c2
5
c
= lim c1 Æ 2
1 1 1 + lim c Æ 2 x-2 0 x-2 2
c2
Ê Ê 1 1 1ˆ 1 ˆ = lim Á - ˜ + lim Á - + c1 Æ 2 Ë c1 - 2 2 ¯ c2 Æ2 Ë 3 c2 - 2 ˜¯ = -• + •,
indeterminate form
Hence, no conclusion can be made about the value of the integral.
example 9 a
Evaluate
1
∫
a − x2 2
−a
dx .
Solution 1
The integrand
2
a - x2 a
Ú -a
1 a2 - x2
is unbounded at x = ± a .
0
Ú c1 Æ- a
c1
c2
1
dx = lim
Ú c2 Æ a
a2 - x2
a2 - x2
0
0
= lim sin -1 c1 Æ- a
x a
1
dx + lim
+ lim sin -1 c1
c2 Æ a
x a
c2 0
dx
2.4
Improper Integrals of the Second Kind
c ˘ c È È ˘ = lim Ísin -1 0 - sin -1 1 ˙ + lim Ísin -1 2 - sin -1 0 ˙ c1 Æ- a Î a ˚ c2 Æ a Î a ˚ Ê aˆ Ê aˆ = - sin -1 Á - ˜ + sin -1 Á ˜ Ë a¯ Ë a¯ = sin -1 1 + sin -1 1 = 2 sin -1 1 = 2◊
p 2
=p
example 10 π 2
∫
Evaluate
0
sin x dx. 1 − cos x
Solution The integrand is unbounded at x = 0. p 2
Ú 0
sin x 1 - cos x
p 2
dx = lim Ú cÆ0
c
sin x 1 - cos x
p 2
= lim Ú (1 - cos x ) cÆ0
= lim
cÆ0
= lim
cÆ0
-
dx 1 2
sin x dx
c p 1 2 (1 - cos x ) 2
1 2
c
p 1 2 2(1 - cos x ) 2 c
ÈQ Ú[ f ( x )]n f ¢( x )dx ˘ Í ˙ [ f ( x )]n +1 ˙ Í Í = n +1 ˙ Î ˚
2.15
2.16
Chapter 2
Improper Integrals
1˘ È = lim 2 Í1 - (1 - cos c) 2 ˙ cÆ0 Í ˙˚ Î =2
2.5
Improper Integral of the thIrd KInd
It is a definite integral in which one or both limits of integration are infinite, and the integrand become infinite at one or more points within or at the end points of the interval of integration. Thus, it is a combination of the first kind and the second kind. •
For example,
1
Ú x 2 dx
is an improper integral of the third kind as the upper limit of
0
integration is infinite and integrand
1
is infinite at x = 0.
x2
example 1 •
Evaluate the improper integral
Ú 0
1 x2
dx.
Solution •
1
1
c2
1
1
Ú x 2 dx = clim Ú x 2 dx + clim Ú x 2 dx 1 Æ0 2 Æ• 0
1
c1
= lim c1 Æ 0
1 x
1
c
+ liim c1
c2 Æ•
1 2 x1
Ê Ê 1 ˆ 1ˆ = lim Á -1 + ˜ + lim Á - + 1˜ c1 Æ 0 Ë c1 ¯ c2 Æ• Ë c2 ¯ = • +1 =•
2.6
Convergence and Divergence of Improper Integrals
2.17
exercIse 2.2 Evaluate the following improper integrals: 1
1.
1
∫ x dx 0
[ans.: •] 1
2.
∫
1
dx
x
0
[ans.: 2] 1
3.
1
∫x
dx
2
0
[ans.: •]
1
4.
∫ log xdx 0
[ans.: –1] 9
5.
0
1
6.
1
∫
(x − 1) 1
∫ 1− x
4
2 3
dx [ans.: 9]
dx
0
[ans.: •] 2
7.
∫ 0
1 x(2 − x)
dx [ans.: p ]
2.6
convergence and dIvergence of Improper Integrals
(i) Direct Comparison Test (a) If f (x) and g(x) are two continuous functions for x ≥ a such that 0 £ f ( x ) £ g( x ), then ∞
∫
∞
∞
∫ g ( x)dx converges and
∫
f ( x)dx ≤ ∫ g ( x)dx .
a
a
a
∞
f ( x)dx converges, if
a
(b) If f (x) and g(x) are two continuous functions for x ≥ a such that f ( x) ≥ g ( x), ∞
then
∫ a
∞
f ( x)dx diverges, if
∫ g ( x)dx a
diverges.
2.18
Chapter 2
Improper Integrals
(ii) Limit Comparison Test If f (x) and g (x) are two continuous functions for x ≥ a such that f ( x) > 0, g ( x) > 0
f ( x) = l , 0 < l < ∞ then x →∞ g ( x ) simultaneously. and lim
∞
∫
∞
f ( x) dx and
a
∫ g ( x) dx
both converge or diverge
a
Note: Convergence of improper integral of second kind can be tested by direct comparison test and limit comparison test similarly.
example 1 •
Ú
Test the convergence of the improper integral
x2
1
Solution f ( x) =
cos x
dx .
cos x 1 and let g ( x) = 2 2 x x
cos x 1 ≤ 2 x2 x
for
∞
∞
1
1
∫ g ( x) dx = ∫
x ≥1
[ Q cos x ≤ 1] b
1 1 dx = lim ∫ 2 dx 2 →∞ b x x 1 b
= lim − b →∞
1 x1
1 = lim − + 1 b →∞ b =1 ∞
Thus,
1
∫x
2
dx is convergent.
1
∞
By direct comparison test,
∫ 1
cos x is convergent. x2
example 2 •
Test the convergence of the improper integral
Úe
- x2
1
Solution 2
f ( x) = e − x and let g ( x) = e − x 2
e− x ≤ e− x
for
Q x 2 ≥ x x ≥1 2 − x ≤ − x
dx .
2.6
Convergence and Divergence of Improper Integrals
•
•
-x Ú g( x) dx = Ú e dx 1
1 b
= lim Ú e - x dx b Æ•
1
= lim (e -1 - e - b ) b Æ•
= e -1 1 = e
•
Thus,
Úe
-x
dx is convergent.
1 •
By direct comparison text,
Úe
- x2
dx is convergent.
1
example 3 •
Test the convergence of the improper integral
Ú 4
3x + 5 x4 + 7
Solution f ( x) =
1 3x + 5 and let g ( x) = 3 4 x x +7
3x + 5 4 f ( x) lim = lim x + 7 1 x Æ• g( x ) x Æ• x3 = lim
3x 4 + 5x3
x4 + 7 5 3+ x = lim 7 x Æ• 1+ 4 x =3 x Æ•
•
•
Ú g ( x ) dx =
Ú x 3 dx
4
1
4
b
= lim Ú b Æ•
4
1 x3
dx
dx .
2.19
2.20
Chapter 2
Improper Integrals
= lim b Æ•
= 0+
∞
1
∫x
3
2 x2 1
4
2( 4 2 )
1 32
=
Thus,
b
1
dx is convergent.
4
∞
By limit comparison test,
3x + 5 is convergent. 4 +7 4
∫x
example 4 1
Test the convergence of the improper integral
Ú 0
1 2
x + x
dx.
Solution f ( x) =
1 x2 + x
1
and let g ( x) =
x 1
1
x]
x
1
1
∫ g ( x) dx = ∫ 0
0
1 x
1
dx = lim ∫ c→0
c
1
dx
x 1
= lim 2 x c c→0
= lim ( 2 − 2 c ) c→0
=2 1
Thus,
∫ 0
1
dx is convergent.
x 1
By direct comparison test,
∫x 0
1 2
+ x
dx is convergent.
example 5 1 − e− x dx. Test the convergence of the improper integral ∫ x3 0 1
2.6
Convergence and Divergence of Improper Integrals
2.21
Solution f ( x) =
1 1 − e− x and let g ( x) = 3 x x3
1 − e− x 3 f ( x) lim = lim x = lim (1 − e − x ) = 1 x →∞ g ( x ) x →∞ x →∞ 1 3 x 1 1 1 ∫0 g ( x) dx = ∫0 x3 dx 1
= lim ∫ c→0
c
1 dx x3 1
1 c→0 2x2 c 1 1 = lim − + 2 c→0 2 2c =∞ = lim −
1
Thus
1
∫x
dx is divergent.
3
0
1 − e− x ∫0 x3 dx is divergent. 1
By limit comparison test,
exercIse 2.3 Test the convergence of the following improper integrals. ∞
1.
2.
3.
sin2 x ∫1 x 2 dx ∞
∫
1 sin2 x
1
x
∞
x −1
∫ 2
x 6 + 16
∫ 3x 1
∞
dx [ans.: convergent]
2
∞
4.
[ans.: convergent]
4
dx
x dx + 5x 2 + 1
log x dx 5. ∫ −x 1 x +e
[ans.: divergent]
[ans.: convergent] [ans.: divergent]
2.22
Chapter 2
Improper Integrals
∞
6.
2 + sin x dx 2 −∞ x + 1
∫
1
7.
∫ ( x + 1) 0
1 1− x2
[ans.: convergent] dx [ans.: convergent]
π 2
8.
∫ log sin x dx
[ans.: convergent]
0
π 2
9.
e − x cos x dx x 0
∫ π 2
10.
∫ 1
tan x x
3 2
[ans.: divergent]
dx [ans.: divergent]
points to remember Improper Integrals of the First Kind It is a definite integral in which one or both limits of integration are infinite.
Improper Integrals of the Second Kind It is a definite integral in which integrand become infinite (or unbounded or discontinuous) at one or more points within or at the end points of the interval of integration.
Improper Integral of the Third Kind It is a definite integral in which one or both limits of integration are infinite, and the integrand become infinite at one or more points within or at the end points of the interval of integration.
Convergence and Divergence of Improper Integrals (i) Direct Comparison Test (a) If f (x) and g(x) are two continuous functions for x ≥ a such that 0 £ f ( x ) £ g( x ), then ∞
∫ a
∞
∞
∫ g ( x)dx converges and
∫
f ( x)dx ≤ ∫ g ( x)dx .
a
a
a
∞
f ( x)dx converges, if
Multiple Choice Questions
2.23
(b) If f (x) and g(x) are two continuous functions for x ≥ a such that f ( x) ≥ g ( x), ∞
then
∫ f ( x)dx
∞
∫ g ( x)dx
diverges, if
a
diverges.
a
(ii) Limit Comparison Test If f (x) and g (x) are two continuous functions for x ≥ a such that f ( x) > 0, g ( x) > 0
f ( x) = l , 0 < l < ∞ then x →∞ g ( x ) simultaneously. and lim
∞
∫
∞
f ( x) dx and
a
∫ g ( x) dx
both converge or diverge
a
multiple choice Questions Select the most appropriate response out of the various alternatives given in each of the following questions: 0
1. The value of
1
Ú
3- x
-•
(a) 0
(b) • •
2. The value of
dx is
1
Ú x 2 dx
(c) 3
(d) 1
(c) •
(d) 2
(c) 3
(d) •
(c) 3
(d) •
(c) 2
(d) •
is
0
(a) 0
(b) 1 •
3. The value of
Ú (1 + 2 x)e
-x
dx is
0
(a) 0
(b) 1 1
4. The value of
1
Ú 10 + 2 z dz
is
-5
(a) 0
(b) 1 1
5. The value of
Ú
6 - y dy is
-•
(a) 0
(b) 1 •
6. The value of
9
Ú (1 - 3z)4 dz
is
2
(a) 0
(b)
3 125
(c)
1 125
(d) •
2.24
Chapter 2
Improper Integrals
•
7. The value of
Ú -•
6 x3 ( x 4 + 1)2
(a) 0
(b) 1 •
8. The value of
dx is
1
Ú x 2 + x - 6 dx
(c) 6
(d) •
(c) 6
(d) •
(c) 2
(d) •
(c) p
(d) •
is
1
(a) 0
(b) 1 1 0
9. The value of
ex
Ú
dx is
x2
-•
(a) 0
(b) 1 p
10. The value of Ú sec 2 x dx is 1
(a) 0
(b) 1 •
11. The value of
dx
Ú x2 + 4
is
0
(a) 0
(b) 1 3
12. The value of
dx
Ú
(a) 0
p 2
(d)
(c)
p 2
(d) •
(c) –2
(d) •
is
9 - x2
0
(b) 1 •
13. The value of
Úx
p 4
(c)
2 -x
e dx is
0
(a) 0
(b) 2 0
14. The value of
Ú2
5x
dx is
-•
(a) 0
(b) •
15. The value of
Ú -•
(a)
p 10
1 5 log 2
dx 25 + 4 x 2 (b)
p 5
2 5 log 2
(c)
1 log 2
(d)
(c)
p 2
(d) p
is
Multiple Choice Questions
•
16. The value of
1
Ú x 2 dx
2.25
[Winter 2015]
is
1
(a) 1 17. The value of (a) p 18. The value of (a) 0
(b) 0 •
(c) –1
1
Ú0 1 + x 2 dx (b) • -x
Ú0 e
(d) does not exist [Winter 2016]
is
p 2
(c) 0
(d) 1 [Summer 2017]
cos 2 x dx is
(b) -
1 5
(c)
1 5
(d)
2 5
answers 1. (b) 2. (c) 3. (c) 4. (d) 5. (d) 6. (c) 7. (a) 8. (d) 9. (b) 10. (d) 11. (d) 12. (c) 13. (b) 14. (b) 15. (a) 16. (a) 17. (b) 18. (c)
3 Gamma and CHAPTER
Beta Functions chapter outline 3.1 3.2 3.3 3.4 3.5 3.6
3.1
Introduction Gamma Function Properties of Gamma Function Beta Function Properties of Beta Functions Beta Function as Improper Integral
IntroductIon
There are some special functions which have importance in mathematical analysis, functional analysis, physics or other applications. In this chapter, we will study two special functions, gamma and beta functions. The beta function is also called the Euler integral of the first kind. The gamma function is an extension of the factorial function to real and complex numbers and is also known as Euler integral of the second kind. Gamma function is a component in various probability distribution functions. It also appears in various areas such as asymptotic series, definite integration, number theory, etc.
3.2
Gamma functIon
Gamma function is defined by the improper integral by n . Hence,
∞
∫
∞
0
e - x x n -1dx, n > 0 and is denoted
n = ∫ e - x x n -1dx, n > 0 0
Alternate form of gamma function ∞ 2 n = 2 ∫ e - x x 2 n -1dx 0
3.2
Chapter 3
Gamma and Beta Functions
Proof: By definition, ∞
n = ∫ e - x x n -1dx 0
Let x = t2,
dx = 2t dt
∞
n = ∫ e - t . t 2 n - 2 . 2t dt 2
0
∞
= 2 ∫ e - t ⋅ t 2 n -1 dt 2
0
Changing the variable t to x,
∞
n = 2 ∫ e - x . x 2 n -1dx 2
0
3.3
ProPertIes of Gamma functIon
Property 1:
n +1 = n n
Proof:
n + 1 = ∫ e - x x n dx
∞
0
Integrating by parts, n + 1 = -e - x x n
∞ 0
∞
- ∫ (-e - x ) nx n -1dx 0
∞
= n ∫ e - x x n-1dx 0
=n n
n +1 = n n Hence, This is known as recurrence or reduction formula for Gamma function. Note: (i) n + 1 = n ! if n is a positive integer
(ii)
n +1 = n n
(iii)
n=
(iv)
n 1- n =
if n is a positive real number
n +1 n
if n is a negative fraction
π sin nπ
Property 2: 1 = π
2
Proof: By alternate form of Gamma function, 1
2. - 1 ∞ 2 1 = 2 ∫ e - x x 2 dx = 2 0 2
∫
∞
0
2
e - x dx
2 2 1. 1 = 2 ∫ e - x dx . 2 ∫ e - y dy 0 0 2 2
∞
= 4∫
∞
0
∞
∫
∞
0
e-( x
2
+ y2 )
Changing to polar coordinates, x = r cosq,
dx dy
y = r sinq dx dy = r dr dq
3.3
x=0 y=0
to to
x→∞ y→∞
This shows that the region of integration is the first quadrant. Draw an elementary radius vector in the region q=p which starts from the pole and extends up to ∞. 2 Limits of r r=0 to r→∞ π θ= Limits of q q=0 to 2
r
ä
Limits of x Limits of y
3.3
Properties of Gamma Function
∞
π 2 1 1 2 ∞ ⋅ = 4 ∫ ∫ e − r . r d r dθ 0 2 2 0
π 2
∞
1 2 = 4 ∫ dθ . ∫ − e− r (− 2r ) d r 2 0 0 =
4 θ −2
= −2 . =π
π 2 0
e− r
2
0
π ( 0 − 1) 2
Example 1
5 Find the value of − . 2 Solution
5 − = 2
n +1 n −
5 +1 2 3 2 =− − 5 5 2 − 2
2 =− . 5
4 . = 15
−
−
q =0
Fig. 3.1
∞
1 = π 2
n=
O r=0
3 +1 4 1 2 = − 3 15 2 − 2
1 +1 8 1 8 π 2 =− =− 1 15 2 15 − 2
∵ e f ( r ) . f ′ (r )dr = e f ( r ) ∫
3.4
Chapter 3
Gamma and Beta Functions
example 2 Given
8 12 = 0.8935 , find the value of - . 5 5
Solution: n +1 n
n=
2 12 7 - +1 +1 - +1 25 5 5 . 5 . 5 == 7 2 12 12 84 5 5 5
-
12 = 5
3 +1 625 8 125 . 5 625 ===(0.8935) = -1.108 3 504 5 504 168 5
example 3 ∞ - x3
∫0 e
Evaluate
dx .
Solution 1
1 -2 x3 = t, x = t 3 , dx = t 3 dt 3 x = 0, t=0
Let when when
x → ∞,
∫
∞
0
e - x dx = ∫ 3
∞
0
t→∞ 2 1 -1 1 1 1 1 ∞ e - t . t 3 dt = ∫ e - t t 3 dt = 0 3 3 3 3
example 4 ∞ - x
∫0 e
Evaluate
1 x4
dx .
Solution x = t , x = t2, dx = 2t dt x = 0, t=0 x → ∞, t→∞
Let when when
∫
∞
0
1
1
∞
e - x x 4 dx = ∫ e - t ( t 2 ) 4 2t dt 0
∞
3
= 2 ∫ e - t t 2 dt = 2 0
5 3 1 1 3 =2. . = p 2 2 2 2 2
3.3
Properties of Gamma Function
example 5 Evaluate
∞
∫0
( x 2 + 4)e -2 x dx . 2
Solution -
1
Let when when
∫
∞
0
( x + 4) e 2
dx = ∫
-2 x 2
∞
0
=
= =
-
1
t -t t 2 dt + 4 e . 2 2 2
1 4 2
∫
∞
0
1
e - t t 2 dt +
2 2
∞
∫0
x ne -
ax
1
π+
8 2
2 2
π =
Solution x = 0, x → ∞,
∫
∞
n
xn e-
ax
0
∞ t2 2t dx = ∫ e - t ⋅ dt 0 a a
=
=
2 a
n +1
2 a n +1
example 7 Evaluate
t2 2t , dx = dt a a t=0 t→∞
ax = t , x =
when when
∞
∫0
x e
- x2
dx . ∫
∞
0
e- x
2
x
e-t t
-
1 2
dt
0
3 2 1 1 .1 1 2 1 + = + 4 2 2 2 2 4 2 2 2 2 2
dx .
Let
∫
∞
1
example 6 Evaluate
1
t 1 1 - 12 t 2 2x = t, x = 2 , dx = ⋅ t dt = dt 2 2 2 2 2 x = 0, t=0 x → ∞, t→∞ 2
dx .
∫
∞
e - t t 2 n +1dt
0
2n + 2
17 π 8 2
3.5
3.6
Chapter 3
Gamma and Beta Functions
Solution 1
x2 = t, x = t 2 , dx =
Let
x = 0,
when when
t=0
x → ∞,
∫
∞
x e - x dx . ∫ 2
0
e- x
2
1 - 12 t dt 2
t→∞
∞ e-t 1 -1 1 -1 dx = ∫ t e . t 2 dt . ∫ 1 . t 2 dt 0 0 2 2 x t4
∞
0
1 ∞ 4 -t
=
1 ∞ - t - 14 . ∞ - t - 34 e t dt ∫ e t d t 0 4 ∫0
=
1 3. 1 1 1 1 = 14 4 4 4 4 4
=
π 1 1 π = .π 2 = ⋅ 4 sin π 4 2 2 4
example 8 Evaluate
∞
∫0
e- x
3
x
∞
dx . ∫ x 4 e - x dx . 6
0
Solution 1 1 -2 x3 = t, x = t 3 , dx = t 3 dt 3 x = 0, t=0
Let when when
x → ∞,
∫
∞
0
e- x
3
x
∞
dx . ∫ x 4 e - x dx = ∫ 6
0
∞
0
t→∞ e - t . 1 - 32 . ∞ 34 - t 2 . 1 - 32 t dt ∫ t e t dt 1 0 3 3 6 t
=
1 ∞ - t - 56 . ∞ - t 2 32 e t dt ∫ e t d t 0 9 ∫0
=
∞ 2 2 -1 1 1 1 ⋅ ⋅ 2 ∫ e - t t 6 dt 0 9 6 2
=
1 1 1 5 ⋅ 9 6 2 6
5
=
1 1 1 π 1 p 1- = ⋅ = p 18 6 6 18 9 sin 6
Q 2 ∞ e - x2 x 2 n -1dx = n ∫0
3.3
Properties of Gamma Function
3.7
example 9 1
∫0 (log x)
Evaluate
5
dx .
Solution log x = - t, x = e-t, dx = - e-t dt x = 0, t→∞ x = 1, t=0
Let when when
∫
1
0
0
(log x)5 dx = ∫ ( -t )5 ( -e - t ) dt ∞
∞
= - ∫ e - t t 5 dt 0
= - 6 = -120
example 10 4
1 Evaluate ∫0 x log dx . x 1
3
Solution 4
1 1 1 3 3 ∫0 x log x dx = ∫0 x . 4 log x dx 1
1 1 = 4 ∫ x 3 log dx 0 x
1 1 log = t , = et , x = e - t , dx = -e - t dt x x x = 0, t→∞
Let when
x = 1,
when
∫
1
0
t=0
4
1 x 3 log dx = 4∫ e -3t t ( -e - t ) dt x ∞ 0
∞
= 4 ∫ e -4t t 2 -1dt 0
= 4⋅ =
1 4
2 ( 4) 2
∞ - kx n -1 n Q ∫0 e x dx = n k
3.8
Chapter 3
Gamma and Beta Functions
example 11 1
∫0
Evaluate
dx 1 x log x
.
Solution
∫
1
0
1
dx
=∫ x
1 x log x
0
-
1 2
1 log x
-
1 2
dx
… (1)
1 1 log = t , = et , x = e-t, dx = -e-t dt x x x = 0, t→∞
Let when
x = 1,
when
∫
1
0
1 1 x log x
t=0 -
1
1
-1
dx = ∫ ( e - t ) 2 . t 0
-
∞
∞
=∫ e
-
t 2
1 2
( - e - t ) dt
t 2 dt
0
=
1 2 1 2
∞ - kx n -1 n Q ∫0 e x dx = n k
1 2
= 2π
example 12 ∞
Evaluate
∫0
xa dx . ax
Solution Let
ax = et, x log a = t, dx =
when when
x = 0, x → ∞,
1 dt log a
t=0 t→∞ a ∞ xa ∞ t . 1 . 1 x d = ∫0 a x ∫0 log a et log a dt =
1 (log a ) a +1
∫
∞
0
e - t t a dt
3.3
Properties of Gamma Function
=
1 a +1 (log a ) a +1
=
a +1 (log a ) a +1
3.9
example 13 ∞
Evaluate
∫0 3
-4 x 2
dx .
Solution 2
3-4 x = e - t , - 4 x 2 log 3 = -t log e, 4 x 2 log 3 = t
Let
x= when when
∫
∞
0
t 2 log 3
, dx =
x = 0,
t=0
x → ∞,
t→∞
∞
3-4 x dx = ∫ e - t . 2
0
= =
. 1 dt 2 log 3 2 t
. 1 dt 4 log 3 t
1
4
1
1
∫ log 3
∞
e-t t
-
1 2
dt
0
π 1 = 4 log 3 2 4 log 3 1
example 14 Prove that
∞
∫0
xe - ax sinbx dx =
Solution
∫
∞
0
2ab . ( a + b2 )2 2
∞
xe - ax sin bx dx = ∫ xe - ax [Imaginary part of eibx]dx 0
∞
= Im. part
∫
= Im. part
∫
= Im. part
2 (a - ib) 2
= Im. part
1 (a 2 - b 2 ) - 2iab
xe - ax . eibx dx
0 ∞
e - ( a -ib ) x . x dx
0
Q
∫
∞
0
e - kx x n -1 dx =
n kn
3.10
Chapter 3
Gamma and Beta Functions
2 2 2ab = Im. part (a - b ) + 2iab = 2 2 2 2 2 2 2 2 ( a - b ) + 4a b ( a + b )
exercIse 3.1 1. Evaluate the following integrals: (i)
∫
∞
xe
-3 x
(ii)
dx
0
(iv)
∫
e
-
x2 4
dx
(iii)
1
(x log x )4 dx
(v)
∫
1
3
∫
1
(viii)
1 dx x 1 x log dx x
log
0
1 4 ∫0 x log x dx 1
0
3
(vi) (ix)
315 ans. : (i) 16 p 8 p (iii) 3 p (v) 2 6 (vii) 625 p (ix) 4 log 5 2. Prove that ∞
∫
0
1
n
x m e - ax dx = na
( m + 1) n
m +1 . n
3. Prove that ∞ ∞ π 2 -x4 -x4 ∫0 x e dx . ∫0 e dx = 8 2 . 4. Prove that ∞
∫
0
x e - x dx . ∫ 2
∞
0
e-x
2
dx =
x
π 2 2
.
5. Prove that
∫
∞
0
∫
∞
0
0
0
(vii)
∫
∞
∞
xe - x dx . ∫ x 2 e - x dx = 8
0
4
π 16 2
.
∫
x
∞
x
dx
7 4
dx
1
0
∫
e-
- log x 2
5-4 x dx.
0
(ii) p (iv)
4! 55
(vi) p 4
3 3 (viiii) 4
4 3
3.4
Beta Function
6. Prove that 1
n x m (log x )n dx = (-1) n + 1. 0 (m + 1)n +1
∫
7. Prove that n
1 n +1 ∫0 x log x dx = (m + 1)n +1 . 1
m
8. Prove that
∫
∞
0
x m -1 cos ax dx =
mπ m cos . m 2 a
9. Prove that
∫
∞
x n -1e - ax sin bx dx =
0
3.4
n n 2 2
(a 2 + b )
b sin n tan-1 . a
Beta functIon
Beta function B(m, n) is defined by 1
B (m, n) = ∫ x m -1 (1 - x) n -1 dx, m > 0, n > 0. 0
B(m, n) is also known as Euler’s integral of first kind.
3.4.1 trigonometric form of Beta function π
B(m, n) = 2 ∫ 2 sin 2 m -1 x cos 2 n -1 x dx 0
Proof:
B(m, n) =
∫
1
0
Let when when
x m -1 (1 - x) n -1 dx
x = sin2q, dx = 2sinq cosq dq x = 0, q =0 π x = 1, q = 2 π 2 B(m, n) = ∫ (sin 2 θ ) m -1 (1 - sin 2 θ ) n -1 ⋅ 2 sin θ cos θ dθ 0
π
= 2 ∫ 2 sin 2 m -1θ cos 2 n -1θ dθ 0
Changing the variable q to x, π 2 2 m -1 x cos 2 n -1 x dx B(m, n) = 2 ∫ sin 0
3.11
3.12
Chapter 3
Gamma and Beta Functions
Putting 2m -1 = p,
Corollary:
m=
2n -1 = q
p +1 , 2
q +1 2
n=
π
p +1 q +1 2 p q B , = 2 ∫0 sin x cos x dx 2 2
3.5
ProPerties of Beta functions
1. Symmetry B(m, n) = B(n, m) B(m, n) =
Proof:
∫
1 - x = t,
1
0
x m -1 (1 - x) n -1 dx
- dx = dt x = 0, t=1 x = 1, t=0
Let when when
0
1
∫t
B (m, n) = ∫ (1 - t ) m -1 t n -1 (-dt ) =
n -1
0
1
(1 - t ) m -1 dt
= B (n, m). 2. Relation between Beta and Gamma functions q=p 2
B(m, n) =
m+n Proof: By alternate form of Gamma function, ∞
m n = 2∫ e
- x2
x
2 m -1
0
= 4∫
∞
0
∫
∞
∞
dx . 2 ∫ e
- y2
y
2 n -1
r
ä
m n
∞
dy
0
e-( x
2
+ y2 )
x 2 m -1 y 2 n -1dx dy
0
Changing to polar coordinates x = r cosq, y = r sinq dx dy = r dr dq O r=0 Limits of x x=0 to x→∞ fig. 3.2 Limits of y y=0 to y→∞ This shows that the region of integration is the first quadrant. Draw an elementary radius vector in the region which starts from pole and extends up to ∞. Limits of r
r=0
Limits of q
q=0
to
r→∞ π q= 2
to
m n=4∫
π 2
0
π
∫
∞
2
e - r (r cos θ ) 2 m -1 (r sin θ ) 2 n -1 r dr dθ
0
π
2 = 4 ∫ 2 (cos θ ) 2 m -1 (sin θ ) 2 n -1 dθ . ∫ e - r r 2 ( m + n ) -1dr 2
0
0
q =0
3.5
Properties of Beta Functions
3.13
1 1 B(m, n) . m + n 2 2
=4.
m n
B(m, n) =
m+n
3. Duplication formula m m+
1 o 2m = 2 m -1 2 2 π 2
B(m, n) = 2 ∫ sin 2 m -1θ cos 2 n -1θ dθ
Proof:
0
Putting n = m, π 2
B(m, m) = 2 ∫ (sin θ cos θ ) 2 m -1 dθ 0
m m 2m Let 2q = t, when when
=
2 22 m -1
∫
π 2
(sin 2θ ) 2 m -1 dθ
0
1 dθ = dt 2 q = 0, π q= , 2
t=0 t=p
m⋅ m 2 π 1 = 2 m-1 ∫ (sin t ) 2 m-1 ⋅ dt 0 2 2 2m = =
=
m m 2m
1 22 m -1 2 2
2 m -1
1 2
2 m -1
1
π 2
⋅ 2 ∫ (sin t )
2 m -1
0
(cos t ) dt
0
∫
π 2
(sin t ) 2 m-1 (cos t )
0
1 B m, 2 1 2 1 m+ 2 m
=
22 m -1
=
1 m π 22 m -1 1 m+ 2
1 2 -1 2
dt
Q 2 a f ( x)dx = 2 a f ( x)dx ∫0 ∫0 if f (2a - x) = f ( x)
3.14
Chapter 3
Gamma and Beta Functions
m m+
1 π 2m = . 2 22 m -1
example 1 Find the value of
4 5 (ii) B , 3 3
3 1 (i) B , 2 2 Solution
3 1 3 1 2 2 B , = 2 2 2 1 1 1 1 = 2 2 2 = π. π 1 2 π = 2
(i)
4 5 4 5 B , = 3 3 3 3 3
(ii)
=
1 1 2 1 1 1.2 2 +1 . +1 = . 2 3 3 2 3 3 3 3
1 1 1 1 π 1- = . π 9 3 3 9 sin 3 1 . 2π 2π = = 9 3 9 3 =
π Q n 1- n = sin nπ
example 2 If B(n, 3) =
1 and n is a positive integer, find the value of n. 60
Solution 1 60 n 3 1 = n + 3 60
B(n, 3) =
n.2 (n + 2) (n + 1) n n
=
1 60
3.5
Properties of Beta Functions
3.15
n3 + 3n2 + 2n = 120 n + 3n2 + 2n - 120 = 0 n = 4, -3.5 But n is a positive integer. Hence, n = 4. 3
example 3 Prove that B(n, n) =
π .
2
2 n -1
1 n+ 2
Solution B(n, n) =
n
.
n . = 2n 2n
n n
n n+
1 2
1 2 n . 1 . π 2n = 22 n -1 2n 1 n+ 2
=
π .
2
n+
[By Duplication formula]
n
2 n -1
n+
1 2
example 4
1 1 π 1-4 n Prove that B(n, n). B n + , n + = 2 . 2 2 n Solution B(n,
n n. 1 1 n) B n + , n + = 2 2 2n
n+
1 n n + 2
1 1 n+ 2 2 2n + 1
2
1 n n+ 1 2 = = 2n . n 2 2n 2n 2n 2
=
example 5 Prove that n 1 - n = 2
1 π π = 21- 4 n 2n 22 n -1 n
n 2 . nπ 21- p cos 2
π
2
3.16
Chapter 3
Gamma and Beta Functions
Solution we know that n 1- n = Replacing n by
π sin nπ
n +1 , 2 n +1 n +1 1= 2 2
n +1 1- n = 2 2
π n +1 sin π 2
π π nπ sin + 2 2
n π n n 1 1- n 2 + = nπ 2 2 2 2 cos 2 n π π n 1- n 2 = nπ 2 2n -1 cos 2 n π 1- n 2 = n nπ 2 1- n 2 cos 2
exercIse 3.2 1. Find the value of (i)
5 3 B , 2 2
1 2 (ii) B , 2 3
π Ans. : (i) 16 2. If B(n, 2) =
(ii)
2π 3
1 and n is a positive integer, find the value of n. 42 [Ans.: n = 6]
3. Prove that 1 Bn + , n + 2
1 n+ 1 1 . 2 π. = 2 22 n n + 1
3.5
Properties of Beta Functions
4. Prove that B(m, n) = B(m, n + 1) + B(m + 1, n). 3 3 -n +n 2 2
5. Prove that
1 = - n 2 p sec np, (-1 < 2n < 1). 4
Problems Based on Definition of Beta Function
example 1 Evaluate
1 5 3 ∫0 x (1- x ) dx .
Solution x = t , x = t2, dx = 2t dt
Let when when
∫ x (1 1
x = 0,
t=0
x = 1,
t=1
x ) dx = ∫ t 6 (1 - t ) 2t dt 1
5
3
0
5
0
1
= 2 ∫ t 7 (1 - t ) dt = 2B(8, 6) 5
0
8 6
=2
=2
14
1 7 ! 5! = 13! 5148
example 2 1
Evaluate
∫0
Solution
x2 1 - x4
dx .
1
dx
1
∫0
1 - x4
when
1 - 34 t dt 4 x = 0,
when
x = 1,
Let x4 = t, x = t 4 , dx =
∫
1
0
x2 1 - x4
dx .
∫
1
0
dx 1 - x4
=∫
1
0
=
.
t=0 t=1 1 2
3 1 1 . 1 - 34 . 1 t 4 dt . ∫0 1 - t 4 t dt 1- t 4
t
1 1 1 -3 1 1 - 14 2 . t 4 (1 - t ) 2 dt t ( t ) d t 1 ∫0 16 ∫0
3.17
3.18
Chapter 3
Gamma and Beta Functions
=
1 3 1 . 1 1 B , B , 16 4 2 4 2
3 1 1 1 1 4 2 . 4 2 1 π . 1 = = 16 5 16 1 1 4 3 4 4 4 4
example 3 1
Evaluate
∫0
1 - y 4 dy .
Solution 1
when
1 - 34 t dt 4 y = 0,
t=0
when
y = 1,
t=1
Let y4 = t, y = t 4 , dy =
∫
1
0
1 1 1 -3 1 - y 4 dy = ∫ (1 - t ) 2 . t 4 dt 0 4
3 1 1 3 1 1 2 4 = B , = 4 2 4 4 7 4 1 1 1 1 2 2 4 1 = = 6 4 3 3 4 4 2
p
1 4
2
3 1 4 4 2
1 p 4 p = = p 6 6 1 1 1p 4 4 sin 4 1 4
2
1 2 1 1 p 4 = = 6 p 2 6 2π 4
example 4 Evaluate
1 3 -3 y
∫0 y (8 - ) 2
4
dy .
π =
p 4
3.5
Properties of Beta Functions
Solution 1 1 -2 Let y3 = 8t, y = 2t 3 , dy = 2 . t 3 dt 3 when y = 0, when y = 2,
∫
2
0
t=0 t=1
1
( ) (8 - 8t ) 1 4
y 4 ( 8 - y 3 ) 3 dy = ∫ 2t 3 -
1
-
0
=
1 3
2 - 32 t dt 3
1 16 1 32 16 5 2 3 dt = ( ) t t 1 B , ∫ 0 3 3 3 3
5 2 16 3 3 16 = = 3 3 7 3
2 2 3 3 4. 1 3 3
2 2 3 3 =8 1 1 3 3
2
example 5 Evaluate
2a
∫0
x 2 2ax - x 2 dx .
Solution
∫
2a
0
5
2a
x 2 2ax - x 2 dx = ∫ x 2 2a - x dx 0
Let x = 2at, dx = 2a dt x = 0,
when
t=0
x = 2a,
when
∫
2a
0
t=1
1
x 2 2ax - x 2 dx = ∫ (2at ) 0
5 2
2a - 2at . 2a dt
1 1 5 7 3 4 = 16a 4 ∫ t 2 (1 - t ) 2 dt = 16a B , 2 2 0
7 3 16a 4 . 5 . 3 . 1 1 . 1 1 = 16a 2 2 = 5 24 2 2 2 2 2 2 4
=
15 π a 4 24
example 6 3
Evaluate
3
∫0
x2 3- x
dx . ∫
1
dx
0
1-
1 x4
.
3.19
3.20
Chapter 3
Gamma and Beta Functions
Solution 3
x2
3
I1 = ∫
dx
3- x
0
Let x = 3t, dx = 3 dt x = 0, x = 3,
when when
t=0 t=1 3
(3t ) 2
1
I1 = ∫
3 - 3t
0
⋅ 3 dt
1 3
= 9 ∫ t 2 (1 - t )
-
1 2
0
5 1 dt = 9 B , 2 2
5 1 9 3 1 1 1 27π =9 2 2 = . . = 3 2 2 2 2 2 8 1 dx I2 = ∫ 1
0
1- x4
1
Let x 4 = t , x = t 4, dx = 4t 3 dt x = 0, x = 1,
when when
t=0 t=1
1
4t 3
0
1- t
I2 = ∫
dt
1
= 4 ∫ t 3 (1 - t )
-
0
4 =4
9 2
1 2 =4
1 2
1 dt = 4 B 4, 2
1 128 2 = 7 . 5 . 3 . 1 1 35 2 2 2 2 2 3!
3
∫
Hence,
x2
3
3- x
0
dx . ∫
0
example 7 dx
a
Prove that ∫0
( an -
1 xn )n
1
=
1
1- x
1 4
=
27π . 128 432 p = 8 35 35
π π cosec . n n
Solution Let xn = an t, x = at n , dx =
dx
a 1n -1 t dt n
3.5
x = 0, x = a,
when when a
dx
0
(a n - x n ) n
∫
1
=
3.21
t=0 t=1
1
1
0
(a n - a n t ) n
=∫
Properties of Beta Functions
1
1 -1 . a t n dt n
1 1 -1 1 1 1 1 1 n n 1 ( ) t t dt = B - + 1, ∫ n 0 n n n
1 1 n n =1. π n sin π 1 n π π = cosec n n
1 = . n
1-
π Q 1- n n = sin nπ
example 8 b
m n m+ n+1 B( m + 1, n + 1) and hence, Prove that ∫ ( x - a) (b - x ) dx = (b - a) a
9
deduce that Solution
∫5 4 ( x - 5)(9 - x) dx =
1 2 4
2
3 π
.
Let (x - a) = (b - a) t, dx = (b - a) dt when x = a, t=0 when x = b, t=1
∫
b a
1
( x - a ) m (b - x) n dx = ∫ [ (b - a )t ] [b - {a + (b - a )t}] (b - a )dt m
n
0
1
= (b - a ) m + n +1 ∫ t m (1 - t ) n dt 0
= (b - a ) m + n +1 B(m + 1, n + 1) 1 1 , n = in the above integral, 4 4 1 1 1 1 9 + +1 1 1 4 4 4 4 ( x 5 ) ( 9 x ) d x = ( 9 5 ) B + 1, + 1 ∫5 4 4
Putting a = 5, b = 9, m =
2
1 1 1 5 5 2 4 = 23 4 4 = 8 4 4 = 3 π 5 3.1 1 2 2 2 2
2
3.22
Chapter 3
example 9
∫0
1 x m -1 (1 -
∫0
Prove that 1 x2
Gamma and Beta Functions
B( m, n) x ) n-1 dx = and hence, evaluate m+ n ( a + bx ) ( a + b) m a n
- 2 x3 + x 4 dx. (1 + x )6
Solution Let
x=
at , a + b - bt
dx =
a (a + b - bt ) - at (-b) a ( a + b) dt = dt (a + b - bt ) 2 (a + b - bt ) 2
x = 0, x = 1,
when when
t=0 t=1
1- x = 1-
at (a + b)(1 - t ) = a + b - bt a + b - bt
a + bx = a +
a ( a + b) bat = a + b - bt a + b - bt
Also,
m -1
at (a + b)(1 - t ) 1 x m -1 (1 - x ) n -1 1 a + b bt a + b - bt m+n ∫0 (a + bx)m + n dx = ∫0 a ( a + b) a + b - bt
=
1 ( a + b) m a n
1
∫t 0
m -1
(1 - t ) n -1 dt =
Putting a = 1, b = 1, m = 3, n = 3 in the above integral 1 x 2 (1 - x ) 2 1 ∫0 (1 + x)6 dx = (1 + 1)3 .13 B(3, 3)
∫
1
0
x 2 - 2 x3 + x 4 1 3 3 1. 4 1 dx = = = (1 + x)6 8 6 8 120 240
example 10 Prove that
1 (1 -
3 4 4 x )
1 1
∫0 (1 + x 4 )2 dx = 4 .
Solution 1
Let x4 = t, x = t 4 , dx = when when
1 - 34 t dt 4 x = 0, x = 1,
1 24
1 7 B , . 4 4
t=0 t=1
n -1
. a ( a + b ) dt (a + b - bt ) 2
1 B(m, n) ( a + b) m a n
3.5
3
3.23
Properties of Beta Functions
3
3
3
-
1 (1 - t ) 4 (1 - x 4 ) 4 1 -3 1 1 t 4 (1 - t ) 4 ∫0 (1 + x 4 )2 dx =∫0 (1 + t )2 . 4 t 4 dt = 4 ∫0 (1 + t )2 dt 1
Let t =
u , 2-u
dt =
2 (2 - u ) - u ( -1) du du = 2 (2 - u ) 2 (2 - u ) t = 0, t = 1,
when when
u=0 u=1 -
3
3
u 4 u 4 3 1 4 1 (1 - x ) 4 1 1 2 - u 2 - u . 2 d x = du 2 ∫0 (1 + x 4 )2 ∫ 4 0 (2 - u ) 2 u 1 + 2 - u 2 1 u = ∫ 4 0
-
3 4
3
( 2 - 2u ) 4 2
2
1 1 du = . 1 4 4 2
∫
1
u
-
3 4
0
3
(1 - u ) 4 du
1 1 1 7 = . 1 B , 4 4 4 4 2
exercIse 3.3 1. Evaluate the following integrals: 2
(i)
∫
1
∫
2
0
(iv)
0
1 - x m dx
(ii)
∫
1
∫
a
0
-
1
x 2 (2 - x ) 2 dx (v)
0
dx 1- x6 x 4 a 2 - x 2 dx
1 2 7 4 ∫3 4 (x - 3)(7 - x) dx = 3 π
(vi)
∫
1 2
0
x 3 1 - 4 x 2 dx
ans. : 1 1 3 1 1 1 (i) B , (ii) B , m m 2 8 8 2 64 2 (iii) 128 (iv) 1155 15 6 1 (v) π a (vi) 32 120
2. Prove that
(i)
1 3 (iii) ∫ 1 - x 4 dx 0 1
2
(ii)
∫
6
5
(x - 5)5 (6 - x )6 dx =
5! 6 ! 12!
3.24
Chapter 3
Gamma and Beta Functions
3. Prove that
∫
1
0
x
-
1 3
(1 - x ) (1 + 2x )
-
2 3
dx =
π 7
.
36
4. Prove that 1 x m -1(1 - x )n -1 B(m, n) ∫0 (1 + x)m + n dx = 2m and hence, evaluate
5 Prove that 1 x n -1 ∫0 (1 + cx)(1 - x)n dx 1 . π , 0 < n < 1. = (1 + c)n sin nπ
x 3 - 2x 2 + x dx. 0 (1 + x )5 1 ans. : 48
∫
1
Problems Based on trigonometric form of Beta function
example 1 Evaluate
π 2 0
∫
cotθ dθ .
∫
Solution
π 2
0
π
1
-
1
cot θ dθ = ∫ 2 (cos θ ) 2 (sin θ ) 2 dθ 0
1 1 +1 - +1 1 2 , 2 = B 2 2 2 3 1 1 3 1 1 4 4 = B , = 2 4 4 2 1 =
1 1 1 1 π p = ⋅ 1= π 2 4 4 2 sin 2 4
example 2 Evaluate
π 4 0
∫
cos3 2θsin 4 4θ dθ .
Solution
∫
π 4
0
π
cos3 2θ sin 4 4θ dθ = ∫ 4 cos3 2θ (2 sin 2θ cos 2θ ) 4 dθ 0
3.5
Properties of Beta Functions
3.25
π 4
1 dθ = d t 2
Let 2q = t, when when
∫
π 4
0
= 16 ∫ cos 7 2θ sin 4 2θ dθ 0
q = 0,
t=0
q= π , 4
t=
π 2
π 2
1 cos 2θ sin 4θ dθ = 16 ∫ sin 4 t . cos 7 t ⋅ dt 0 2 3
4
5 4 1 5 = 8. B , 4 = 4 2 2 2 13 2 3 .1 1 . 3! 128 2 2 2 = = 4⋅ 11 9 7 5 3 1 1 1155 ⋅ ⋅ ⋅ ⋅ ⋅ 2 2 2 2 2 2 2
example 3 2π
Evaluate
∫0
sin 2θ (1 + cosθ ) 4 dθ .
Solution
2π
I = ∫ sin 2 θ (1 + cos θ ) 4 dθ 0
2
4
2π θ θ θ = ∫ 2 sin cos 2 cos 2 dθ 0 2 2 2 2π
= 26 ∫ sin 2 0
Let θ = t , dθ = 2dt 2 when when
q = 0, q = 2p,
θ θ cos10 dθ 2 2
t=0 t=p π
I = 26 ∫ sin 2 t cos10 t . 2dt 0
π
= 27 . 2 ∫ 2 sin 2 t cos10 t dt 0
Q f ( x ) dx = 2 ∫ f ( x ) d x ∫0 0 if f (2a - x) = f ( x)
3 11 1 3 11 7 2 2 = 28 . B , =2 2 2 2 7
2a
a
3.26
Chapter 3
Gamma and Beta Functions
=
27 . 1 1 9 . 7 . 5 . 3 . 1 1 21p . = 6! 2 2 2 2 2 2 2 2 8
example 4 Evaluate
π 4 ( cosθ π 4
1 + sinθ ) 3 dθ .
∫
Solution 1
∫ = Let
∫
π 4 π 4
π 4 π 4
1 3
(cos θ + sin θ ) dθ =
∫
π 4 π 4
1
1
1 6
π π 3 2 sin cos θ + cos sin θ dθ = 4 4
π + θ = t, 4
q = -
when
q= π 4 π 4
∫
π 4 π 4
π 3 2 sin + θ dθ 4 1 6
dq = dt
when
∫
1 1 3 cos θ + sin θ dθ 2 2 2
π , 4
t=0
π , 4
1
t= π
1
π 2
1
(cos θ + sin θ )3 dθ = 2 6 ∫ 2 (sin t ) 3 dt 0
1
= 2
1 6
π 2 0
∫
1 3
26 4 1 (sin t ) (cos t )0 dt = B , 2 6 2
2 1 2 π 1 3 6 π 3 2 = 5 ⋅ = 5 = 5 7 1 1 26 26 26 6 6 6 1
2 3 1 6
example 5 Prove that Solution
π 2 0
∫
tan n x dx =
π 2 0
∫
π nπ sec . 2 2
tan n x dx =
π 2 0
∫
(sin x) n (cos x) - n dx
1 n + 1 -n + 1 1 , = B = 2 2 2 2
n + 1 -n + 1 2 2 1
3.5
=
3.27
1 n +1 n +1 12 2 2 π Q n 1- n = sin nπ
π n +1 π sin 2
= 1⋅ 2
=
Properties of Beta Functions
π ⋅ 2
1 π nπ sin + 2 2
=
π π 1 nπ . = sec ⋅ 2 2 cos n π 2 2
example 6 π
∫0 x sin
Evaluate
7
x cos 4 x dx .
Solution π
∫
0
x sin 7 x cos 4 x dx =
∫
π
0
(π - x) sin 7 (π - x) cos 4 (π - x)dx Q a f ( x)dx = a f (a - x)dx ∫0 ∫0 π
π
0
0
= π ∫ sin 7 x cos 4 x dx - ∫ x sin 7 x cos 4 x dx π
π
0
0
2 ∫ x sin 7 x cos 4 x dx = π ∫ sin 7 x cos 4 x dx π π = π ∫ 2 sin 7 x cos 4 x dx + ∫ 2 sin 7 (π - x) cos 4 (π - x)dx 0 0
Q 2 a f ( x)dx = a f ( x)dx + a f (2a - x)dx ∫0 ∫0 ∫0 π
= 2π 2 sin 7 x cos 4 x dx = 2π ⋅ ∫ 0
5 5 3! 2 2=π 11 9 7 5 5 13 ⋅ ⋅ ⋅ 2 2 2 2 2 2
4 =π
∫
π
0
x sin 7 x cos 4 x dx =
16π 1155
1 B 4, 2
5 2
3.28
Chapter 3
Gamma and Beta Functions
exercIse 3.4 1. Evaluate the following integrals: π 2 0
(i)
∫
(iii)
∫
(v)
∫
π 3 π 6
2π
0
tan θ dθ
(
(ii)
)
1 4
π 6 0
∫
dθ
(iv)
∫
sin2 θ (1 + cos θ )4 dθ
(vi)
∫
3 sin θ + cos θ
cos6 3θ sin2 6θ dθ
π 2 π 2 π
0
cos3 θ (1 + sin θ )2 dθ
x sin5 x cos6 x dx
π ans. : (i) 2 5 3 (iii) 2 4 π 8 9 8 21π (v ) 8 2. Prove that π 2 0
∫ 3.6
(sin x )2n dx =
1 ⋅ 3 ⋅ 5 L (2n - 1) π ⋅ . 2 2n (n !)
Beta functIon as ImProPer InteGral B(m, n) =
∫
∞
0
x m -1 dx (1 + x) m + n
Proof: Let x = tan2q, dx = 2 tanq sec2q dq when
x = 0,
when
x→Ç,
∫
∞
0
x m -1 dx = (1 + x) m + n
q=0 π θ= 2 (tan 2 θ )m -1 ⋅ 2 tan θ sec 2 θ dθ (1 + tan 2 θ )m + n
π 2 0
∫
π
= 2∫2 0
(tan θ )2 m -1 sec 2 θ dθ (sec θ )2 m + 2 n
π
= 2 ∫ 2 (sin θ )2 m -1 (cos θ )2 n -1 dθ 0
= B(m, n)
7π 384 8 (iv ) 5 8π (iv ) 693 (ii)
3.6
Beta Function as Improper Integral
3.29
example 1 Prove that ∞
of
∫0
x m-1 1 dx = n m B( m, n) and hence, find the value m+ n ( a + bx ) a b
∞
∫0
x5 . ( 2 + 3 x )16
Solution a dt b
Let bx = at, dx = when
x = 0,
t=0
when
x → ∞,
t→∞ m -1
∫
m -1
x dx = (a + bx) m + n
∞
0
=
∞
∫
0
a t a b ⋅ dt m+n b (a + at )
1 n a bm
∫
∞
0
1 t m -1 dt = n m B(m, n) m+n a b (1 + t )
Putting a = 2, b = 3, m = 6, n = 10 in the above integral,
∫
∞
0
x5 1 dx = 10 6 B(6, 10) 2 ⋅3 (2 + 3 x)16 =
1 5!10 ! 1 6 10 = 10 6 2 ⋅ 3 15! 210 ⋅ 36 16
example 2 ∞
Prove that
∫0
Solution
∫
∞
0
x 8 (1 - x 6 ) dx = 0 . (1 + x ) 24 x8 (1 - x 6 ) dx = (1 + x) 24 =
∫
∞
∫
∞
0
0
∞ x8 x14 dx - ∫ dx 24 0 (1 + x ) 24 (1 + x) ∞ x 9 -1 x15 -1 x d ∫0 (1 + x)15+ 9 dx (1 + x)9 +15
= B(9, 15) - B(15, 9) =0
3.30
Chapter 3
Gamma and Beta Functions
example 3 ∞
∫0
Prove that
x2 5π 2 dx = . 4 3 128 (1 + x )
Solution 1
Let x4 = t, x = t 4 , dx =
1 - 43 t dt 4
when
x = 0,
when
x → ∞,
∫
x2 dx = (1 + x 4 )3
∞
0
t=0
∞
∫
0
t→∞ 1 2
t 1 -3 ⋅ t 4 dt 3 (1 + t ) 4 -
1
3
1 ∞ t 4 1 ∞ = ∫ dt = ∫ 3 4 0 (1 + t ) 4 0 1 3 = B , 4 4
t4
-1
(1 + t )
3 9 + 4 4
dt
3 9 9 1 4 4 = 4 4 3
3 5 1 1 ⋅ ⋅ 1 4 4 4 4 5 1 1 1= ⋅ = 2! 4 128 4 4 π Q 1- n n = nπ sin
5 π ⋅ 128 sin π 4 5π 2 = 128 =
example 4 Prove that
∞
∫0 sech
6
8 . 15
x dx =
Solution
∫
∞
0
sech x dx = 6
∫
∞
0
= 26 ⋅
6
2 x dx e + e- x 1 ∞ 1 dx x ∫ ∞ 2 (e + e - x ) 6
e x + e- x Q cosh x = 2 Q a f ( x)dx = 2 a f ( x)dx ∫0 ∫- a if f (- x) = f ( x)
3.6
when when
∫
∞
0
= 25 ∫
∞
∫
∞
sec h 6 x dx = 25
0 ∞
= 24 ∫0 = 16 ⋅
example 5 ∞
∫0
Solution
∞
∫
0
e2x = t,
∫
0
33
= 16 ⋅
6
8 2! 2! = . 15 5!
1 ∞ e2( m + n) x + e2( n - m) x dx 2 ∫- ∞ (1 + e 2 x ) 2 n dx =
x → - ∞, x → ∞, ∞
t 3-1 dt = 24 B(3, 3) (1 + t )3+ 3
e 2 mx + e - 2 mx 1 ∞ (e 2 mx + e -2 mx ) e 2 nx dx dx = ∫ x - x 2n 2 -∞ (e 2 x + 1) 2 n (e + e ) Q a f ( x)dx = 2 a f ( x)dx ∫0 ∫- a if f (- x) = f ( x)
2e2x dx = dt,
when when
t3 1 ⋅ dt 6 2t (t + 1)
e 2 mx + e - 2 mx 1 dx = B( n + m, n - m), n > m. x - x 2n 2 (e + e )
=
Let
3.31
e6 x dx - ∞ (e 2 x + 1)6 1 2e2x dx = dt, dx = dt 2t x → - ∞, t=0 x → ∞, t→∞
Let e2x = t,
Prove that
Beta Function as Improper Integral
1 dt 2t t=0 t → ∞,
e 2 mx + e -2 mx 1 ∞ t m+n + t n-m 1 x = d ⋅ dt (e x + e - x ) 2 n 2 ∫0 (1 + t ) 2 n 2t =
∞ 1 ∞ t ( m + n ) -1 t ( n - m ) -1 t d dt + ∫0 (m+ n)+(n- m) (n- m)+(n+ m) ∫ 0 4 (1 + t ) (1 + t )
1 [ B(m + n, n - m) + B(n - m, n + m)] 4 1 = B (n + m, n - m) [Q B(m, n) = B(n, m)] 2 =
3.32
Chapter 3
Gamma and Beta Functions
example 6 Prove that
deduce that
π
∫0 π
∫0
2n-1
sin n-1 x dx = ( a + b cosx ) n
sinx
dx =
3 (5 + 3 cosx ) 2
n n B , and hence, 2 2
n 2 2 2 (a - b )
3 4
2
2 2π
.
Solution Let tan
x x 2 = t , = tan -1 t , dx = dt 2 2 1+ t2
when
1- t2 2t , sin x = 2 1+ t 1+ t2 x = 0, t=0
when
x = p,
cos x =
∫
π
0
n -1
sin x dx = (a + b cos x) n
∫
2t 1+ t2
∞
0
t→∞
1 - t 2 a + b 1 + t 2
t = 0, t → ∞,
when when
⋅
n
n 2 dt = 2 2 1+ t
a+b ⋅ u
(a - b) t2 = (a + b)u, t =
Let
n -1
, dt =
a-b
∫
t n -1
∞
0
(a + b) + (a - b) t 2
a+b 2 a-b
⋅
1 u
1 2
n
dt
du
u=0 u→∞ n -1
(a + b)u 2 ( a - b) π sin n -1 x ∞ a+b 1 ⋅ du ∫0 (a + b cos x)n dx = 2n ∫0 n [(a + b) + (a + b) u ] 2 a - b u 12 n
2n -1
=
n 2
( a + b) ( a - b) Putting a = 5, b = 3, n =
∫
0
sin x (5 + 3 cos x)
∫
0
-1
(1 + u )
3 in the above integral, 2 3
π
n 2
u2
∞
3 2
dx =
22
-1 3 2 4
(52 - 3 )
3 3 B , 4 4
n n + 2 2
du =
2n -1 n 2 2
(a 2 - b )
n n B , 2 2
3.6
Beta Function as Improper Integral
2
3 3 3 2 4 2 = 3 4 4= 2 3 1 1 23 ⋅ 2 2 2
=
3 4
2
2 2π
example 7 Prove that
π 2 0
∫
cos 2 m-1θ sin 2 n-1θ Β( m, n) dθ = 2 m 2 n . 2 2 2 2 m+ n ( a cos θ + b sin θ ) 2a b
Solution π 2 0
cos 2 m -1 θ sin 2 n -1 θ dθ (a 2 cos 2 θ + b 2 sin 2 θ )m + n
π 2 0
cos 2 m-1 θ sin 2 n -1 θ (cosθ )-2 m- 2 n dθ (a 2 + b 2 tan 2 θ )m+ n
∫ = =
b2 tan 2 θ = t, 2 a
Let
∫
1 a
2( m+ n )
tan q = q = 0, π q= , 2
when when π 2 0
∫
π 2 0
∫
(tan θ )2 n -1 sec 2 θ b2 2 1 + a 2 tan θ
a t, b
sec2 q dq =
=
Prove that B(m, n) =
∫0
a 1 ⋅ dt 2b t
t→∞
=
1 x m -1 +
dθ
t=0
cos 2 m -1 θ sin 2 n -1 θ 1 dθ = 2 ( m + n ) (a 2 cos 2 θ + b 2 sin 2 θ ) m + n a
example 8
m+ n
x n-1
(1 + x ) m+ n
∫
1 2a 2 m b 2 n
1 2m
2a b 2 n
dx .
a 12 t ∞ b
2 n -1
(1 + t ) m + n
0
∫
∞
0
⋅
a 1 dt ⋅ 2b t
t n -1 dt (1 + t ) m + n
B(m, n)
3.33
3.34
Chapter 3
Gamma and Beta Functions
Solution
x m -1 ∫0 (1 + x)m + n dx ∞ 1 x m -1 x m -1 = ∫ dx + ∫ dx m + n 0 (1 + x ) 1 (1 + x ) m + n ∞
B(m, n) =
we have
I=
Consider, 1 , y
∫
∞
1
x m -1 dx (1 + x) m + n
when
1 dy y2 x =1,
y=1
when
x → ∞,
y=0
Let
x=
dx = -
I=
0
∫
1
1 y
m -1
1 1 + y
m+n
1 - y 2 dy =
y n -1 ∫0 ( y + 1)m + n dy 1
Substituting in Eq. (1), B(m, n) =
1 x m -1 y n -1 x d + ∫0 (1 + x)m + n ∫0 (1 + y)m + n dy
B(m, n) =
1 x m -1 x n -1 x d + ∫0 (1 + x)m + n ∫0 (1 + x)m + n dx
1
Replacing y by x,
=
1
∫
1
0
x m -1 + x n -1 dx (1 + x) m + n
example 9
Prove that
π 2 0
∫
dθ 1 1 - sin 2θ 2
=
1 4
4 π
Solution Let tan
2
2 θ θ dt = t, = tan -1 t , dθ = 1 + t2 2 2
.
… (1)
3.6
sin q =
Beta Function as Improper Integral
2t 1+ t2
q = 0, π θ= , 2
when when π 2 0
∫
dθ 1 1 - sin 2 θ 2
1
1 - 43 u du 4
Let t4 = u, t = u 4 , dt = when when π 2 0
∫
dθ 1-
1 sin 2 θ 2
=
∫
t=0 t=1 1
1
0
1-
= 2∫
1 2t 2 1 + t 2
1
1
t = 0,
(1 + t ) u=0
t = 1,
u=1
= 2∫
1
0
(1 + u ) 2 1
1 = 2
exercIse 3.5 1. Evaluate
∫
∞
0
∫
0
x m -1 + x n -1 dx = 2B (m, n). (1 + x )m + n
3. Prove that
∫
∞
0
0
u4
1
-1 1
(1 + u ) 2
1 du = 4
1 1 1 B , 4 4 4
1 1 1 4 4 = = 4 1 2
2 dt 1+ t2
∫
1
0
1
-1
u4 + u4 1
(1 + u ) 4
+
-1 1 4
du
[From Ex. 8]
1 4
2
4 π
dy . 1+ y 4
2. Prove that ∞
∫
1
⋅
1 -3 ⋅ u 4 du 4
1
1
2
dt
1 4 2
0
=
3.35
x π dx = . 2 (4 + 4 x + x ) 4 2
π ans. : 2 2
3.36
Chapter 3
Gamma and Beta Functions
4. Prove that
∫
∞
0
dx 1 n n = B , (e x + e - x )n 4 2 2
and hence, evaluate
∫
5. Prove that
∞
0
sech 8 x dx.
dx = B(p + q, 1 - q), if - p < q < 1. (x - 1)q
∞
1
∫
p +1
x
Points to remember Gamma Function ∞
n = ∫ e - x x n -1dx, n > 0 0
∞
n = 2 ∫ e - x x 2 n -1dx 2
0
Properties of Gamma Function n +1 = n n
Property 1: (i) (ii)
n + 1 = n!
(iii)
n=
(iv)
n 1- n =
if n is a positive integer if n is a positive real number
n +1 = n n
n +1 n
Property 2:
if n is a negative fraction
π sin nπ
1 = π 2
Beta Function 1
B (m, n) = ∫ x m -1 (1 - x) n -1 dx, m > 0, n > 0. 0
Trigonometric Form of Beta Function π
B(m, n) = 2 ∫ 2 sin 2 m -1 x cos 2 n -1 x dx 0
16 ans. : 35
Multiple Choice Questions
3.37
Properties of Beta Functions Symmetry B(m, n) = B(n, m) Relation between Beta and Gamma Functions m n
B(m, n) =
m+n
Duplication Formula m m+
1 o 2m = 2 m -1 2 2
Beta Function as Improper Integral B(m, n) =
∫
∞
0
x m -1 dx (1 + x) m + n
multiple choice Questions Select the most appropriate response out of the various alternatives given in each of the following questions: 1. The value of the integral I =
2p
(b) p
(a) 1
2 ∞ Ê -x ˆ ˜ ÁË 8 ¯
1
Úe
dx is
0
(c) 2
(d) 2p
2. Match the items in columns I and II for the following special functions I II 1 (P) b (p, q) (i) 2 ∞
(Q)
p q p+q
(R) (S)
p p sin pp
(ii)
y p -1
Ú (1 + y)( p + q) dy 0
(iii) b (p, q) (iv) p 1 - p
(a) P-(iv), Q-(iii), R-(i), S-(ii) (b) P-(ii), Q-(iii), R-(i), S-(iv) (c) P-(iii), Q-(ii), R-(i), S-(iv) (d) P-(ii), Q-(iii), R-(iv), S-(i)
3.38
Chapter 3
Gamma and Beta Functions
∞
3. The value of
3
y e - y dy is
Ú 0
p 2
(a)
p 3
(b)
(c)
4. The value of B (m + 1, n) is n (a) B (m, n) m+n (c)
m B (m, n) m+n
5. The value of (a)
3 325
7. If B (n, 2) = (a) 3 8. The value of
(a)
Ú
p 2 2
6. The value of (a)
p 2 0
p 2
p
(d)
p 6
(b)
n B (m, n) m +1
(d)
m B (m, n) m +1
(c)
p 2 4
(d)
p 4
3 625
(d)
6 325
cot q dq is
(b) 1 4 x 0
Ú
(b)
p 2 3
È Ê 1ˆ˘ Ílog ÁË ˜¯ ˙ dx is x ˚ Î 6 625
(c)
1 and n is a positive integer, then the value of n is 6 (b) - 2 (c) 2 (d) - 3 •
Ú0
t 2 dt 1 + t4
(b)
is
p 2
(c)
p 2
(d)
p 4
9. The value of B (m, m) is 1ˆ Ê (a) 21 - 2m B ÁË m, ˜¯ 2
1ˆ Ê (b) 21 - 2m B ÁË m + 1, ˜¯ 2
1 ˆ 3ˆ Ê Ê (c) 21 - 2m B ÁË m + , 1˜¯ (d) 21 - 2m B ÁË m, ˜¯ 2 2 10. Gamma function is discontinuous for (a) all p < 0 (b) any p > 0 (c) p = 0 only (d) p = 0 and negative integers
Multiple Choice Questions
3.39
11. Beta function B (p, q) is convergent for (a) p > 0, q < 0 (b) p > 0, q > 0 (c) p < 0, q > 0 (d) p < 0, q < 0 answers 1. (a) 9. (a)
2. (b) 10. (d)
3. (b) 11. (b)
4. (c)
5. (a)
6. (b)
7. (c)
8. (a)
4 Applications CHAPTER
of Definite Integrals chapter outline 4.1 4.2 4.3 4.4
4.1
Introduction Volume using Cross-sections Length of Plane Curves Area of Surface of Solid of Revolution
IntroductIon
In this chapter, we will explore few applications of definite integrals. Applications of definite integrals in finding volume by cross-sections, length of plane curves and areas of surfaces of revolution are discussed in detail.
4.2
Volume usIng cross-sectIons
The volume of a solid of known integrable cross-section area A(x) formed by a plane perpendicular to the x-axis at any point between x = a to x = b is b
V = ∫ A( x)dx a
Note: If the cross-section A(y) is perpendicular to the y-axis at any point between y = c to y = d, then the volume of the solid is d
V = ∫ A( y )dy c
4.2
Chapter 4 Applications of Definite Integrals
example 1 Find the volume of the solid generated by rotating the plane region 1 bounded by y = , x = 1 and x = 3 about the x-axis. x Solution The volume is generated by rotating the dotted region about the x-axis. The cross-section PQ of the generated volume is a circle of radius y perpendicular to x-axis. y Area of cross-section, A = p y2 = p◊
1
y=
x2
For the region shown, x varies from 1 to 3. Volume,
V =Ú
3
1
1 x
P (x, y) x=1
p x2
dx 3
=p -
1 x1
O
B
A
x=3 x
Q
Ê 1 ˆ = p Á - + 1˜ Ë 3 ¯ =
2 p 3 Fig. 4.1
example 2 Find the volume of the solid that lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections perpendicular to the x-axis between these planes run from one side of the parabola x = y 2 to the other. The cross-sections are squares with bases in the xy-plane. Solution The cross-section of the solid is a square PQRS with side runs from y = − x to y= x Length PQ = 2 y = 2 x Area of cross-section, A = (2 x ) 2 = 4 x
4.2 Volume Using Cross-sections 4.3
y
For the region shown, x varies from 0 to 4.
S P(x,y)
4
V = Ú 4 x dx
Volume,
0
=4
x2 2
4
S
P(x,y)
R
Q
O
x R
x=4
0
= 2(16 - 0) = 32
Q
Fig. 4.2
example 3 Find the volume of a right circular cone of base radius r and height h. Solution The cone is generated by rotating the line AB about the y-axis. Equation of the line AB passing through the points A(r, 0) and B (0, h) is
y B(0,h)
h-0 (x - r) 0-r yˆ Ê x = r Á1 - ˜ Ë h¯
y-0 =
x
P
The cross-section PQ of the cone is a circle of radius x perpendicular to the y-axis.
O
Q(x,y)
x
Area of cross-section, A = p x 2 yˆ Ê = p r 2 Á1 - ˜ Ë h¯
2
Fig. 4.3
For the region shown, y varies from 0 to h. 2
Volume,
h yˆ Ê V = Ú p r 2 Á 1 - ˜ dy o Ë h¯ 3
yˆ Ê ÁË 1 - h ˜¯ = p r2 Ê 1ˆ 3Á - ˜ Ë h¯
p r2h (0 - 1) =3 1 = pr 2 h 3
h
o
A(r,0)
x
4.4
Chapter 4 Applications of Definite Integrals
example 4 Find the volume of the solid with a circular base of radius 5 and whose cross-sections perpendicular to the base and parallel to the x-axis are equilateral triangles. Solution Equation of the circular base of radius 5 is, x 2 + y 2 = 25 The cross-section of the solid is an equilateral triangle PQR with base 2x. Area of cross-section,
1 ¥ (base) ¥ (height ) 2 1 = (2 x )(h ) 2 1 = (2 x )( x tan 60°) 2 = x2 3
A=
= (25 - y 2 ) 3
[from Eq. (1)]
For the region shown, y varies from −5 to 5. y Q Q h R
P (x, y ) x
O
R
Fig. 4.4
Volume,
V =Ú
5
3 (25 - y 2 ) dy
-5
= 2Ú
5
0
3 (25 - y 2 ) dy
y3 = 2 3 25 y 3
5
0
125 ˆ Ê = 2 3 Á 125 Ë 3 ˜¯ 500 3 = 3
60° 2x
P
…(1)
4.2 Volume Using Cross-sections 4.5
example 5 Use the method of slicing to find the volume of solid with semicircular È p p˘ base defined by y = 5 cos x on the interval Í - , ˙ . The cross-sections Î 2 2˚ of the solid are squares perpendicular to the x-axis with base running from x-axis to the curve. [Winter 2015] Solution The cross-section of the solid is a square with side runs from y = 0 (x-axis) to y = 5 cos x . Length of one side of the square = 5 cos x 2
(
Area of cross-section, A = 5 cos x
)
= 25 cos2 x p p È p p˘ In the given interval Í - , ˙ , x varies from - to . 2 2 2 2 Î ˚ p 2
Volume,
V =
Ú -
25 cos2 x dx
p 2 p 2
Ê 1 + cos 2 x ˆ = 25 Ú Á ˜¯ dx Ë 2 p -
2
25 sin 2 x = x+ 2 2
p 2 -
p 2
=
˘ 25 ÈÊ p p ˆ + + 0˙ Í 2 ÎÁË 2 2 ˜¯ ˚
=
25 p 2
[Qsin p = 0]
exercIse 4.1 1. Find the volume of the solid that lies between planes perpendicular to the y-axis at y = 0 and y = 2. The cross-sections perpendicular to the
4.6
Chapter 4 Applications of Definite Integrals
y-axis are circular disks with diameters running from the y-axis to the parabola x = 5y 2 . [Ans.: 8p ] 2. Find the volume of the solid that lies between the planes perpendicular to the x-axis at x = −1 and x = 1. The cross-sections perpendicular to the x-axis are circular disks whose diameters run from the parabola y = x2 to the parabola y = 2 − x2. 16 Ans.: 15 π 3. Show that the volume of a sphere of radius r is
4 3 πr . 3
4. Find the volume of the solid whose base is the region bounded between the curves y = x and y = x2, and whose cross-sections perpendicular to the x-axis are squares. 1 Ans.: 30 5. Find the volume of the solid whose base is a triangle with vertices (0, 0), (2, 0) and (0, 2) and whose cross-sections perpendicular to the base and parallel to the y-axis are semicircles. π Ans.: 3
4.3
length oF PlAne curVes
The process of determining the length of an arc of a plane curve is known as rectification of curves.
4.3.1
length of Arc in cartesian Form
We know from differential calculus that for the curve y y = f (x), ds Ê dy ˆ = 1+ Á ˜ Ë dx ¯ dx
y = f (x)
2
The length of the arc of the curve y = f (x) between x = a and x = b is given by, O x=a x=b x b ds Fig. 4.5 dx s=∫ a dx
=∫
b a
2
dy 1 + dx dx
4.3 Length of Plane Curves 4.7
Similarly, the length of the arc of the curve x = f (y) between y = c and y = d is given by,
s=Ú
d
=Ú
d
c
ds dy dy 2
Ê dx ˆ 1 + Á ˜ dy Ë dy ¯
c
example 1 y Show that the length of the arc of the curve 4 ax = y 2 - 2 a 2 log - a 2 a y2 a - - x. from (0, a) to any point (x, y) is given by 2a 2 Solution y - a2 a dx a 1 4a = 2 y - 2a2 ◊ ◊ dy y a 4ax = y 2 - 2a 2 log
… (1)
dx y a y2 - a2 = = 2 ay dy 2 a 2 y For the required arc, y varies from a to y. 2
Length of the arc,
s=Ú
y a
Ê dx ˆ 1 + Á ˜ dy Ë dy ¯ 2
=∫ =∫
y a
y a
y2 − a2 1+ dy 2ay 2
( y2 + a2 ) dy (2ay ) 2
=
1 y y2 + a2 dy 2 a ∫a y
=
1 y y dy + a 2 2a ∫a
=
1 y2 + a 2 log y 2a 2 a
∫ y
y a
1 dy y
4.8
Chapter 4 Applications of Definite Integrals
=
y a2 1 y2 2 + a log − a 2 2a 2
a2 a2 1 y2 y2 + − 2ax − − 2a 2 2 2 2 1 2 ( y − 2ax − a 2 ) = 2a y2 a = −x− 2 2a 2 y a = − −x 2a 2 =
… [From Eq. (1)]
example 2 Find the length of the arc of the curve y = ex from the point (0, 1) to (1, e). Solution
y y = ex
y = ex dy = ex dx For the required arc, x varies from 0 to 1.
B(1, e) A
Length of the arc AB,
1
s=∫
0
=∫
1
0
Putting
2
dy 1 + dx dx 1 + e 2 x dx
1 + e2x = t2, 2e2xdx = 2t dt t dx = 2 dt t −1
when
x = 0,
t= 2
when
x = 1,
t = 1 + e2
Length of the arc,
s=∫
=∫ =∫
1+ e2 2
t⋅
t dt t −1 2
t2 −1+1 dt 2 t2 −1 1+ e2 1 1 + 2 dt 2 t −1 1+ e2
(0, 1) x
O Fig. 4.6
4.3 Length of Plane Curves 4.9
1 t −1 = t + log 2 t +1
1+ e2
2
1 1 + e2 − 1 2 − 1 = 1 + e 2 − 2 + log − log 2 2 + 1 1 + e2 + 1 2 ( 2 − 1)2 1 ( 1 + e 2 − 1) = 1 + e − 2 + log − log − 2 1 + e2 − 1 2 1 1 = 1 + e 2 − 2 + log ( 1 + e 2 − 1) − log e 2 − log ( 2 − 1) 2 2
= 1 + e 2 − 2 + log ( 1 + e 2 − 1) − 1 − log ( 2 − 1)
[Q log e2 = 2 log e = 2]
example 3 Find the length of the arc of the curve y = log sec x from x = 0 to x =
p . 3
Solution y = log sec x dy 1 = ⋅ sec x tan x = tan x dx sec x p For the required arc, x varies from 0 to . 3 2
dy s = ∫ 3 1 + dx 0 dx p
Length of the arc,
p
= ∫ 3 1 + tan 2 x dx 0
p
= ∫ 3 sec x dx 0
p
= log(sec x + tan x) 03 = log ( 2 + 3 )
example 4 Ê e x - 1ˆ ˜ from x = 1 to x = 2. Ë e x + 1¯
Find the length of the arc of the curve y = log Á
4.10
Chapter 4 Applications of Definite Integrals
Solution y = log
ex − 1 ex + 1
y = log (e x − 1) − log (e x + 1) dy ex ex 2e x = x − x = 2x dx e − 1 e + 1 e − 1 For the required arc, x varies from 1 to 2. Length of the arc,
s=∫
2
=∫
2
=∫
2
1
1
1
2
dy 1 + dx dx 1+
4e 2 x dx (e 2 x − 1) 2
(e 2 x − 1) 2 + 4e 2 x dx (e 2 x − 1) 2
2x 2 e + 1 = ∫ 2 x dx 1 e − 1 x −x 2 e + e =∫ x dx 1 e − e− x 2
= log (e x − e − x ) 1
= log (e 2 − e −2 ) − log(e − e −1 ) e 2 − e −2 e − e −1 = log (e + e −1 ) = log
1 = log e + e
example 5 Show that the length of the arc of the curve ay2 = x3 from the origin to the point whose abscissa is b is Solution ay 2 = x 3
3 8a ÈÍÊ 9b ˆ 2 ˘˙ 1+ -1 . ˙ 7 ÍÁË 4 a ˜¯ Î ˚
4.3 Length of Plane Curves 4.11
2ay
dy = 3x 2 dx
y
dy 3 x 2 = = dx 2ay
3x 2 3
x a
2a
P
3 x 2 a
=
O
For the arc OP, x varies from 0 to b. Length of the arc OP,
s=∫
b
=∫
b
0
0
x=b
x
2
dy 1 + dx dx
1+
9x dx 4a
Fig. 4.7 b
3
2 9 x 2 4a = 1 + 3 4a 9 0
8a 9b = 1 + − 1 27 4a 3 2
example 6 x For the catenary y = c cosh , prove that the length of the arc s, meac sured from its vertex to any point (x, y), is (i) s = c sinh
x c
(ii) s2 = y 2 - c 2
y
Solution (i)
y = c cosh
x c P(x, y)
dy x = sinh dx c For the arc AP, x varies from 0 to x. Length of the arc AP,
(iii) s = c tan y
s=∫
x
0
=∫
x
0
A (0, c) 2
dy 1 + dx dx 1 + sinh 2
x dx c
y x
Fig. 4.8
4.12
Chapter 4 Applications of Definite Integrals
x
= ∫ cosh 0
x dx c x
= c sinh
x c0
= c sinh
x c
s 2 = c 2 sinh 2
(ii)
x c
x = c 2 cosh 2 − 1 c x = c 2 cosh 2 − c 2 c = y 2 − c2 (iii) The tangent at point P(x, y) makes an angle y with the x-axis. dy tan y = dx x = sinh c s = c s = c tan y
[From (i)]
example 7 2
Ê 1 ˆ Prove that the length of the arc of the curve y = x Á 1 - x˜ from the Ë 3 ¯ 4 origin to the point P(x, y) is given by s2 = y 2 + x 2 . Hence, rectify the 3 loop. y 2
Solution
P(x, y)
1 y 2 = x 1 − x 3
y=
2
1 1 3 x x 1 − = x 2 − x 2 3 3
dy 1 − 12 1 3 12 (1 − x) = x − ⋅ x = dx 2 3 2 2 x
A (3, 0) x
O
Fig. 4.9
4.3 Length of Plane Curves 4.13
For the arc OP, x varies from 0 to x. Length of the arc OP,
s=∫
x
=∫
x
=∫
x
=∫
x
0
0
0
0
=
2
dy 1 + dx dx 1+
(1 − x) 2 dx 4x
(1 + x) 2 dx 4x 1+ x dx 2 x
1 1 x − 12 ( x + x 2 ) dxx ∫ 2 0 x 1 2
3 2
x 1 x + 1 3 2 2 2 0 x = x 1 + 3 =
x s 2 = x 1 + 3
2
2
4 x = x 1 − + x 2 3 3 4 = y 2 + x2 3 2
1 2 The points of intersection of the curve y = x 1 − x and x-axis are obtained as, 3 1 0 = x 1 − x 3
2
x = 0, 3, 3 and y = 0, 0, 0 Hence, A: (3, 0) is the point of intersection. 3 Length of the upper half of the loop = 3 1 + = 2 3 3 Length of the complete loop = 4 3
4.14
Chapter 4 Applications of Definite Integrals
example 8 Show that the length of the loop of the curve 9ay 2 = ( x - 2 a )( x - 5a)2 is 4 3a . Solution The points of intersection of the curve 9ay 2 = ( x − 2a )( x − 5a ) 2 and x-axis are obtained as, y 0 = ( x − 2a)( x − 5a) 2 x = 2a, 5a and y = 0, 0 Hence, A: (2a, 0) and B: (5a, 0) are the points of intersection. A O (2a, 0)
9ay 2 = ( x − 2a )( x − 5a ) 2
dy 18ay = ( x − 2 a ) 2 ( x − 5a ) + ( x − 5 a ) 2 dx = ( x − 5a )(3 x − 9a ) dy ( x − 5a )( x − 3a ) = dx 6ay
B (5a, 0)
Fig. 4.10
For the upper half of the loop, x varies from 2a to 5a. Length of the loop of the curve, s = 2 (Length of upper half of the loop) = 2∫ = 2∫ = 2∫ = 2∫
5a 2a
5a 2a
5a 2a
5a 2a
5a
= 2∫
2a
= 2∫
2a
2
dy 1 + dx dx 1+
( x − 5a ) 2 ( x − 3a ) 2 dx 36a 2 y 2
1+
(xx − 5a ) 2 ( x − 3a ) 2 dx 4a ( x − 2a )( x − 5a ) 2
1+
( x − 3a ) 2 dx 4a ( x − 2a )
( x − a)2 dx 4a ( x − 2a ) x−a
5a
dx 2 a ⋅ x − 2a 1 5 a ( x − 2a ) + a dx = ∫ a 2a x − 2a =
1 − 2 x − 2 a + a x − 2 a ( ) dx ∫ a 2a
1
5a
x
4.3 Length of Plane Curves 4.15
3 1 1 2 = ( x − 2a ) 2 + 2a ( x − 2a ) 2 a 3
5a
2a
3 1 1 2 2 2 a a a = ( 3 ) + 2 ( 3 ) a 3
= 2 3 a + 2a 3 = 4 3a
example 9 In the evolute 27ay 2 = 4( x - 2 a )3 of the parabola y 2 = 4 ax, show that the length of the arc from one cusp to the point where it meets the parabola is 2 a(3 3 - 1) . Solution
y (8a, ≥32a)
(i) The points of intersection of the parabola y 2 = 4ax with its evolute 27 ay 2 = 4( x − 2a )3 are obtained as, 27a ⋅ 4 ax = 4( x − 2a)3 x 3 − 6 ax 2 − 15a 2 x − 8a3 = 0
B
A O
x
(2a, 0)
( x + a ) 2 ( x − 8a ) = 0 x = − a, 8 a
C (8a, −≥32a)
But x = –a does not lie on the parabola. x = 8a and y = ± 32a
(
Fig. 4.11
)
Hence, B: 8a, 32a is the point of intersection. (ii) The points of intersection of the evolute 27 ay 2 = 4( x − 2a )3 with the x-axis are obtained as, x = 2a and y = 0 Hence A: (2a, 0) is the point of intersection. Now, 27 ay 2 = 4( x − 2a )3 y=
2 3 3a
3
( x − 2a ) 2
1 dy 2 3 = ⋅ ( x − 2a ) 2 = dx 3 3a 2 For the arc AB, x varies from 2a to 8a.
x − 2a 3a
4.16
Chapter 4 Applications of Definite Integrals
Length of the arc AB, s = ∫ =∫
8a 2a 8a 2a
=
1
=
1
2
dy 1 + dx dx 1+
∫ 3a
8a 2a
x − 2a dx 3a x + a dx
3 2 ( x + a) 2 3a 3
8a
⋅
2a
3 2 ⋅ (9a ) − (3a ) 2 3a 3 3 3 2 = (3a ) 2 (3 2 − 1) 3 3a
=
3 2
1
= 2a (3 3 − 1)
example 10 Find the length of the parabola x2 = 4y which lies inside the circle x2 + y2 = 6y. Solution The equation of the circle is x2 + y2 = 6y 2 x + y2 – 6y = 0
y
The centre of the circle is (0, 3) and radius is 3. The points of intersection of parabola x2 = 4y and circle x2 + y2 = 6y are obtained as, 4 y + y2 = 6 y y2 − 2 y = 0 y ( y − 2) = 0 y = 0, 2 when
(
)
x=0
y = 2,
x = ±2 2
)
B (2≥2, 2)
O Fig. 4.12
y = 0,
(
A (−2≥2, 2)
Hence A : −2 2 , 2 and B : 2 2 , 2 are the points of intersection. Now,
x2 = 4 y
x
4.3 Length of Plane Curves 4.17
dy x = dx 2 For the arc, OB, x varies from 0 to 2 2. Length of the arc OB,
s = 2(Length of the arc OB) = 2∫
2 2
= 2∫
2 2
0
0
=∫
2 2
0
=
2
dy 1 + dx dx 1+
x2 dx 4
x 2 + 4 dx
(
x 2 x + 4 + 2 log g x + x2 + 4 2
(
)
2 2
0
)
= 2 ⋅ 12 + 2 log 2 2 + 12 − 2 log 2 = 2 6 + log
(
)
2+ 3
example 11 Show that the length of the parabola y2 = 4ax from the vertex to the end of the latus rectum is a È 2 + log (1 + 2 )˘ . Find the length of the arc Î ˚ cut off by the line 3y = 8x. Solution (i) The points of intersection of the parabola y2 = 4ax and its latus rectum x = a are obtained as, y 2 = 4a ⋅ a = 4a 2
y
P (a, 2a) A
y = ± 2a and x = a
Hence, P: (a, 2a) and Q : (a, − 2a ) are the points of intersection.
3y = 8x
x=a
O
(a, 0)
x
2
Now,
y 4a dx y = dy 2a x=
For the arc OP, y varies from 0 to 2a.
Q (a, −2a) y 2 = 4ax
Fig. 4.13
4.18
Chapter 4 Applications of Definite Integrals
2
s=∫
Length of the arc OP,
dx 1 + dy dy
2a
0
=∫
2a
1+
0
y2 dy 4a 2
=
1 2a 2 y + 4 a 2 dy 2a ∫ 0
=
1 y 2a 2
y 2 + 4a 2 +
(
4a 2 log y + y 2 + 4a 2 2
)
2a
0
1 a ⋅ 2a 2 + 2a 2 log ( 2a + 2a 2 ) − 2a 2 log 2a = 2a 2a + 2a 2 = a 2 + log 2a = a 2 + log (1 + 2 ) (ii) The points of intersection of the parabola y2 = 4ax and the line 3y = 8x are obtained as, 3y y 2 = 4a 8 3a y y − = 0 2
y = 0,
3a 9a and x = 0, 2 16
9a 3a Hence, A : , is the point of intersection. 16 2 For the arc OA, y varies from 0 to 3a . 2
Length of the arc OA, s =
3a 2
1 2a ∫0
1 y = 2a 2
y 2 + 4a 2 dy
(
4a 2 y 2 + 4a 2 + log y + y 2 + 4a 2 2
)
3a 2
0
3a 9a 2 1 3a 9a 2 = + 4a 2 + 2a 2 log + + 4a 2 − log 2a 2 4 2a 4 4 1 3a 5a 3a 5a 2 g + − log 2a = ⋅ + 2a log 2 2a 4 2 2
4.3 Length of Plane Curves 4.19
1 15a 2 + 2a 2 log 2 2a 8 15 = a log 2 + 16 =
exercIse 4.2 1. Find the length of the arc of following curves: xˆ Ê (i) y = log Á tanh ˜ from x = 1 to x = 2 Ë 2¯ (ii) 24 xy = x 4 + 48 from x = 2 to x = 4 3
(iii) x = 3y 2 - 1 from y = 0 to y = 4 (iv) y = x (2 - x) from x = 0 to x = 2
È 1ˆ 17 8 Ê (82 82 - 1), (iv) 1 log (2 + 5 )+ 5 ˘˙ Í Ans.: (i) log ÁË e + ˜¯ , (ii) , (iii) e 6 243 2 Î ˚ 2. Find the length of the curve y 2 = (2 x - 1)3 cut off by the line x = 4.
1022 ˘ È Í Ans.: 27 ˙ Î ˚ 3. Find the arc of the parabola y 2 = 4 a(a - x) cut off by the y-axis.
(
)
È Ans.: a È2 2 - log 3 - 2 2 ˘ ˘ ÍÎ Î ˚ ˙˚ 4. Find the length of the arc of the parabola y2 = 8x cut off by its latus rectum. Find the length of the arc cut off by the line 3y = 8x È 15 ˆ ˘ Ê Í Ans.: 4 ÈÎ 2 + log(1 + 2)˘˚ , 2 ÁË log 2 + ˜¯ ˙ 16 ˚ Î 5. Find the length of the arc of the parabola x2 = 4ay measured from the vertex to one extremity of the latus rectum. È Ans.: log È x + (1 + x 2 ) ˘ ˘ Î ˚ ˚˙ ÎÍ 6. Show that if s is the arc of the curve 9y 2 = x(3 - x)2 measured from the origin to the point P(x, y), then 3s 2 = 3y 2 + 4 x 2 . 7. Find the length of the loop of the curve (i)
3ay 2 = x(x - a)2
4.20
Chapter 4 Applications of Definite Integrals
(ii) 9y 2 = (x + 7)(x + 4)2 (iii) 9ay 2 = x(x - 3a)2 (iv) ay 2 = x 2 (a - x) 4 4 ˘ È a, (ii) 4 3, (iii) 4 3a, (iv) a˙ Í Ans.: (i) 3 3 ˚ Î
4.3.2
length of Arc in Parametric Form
When the equation of the curve is given in parametric form x = f1(t), y = f2(t), we have, from differential calculus, 2
ds dx dy = + dt dt dt
t2
t1
y=d
x = f (y)
2
The length of the arc of the curve between the points t = t1 and t = t2 is given by,
s=∫
y
2
y=c x
O Fig. 4.14
2
dx dy + dt dt dt
example 1 Find the length of the curve x = a(cos q + q sinq ), y = a(sin q – q cos q ), from q = 0 to q = 2p. Solution x = a(cos q + q sin q )
dx = a ( − sin q + sin q + q cos q ) = a q cos q dq y = a (sin q − q cos q ) dy = a (cos q − cos q + q sin q ) = a q sin q dq For the required arc, q varies from 0 to2p. Length of the curve,
s=∫
2p
=∫
2p
0
0 2p
2
(a q cos q ) 2 + (a q sin q ) 2 dq
= a ∫ q dq 0
2
dx dy + dq dq dq
4.3 Length of Plane Curves 4.21
q2 =a 2
2p
0
= 2ap 2
example 2 q qˆ q q Ê Find the length of the curve x = eq ÊÁ sin + 2 cos ˆ˜ , y = eq Á cos - 2 sin ˜ Ë Ë 2 2¯ 2 2¯ measured from q = 0 to q = p.
Solution q q x = eq sin + 2 cos 2 2 dx q q q 5 q q 1 = eq sin + 2 cos + eq cos − sin = eq cos dq 2 2 2 2 2 2 2 q q y = eq cos − 2 sin 2 2 5 q q q q q dy 1 = eq cos − 2 sin + eq − sin − cos = − eq sin 2 2 2 2 dq 2 2 2 For the required arc, q varies from 0 to p. Length of the curve
s=∫
p
=∫
p
=∫
p
0
0
0
2
2
dx dy + dq dq dq 25 2q q 25 q e cos 2 + e 2q sin 2 dq 4 2 4 2 25 2q e dq 4
5 p q e dq 2 ∫0 5 p = eq 0 2 5 p = (e − 1) 2 =
example 3 Find the length of the cycloid from one cusp to the next cusp x = a(q + sin q ), y = a(1 - cos q ).
4.22
Chapter 4 Applications of Definite Integrals
Solution
y
x = a (q + sin q ) dx = a (1 + cos q ) dq y = a (1 − cos q ) dy = a sin q dq For the arc OB, x varies from 0 to ap, hence q varies from 0 to p. Length of the arc AB,
A
−ap
B
O
ap
x
Fig. 4.15
s = 2 (Length of arc OB)
= 2∫
p
= 2∫
p
2
0
a 2 (1 + cos q ) 2 + a 2 sin 2 q dq
0
= 2a ∫
p
0
= 4a
2
dx dy + dq dq dq
∫
2(1 + cos q ) dq
p
0
cos
q dq 2 p
= 4a 2 sin
q 20
= 8a
example 4 Find the length of one arc of the cycloid x = a(q – sinq ), y = a(1 + cosq ). y
Solution
A
x = a (q − sin q ) dx = a (1 − cos q ) dq y = a (1 + cos q ) O dy = − a sin q Fig. 4.16 dq For the arc AB, x varies from 0 to 2ap, hence q varies from 0 to 2p. Length of the arc AB,
s=∫
2p
0
2
2
dx dy + dq dq dq
B
2ap
x
4.3 Length of Plane Curves 4.23
=∫
2p
a 2 (1 − cos q ) 2 + a 2 sin 2 q dq
0
= a∫
2p
= a∫
2p
0
0
2 − 2 cos q dq 2 ⋅ 2 sin 2
2p
= 2a ∫ sin 0
q dq 2
q dq 2 2p
= 2a −2 cos
q 20
= −4a (cos p − cos 0) = 8a
example 5 È Ê t ˆ˘ Find the length of the tractrix x = a Ícos t + log tan Á ˜ ˙ , y = a sin t from Ë 2¯ ˚ Î p t = to any point t. 2 Solution t x = a cos t + log tan 2 1 dx t 1 sec 2 ⋅ = a − sin t + 2 2 dt t tan 2 1 = a − sin t + t t 2 sin cos 2 2 1 = a − sin t + sin t (1 − sin 2 t ) sin t cos 2 t =a sin t y = a sin t dy = a cos t dt =a
4.24
Chapter 4 Applications of Definite Integrals
For the required arc, t varies from
p to t. 2 2
Length of the curve,
2
t dx dy s = ∫p + dt dt dt 2
cos 4 t + a 2 cos 2 t dt sin 2 t
t
= ∫p a 2 2
t
= a ∫p cos t cot 2 t + 1 dt 2 t
= a ∫p cot t dt 2
t
= a log sin t p 2
= a log sin t
example 6 For the curve x = a (2 cos t - cos 2t ), y = a (2 sin t - sin 2t ), show that the length of the arc of the curve measured from t = 0 to the point where the tangent makes an angle y with the tangent, at t = 0 is given by y s = 16 a sin 2 . 6 Solution x = a (2 cos t − cos 2t ) dx = a ( −2 sin t + 2 sin 2t ) = 2a (sin 2t − sin t ) dt y = a ( 2 sin t − sin 2t ) dy = a ( 2 cos t − 2 cos 2t ) = 2a (cos t − cos 2t ) dt For the required arc, t varies from 0 to t. Length of the curve,
s=∫
t
=∫
t
2
2
dx dy + dt dt dt
0
4a 2 [(sin 2t − sin t ) 2 + (cos t − cos 2t ) 2 dt
0
= 2a ∫
t
0 t
2 − 2(sin 2t sin t + cos t cos 2t ) dt
= 2a ∫
2 [1 − cos(2t − t )] dt
= 2a ∫
2 (1 − cos t ) dt
0 t 0
4.3 Length of Plane Curves 4.25
= 2a ∫
t
0
2 ⋅ 2 sin 2
t
= 4a ∫ sin 0
t dt 2
t dt 2
t = 8a − cos 2
t 0
t = 8a 1 − cos 2 = 16a sin 2
t 4
… (1)
dy dy dt 2a (cos t − cos 2t ) = = dx dx 2a (sin 2t − sin t ) dt
3t t sin 2 2 = tan 3t = 3t t 2 2 cos sin 2 2 2 sin
At t = 0, y = 0,
dy =0 dx
Hence, the tangent is x-axis at t = 0. At the point where tangent makes an angle y with the tangent at t = 0, i.e., x-axis, we get dy = tany dx tan
3t = tan y 2
3t 2 2y t= 3
y =
Putting t in Eq. (1),
2y 12 2 y = 16a sin 6
s = 16a sin 2
4.26
Chapter 4 Applications of Definite Integrals
example 7 2
2
x 3 y 3 Find the total length of the curve ÊÁ ˆ˜ + ÊÁ ˆ˜ = 1. Hence, deduce the Ë a¯ Ë b¯
total length of the curve
2 x3
+
divides the length of the curve ratio 1:3.
2 y3
=
2 a3.
+
2 y3
2 x3
Also show that the line q = =
2 a3
p 3
in the first quadrant in the
Solution
y
(i) The parametric equations of the curve 2 3
(0, b) B
2 3
q= p 3
x y + = 1 are given by, a b
C p 3
x = a cos q , y = b sin q 3
3
(−a, 0)
dx = −3a cos 2 q sin q dq dy = 3b sin 2 q cos q dq
O
(0, −b)
For the arc AB, x varies from a to 0, hence q p varies from 0 to . 2 Total length of the curve, s = 4 (Length of the arc AB) 2
Fig. 4.17
2
p dx dy = 4∫ 2 + dq 0 dq dq p
= 4∫ 2 9a 2 sin 2 q cos 4 q + 9b 2 sin 4 q cos 2 q dq 0
p
= 12∫ 2 sin q cos q a 2 + (b 2 − a 2 ) sin 2 q dq 0
Putting a + (b − a ) sin q = t , 2(b 2 − a 2 ) sin q cos q dq = 2t dt 2
2
2
2
2
sin q cos q dq =
when q = 0, p when q = , 2
t dt b − a2 2
t=a t=b b
s = 12∫ t ⋅ a
t dt b − a2 2
A (a, 0)
x
4.3 Length of Plane Curves 4.27
12 t 3 = 2 b − a2 3
b
a
4(b − a ) b2 − a 2 4(a 2 + ab + b 2 ) = a+b 3
3
=
(ii) Putting b = a, 2
2
2
Total length of the curve ( x 3 + y 3 = a 3 ) = 2
2
4( a 2 + a 2 + a 2 ) = 6a 2a
2
(iii) Length of the curve x 3 + y 3 = a 3 in the first quadrant =
3 6a = a 2 4
p
Length of the arc AC = ∫ 3 3a sin q cos q dq 0
=
3a p3 sin 2q dq 2 ∫0
3a − cos 2q = 2 2
p 3 0
9a = 8 Length of the arc BC = length of the arc AB − length of the arc AC 3a 9a 3a − = 2 8 8 Length of the arc BC 1 = Length of the arc AC 3 =
example 8 Show that the length of the arc of the curve x sin q + y cos q = f ¢(q), x cos q – y sin q = f ¢¢(q) is given by s = f (q) + f ¢¢(q) + C. Solution x sin q + y cos q = f ′ (q )
… (1)
x cos q − y sin q = f ′′ (q )
… (2)
Multiplying Eq. (1) by sin q and (2) by cos q and adding, x = sin q f ′(q ) + cos q f ′′(q )
… (3)
4.28
Chapter 4 Applications of Definite Integrals
Multiplying Eq. (1) by cos q and (2) by sin q and subtracting, y = cos q f ′(q ) − sin q f ′′(q ) dx = cos q f ′(q ) + sin q f ′′(q ) − sin q f ′′(q ) + cos q f ′′′(q ) dq = cos q [ f ′(q ) + f ′′′(q )] dy = cos q f ′′(q ) − sin q f ′(q ) − cos q f ′′(q ) − sin q f ′′′(q ) dq = − sin q [ f ′(q ) + f ′′′(q )] 2
Length of the arc,
s=∫
2
dx dy + dq dq dq
= ∫ (cos 2 q + sin 2 q ) [ f ′(q ) + f ′′′(q )]2 dq = ∫ [ f ′(q ) + f ′′′(q )] dq = f (q ) + f ′′(q ) + C
exAmPle 9 Show that for the curve 8a 2 y 2 = x 2 (a 2 - x 2 ), arc length a s= (2q + sin q cos q) where x = a sinq and that the perimeter of 2 2 pa one of the loop is . 2 y
Solution When
x = a sinq 8a 2 y 2 = a 2 sin 2 q ( a 2 − a 2 sin 2 q )
= a 4 sin 2 q cos 2 q a a y= sinq cosq = sin 2q 2 2 4 2 dx = a cos q dq dy a = cos 2q dq 2 2 For the upper half of the loop OA, x varies from 0 to a, hence q varies from 0 to
p . 2
B (−a, 0)
O
Fig. 4.18
A (a, 0) x
4.3 Length of Plane Curves 4.29
Length of one loop,
s = 2( Length of upper half of the loop OA) = 2∫
2
p 2
2
dx dy + dq dq dq
0 p
= 2 ∫ 2 a 2 cos 2 q + 0
=
a
=
a
=
a
=
a
2 2 2
p 2
∫
0 p 2
∫
0
∫
p 2
0
∫ 2
p 2
0
a2 cos 2 2q dq 8
8 cos 2 q + ( 2 cos 2 q − 1) 2 dq 4 cos 4 q + 4 cos 2 q + 1 dq ( 2 cos 2 q + 1) dq ( 2 + cos 2q ) dq p 2
sin 2q = 2q + 2 2 a = (p ) 2 pa = 2 a
0
example 10 Show that the length of one complete wave of the curve y = b cos
x is a
equal to the perimeter of the ellipse whose semi-axes are a 2 + b2 and a. Solution y = b cos
(i)
x a
dy b x = − sin dx a a For one complete wave
x varies from 0 to 2p i.e., x varies from 0 to 2pa. a 2p a
Length of one complete wave, s1 = ∫0
2
dy 1 + dx dx
4.30
Chapter 4 Applications of Definite Integrals
=∫
2p a
1+
0
x 1 2p a 2 a + b 2 sin 2 dx ∫ 0 a a
= Putting
b2 x sin 2 dx a a2
x = t, a
dx = a dt when x = 0, when x = 2pa,
t=0 t = 2p s1 =
1 2p 2 a + b 2 sin 2 t ⋅ a dt a ∫0
=∫
2p
a 2 + b 2 sin 2 t dt
0
Q a + b sin t dt
p
= 2∫
2
0
= 4∫
p 2
2
2
∫
2a
0
a 2 + b 2 sin 2 t dt
0
f (t ) dt = 2 ∫ f (t ) dt 0 if f ( 2a − t ) = f (t ) (1) a
(ii) Now, parametric equations of the given ellipse are x = a 2 + b 2 cost and y = a sin t y
dx dy = − a 2 + b 2 sin t , = a cos t dt dt
B (0, a)
For the arc AB, x varies from a + b p to 0, hence t varies from 0 to . 2 2
2
A
x
(√a 2 + b 2, 0(
Perimeter of the ellipse, s2 = 4 (Length of the arc AB) 2
2
p dx dy = 4∫ 2 + dt 0 dt dt
Fig. 4.19
p
= 4∫ 2 (a 2 + b 2 ) sin 2 t + a 2 cos 2 t dt 0
p
= 4∫ 2 a 2 + b 2 sin 2 t dt 0
From Eqs (1) and (2), Length of one complete wave = perimeter of the ellipse.
… (2)
4.3 Length of Plane Curves 4.31
example 11 x 2 y2 Show that the perimeter of the ellipse + 2 = 1 is 2 a b È e2 12 ◊ 3 4 12 ◊ 32 ◊ 5 6 ˘ 2p a Í1 - 2 - 2 2 e - 2 2 2 e -º˙ , where e is the eccentricity 2 ◊4 2 ◊ 4 ◊6 Î 2 ˚ of the ellipse. Solution The parametric equations of the given ellipse are x = a cosq and y = b sinq . dx = − a sin q dq dy = b cos q dq
y B (0, b) A (a, 0) x
O
For the arc AB, x varies from a to 0, hence q p varies from 0 to . 2
Fig. 4.20
Perimeter of the ellipse = 4(Length of the arc AB) 2
2
p dx dy = 4 ∫ 2 + dq 0 dq dq p
= 4∫ 2 a 2 sin 2 q + b 2 cos 2 q dq 0
p
= 4∫ 2 a 2 sin 2 q + a 2 (1 − e 2 ) cos 2 q dq 0
p
b2 Q e = 1 − 2 a
1
= 4a ∫ 2 (1 − e 2 cos 2 q ) 2 dq 0
= 4a ∫
p 2
0
11 2 − 1 1 1 + ( −e 2 cos 2 q ) + 2 ( −e 2 cos 2 q ) 2 2 2! 11 1 2 − 1 2 − 2 2 ( −e 2 cos 2 q )3 + K dq + 3!
p 1 4 1⋅ 3 6 1 = 4a ∫ 2 1 − e 2 cos 2 q − e cos 4 q − e cos 6 q K dq 0 2 2⋅ 4 2⋅ 4⋅6
4.32
Chapter 4 Applications of Definite Integrals
1 p 1 43 1 p 1⋅ 3 6 5 ⋅ 3 ⋅1 p p 1 = 4a − e 2 ⋅ ⋅ − e ⋅ ⋅ − e ⋅ ⋅ − K 2 2 2⋅ 4 4 2 2 2⋅ 4⋅6 6⋅ 4⋅ 2 2 2 2
e 2 12 ⋅ 3 12 ⋅ 32 ⋅ 5 = 2p a 1 − 2 − 2 2 e 4 − 2 2 2 e6 ⋅ K 2 ⋅4 2 ⋅4 ⋅6 2
exercIse 4.3 1. Find the length of the following curves: (i) x = a (2 cos q + cos 2q ), y = a (2 sin q + sin2q ), from q = 0 to any point q. (ii) x = a (q - sin q ), y = a (1 - cos q ) from q = 0 to q = 2p q (iii) x = aeq sin q , y = ae cos q from q = 0 to q =
p 2
(iv) x = log(sec q + tan q ) - sinq , y = cos q from q = 0 to any point q (v)
x = a (t - tanh t ), y = a sech t from t = 0 to any point t.
Êa+b ˆ Êa+b ˆ (vi) x = (a + b) cosq - bcos Á q ˜ , y = (a + b) sinq - b sin Á q ˜ from Ë b ¯ Ë b ¯ pb q = to any point q. a (vii) x = a sin2q (1 + cos 2q ), y = a cos 2q (1 - cos 2q ), from q = 0 to any point q q È ˘ (ii) 8a Í Ans.: (i) 8a sin 2 , ˙ Í ˙ p Í ˙ 2 e a (iii) 2( 1) , (iv) log sec q Í ˙ Í 4b Ê aq ˆ ˙ (v) log cosh t (vi) (a + b)cos Á ˜ ˙ Í Ë 2b ¯ ˙ a Í Í ˙ 4 a sin q (vii) Í ˙ 3 Î ˚ 1 3 t is of length 4 3. 3 3. Show that the length of the arc of the curve x = a(3 sinq - sin3 q ), 3 y = a cos3q measured from (0, a) to any point (x, y) is a(q + sin q cos q ). 2 4. If ‘s’ be the length of the arc of the curve x = a(q + sinq cosq ), p y = a(1 + sinq )2, measured from the point q = - to a point q, show that 2 s4 varies as y3. 2. Prove that the loop of the curve x = t2, y = t -
4.3 Length of Plane Curves 4.33
4.3.3
length of Arc in Polar Form
For the curve r = f (q ), we have, from differential calculus, ds dr = r2 + dq dq
2
q=p 2
The length of the arc of the curve r = f (q) between the points q = q 1 and q = q 2 is given by,
P (r, q ) r
2
s=Ú
q2
q1
Ê dr ˆ r 2 + Á ˜ dq Ë dq ¯
q q1 q2
Similarly, the length of the arc of the curve q = f (r) between the points r = r1 and r = r2 is given by,
q =0 Fig. 4.21
2
s=Ú
r2
r1
Ê dq ˆ 1 + r 2 Á ˜ dr Ë dr ¯
example 1 Find the length of the spiral r = e2q from q = 0 to q = 2p. Solution r = e 2q dr = 2e 2q dq For the required length of the spiral, q varies from 0 to 2p.
Length of the spiral,
2p
s=∫
0
=∫
2
dr r 2 + dq dq
2p
e 4q + 4e 4q dq
0
= 5∫
2p 0
e 2q = 5 2 =
e 2q dq 2p
0
5 4p (e − 1) 2
4.34
Chapter 4 Applications of Definite Integrals
example 2 Find the length of the arc of the equiangular spiral r = aeq cot a from the point corresponding to q = 0 to the point corresponding to q = tan a. Solution r = aeq cot a
dr = a cot a eq cot a dq For the required arc, q varies from 0 to tan a. Length of the arc,
s=∫
tan a
=∫
tan a
0
0
2
dr r 2 + dq dq a 2 e 2q cot a + a 2 cot 2 a e 2q cot a dq
= a 1 + cot 2 a ∫ = a cosec a
tan a 0
eq cot a dq
tan a
eq cot a cot a
0
a cosec a tan a cot a (e − 1) cot a = a sec a (e − 1) = a (e − 1) sec a =
example 3 Find the length of the cissoid r = 2 a tan q sin q from q = 0 to q = Solution r = 2a tan q sin q
dr = 2a (sec 2 q sin q + tan q cos q ) dq = 2a sin q (sec 2 q + 1)
For the required arc length of the cissoid, q varies from 0 to 2
Length of the curve,
p dr s = ∫ 4 r 2 + dq 0 dq
p . 4
p . 4
4.3 Length of Plane Curves 4.35
p
= ∫ 4 4a 2 tan 2 q sin 2 q + 4a 2 sin 2 q (sec 2 q + 1) 2 dq 0
p
= ∫ 4 4a 2 sin 2 q (sec 2 q − 1 + sec 4 q + 2 sec2 q + 1) dq 0
p
= ∫ 4 4a 2 sin 2 q sec 2 q (sec 2 q + 3) dq 0
p
= ∫ 4 4a 2 tan 2 q (tan 2 q + 4) dq 0
p
= ∫ 4 2a tan q tan 2 q + 4 dq 0
Putting
tan 2 q + 4 = t 2 ,
2 tan q sec 2 q dq = 2t dt
t dt t dt t dt = = 2 2 2 sec q 1 + tan q t − 3 q = 0, t = 2 p q = , t= 5 4
tan q dq = When When
t2 dt 2 t −3 5 3 dt = ∫ 2a 1 + 2 2 t − 3 5
s = ∫ 2a ⋅
= 2a t +
2
3 2 3
log
t+ 3
3 log = 2a 5 + 2 = 2a 5 − 2 + = 2a 5 − 2 +
5
t− 3 2
5− 3 5+ 3
3 log 2
(
−2−
3 2 − 3 log 2 2 + 3
)
(
2 2 − 3 5 − 3 − log 5 − 3 4 − 3
) 2
3 log 4 − 15 − log 7 − 4 3 2
(
)
(
)
Note: Only positive values of t are considered since q lies in the first quadrant.
example 4 Find the length of the whole arc of the cardioid r = a (1 + cos q ) and p show that the upper half is bisected by the line q = . 3
4.36
Chapter 4 Applications of Definite Integrals
Solution
q= p 2 C
r = a (1 + cos q ) dr = − a sin q dq (i) For the arc BACDO, q varies from 0 to p.
q= p 3 A
D B
O
q=p
q=0
Length of the whole arc of the curve, s = 2( Length of arc BACDO ) dr = 2 r + dq
Fig. 4.22
2
2
= 2∫
p
a 2 (1 + cos q ) 2 + ( − a sin q ) 2 dq
0 p
= 2∫ a 2 + 2 cos q dq 0
p
= 2∫ a 2 ⋅ 2 cos 2 0
p
= 4a ∫ cos 0
= 4a 2sin
q dq 2
q dq 2
q 2
p
0
= 8a Length of the upper half of the cardioid = 4a p intersects the cardioid at point A. 3 p q q Length of the arc, BA = ∫ 3 2a cos dq = 2a 2sin 0 2 2
(ii) Let q =
p 3
= 2a
0
Hence, the upper half of the cardioid is bisected by the line q =
p . 3
example 5 Find the length of the cardioid r = a (1 - cos q ) lying outside the circle r = a cos q . Solution The points of intersection of cardioid r = a (1 − cos q ) and the circle r = a cos q is obtained as,
4.3 Length of Plane Curves 4.37
a (1 − cos q ) = a cos q 1 = 2 cos q 1 cos q = 2 q =±
Hence at A,
q =
q= p 3
q= p 2 A
p 3
q=p B
p 3
O
q=0
r = a (1 − cos q ) Fig. 4.23
dr = a sin q dq
For the arc of the cardioid lying outside the circle, q varies from
p to p. 3
Length of the cardioid lying outside the circle, s = 2 (Length of arc AB) 2
dr r2 + dq dq
p
= 2 ∫p
3 p
= 2 ∫p
a 2 (1 − cos q ) 2 + ( a sin q ) 2 dq
3 p
= 2∫p a 2 − 2 cos q dq 3
p
= 2∫p a 2 ⋅ 2 sin 2 3
q dq 2
q dq 2 3 p q = 4a −2 cos 2p p
= 4a ∫p sin
3
3 = −8a − 2 = 4a 3
example 6 Show that the length of the arc of that part of cardioid r = a (1 + cos q ) which lies on the side of the line 4r = 3a sec q away from the pole is 4a.
4.38
Chapter 4 Applications of Definite Integrals
Solution The points of intersection of cardioid r = a (1 + cos q ) and the line 4r = 3a sec q are obtained as, 3a a (1 + cos q ) = secq 4 4 (1 + cos q ) cos q = 3
q= p 2
A
B (2a, 0) q = 0
O
4r = 3a sec q
4 cos q + 4 cos 2 q − 3 = 0 (2 cos q + 3) (2 cos q − 1) = 0 cos q =
q= p 3
C Fig. 4.24
1 −3 and cos q = (does not exist) 2 2
q =±
p 3
Hence at A, q =
p 3
r = a (1 + cos q ) dr = − a sin q dq
For the arc BA, q varies from 0 to Length of the arc CBA,
p . 3
s = 2 (length of arc BA) 2
p dr = 2∫ 3 r 2 + dq 0 dq p
= 2∫ 3 a 2 (1 + cos q ) + ( − a sin q ) dq 2
0
p
= 2∫ 3 a 2 + 2 cos q dq 0
p
= 2∫ 3 a 2 ⋅ 2 cos 2 0
p
= 4a ∫ 3 cos 0
q dq 2
q = 4a 2 sin 2 = 4a
p 3 0
q dq 2
2
4.3 Length of Plane Curves 4.39
example 7 Find the total length of the curve r = a sin 3
q . 3
q= p 2
Solution r = a sin 3
q 3
B
q q 1 dr = a ⋅ 3sin 2 cos ⋅ dq 3 3 3 q q = a sin 2 cos 3 3 For the arc OABCD, q varies from 0 to
A C
3p 2
0
= 2∫
3p 2
0
Fig. 4.25
3p 2
2
dr r 2 + dq dq a 2 sin 6
q q q + a 2 sin 4 cos 2 dq 3 3 3
q dq 3 3p 2q = a∫ 2 1 − cos dq 0 3 = 2∫
0
a sin 2
3 2q = a q − sin 2 3
3p 2 0
3p 3 = a⋅ = pa 2 2
example 8 Find the perimeter of the lemniscate r 2 = a 2 cos 2q . Solution
r 2 = a 2 cos 2q 2r
dr = a 2 ( − sin 2q ) ⋅ 2 dq
dr a2 = − sin 2q dq r p For the arc OBA, q varies from 0 to . 4
q=0
D
3p . 2
Length of the curve = 2 (Length of the arc OABCD) = 2∫
O
4.40
Chapter 4 Applications of Definite Integrals
Perimeter of the curve
q= p 2
= 4 (Length of the arc OBA)
q= p 4
2
p dr = 4 ∫ 4 r 2 + dq 0 dq
= 4∫
A
p 4
a4 a cos 2q + 2 sin 2 2q dq r 2
0 p
= 4 ∫ 4 a 2 cos 2q + 0
p
= 4 ∫ 4 a 2 cos 2q + 0
p
a 4 sin 2 2q dq a 2 cos 2q
2
0
p
= 4a ∫ 4 0
1 cos 2q
Putting
2q = t , 2dq = dt
When q = 0,
t=0
dq
p p , t= 2 4
Perimeter of the curve =
4a p2 dt 2 ∫0 cos t −1
p
= 2a ∫ 2 sin 0 t ⋅ (cos t ) 2 dt 0
1 1 = aB , 2 4 a =
a =
O
a4 sin 2 2q dq r2
a4 dq a cos 2q
= 4∫ 4
When q =
B
1 1 2 4 3 4 1 1 2 4 1 1 1− 4 4
2
Fig. 4.26
q=0
4.3 Length of Plane Curves 4.41
1 1 a 2 4 = p p sin 4
=
2
1 a p 4
2
p Q n 1 − n = sin np
p 2 a 1 2p 4
=
2
example 9 2a = 1 + cos q , the arc intercepted between r the vertex and the extremity of the latus rectum is a ÈÎ 2 + log(1 + 2)˘˚ . Show that for the parabola
Solution The latus rectum is the line q =
q= p 2 B
p . 2
2a = 1 + cos q r r=
2a = 1 + cos q
2a q 2 cos 2
= a sec 2
2
q 2
A
O
dr q q = a sec 2 ⋅ tan dq 2 2 For the arc AB, q varies from 0 to
Fig. 4.27
p . 2 2
Length of the arc AB,
p dr s = ∫ 2 r 2 + dq 0 dq p
= ∫ 2 a 2 sec 4 0
p
= ∫ 2 a sec 2 0
q q q + a 2 sec 4 tan 2 dq 2 2 2
q q 1 + tan 2 dq 2 2
q=0
4.42
Chapter 4 Applications of Definite Integrals
Putting
tan
q = t, 2
1 2q q sec dq = dt , sec 2 dq = 2dt 2 2 2 When
q = 0,
When
q =
t=0
p , t=1 2 1
s = ∫ 2a 1 + t 2 dt 0
= 2a
(
t 1 1 + t 2 + log t + 1 + t 2 2 2
(
)
1
0
)
1 1 2 + log 1 + 2 = 2a 2 2
(
)
= a 2 + log 1 + 2
example 10 Show that the whole length of the limacon r = a cos q + b (a < b) is equal to that of an ellipse whose semi-axes are equal in length to the maximum and minimum radii vectors of the limacon. Solution
q= p 2 B
r = a cos q + b dr = − a sin q dq
For the arc ABC, q varies from 0 to p.
q=p
C (b−a, 0)
O
whole length of the limacon
q=0 A (b + a, 0)
= 2(length of the arc ABC ) = 2∫
p
0 p
= 2∫
0
= 2∫
p
0
Fig. 4.28
2
dr r 2 + dq dq (a cos q + b) 2 + ( − a sin q ) 2 dq
a 2 + b 2 + 2ab cosq dq
… (1)
Maximum radius vector of the limacon = a (1) + b = b + a
[Q Maximum value of cosq
= 1]
4.3 Length of Plane Curves 4.43
Minimum radius vector of the limacon = a ( −1) + b = b − a
[Q Minimum value of cosq The parametric equations of the ellipse with above radii vectors as semi-axes are given as, x = (b + a ) cos q and y = (b − a ) sin q
= −1]
q= p 2 B (0, b- a)
dx = −(b + a) sin q dq dy = (b − a ) cos q dq p For the arc AB, q varies from 0 to . 2
O
A (b+a, 0) q =0
Fig. 4.29
whole length of the ellipse = 4(length of the arc AB) 2
2
p dx dy = 4 ∫ 2 + dq 0 dq dq p
= 4∫ 2 [ −(b + a ) sin q ]2 + [(b − a ) cosq ]2 dq 0
p
= 4∫ 2 (a 2 + b 2 + 2ab sin 2 q − 2ab cos 2 q ) dq 0
p
= 4∫ 2 (a 2 + b 2 − 2ab cos 2q ) dq 0
Putting
When When
2q = t , dt dq = 2 q = 0, t = 0 p q = , t=p 2
whole length of the ellipse = 2∫
p
= 2∫
p
= 2∫
p
0
0
0
a 2 + b 2 − 2ab cos t dt a 2 + b 2 − 2ab cos (p − t ) dt a 2 + b 2 + 2ab cos t dt
Q a f ( x) dx = a f (a − x) dx ∫0 ∫0 From Eqs. (1) and (2), whole length of the limacon = whole length of the ellipse
… (2)
4.44
Chapter 4 Applications of Definite Integrals
example 11 Find the length of the arc of the hyperbolic spiral rq = a from the point r = a to r = 2a. Solution rq = a r
dq +q = 0 dr
Length of the arc,
s=∫ =∫ =∫ =∫
Putting
2
dq 1 + r 2 dr dr
2a a 2a
1 + q 2 dr
a 2a
1+
a
a2 dr r2
r 2 + a2 dr r2
2a a
r 2 + a2 = t 2 , t dt
2r dr = 2t dt , dr =
t − a2 2
When
r = a,
t=a 2
When
r = 2a,
t=a 5
s=∫
t − a t − a2 2 2 5t −a +a dt 2 t 2 − a2 a 5 5 dt dt + a 2 ∫ a 2 t 2 − a2 2 2
a 2 a
=∫
a
=∫
a
a
a 2
=a
(
=a
(
2
2
2
a 5
= t
t dt
t
a 5
+
a2 t−a log 2a t+a
a 5
a 2
a a 5−a a 2 − a − log log 2 a 5+a a 2 + a 5 −1 2 − 1 a 5 − 2 + log − log 2 5 +1 2 + 1
)
5− 2 +
)
4.3 Length of Plane Curves 4.45
=a
(
=a
(
=a
(
)
a log 2 a 5 − 2 + log 2
5 − 1 2 + 1 5 + 1 2 − 1
)
5 −1
5− 2 +
5− 2
)
( 2( + a log
) 2 + 1)
5 +1
(
2
)
2 2 +1 2 −1
5 +1
exercIse 4.4 1. Find the perimeter of the following curves: (i)
r = a cos q
(ii)
r = a(q 2 - 1)
(iii)
Êq ˆ r = a cos3 Á ˜ Ë 3¯
(iv)
r = ae mq
(v)
r = aq
(vi)
r = a sec2
q 2
(vii) r = 4 sin2 q È ˘ ÍAns. : ˙ Í ˙ 8a (i) p a (ii) Í ˙ 3 Í ˙ 2 Í ˙ 3p a 1+ m (iii) (iv) (r2 - r1 ) Í ˙ 2 m Í ˙ a Í (v) Èq 1 + q 2 + sinh-1 q ˘ (vi) 2a È 2 + log( 2 + 1)˘ ˙ Î ˚˙ ˚ 2Î Í 4 Í ˙ (vii) 8 + log( 3 + 2) Í ˙ 3 Î ˚ 2. Find the perimeter of the cardioid r = a(1 - cosq ) and prove that the line 2p q= bisects the upper half of the cardioid. 3 [Ans.: 8a] 3. Find the length of the cardioid r = a(1 + cosq ) which lies outside the circle r + a cosq = 0. È Ans.: 4 3a ˘ Î ˚ q cota 4. Prove that the length of the spiral r = ae as r increases from r1 to r2 is given by (r2 - r1)sec a .
4.46
Chapter 4 Applications of Definite Integrals
5. Find the length of the cardioid r = a(1 - cos q ) lying inside the circle r = a cos q . È Ê 3ˆ˘ Í Ans. : 8a Á 1 ˙ 2 ˜¯ ˙˚ Ë ÍÎ 6. Find the length of the spiral r = ae mq lying inside the circle r = a. a È 2 ˘ Í Ans.: m 1 + m ˙ Î ˚ 7. Find the length of the arc of parabola
l = 1 + cos θ cut off by its latus r
rectum.
(
)
Ans. : l 2 + log 1 + 2
4.4 AreA oF surFAce oF solId oF reVolutIon Let y = f (x) be a curve included between two lines x = a and x = b. Let P(x, y) be any point on the curve. when the chord PQ is revolved about the x-axis, a solid of revolution is generated. The elementary surface area d S is y = f (x) approximately equal to the circumference of the y B circle multiplied by the PQ. Q A P d S = 2p yPQ = 2p y d s The total surface area of the solid of revolution about x-axis is given by, S = ∫ 2π y ds
O
x=a
x=b
x
Fig. 4.30
4.4.1 Area of surface of solid of revolution in cartesian Form Area of surface generated by revolving the arc of the curve y = f (x) about the x-axis is given by, b
S = ∫ 2p y ds a
b
= ∫ 2p y a
ds dx dx 2
b dy = ∫ 2p y 1 + dx a dx
4.4 Area of Surface of Solid of Revolution 4.47
Similarly, the area of the surface generated by revolving the arc of the curve x = f (y) about y-axis is given by, d
S = ∫ 2p x ds c
d
= ∫ 2p x c
ds dy dy 2
=∫
d c
dx 2p x 1 + dy dy
example 1 Find the area of the surface of revolution generated by revolving the curve x = y3 from y = 0 to y = 2. Solution x = y3
y
dx = 3y2 dy
A(8, 2) 2
dx ds = 1 + = 1 + 9 y4 dy dy The area of the surface is generated by revolving the region about the y-axis. For the region shown, y varies from 0 to 2. Surface area,
2
S = ∫ 2p x 0
x
O Fig. 4.31
ds dy dy
2
= ∫ 2p y 3 1 + 9 y 4 dy 0
=
1 2p 2 (1 + 9 y 4 ) 2 (36 y 3 )dy ∫ 36 0 2
=
3 4 2
p (1 + 9 y ) 3 18 2
(
[ f ( y )]n +1 n Q ∫ [ f ( y )] f ′( y ) dy = n +1 0
)
π = 145 145 − 1 27
4.48
Chapter 4 Applications of Definite Integrals
example 2 Find the area of the surface of revolution of the solid generated by x 2 y2 revolving the ellipse + = 1 about the x-axis. 16 4 Solution x2 + 16 2x 2 y + 16 4
y2 =1 4 dy =0 dx x dy =− dx 4y
y B (0, 2)
ds dy = 1+ dx dx = 1+
O
(− 4, 0) 2
A x (4, 0)
(0, −2)
x2 16 y 2
Fig. 4.32
x 2 + 16 y 2 4y The area of the surface of solid is generated by revolving the upper half of the ellipse about the x-axis. For the region above the x-axis, x varies from - 4 to 4. Due to symmetry about the y-axis, considering the region in the first quadrant where x varies from 0 to 4, 4 ds Surface area, S = 2 ∫ 2π y dx 0 dx =
4
= 4p ∫ y 0
= p∫
4
x 2 + 16 y 2 dx 4y
x 2 + 64 − 4 x 2 dx
0
2
= p 3∫
4
0
8 2 − x d x 3
x 64 1 64 x 3 − x 2 + ⋅ sin −1 =p 3 2 3 2 3 8 64 32 3 = p 3 2 − 16 + sin −1 3 3 2 4p = 8p 1 + 3 3
4
0
4.4 Area of Surface of Solid of Revolution 4.49
example 3 The part of the parabola y2 = 4ax cut off by the latus rectum revolves about the tangent at the vertex. Find the surface area of the revolution. Solution The points of intersection of the parabola y2 = 4ax and its latus rectum x = a are obtained as, y y2 = 4a . a = 4a2 A y = ± 2a and x = a (a, 2a)
Hence, A: (a, 2a) and B: (a, -2a) are the points of intersection. Now,
y2 4a dx 2 y y = = dy 4 a 2 a x=
O
C (a, 0)
x
(a, −2a)
dx ds = 1+ dy dy
2
B Fig. 4.33
y2 4a 2
= 1+
The surface area is generated by revolving the region about the tangent at the vertex i.e., y-axis. For the region shown, y varies from -2a to 2a. Due to symmetry about x-axis, considering the region in the first quadrant where y varies from 0 to 2a, Surface area,
S = 2∫ = 2∫
Putting when when
2a 0 2a 0
2π x
ds dy dy
2π ⋅
y2 y2 1 + 2 dy 4a 4a
y = 2a tan q, dy = 2a sec2q dq y = 0, q = 0 p y = 2a, q = 4 S = 4p ∫
p 4
4a 2 tan 2 q 1 + tan 2 q 2a sec 2 q dq 4a
0
= 8p a 2 ∫
p 4 0
tan 2 q sec3 q dq
p
= 8p a 2 ∫ 4 (sec5 q − sec3 q ) dq 0
4.50
Chapter 4 Applications of Definite Integrals
= 8p a 2
1 3 3 tan q sec3 q + tan q sec q + log(sec q + tan q ) 4 8 8 p 4 1 1 − tan q sec q − log(sec q + tan q ) 2 2 0
[Using reduction formula] 3 3 1 1 1 = 8π a 2 2 2 + 2 + log( 2 + 1) − 2 − log 8 8 2 2 4 = π a 2 3 2 − log
(
(
)
2 +1
)
2 +1
example 4 Find the surface area generated by revolving the loop of the curve 9ay2 = x(3a - x)2 about the x-axis. Solution The points of intersection of the curve 9ay2 = x(3a - x)2 and x-axis are obtained as, 0 = x(3a - x)2 x = 0, 3a, 3a
and y = 0, 0, 0
y
Hence, A : (3a, 0) is the point of intersection. Now,
9ay 2 = x(3a − x) 2 dy 18ay = (3a − x) 2 − 2 x(3a − x) dx dy (3a − x) 2 − 2 x(3a − x) = 18ay dx (3a − x)(a − x) = 6ay ds dy = 1+ dx dx = 1+ = =
A (3a, 0)
O
2
(3a − x) 2 (a − x) 2 36a 2 y 2
36a 2 y 2 + (3a − x) 2 (a − x) 2 36a 2 y 2 1 4ax(3a − x) 2 + (3a − x) 2 (a − x) 2 6ay
Fig. 4.34
x
4.4 Area of Surface of Solid of Revolution 4.51
1 (3a − x) 2 (a + x) 2 6ay (3a − x)(a + x) = 6ay =
The surface area is generated by revolving the loop about the x-axis. For the loop, x varies from 0 to 3a. ds dx dx 3a (3a − x)(a + x) dx = 2p ∫ y ⋅ 0 6ay p 3a (3a 2 + 2ax − x 2 ) dx = 3a ∫0 3a
Surface area, S = ∫ 2p y 0
=
x3 p 3a 2 x + ax 2 − 3a 3
= 3p a
3a
0
2
example 5 Find the area of the surface of revolution of a quadrant of a circular arc as obtained by revolving it about a tangent at one of its ends. Solution Let x2 + y2 = a2 be the equation of the circle and let AC be the tangent at A. x2 + y 2 = a2 dy 2x + 2 y =0 dx dy x =− dx y
y C B P(x, y) 2
ds x2 dy = 1+ = 1+ 2 dx y dx
x
Fig. 4.35
a x +y = 2 y y 2
=
A (a, 0)
2
The surface area is generated by revolving the quadrant of circular arc APB about the line AC. If P(x, y) is any point on the circle, the distance of P from the tangent at A = a - x. For the region shown, x varies from 0 to a. a
Surface area, S = ∫ 2p (a − x) 0
ds dx dx
4.52
Chapter 4 Applications of Definite Integrals
a a = 2p ∫ (a − x) dx 0 y a a−x = 2p a ∫ dx 0 a2 − x2 a a x = 2p a ∫ − dx 0 a2 − x2 a2 − x2 −1 a a 1 = 2p a ∫ + (a 2 − x) 2 ( −2 x) dx 2 2 0 2 a −x
= 2π a a sin −1
x + a2 − x2 a
a
[ f ( x)]n +1 n Q ∫ [ f ( x)] f ′( x)dx = n +1
0
π = 2π a a − a 2 2 = π a (π − 2)
exercIse 4.5 1. Find the surface area of the solid generated by revolving the arc of the parabola y2 = 4ax bounded by its latus rectum about the x-axis. È ˘ 8a 2 p 2 2 -1 ˙ Í Ans.: 3 Î ˚
(
)
2. Find the area of the curved surface generated when one loop of the curve x2(a2 - x2) = 8a2y2 is revolved about the x-axis. È p a2 ˘ Í Ans.: ˙ 4 ˚ Î 3. Prove that the surface area of the solid obtained by revolving the ellipse È ˘ Ê 1ˆ b2x2 + a2y2 = a2b2 about the x-axis is 2p ab Í 1 - e 2 + Á ˜ sin-1 e ˙ , e being Ë ¯ e Î ˚ the eccentricity of the ellipse. 4. Show that the surface area of the solid obtained by revolving the arc of the curve y = sin x from x = 0 to x = p about the x-axis is p 2 È 2 + log 2 + 1 ˘ . Î ˚
(
)
5. Show that the area of the surface formed by rotating the curve y2 = x3 128p from x = 0 to x = 4 about the y-axis is 1 + 125 10 . 1215
(
)
4.4 Area of Surface of Solid of Revolution 4.53
6. Find the area of the curved surface of the cup formed by the revolution of the smaller part of the parabola y2 = 4ax cut off by the line x = 3a about its axis. 56 2 ˘ È Í Ans.: 3 p a ˙ Î ˚ 7. The arc of the parabola y2 = 4ax between its vertex and an extremity of its latus rectum revolves about its axis. Find the surface area traced out. 8 È 2˘ Í Ans.: 3 (2 2 - 1)p a ˙ Î ˚ 8. The arc of the curve a2y = x3 between x = 0 and x = a is revolved about the x-axis. Find the area of the surface so generated. È ˘ p a2 10 10 - 1 ˙ Ans.: Í 27 Î ˚
(
)
9. Find the surface area of the solid formed by the revolution of the loop of the curve 3ay2 = x (x - a)2 about the x-axis. È p a2 ˘ Í Ans.: ˙ 3 ˚ Î 10. Find the surface area of the solid generated by revolving the area bounded by the circle x2 + y2 = a2 about the line y = a. [Ans.: 4p 2a2]
4.4.2 Area of surface of solid of revolution in Parametric Form When the equation of the curve is given in parametric form x = f1(t), y = f2(t) with t1 ≤ t ≤ t2, the area of surface of solid of revolution about the x-axis is given by, t2
S = ∫ 2p y t1
ds dt dt 2
2
t2 dx dy = ∫ 2p y + dt t1 dt dt
Similarly, the area of surface of solid of revolution about the y-axis is given by, t2
S = ∫ 2p x t1
ds dt dt 2
2
t2 dx dy = ∫ 2p x + dt t1 dt dt
4.54
Chapter 4 Applications of Definite Integrals
example 1 Prove that the surface generated by the revolution of the tractrix 1 t x = a cos t + a log tan 2 , y = a sin t about its asymptote is equal to 2 2 the surface of the radius a. y Solution A (0, a)
1 t x = a cos t + a log tan 2 2 2 dx t 1 1 = − a sin t + a ⋅ ⋅ sec 2 ⋅ t dt 2 2 tan 2 a = − a sin t + sin t a cos 2 t = sin t y = a sin t dy = a cos t dt 2
ds dx dy = + dt dt dt
O
x
B (0, −a) Fig. 4.36
2
a 2 cos 4 t + a 2 cos 2 t sin 2 t a cos t = sin t =
The surface area is generated by revolving the tractrix about its asymptote, i.e., x-axis. For the region shown, x varies from – • to •, hence t varies from 0 to p. Due to symmetry about the y-axis, considering the region in the second quadrant where t p varies from 0 to , 2 p ds S = 2∫ 2 2p y dt Surface area, 0 dt p
= 4p ∫ 2 a sin t ⋅ 0
a cos t dt sin t
p
= 4p a 2 ∫ 2 cos t dt 0
p
= 4p a 2 sin t 02 = 4p a 2
4.4 Area of Surface of Solid of Revolution 4.55
example 2 Find the surface area of the solid generated by revolving the astroid 2 x3
+
2 y3
=
2 a3
about the x-axis.
Solution
q= p 2
The parametric equations of the astroid are x = a cos3 q , y = a sin 3 q dx = −3a cos 2q sinq , dq dy = 3a sin 2q cosq dq 2
ds dx dy = + dq dq dq
B (0, a)
(−a, 0)
O
A(a, 0)
q=0
2
(0, a)
= 9a 2 cos 4 q sin 2 q + 9a 2 sin 4 q cos 2 q Fig. 4.37 = 3a siin q cos q The surface area is generated by revolving the upper half of the astroid about the x axis. For the region shown, x varies from –a to a, hence q varies from p to 0. Due to symmetry about the y-axis, considering the region in the first quadrant, where p q varies from 0 to , 2 p
Surface area,
S = 2∫ 2 2p y 0
ds dq dq
p
= 4p ∫ 2 a sin 3 q ⋅ 3a sin q cos q dq 0
p
= 12p a 2 ∫ 2 sin 4 q cos q dq 0
sin 5 q = 12p a 5 2
p 2
0
[ f (q )]n + 1 n Q [ f ( q ) ] f ( q ) d q = ′ ∫ n +1
12 = p a2 5
example 3 Find the surface area of the solid formed by revolving one arch of the cycloid x = a(q - sin q ), y = a(1 - cos q ) about the y-axis.
4.56
Chapter 4 Applications of Definite Integrals
Solution x = a(q − sin q ) y
dx = a (1 − cos q ) dq y = a(1 − cos q ) dy = a sin q dq 2
ds dx d y = + dq dq dq
O
(2ap, 0) q = 2p
q =0
2
x
Fig. 4.38
= a 2 (1 − cos q ) 2 + a 2 sin 2 q = 2a 2 (1 − cosq q) q 2 The surface area is generated by revolving one arch of the curve about the y-axis. For the region shown, q varies from 0 to 2p. = 2a sin
Surface area, S = ∫
2p
0
2p x
ds dq dq
q dq 2 2p q q q = 4p a 2∫ q sin − 2 sin 2 cos dq 0 2 2 2 = 2p ∫
2p
0
a (q − sin q )2a sin
q q 4 q = 4p a q −2 cos − 1 −4 sin − sin 3 2 2 3 2
2p
2
0
[ f (q )]n +1 n Q [ f ( q ) ] f ( q ) d q = ′ ∫ n +1 = 4p a2(4p ) = 16p 2 a2
example 4 A circular arc of radius a revolves round its chord. Show that the surface of the spindle generated is 4p a 2(sin a - a cos a ), where 2a is the angle subtended by the arc at the centre. Find the surface area if the circular arc is a quadrant of circle.
4.4 Area of Surface of Solid of Revolution 4.57
Solution Taking the centre of the circle as origin and radius as a, the equation of the circle is x2 + y2 + a2. The parametric equations of the circle are: x = a cos q , y = a sin q y
dx = − a sin q , dq
dy = a cos q dq
2
ds dx d y = + dq dq dq
A M a
2
P(x, y)
q O
L
N C(a, 0)
x
a = a sin q + a cos q = a The arc ACB is revolved about the chord AB. If P(x, y) is any point on the circle and M B is the foot of perpendicular from P on AB, then Fig. 4.39 PM = ON − OL = x − a cosa For the region shown, q varies from –a to a. Due to symmetry about the x-axis, considering the region in the first quadrant where q varies from 0 to a, 2
2
a
2
Surface area, S = 2∫ 2p ( PM ) 0
2
ds dq dq
a
= 4p ∫ ( x − a cos a ) adq 0
a
= 4p a∫ (a cos q − a cos a ) dq 0
= 4p a 2 sin q − q cos a
a 0
= 4p a (sin a − a cos a ) 2
when circular arc is quadrant of a circle, a =
p 4
1 p 1 S = 4p a 2 − 2 4 2 =
p a2 2
(4 − p )
example 5 Show that the total surface area of the solid generated by the revolution Ê 1 - e2 1+ eˆ log of an ellipse about its minor axis is 2p a 2 Á 1 + ˜ , where a Ë 2e 1- e¯ is the semi-major axis and e is the eccentricity.
4.58
Chapter 4 Applications of Definite Integrals
Solution y
The parametric equations of the ellipse are, x = a cos q , y = b sin q dx = − a sin q , dq
dy = b cos q dq
2
ds dx d y = + dq dq dq
B (0, b) A (a, 0) x
O
(−a, 0)
(0, −b)
2
Fig. 4.40
= a 2 sin 2 q + b 2 cos 2 q The surface area of the solid is generated by the revolution of the ellipse about its minor p p axis. For the region shown, y varies from –b to b, hence q varies from − to . Due 2 2 to symmetry about x-axis, considering the region in the first quadrant where q varies p from 0 to , 2 p
S = 2 ∫ 2 2p x
Surface area,
0
ds dq dq
p
= 4p ∫ 2 a cos q a 2 sin 2 q + b 2 cos 2 q dq 0
p
= 4ap ∫ 2 cos q a 2 sin 2 q + b 2 (1 − sin 2 q ) dq 0
p
= 4ap ∫ 2 cos q b 2 + a 2 e 2 sin 2 q dq
where e =
0
Putting when
sinq = t cosq dq = d t q = 0, q =
when S = 4ap ∫
1
0
1−
b2 a2
t=0
p , t =1 2
b 2 + a 2 e 2 t 2 dt
= 4ap ⋅ ae∫
1
0
2
b t 2 + dt ae
t 2 b2 b2 b2 = 4p a e t + 2 2 + 2 2 log t + t 2 + 2 2 2 ae 2a e ae
1
2
0
1 b2 b 1 b a 2 e 2 + b 2 − 2 2 log = 4p a 2e a 2 e 2 + b 2 + 2 2 log 1 + ae ae 2a e 2a e 2ae 2
4.4 Area of Surface of Solid of Revolution 4.59
1 b2 a b2 b = 4p a 2e ⋅ a + 2 2 log 1 + − 2 2 log ae 2a e ae 2a e 2ae
Q b = a 1 − e 2
b2 a(1 + e) = 2p a 2 + log e b 1+ e b2 = 2p a 2 + log 2e 1− e
Q b = a 1 − e 2
1 − e2 1+ e = 2p a 2 1 + log 2e 1− e
exercIse 4.6 1. Find the surface area of the reel formed by the revolution of the cycloid x = a(q + sinq ), y = a(1 - cosq ) about (i) the tangent at the vertex, (ii) y-axis, and (iii) base. È 32 2 4ˆ 64 2 ˘ Ê p a , (ii) 4p a 2 Á 2p - ˜ , (iii) pa ˙ Í Ans.: (i) Ë ¯ 3 3 3 Î ˚ 2. Find the surface area of the solid generated by the revolution of the loop t3 of the curve x = t 2 , y = t about x-axis. 3 [Ans.: 3p ] 3. Show that the area of the surface of the solid generated by revolving the curve x = a(u - tanh u), y = a sechu, about the x-axis is equal to the area of the surface of a sphere of radius a. 4. Find the area of the surface of revolution generated by revolving the cardioid x = 2cosq - cos 2q, y = 2sinq - sin 2q, about the x-axis. 128p ˘ È Í Ans.: 5 ˙ Î ˚ 5. Find the area of the surface generated by revolving the curve x = 3t(t - 2), 3
y = 8t 2 with 0 £ t £ 1 about the y-axis. [Ans.: 39p ] 6. Find the area of the surface generated by revolving the curve x = a cos2 t, y = a sin2 t about x-axis. 12p 2 ˘ È Í Ans.: 5 a ˙ Î ˚
4.60
Chapter 4 Applications of Definite Integrals
7. Show that the ratio of the areas of the surface formed by revolving the arch of the cycloid x = a(q + sinq ), y = a(1 + cosq ) between two consecutive cusps about the x-axis to the area enclosed by the cycloid and x-axis is 64 . 9
4.4.3 Area of surface of solid of revolution in Polar Form For the curve r = f (q), bounded between the radii vectors at q = q 1 and q = q 2, the area of surface of the solid of revolution about the initial line q = 0 is given by, q2
S = ∫ 2p y q1
ds dq dq 2
q2 dr = ∫ 2p r sin q r 2 + dq q1 dq
Similarly, the area of surface of solid of revolution about the line q = q2
S = ∫ 2p x q1
ds dq dq
p is given by, 2
2
q2 dr = ∫ 2p r cos q r 2 + dq q1 dq
example 1 q
The curve r = e 2 is revolved about the initial line. Prove that the area of surface of revolution traced out by the part between the points q = 0 and q = p is equal to p 5 (ep + 1). 2 Solution q
r = e2 d r 1 q2 = e dq 2 2
ds 1 5 q2 dr e = r 2 + = eq + eq = dq dq 4 2 The surface area is generated by revolving the curve about the initial line. For the region shown, q varies from 0 to p.
4.4 Area of Surface of Solid of Revolution 4.61
ds dq dq p ds dq = ∫ 2p r sin q 0 dq q p 5 q2 = 2p ∫ e 2 sin q e dq 0 2 p
S = ∫ 2p y
Surface area,
0
p
= p 5 ∫ eq sin q dq 0
p
=p 5 =
eq (sin q − cos q ) 2 0
p 5 (ep + 1). 2
example 2 Find the area of the surface of the solid generated by revolving upper half of the cardioid r = a(1 - cos q ) about the initial line. q= p 2
Solution r = a(1 − cos q )
dr = a sin q dq
q =p
ds dr = r2 + dq dq
= a 2 (1 − cos q ) 2 + a 2 sin 2 q 2
q q q = a 2 2 sin 2 + a 2 2 sin cos 2 2 2 = 4a 2 sin 2 = 2a sin
O
q=0
2
Fig. 4.41 2
q 2
q 2
The area of the surface of the solid is generated by revolving the upper half of the cardioid about the initial line q = 0. For the region shown, q varies from 0 to p. p
Surface area,
S = ∫ 2p y 0
ds dq dq
4.62
Chapter 4 Applications of Definite Integrals
ds dq dq
p
= ∫ 2p r sin q 0
q dq 2 p q q q q = 4p a 2 ∫ 2 sin 2 2 sin cos sin dq 0 2 2 2 2 p
= 4p a 2 ∫ (1 − cos q ) sin q sin 0
q q cos dq 2 2 p q 1 q = 32p a 2 ∫ sin 4 ⋅ cos dq 0 2 2 2 p
= 16p a 2 ∫ sin 4 0
sin 5
= 32p a 2
q 2
p
[ f (q ) n + 1 ] n ] Q ∫ [ f (q ) f ′ (q ) dq = n +1
5 0
32 = p a2 5
example 3 p p The arc of cardioid r = a(1 + cos q ) included between q = - to 2 2 p is rotated about the line q = . Show that the area of the surface 2 48 2 generated is p a 2. 5 q= p 2 Solution r = a (1 + cos q ) dr = − a sin q dq
O
ds dr = r2 + dq dq
2
Fig. 4.42
= a 2 (1 + cos q ) 2 + a 2 sin 2 q 2
q q q = a 2 2 cos 2 + a 2 2 sin cos 2 2 2 = 4a 2 cos 2 = 2a cos
q 2
q 2
q=0
2
4.4 Area of Surface of Solid of Revolution 4.63
p The area of the surface is generated by revolving the cardioid about the line q = . 2 p p For the region shown, q varies from − to . Due to symmetry about the initial line 2 2 p considering the region in the first quadrant where q varies from 0 to , 2 p ds dq Surface area, S = 2∫ 2 2p y 0 dq p
= 2∫ 2 2p r cos q 0
ds dq dq
p
= 8p a 2 ∫ 2 (1 + cos q ) cosq cos 0
q dq 2
p q q q = 8p a 2∫ 2 2 cos 2 1 − 2 sin 2 cos dq 0 2 2 2 p q q q = 8p a 2∫ 2 2 1 − sin 2 1 − 2 sin 2 cos dq 0 2 2 2
Putting
when
sin
q = t, 2
1 q cos dq = dt 2 2 q = 0,
when
q =
t=0
p 1 , t= 2 2 1
S = 32 p a 2 ∫ 2 (1 − t 2 )(1 − 2t 2 )dt 0
1
= 32 p a 2 ∫ 2 (1 − 3t 2 + 2t 4 ) dt 0
2 = 32 p a t − t + t 5 5 2
3
1 2
0
1 2 1 1 = 32 p a 2 − + ⋅ 5 4 2 2 2 2 =
48 2 2 pa 5
example 4 Find the surface area of the solid formed by the revolution of the loop about the tangent at the pole of the curve r 2 = a 2 cos 2q .
4.64
Chapter 4 Applications of Definite Integrals
Solution
q= p 2
r 2 = a 2 cos 2q 2r
q= p 4
dr = −2a 2 sin 2q dq a 2 sin 2q dr =− dq a cos 2q sin 2q = −a cos 2q
M
p −q 4
q
2
q=− p 4
= a 2 cos 2q + a 2
=
q=0
O
ds dr = r2 + dq dq
=
r sin ( p −q ) 4 P (r, q )
Fig. 4.43
sin 2 2q cos 2q
a 2 cos 2 2q + a 2 sin 2 2q cos 2q a cos 2q
The surface area is formed by the revolution of the loop about the tangent at the pole p p . If P(r, q) is any point on the curve, its distance from the line q = is 4 4 1 p p r(cos q − sin q ). For the region shown, q varies from − to r sin − q , i.e., 4 4 2 p . Due to symmetry about the initial line, considering the region in the first quadrant 4 p where q varies from 0 to , 4 i.e., q =
p
Surface area,
S = 2∫ 4 2p 0
=
4p
∫ 2
p 4
0
1 2
r (cosq − sin q )
ds dq dq
a cos 2q (cos q − sin q ) p
= 2 2p a 2 ∫ 4 (cos q − sin q ) dq 0
= 2 2p a 2 sin q + cos q = 2 2 p a 2 ( 2 − 1)
p 4 0
a cos 2q
dq
Points to Remember 4.65
exercIse 4.7 1. Find the area of the surface of the solid generated by revolving the curve r 2 = a 2 cos 2q about the initial line. È 1 ˆ˘ 2Ê Í Ans.: 4pa Á 1 ˜˙ Ë 2¯˚ Î 2. Find the area of the surface of the solid generated by revolving the curve r = 2a cosq about the initial line. ÈÎ Ans.: 4p a 2 ˘˚ 3. Find the area of the surface of the solid generated by revolving the curve r = 4 cos q about the initial line. [Ans.: 16p ]
Points to remember Volume Using Cross-sections The volume of a solid of known integrable cross-section area A(x) formed by a plane perpendicular to the x-axis at any point between x = a to x = b is b
V = ∫ A( x)dx a
If the cross-section A(y) is perpendicular to the y-axis at any point between y = c to y = d, then the volume of the solid is d
V = ∫ A( y )dy c
Length of Plain Curves Length of Arc (i) Cartesian form (a) s = ∫
b a
2
dy 1 + dx dx 2
d dx (b) s = ∫c 1 + dy dy
(ii) Parametric form s=∫
t2
t1
2
2
dx dy + dt dt dt
4.66
Chapter 4 Applications of Definite Integrals
(iii) Polar form 2
dr r 2 + dq dq
q2
(a) s = ∫q
1
2
dq 1 + r 2 dr dr
r2
(b) s = ∫r
1
Area of Surface of Solid of Revolution (i) Cartesian form (a) S =
2
dy ∫a 2p y 1 + dx dx b
(revolution about x-axis)
2
(b) S =
∫
d
c
dx 2p x 1 + dy dy
(revolution about y-axis)
(ii) Parametric form (a) S =
(b) S =
∫
t2
∫
t2
t1
t1
2
2
dx dy 2p y + dt dt dt 2
(revolution about x-axis)
2
dx dy 2p x + dt dt dt
(revolution about y-axis)
(iii) Polar form (a) S = (b) S =
∫
q2
∫
q2
q1
q1
2
dr 2p r sin q r 2 + dq dq 2
dr 2p r cos q r 2 + dq dq
(revolution about q = 0) p revolution about q = 2
multiple choice Questions Select the most appropriate response out of the various alternatives given in each of the following questions: 1. The length of the arc of the curve y = log sec x from x = 0 to x = (a) log (2 + 3 )
(b) log ( 2 + 3)
(c) log ( 2 + 1)
(d) log ( 3 + 1)
p is 3
Multiple Choice Questions 4.67
2
( x 2 + 2) 3 2. The length of the curve y = from x = 0 to x = 3 is 3 (a) 10 (b) 12 (c) 3p (d) 6p 3. The whole length of the curve r = 2a sin q is equal to (a) p a (b) 2p a (c) 3p a (d) 4p a 4. The length of the arc of the curve 6xy = x4 + 3 from x = 1 to x = 2 is 17 13 19 (a) (b) (c) (d) none of these 12 12 12 5. The arc of the sine curve y = sin x from x = 0 to x = p revolved about the x-axis. The area of the surface of the solid generated is 2p 2 (a) 2p [ 2 + log ( 2 + 1)] (b) [ 2 + log ( 2 + 1)] 3 p p2 (c) [ 2 + log ( 2 + 1)] (d) [ 2 + log ( 2 + 1)] 3 3 6. The area of the surface of the solid generated by revolving the curve r = 2a cos q about the initial line is (a) 2p a2 (b) 4p a2 (c) p a2 (d) 8p a2 7. The area of the surface of the solid generated by revolving the curve x = t3 – 3t, y = 3t2, 0 £ t £ 1 about the x-axis is 24p 36p 48p 40p (a) (b) (c) (d) 5 5 5 5 8. The area of the surface of the solid generated by the revolution of the line segment y = 2x from x = 0 to x = 2 about the x-axis is equal to (a) p 5
(b) 2p 5
(c) 4p 5
(d) 8p 5
Answers 1. (a)
2. (b)
3. (b)
4. (b)
5. (a)
6. (b)
7. (d)
8. (d)
UNIT-2 Chapter 5. Sequences and Series Chapter 6. Taylor’s and Maclaurin’s Series
CHAPTER
Sequences and Series
5
chapter outline 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17
5.1
Introduction Sequence Infinite Series The nth Term Test for Divergence Geometric Series Telescoping Series Combining Series Harmonic Series p-Series Comparison Test D’Alembert’s Ratio Test Raabe’s Test Cauchy’s Root Test Cauchy’s Integral Test Alternating Series Absolute and Conditional Convergent of a Series Power Series
IntroductIon
In this chapter, we will learn about the convergence and divergence of sequence and series. There are various methods to test the convergence and divergence of an infinite series. We will study Comparison Test, D’Alembert’s ratio test, Cauchy’s root test and Cauchy’s integral test. We will also study alternating series, absolute and uniform convergence of the series and power series.
5.2
Chapter 5
5.2
Sequences and Series
Sequence
An ordered set of real numbers as u1, u2, u3, ……..un, …… is called a sequence and is denoted by {un}. If the number of terms in a sequence is infinite, it is said to be an infinite sequence, otherwise it is a finite sequence and un is called the nth term of the sequence.
5.2.1
Limit of a Sequence
A sequence {un} tends to a finite number l as n Æ • if for every Œ > 0 there exists an integer m such that, | un - l | < Œ for all n > m, i.e., lim un = l. n Æ•
5.2.2
continuous Function theorem for Sequences
Let {un} be a sequence of real numbers. If un Æ l and if f is a function that is continuous at l and defined at all un, then f(un) Æ f(l).
5.2.3
convergence, divergence and oscillation of Finite Series
(i) If the sequence {un} has a finite limit, i.e., lim un is finite, the sequence is said to n Æ• be convergent.
e.g.
Ï ¸ Ô 1 Ô {un } = Ì ˝ Ô1 + 1 Ô Ó n˛ lim un = 1 n Æ•
Since limit is finite, the sequence is convergent. (ii) If the sequence {un} has infinite limit, i.e., lim un is infinite, the sequence is said n Æ• to be divergent. e.g.
{un} = {2 n + 1} lim un = •
n Æ•
Since limit is infinite, the sequence is divergent. (iii) If the limit of the sequence {un} is not unique, the sequence is said to be oscillatory. e.g.
{un} = ( -1)n +
1
2n lim un = 1, if n is even n Æ•
= –1, if n is odd Since limit is not unique, the sequence is oscillatory.
5.2
5.2.4
Sequence
5.3
Monotonic Sequence
A sequence is said to be monotonically increasing if un +1 ≥ un for each value of n and is monotonically decreasing if un +1 £ un for each value of n. The sequence is called alternating sequence if the terms are alternate positive and negative. For example, (i) 1, 2, 3, 4, … is a monotonically increasing sequence. 1 1 1 (ii) 1, , , , … is a monotonically decreasing sequence. 2 3 4 (iii) 1, –2, 3, – 4, … is an alternating sequence.
5.2.5
Bounded Sequence
A sequence {un} is said to be a bounded sequence if there exists numbers m and M such that m < un < M for all n. Note 1: Every convergent sequence is bounded but the converse is not true. Note 2: A monotonic increasing sequence converges if it is bounded above and diverges to + • if it is not bounded above. Note 3: A monotonic decreasing sequence converges if it is bounded below and diverges to – • if it is not bounded below. Note 4: If sequence {un} and {vn} converges to l1 and l2 respectively then (i) Sequence {un + vn} converges to l1 + l2 (ii) Sequence {un . vn} converges to l1◊ l2 Ïu ¸ l (iii) Sequence Ì n ˝ converges to 1 provided l2 π 0 v l Ó n˛ 2
example 1 ÏÔ n2 + n ¸Ô Test the convergence of the sequence Ì 2 ˝. ÓÔ 2 n - n ˛Ô
Solution n2 + n
Let
un =
2n2 - n
lim un = lim
n Æ•
Hence, {un} is convergent.
n2 + n
2n2 - n 1 1+ n = lim 1 n Æ• 2n 1 = 2 n Æ•
5.4
Chapter 5
Sequences and Series
example 2 Test the convergence of the sequence {tanh n}. Solution Let
un = tanh n lim un = lim tanh n n Æ•
n Æ•
= lim n Æ•
sinh n cosh n en - e- n
= lim n Æ•
en + e- n e2 n - 1
= lim
e2 n + 1 1 1 - 2n e = lim 1 n Æ• 1 + 2n e =1 n Æ•
Hence, {un} is convergent.
example 3 Test the convergence of the sequence {2n}. Solution Let un= 2n
lim un = lim 2 n
n Æ•
n Æ•
=•
Hence, {un} is divergent.
example 4
{
}
Test the convergence of the sequence 2 - ( -1)n . Solution un= 2 –(–1)n
Let
lim un = lim 2 - ( -1)n n Æ•
n Æ•
=2–1=1
,
= 2 –(–1) = 3 ,
if n is even if n is odd
Since limit is not unique, the sequence {un} is oscillatory.
5.2
Sequence
5.5
example 5
1 1 1 Show that the sequence {un} whose nth term is un = 1 + + 2 + L + n , 3 3 3 is monotonic increasing and bounded. Is it convergent?
Solution 1 1 1 + 2 +L+ n 3 3 3 1 1 1 1 = 1 + + 2 + L + n + n +1 3 3 3 3
un = 1 + un +1
1 un +1 - un =
3
n +1
>0
Hence, {un} is monotonic increasing sequence. un = 1 +
Also,
1 1 1 + +L+ n 3 32 3
1 ˆ Ê 1 Á 1 - n +1 ˜ Ë 3 ¯ = 1 13 =
3Ê 1 ˆ 3 1 - n +1 ˜ < Á 2Ë 3 ¯ 2
3 . 2 Since {un} is monotonic increasing and bounded above, it is convergent. {un} is bounded above by
example 6
1 2 1 + + …+ , 1! 2 ! n! n Œ N, is monotonic increasing and bounded. Is it convergent?
Show that the sequence {un} whose nth term is un =
Solution
1 1 + + …+ 1! 2 ! 1 1 un +1 = + + … + 1! 2 ! 1 un +1 - un = >0 (n + 1)! un+1 > un un =
Hence, {un} is a monotonic increasing sequence.
1 n! 1 1 + n ! (n + 1)!
5.6
Chapter 5
Sequences and Series
Also,
1 1 1 1 + + +L+ 1! 2 ! 3! n! 1 1 1 = 1+ + +L+ 2 ! 3! n! 1 1 1 < 1+ + 2 +L+ n 2 2 2 1 ˆ Ê 1 Á 1 - n +1 ˜ Ë 2 ¯ < 1 12 1 ˆ Ê < 2 Á 1 - n +1 ˜ Ë 2 ¯
un =
[Using sum of G.P]
ny ˙ ÍQ Ì1 + ny + 2! ÍÎ ÓÔ ˙˚ ˛Ô
1 ny
0< x
n(n - 1) y 2 +L 2!
n
1 2
1 +
3
1 +K+
n
> 1+
1 n
1 +
n
1 +K+
n
n n
Sn > n
and
lim n = •
n Æ•
Thus, the series is divergent. Hence, lim un = 0 is a necessary but not sufficient condition for convergence n→∞ of Sun. If lim un π 0 or lim un does not exist, then Sun is divergent. n Æ•
5.5
n Æ•
geoMetrIc SerIeS
Consider the geometric series a + ar + ar 2 + L + ar n -1 + L 2
S n = a + ar + ar + L + ar a (1 − r n ) , 1− r a (r n − 1) = , r −1
=
n −1
if r < 1 if r > 1
lim r n = 0
(i) When | r | < 1,
n Æ•
lim Sn =
n Æ•
a is finite. 1- r
Hence, the series is convergent. (ii) When r > 1,
lim r n Æ ∞
nÆ∞
a(r n - 1) Æ∞ nÆ∞ r -1
lim Sn = lim
nÆ∞
Hence, the series is divergent. (iii) When r = 1, Sn = a + a + a + L = na lim Sn Æ ∞
nÆ∞
Hence, the series is divergent.
... (1)
5.5
Geometric Series
5.11
(iv) When r = –1, Sn = a - a + a - L ( -1)n -1 a
= 0, if n is even = a, if n is odd Hence, the series is oscillatory. (v) When r < –1, let r = –k where k > 0 a[1 - ( - k )n ] n Æ• 1+ k
lim Sn = lim
n Æ•
a[1 - ( -1)n k n ] n Æ• 1+ k = – •, if n is even = + •, if n is odd = lim
Hence, the series is oscillatory. From all the above cases, we conclude that the geometric series (1) is (i) convergent if | r | < 1 (ii) divergent if r ≥ 1 (iii) oscillatory if r £ –1
example 1 Prove that 1 +
2 4 8 16 + + + + L converges and find its sum. 3 9 27 81 [Winter 2014]
Solution The given series is geometric series with a = 1 and r = 1+
• • 2 4 8 Ê 2ˆ + + + L = Â ar n -1 = Â Á ˜ Ë ¯ 3 9 27 n =1 n =1 3
r =
2 m
un -l m
un < l+ Πvn
for all n > m
Neglecting the first m terms of Sun and Svn,
l - Œ
(l - Œ) lim (v1 + v2 + v3 + L + vn )
n Æ•
n Æ•
lim (u1 + u2 + u3 + L + un ) Æ •
[From Eq. (2)]
n Æ•
Hence, Sun is also divergent. Note The following standard limits can be used to solve the problems: log n (i) lim (vi) lim x n = 0 if x < 1 =0 n Æ• n n Æ• n
Ê 1ˆ (ii) lim Á 1 + ˜ = e n Æ• Ë n¯
(vii) lim x n = • if x > 1 n Æ•
1
(iii) lim (n) n = 1
(viii) lim n Æ•
n Æ• 1
(iv) lim (n !) n = • n Æ•
xn = 0 for all x n!
Ê a n - 1ˆ = log a (ix) lim Á nÆ0 Ë n ˜¯ 1
1
Ê n!ˆ n 1 (v) lim Á ˜ = n Æ• Ë n ¯ e
an -1 = log a (x) lim 1 n Æ• n
example 1 •
Test the convergence of the series
 n=1 n
n
2
. +1
5.22
Chapter 5
Sequences and Series
Solution un =
Let
=
n n +1 1 2
3
1ˆ Ê n 2 Á1 + 2 ˜ Ë n ¯ 1
vn =
Let
3
n2 1
un = lim n Æ• vn n Æ• lim
1+
1 n2
[finite and non-zero]
=1
and Svn = Â
1 3 n2
is convergent as p =
3 > 1. 2
Hence, by comparison test, Sun is also convergent.
example 2 •
Test the convergence of the series
2n - 1
 n(n + 1)(n + 2) . n =1
Solution Let
un =
2n - 1 n(n + 1)(n + 2)
1ˆ Ê ÁË 2 - n ˜¯ = Ê 1ˆ Ê 2ˆ n2 Á 1 + ˜ Á 1 + ˜ Ë n¯ Ë n¯
Let
vn =
1 n2
1ˆ Ê ÁË 2 - n ˜¯ un lim = lim n Æ• vn n Æ• Ê 1ˆ Ê 2ˆ ÁË 1 + n ˜¯ ÁË 1 + n ˜¯
= 2 [finite and non-zero]
5.10
and the series Svn = Â
1 n2
Comparison Test
5.23
is convergent as p = 2 > 1.
Hence, by comparison test, Sun is also convergent.
example 3 •
1
Â
Test the convergence of the series
n =1 1 + 2
2
+ 32 + L + n2
.
[Winter 2015] Solution 1
un =
Let
2
1 + 2 + 3 + L + n2 6 = n(n + 1)(2 n + 1) 6 = 1ˆ Ê 1ˆ Ê n3 Á 1 + ˜ Á 2 + ˜ Ë n¯ Ë n¯ 1
Let
vn =
lim n Æ•
n3
un 6 = lim 1ˆ vn nÆ• Ê 1 ˆ Ê ÁË 1 + n ˜¯ ÁË 2 + n ˜¯
=6 and Svn = Â
2
1 n3
[finite and non-zero]
is convergent as p = 3 > 1.
Hence, by comparison test, Sun is also convergent.
example 4 •
Test the convergence of the series
Â
n =1 ( n + 1)
Solution Let
un =
n+2
n+2 (n + 1) n
. n
5.24
Chapter 5
Sequences and Series
=
Let
1+
2 n
1 1 n 2 1 + n
vn =
1 1
n2
Ê 2ˆ ÁË 1 + n ˜¯ un lim = lim n Æ• vn n Æ• Ê 1ˆ ÁË 1 + n ˜¯
=1 and Svn = Â
1 1 2 n
is divergent as p =
[finite and non-zero]
1 < 1. 2
Hence, by comparison test, Sun is also divergent.
example 5 •
Is the series
Â
2n + 1
n =1 ( n + 1)
2
convergent or divergent?
Solution Let
un =
=
2n + 1 (n + 1)2 1ˆ Ê nÁ2 + ˜ Ë n¯ Ê 1ˆ n Á1 + ˜ Ë n¯
2
2
1ˆ Ê 2+ ˜ Á 1 Ë n¯ = n Ê 1ˆ2 ÁË 1 + n ˜¯
Let
vn =
1 n
[Summer 2015]
5.10
lim
n Æ•
Comparison Test
5.25
1ˆ Ê ÁË 2 + n ˜¯
un = lim vn nÆ• Ê 1 ˆ 2 ÁË 1 + n ˜¯
=
2 =2 1
[finite and non-zero]
1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.
and Svn = Â
example 6
•
Â
Test the convergence of the series
n =1
Solution
(2 n (3n
2
1 - 1) 3
1 + 2 n + 5) 4
3
.
1
(2 n2 - 1) 3
un =
Let
1
(3n3 + 2 n + 5) 4 2 n3
=
1
1 ˆ3 Ê ÁË 2 - 2 ˜¯ n 1
3 n4
2 5 ˆ4 Ê ÁË 3 + 2 + 3 ˜¯ n n 1
1 ˆ3 Ê ÁË 2 - 2 ˜¯ n
= 1 n 12
vn =
Let
1
5 ˆ4 2 Ê ÁË 3 + 2 + 3 ˜¯ n n
1 1
n 12 1
un = lim n Æ• vn n Æ• lim
1 ˆ3 Ê ÁË 2 - 2 ˜¯ n 1
2 5 ˆ4 Ê ÁË 3 + 2 + 3 ˜¯ n n
1
(2 ) 3 =
1 (3) 4
[finite and non-zero]
[Winter 2013]
5.26
Chapter 5
and Svn = Â
1 1 n 12
Sequences and Series
is divergent as p =
1 < 1. 12
Hence, by comparison test, Sun is also divergent.
example 7
1 2 3 4 + + + +L. 1◊ 3 3 ◊ 5 5 ◊ 7 7 ◊ 9
Test the convergence of the series Solution Let
n (2 n - 1)(2 n + 1)
un =
1 1ˆ Ê 1ˆ Ê nÁ2 - ˜ Á2 + ˜ Ë n¯ Ë n¯
=
Let
1 n
vn =
lim
n Æ•
un 1 = lim 1ˆ Ê 1ˆ vn nÆ• Ê ÁË 2 - n ˜¯ ÁË 2 + n ˜¯
=
1 [finite and non-zero] 4
1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.
and Svn = Â
example 8
1
Test the convergence of the series
1
3p
+
5p
1 +
Solution Let
un = =
Let
vn =
1 (2 n + 1) p 1 1ˆ Ê np Á 2 + ˜ Ë n¯ 1 np
p
7p
+ L.
5.10
Comparison Test
un 1 = lim p n Æ• vn n Æ• 1ˆ Ê 2 + ÁË n ˜¯ lim
= and Svn = Â
1
[finite and non-zero]
2p
1
is convergent if p > 1 and divergent if p £ 1. np Hence, by comparison test, Sun is also convergent if p > 1 and divergent if p £ 1.
example 9 Test the convergence of the series 2 + 3 + 4 + 5 + L + n + 1 + L. 1 8 27 64 n3 Solution Let
un =
Let
vn =
n +1
n3 1 Ê 1ˆ = 2 Á1 + ˜ n¯ n Ë
lim
n Æ•
1 n2
un Ê 1ˆ = lim 1 + vn nÆ• ÁË n ˜¯
= 1 [finite and non-zero] and Svn = Â
1
is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.
example 10 Test the convergence of the series Solution Let
un =
1 1+ 2
2 +
1+ 2 3
n
1+ n n +1 n = Ê ˆ 3 1 1˜ Á 2 + 1+ n Á 3 n˜ 2 Ën ¯
3 +
1+ 3 4
+L.
5.27
5.28
Chapter 5
Sequences and Series
1
=
Let
ˆ 1 Ê 1 1 n2 Á 3 + 1+ ˜ Á n˜ Ë n2 ¯ 1
vn =
1
n2
lim n Æ•
and Svn =
1 1 n2
un 1 = lim vn nÆ• Ê ˆ Á 1 + 1+ 1 ˜ Á 3 n˜ Ë n2 ¯ = 1 [finite and non-zero]
is divergent as p =
1 < 1. 2
Hence, by comparison test, Sun is also divergent.
example 11
1
Test the convergence of the series
2
3
2 +
Solution n
Let
un =
(n + 1)3 1
n2
=
3
( n + 1) 2 1
n2
= 3 n2
3
Ê 1ˆ 2 ÁË 1 + n ˜¯ 1
=
3
Ê 1ˆ 2 n Á1 + ˜ Ë n¯
Let
vn =
1 n
3
3
3 +
43
+L.
5.10
lim n Æ•
un = lim vn nƕ
=1 1 and Svn = Â is divergent as p = 1. n
Comparison Test
5.29
1 3
Ê 1ˆ 2 ÁË 1 + n ˜¯ [finite and non-zero]
Hence, by comparison test, Sun is also divergent.
example 12 Test the convergence of the series
14 13
24 +
23
34 +
33
+L.
Solution nth term of the numerator = a + (n – 1)d = 14 + (n – 1)10 = 10n + 4 nth term of the denominator = n3 10 n + 4 un = n3 1 Ê 4ˆ = 2 Á 10 + ˜ n¯ n Ë Let
vn =
1 n2
un 4ˆ Ê = lim Á 10 + ˜ n Æ• vn n Æ• Ë n¯ = 10 [finite and non-zero] lim
and Svn =
1
is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.
example 13 Test the convergence of the series
1 2
a ◊1 + b
Solution Let
un =
n
a ◊ n2 + b 1 = bˆ Ê nÁa + 2 ˜ Ë n ¯
2 +
2
a◊2 + b
3 +
a ◊ 32 + b
+ L.
5.30
Chapter 5
Sequences and Series
Let
vn = lim
n Æ•
1 n
un 1 = lim bˆ vn nÆ• Ê ÁË a + 2 ˜¯ n 1 [finite and non-zero] = a
1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.
and Svn = Â
example 14 1
Test the convergence of the series Solution
2
2
1 +m
+
2
2 +m
3 +
2
3 +m
+ L.
n
Let
un =
2
n +m 1 = mˆ Ê n Á1 + 2 ˜ Ë n ¯
Let
vn =
1 n
un 1 = lim n Æ• vn n Æ• Ê mˆ ÁË 1 + 2 ˜¯ n = 1 [finite and non-zero] lim
1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.
and Svn = Â
example 15 Test the convergence of the series Solution Let
un =
2 ◊ 13 + 5 4 ◊ 15 + 1
2 n3 + 5 4 n5 + 1
+
2 ◊ 23 + 5 4 ◊ 25 + 1
+L +
2 ◊ n3 + 5 4 ◊ n5 + 1
+ L.
5.10
Comparison Test
5.31
5ˆ Ê ÁË 2 + 3 ˜¯ n = 1ˆ Ê n2 Á 4 + 5 ˜ Ë n ¯
Let
vn =
1 n2
5ˆ Ê 2+ 3˜ Á Ë u n ¯ lim n = lim n Æ• vn n Æ• Ê 1ˆ ÁË 4 + 5 ˜¯ n
= and Svn = Â
2 [finite and non-zero] 4
1
is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.
example 16 Test the convergence of the series
1◊ 2 2
3 ◊4
2
+
3◊ 4 2
5 ◊6
2
+
5◊6
+L. 72 ◊ 82 [Winter 2013]
Solution Let
un =
=
(2 n - 1) ◊ 2 n
[Using A.P.]
(2 n + 1)2 (2 n + 2)2 1ˆ Ê ÁË 2 - n ˜¯ ◊ 2 2
1ˆ Ê 2ˆ Ê n Á2 + ˜ Á2 + ˜ Ë n¯ Ë n¯
2
2
Let
vn =
1 n2
1ˆ Ê 2Á2 - ˜ Ë un n¯ lim = lim 2 2 n Æ• vn n Æ• 1ˆ Ê 2ˆ Ê ÁË 2 + n ˜¯ ÁË 2 + n ˜¯
=
1 [finite and non-zero] 4
5.32
Chapter 5
and Svn = Â
Sequences and Series
1
is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.
example 17 Test the convergence of the series
1 3 5 7 + + + +L. 1◊ 2 ◊ 3 2 ◊ 3 ◊ 4 3 ◊ 4 ◊ 5 4 ◊ 5 ◊ 6 [Winter 2014]
Solution Let
un =
(2n + 1) n(n + 1)(n + 2)
[Using A.P.]
1ˆ Ê ÁË 2 + n ˜¯ = Ê 1ˆ Ê 2ˆ n2 Á 1 + ˜ Á 1 + ˜ Ë n¯ Ë n¯ 1 vn = 2 n
Let
1ˆ Ê ÁË 2 + n ˜¯ un lim = lim n Æ• vn n Æ• Ê 1ˆ Ê 2ˆ ÁË 1 + n ˜¯ ÁË 1 + n ˜¯
= 2 [finite and non-zero] and Svn = Â
1 n2
is convergent as p = 2 > 1.
Hence, by comparison test, Sun is also convergent.
example 18 •
Test the convergence of the series
Â(
n =1
)
n4 + 1 - n4 - 1 . [Summer 2016]
Solution Let
un = n 4 + 1 - n 4 - 1 =
( (
) ( - 1)
n4 + 1 - n4 - 1 4
n +1 + n
4
n4 + 1 + n4 - 1
)
5.10
=
Comparison Test
5.33
(n 4 + 1) - (n 4 - 1) n4 + 1 + n4 - 1 2
=
n4 + 1 + n4 - 1 1 2 = 2◊ n Ê 1 1 ˆ Á 1+ 4 + 1- 4 ˜ n n ¯ Ë Let
vn = lim
n Æ•
1 n2
un 2 = lim vn nÆ• Ê 1 1 ˆ Á 1+ 4 + 1- 4 ˜ n n ¯ Ë =2
and
1
 vn =  n2
[finite and non-zero]
is convergent as p = 2 > 1.
Hence, by comparison test, Sun is also convergent.
example 19 •
Check for convergence of the series
 n =1
5n3 - 3n n2 (n - 2)(n2 + 5)
.
[Winter 2016] Solution un =
5n 3 - 3n n (n - 2)(n2 + 5) 2
3ˆ Ê n3 Á 5 - 2 ˜ Ë n ¯ = 2ˆ Ê 5ˆ Ê n5 Á 1 - ˜ Á 1 + 2 ˜ Ë n¯ Ë n ¯ 3ˆ Ê ÁË 5 - 2 ˜¯ n = 5ˆ 2 ˆ Ê Ê n2 Á 1 - ˜ Á 1 + 2 ˜ ¯ Ë Ë n n ¯
5.34
Chapter 5
Let
Sequences and Series
vn =
1 n2
3ˆ Ê ÁË 5 - 2 ˜¯ un n lim = lim n Æ• vn n Æ• Ê 2ˆ Ê 5ˆ ÁË 1 - n ˜¯ ÁË 1 + 2 ˜¯ n =5 and
1
 vn =  n2
[finite and non zero]
is convergent as p = 2 > 1.
Hence, by comparison text,
 un
is also convergent.
example 20 •
Test the convergence of the series Solution Let
È1
Ê n + 1ˆ ˘ ˜ . n ¯ ˙˚
Â Í n - log ÁË
n =1 Î
un =
1 Ê n + 1ˆ - log Á Ë n ˜¯ n
=
1 Ê 1ˆ - log Á 1 + ˜ Ë n n¯
=
1 Ê1 1 1 1 ˆ - Á - 2 + 3 - 4 + L˜ ¯ n Ë n 2n 3n 4n
=
1
Let
vn = lim n Æ•
1
3
+
1
1 n2
un 1 Ê1 1 ˆ = lim + - L˜ ¯ vn nÆ• ÁË 2 3n 4n2
= and Svn = Â
-
1
-L 2n 3n 4n4 1 Ê1 1 1 ˆ = 2Á + 2 - L˜ ¯ n Ë 2 3n 4n 2
1 [finite and non-zero] 2
is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.
5.10
example 21
•
Test the convergence of the series
 sin n .
Comparison Test
5.35
1
n =1
Solution un = sin
Let
=
1 n
1 1 1 + -L 3 n 3! n 5! n5
1Ê 1 1 ˆ + - L˜ 1¯ n ÁË 3! n2 5! n 4 1 vn = n =
Let
un 1 1 Ê ˆ = lim Á 1 + - L˜ 2 4 n Æ• vn n Æ• Ë ¯ 3! n 5! n lim
= 1 [finite and non-zero] 1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.
and Svn = Â
example 22 È Test the convergence of the series Â Í n3 + 1 n =1 Í Î Solution •
Let
(
)
1 3
˘ - n ˙. ˙˚
[Summer 2017]
1 È ˘ 3 un = Í(n + 1) 3 - n ˙ ÍÎ ˙˚
1
1 ˆ3 Ê = n Á1 + 3 ˜ - n Ë n ¯ È 1Ê1 ˆ Í 1 1 3 ÁË 3 - 1˜¯ = n Í1 + ◊ 3 + ÍÎ 3 n 2!
=
1Ê1 ˆÊ1 ˆ -2 -1 2 3 ÁË 3 ˜¯ ÁË 3 ˜¯ Ê 1ˆ ÁË 3 ˜¯ + 3! n
1 1 1 1 5 1 ◊ - ◊ + ◊ -L 3 n2 32 n5 34 n8
˘ 3 ˙ Ê 1ˆ ÁË 3 ˜¯ + L˙˙ - n n ˚
5.36
Chapter 5
Sequences and Series
=
Let
vn =
1 Ê1 1 1 5 1 ˆ - 2 ◊ 3 + 4 ◊ 6 - L˜ 2 Á ¯ n Ë3 3 n 3 n
1 n2
un 5 1 Ê1 1 1 ˆ = lim Á - 2 ◊ 3 + 4 ◊ 6 - L˜ n Æ• vn n Æ• Ë 3 ¯ 3 n 3 n lim
=
1 [finite and non-zero] 3
and the series Svn = Â
1 n2
is convergent as p = 2 > 1.
Hence, by comparison test, Sun is also convergent.
example 23 1 1˘ È 3 - n3 ˙. Í ( n 1) + ˚ ÂÎ •
Test the convergence of the series
n =1
Solution 1
Let
1
un = (n + 1) 3 - n 3 1 È ˘ Ê 1ˆ 3 ˙ Í = + 1 1 ÍÁË ˙ n ˜¯ ˙˚ ÎÍ ÈÏ ¸ ˘ 1Ê1 ˆ 1Ê1 ˆÊ1 ˆ - 1˜ - 1˜ Á - 2˜ 1 ÍÔ Á Á Ô ˙ 1 3Ë3 ¯ 1 3Ë3 ¯Ë3 ¯ 1 Ô Ô 3 Í = n Ì1 + + ◊ 2+ ◊ 3 + L˝ - 1˙ Í Ô 3n 2! 3! n n Ô ˙˙ ÍÔ ˛Ô ˚ ÎÓ 1 n3
=
=
Let
vn =
1 2 3n 3
-
1 5 9n 3
+
5 8 81n 3
-L
1 Ê1 1 5 ˆ + - L˜ 2 Á Ë 3 9n 81n2 ¯ n3 1 2
n3
5.10
Comparison Test
5.37
un 5 Ê1 1 ˆ = lim Á + - L˜ 2 n Æ• vn n Æ• Ë 3 ¯ 9n 81n lim
= and Svn = Â
1 2 n3
1 3
[finite and non-zero] 2 < 1. 3
is divergent as p =
Hence, by comparison test, Sun is also divergent.
example 24 ∞
Test the convergence of the series
∑ n =1
n2 + n + 1 − n2 − n + 1 . n
Solution Let
un =
n2 + n + 1 - n2 - n + 1 n 1
1
È Ê 1 1 ˆ ˘2 È Ê 1 1 ˆ ˘2 = Í1 + Á + 2 ˜ ˙ - Í1 + Á - + 2 ˜ ˙ Î Ë n n ¯˚ Î Ë n n ¯˚
Expanding using binomial expansion, È ˘ 1Ê1 ˆ - 1˜ 2 Á Í 1 1 1 ˙ Ë ¯ Ê ˆ 2 2 Ê1 1 ˆ ˙ L un = Í1 + Á + 2 ˜ + + + ÁË n ˜ ˙ Í 2Ën n ¯ 2! n2 ¯ ˙ Í Î ˚ ˘ È 1Ê1 ˆ 2 ÁË 2 - 1˜¯ ˙ Í 1 1 1 1 1 2 Ê ˆ Ê ˆ ˙ L - Í1 + Á - + 2 ˜ + + + ÁË n ˜ ˙ Í 2Ë n n ¯ 2! n2 ¯ Í ˙ ˚ Î
˘ È 1 1 1Ê 1 2 1ˆ = Í1 + + 2 - Á 2 + 3 + 4 ˜ + L˙ Ë ¯ 8 n n n ˚ Î 2n 2n È ˘ 1ˆ 1 1 1Ê 1 2 - Í1 + 2 - Á 2 - 3 + 4 ˜ + L˙ Ë ¯ 8 n n n Î 2n 2n ˚
5.38
Chapter 5
Sequences and Series
Let
=
1 1 - 3 +L n 2n
=
1Ê 1 ˆ 1 - 2 + L˜ Á ¯ n Ë 2n
vn =
lim
n Æ•
1 n
un 1 Ê ˆ = lim 1 + L˜ ¯ vn nÆ• ÁË 2 n2
=1 and Svn = Â
[finite and non-zero]
1 is divergent as p = 1. n
Hence, by comparison test, Sun is also divergent.
example 25 Test the convergence of the series
Ê n2 + 1 - n ˆ ˜. ˜¯ np Ë
 ÁÁ
Solution Let
un =
n2 + 1 - n np
1 È ˘ 2 1 Ê ˆ Í n Á 1 + 2 ˜ - 1˙ ÍË ˙ n ¯ ÍÎ ˙˚ = p n ÈÏ ¸ ˘ 1Ê1 ˆ 1Ê1 ˆÊ1 ˆ - 1˜ Á - 2˜ - 1˜ ÍÔ Á Á Ô ˙ Ë ¯ Ë ¯ Ë ¯ n Ô 1 2 2 1 1 2 2 2 Ô ◊ 6 + L˝ - 1˙ ◊ 4 + = p Í Ì1 + 2 + 2! 3! n Í Ô 2n n n Ô ˙˙ ÍÔ Ô˛ ˚ ÎÓ
=
n Ê 1 1 1 ˆ - 4+ - L˜ p Á 2 6 Ë ¯ n 2n 8n 16 n
=
1 Ê1 1 1 ˆ - L˜ ÁË 2 - 2 + 4 ¯ 8n 16 n n p +1
5.10
Let
n lim
5.39
1
vn =
n Æ•
Comparison Test
p +1
un 1 1 Ê1 ˆ = lim + - L˜ ¯ vn nÆ• ÁË 2 8n2 16 n 4 1 [finite and non-zeero] = 2 1
and Svn = Â n
p +1
is convergent if p + 1 > 1, i.e., p > 0 and divergent if p + 1 £ 1,
i.e., p £ 0. Hence, by comparison test, Sun is also convergent if p > 0 and divergent if p £ 0.
example 26 1 1 1 1 1 + + + + +L, x x -1 x +1 x - 2 x + 2
Test the convergence of the series where x is a positive fraction. Solution
Since it is an infinite series, by ignoring the first term, the series can be rewritten as Ê 1
1 ˆ
Ê 1
1 ˆ
 un = ÁË x - 1 + x + 1˜¯ + ÁË x - 2 + x + 2 ˜¯ + L =
2x
+
2x
x -1 x - 22 2x 2
2
2
+L
=Â
x 2 - n2 2x
un =
x 2 - n2 2x = Ê x2 ˆ n2 Á 2 - 1˜ ¯ Ën
Let
vn = lim
n Æ•
1 n2
un 2x = lim 2 n Æ• vn Êx ˆ Á 2 - 1˜ Ën ¯ = –2x
[finite and non-zero]
5.40
Chapter 5
and Svn =
Sequences and Series
1
is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.
exercISe 5.2 1. Test the convergence of the following series: (i)
∑n
(iii)
∑(
(v) (vii)
2
1 +1
)
n4 + 1 − n4 − 1
n
∑(
(iv)
∑
(vi)
∑
(viii)
∑
p
∑ (n + 1)
q
∑ tan
(ii)
−1
1 n
n +1− n
)
np n +1+ n
1 tan n n
1
1 n
b a + n
Ans. : (ii) Divergent (iii) Convergent (i) Convergent 1 1 (iv) Convergent if p < − Divergent if p ≥ − 2 2 (v) Convergent if p − q + 1 < 0, Divergent if p − q + 1 ≥ 0 (vi) Convergent (vii) Divergent (viii) Convergent if a > 1, Divergent if a ≤ 1 2. Test the convergence of the series
(1 + a)(1 + b) (2 + a)(2 + b) (3 + a)(3 + b) + + +K. 1⋅ 2 ⋅ 3 2⋅3⋅4 3⋅ 4 ⋅5 [ans. : Divergent]
5.11
d’aLeMBert’S ratIo teSt
If Sun is a positive-term series and lim
n Æ•
(i) Sun is convergent if l < 1 (ii) Sun is divergent if l > 1
un +1 = l then un
5.11
D’Alembert’s Ratio Test
5.41
Proof
un +1 = l < 1. n Æ• un
Case I If lim
Consider a number l < r < 1 such that
un +1 < r for all n > m un
… (1)
Neglecting the first m terms, •
Â
un = um +1 + um + 2 + um + 3 + K •
n = m +1
Ê u ˆ u u = um +1 Á 1 + m + 2 + m + 3 + m + 4 + K˜ um +1 um +1 um +1 Ë ¯
Ê u ˆ u u u u u = um +1 Á 1 + m + 2 + m + 3 ◊ m + 2 + m + 4 ◊ m + 3 ◊ m + 2 + K˜ um +1 um + 2 um +1 um + 3 um + 2 um +1 Ë ¯ < um +1 (1 + r + r ◊ r + r ◊ r ◊ r + K) 2
[ Using Eq. (1)]
3
= um +1 (1 + r + r + r + K) = um + 1 ◊ •
Â
\
un
1, n Æ• un lim
un +1 > 1 for all n > m un
... (2)
5.42
Chapter 5
Sequences and Series
Neglecting the first m terms, •
Â
un = um +1 + um + 2 + um + 3 + um + 4 + K •
n = m +1
Ê u ˆ u u = um +1 Á 1 + m + 2 + m + 3 + m + 4 + K˜ u u u Ë ¯ m +1 m +1 m +1 Ê u ˆ u u u u u = um +1 Á 1 + m + 2 + m + 3 ◊ m + 2 + m + 4 ◊ m + 3 ◊ m + 2 + K˜ um +1 um + 2 um +1 um + 3 um + 2 um +1 Ë ¯ > um +1 (1 + 1 + 1 + 1 + K) \ (um +1 + um + 2 + K to n terms) > um +1 (1 + 1 + 1K to n terms) Sn > um +1 .n lim Sn > lim num +1 Æ •
n Æ•
[Q um +1 is positive]
n Æ•
•
Â
Thus, the series
un is divergent.
n = m +1
The nature of a series remains unchanged if we neglect a finite number of terms in the •
beginning. Hence, the series
 un
is divergent.
n=1
un +1 = 1 the ratio test fails, i.e. no conclusion can be drawn about the n Æ• un
Note 1: If lim
convergence or divergence of the series.
Note 2: It is convenient to use D’Alembert’s ratio test in the following form: u If Sun is a positive term series and lim n = l, then n Æ• un +1 (i) Sun is convergent if l > 1 (ii) Sun is divergent if l < 1 (iii) The ratio test fails if l = 1
example 1 •
Test the convergence of the series
 n£0
Solution Let
32 n un =
23 n
32 n 23 n
.
[Summer 2014]
5.11
32( n +1)
un +1 =
=
23( n +1)
D’Alembert’s Ratio Test
5.43
32 n + 2 23 n + 3
un 32 n 23n + 3 = 3n ◊ 2 n + 2 un +1 2 3 23 = lim n Æ•
2
3
=
8 9
un 8 8 = lim = < 1 un +1 nƕ 9 9
Hence, by D’Alembert’s ratio test, the series is divergent.
example 2 5n -1 . n =1 n ! •
Test the convergence of the series Solution Let
Â
5n -1 n! 5n = (n + 1)!
un =
un +1
un 5n -1 (n + 1)! = ◊ un +1 n! 5n n +1 5 n +1 = lim Æ • >1 n Æ• 5 =
lim n Æ•
un un +1
Hence, by D’Alembert’s ratio test, the series is convergent.
example 3 •
Test the convergence of the series
 n =1 n
2n 3
. +1
Solution Let
un = un +1 =
2n n3 + 1 2 n +1 (n + 1)3 + 1
[Winter 2016]
5.44
Chapter 5
Sequences and Series
un 2 n (n + 1)3 + 1 = 3 ◊ un +1 n + 1 2 n +1 3
1 Ê 1ˆ ÁË 1 + n ˜¯ + 3 1 n = ◊ 1 2 1+ 3 n 3
1 Ê 1ˆ ÁË 1 + n ˜¯ + 3 1 un n lim = lim ◊ 1 n Æ• un +1 n Æ• 2 1+ 3 n 1 = 1
Hence, by D’Alembert’s ratio test, the series is convergent.
5.11
D’Alembert’s Ratio Test
example 5 Test the convergence of the series
n !(2) n ∑ nn .
Solution Let
un = un +1 =
n !(2)n nn (n + 1)!(2)n +1 (n + 1)n +1
un n! 2 n (n + 1)n +1 = n ◊ un +1 (n + 1)! 2 n +1 n =
1 Ê n + 1ˆ 2 ÁË n ˜¯
n
un 1 Ê 1ˆ = lim Á 1 + ˜ n Æ• un +1 2 nÆ• Ë n ¯
n
lim
e >1 2 Hence, by D’Alembert’s ratio test, the series is convergent. =
example 6 n3 . Test the convergence of the series  n =1 ( n - 1)! Solution •
Let
un =
n3 (n - 1)!
un +1 =
(n + 1)3 n!
un n3 n! = ◊ un +1 (n - 1)! (n + 1)3 = =
n3 n(n - 1)! ◊ (n - 1)! (n + 1)3 n Ê 1ˆ ÁË 1 + n ˜¯
3
5.45
5.46
Chapter 5
Sequences and Series
lim n Æ•
un n = lim Æ • >1 3 n Æ• un +1 Ê 1ˆ ÁË 1 + n ˜¯
Hence, by D’Alembert’s ratio test, the series is convergent.
example 7 ∞
Test the convergence of the series Solution Let
(n + 1) n . ∑ n! n =1 (n + 1)n n! (n + 2)n +1 = (n + 1)!
un = un +1
un (n + 1)n (n + 1)! = ◊ un +1 n! (n + 2)n +1 = =
lim
n Æ•
(n + 1)n (n + 1)(n !) ◊ n! [(n + 1) + 1]n+1 1 1 ˆ Ê ÁË 1 + n + 1˜¯
n +1
un 1 = lim n +1 un +1 nÆ• Ê 1 ˆ ÁË 1 + n + 1˜¯
1 1 2 Hence, by D’Alembert’s ratio test, the series is convergent. =
example 9 •
Test the convergence of the series Solution
Â
n =1 3
un =
Let
5n + a
un +1 =
n
+b
, a > 0, b > 0.
5n + a 3n + b 5n +1 + a 3n +1 + b
un 5n + a 3n +1 + b = lim n ◊ n Æ• un +1 n Æ• 3 + b 5n +1 + a lim
a b 3+ n n 3 5 ◊ = lim b a n Æ• 1+ n 5 + n 5 3 3 = 1 1 un +1 nÆ• 1+ n Hence, by D’Alembert’s ratio test, the series is convergent. lim
n Æ•
example 13 Test the convergence of 1 +
2 p 3p 4 p + + + … ,( p > 0). 2! 3! 4!
Solution un =
Let
un +1 =
lim n Æ•
un un +1
np n! (n + 1) p (n + 1)!
np n! = lim n Æ• ( n + 1) p (n + 1)! = lim n Æ•
= lim
n Æ•
np (n + 1)
◊ p
(n + 1) Ê 1ˆ ÁË 1 + n ˜¯
p
(n + 1)! n!
Æ • >1
Hence, by D’Alembert’s ratio test, the series is convergent.
example 14 Test the convergence of the series
1 2 3 4 + + + + L. 2 3 1 + 2 1 + 2 1 + 2 1 + 24
5.50
Chapter 5
Sequences and Series
Solution n
Let
un =
un +1 = un un +1
1 + 2n n +1
1 + 2 n +1 n 1 + 2 n +1 = ◊ 1 + 2n n + 1
1 +2 2n = Ê 1 ˆ Ê 1ˆ ÁË n + 1˜¯ ÁË 1 + n ˜¯ 2 1 +2 un 2n lim = lim n Æ• un +1 n Æ• Ê 1 ˆ Ê 1ˆ ÁË n + 1˜¯ ÁË 1 + n ˜¯ 2
=2>1
ÈQ lim 2 n Æ • ˘ ÍÎ nÆ• ˙˚
Hence, by D’Alembert’s ratio test, the series is convergent.
example 15 •
Test the convergence of the series
2 ◊ 4 ◊ 6 K 2n
 5 ◊ 8 ◊11K(3n + 2).
n =1
Solution Let
2 ◊ 4 ◊ 6 K 2n 5 ◊ 8 ◊ 11K (3n + 2) 2 ◊ 4 ◊ 6 K 2 n(2 n + 2) = 5 ◊ 8 ◊ 11K (3n + 2)(3n + 5)
un = un +1
un 2 ◊ 4 ◊ 6 K 2n 5 ◊ 8 ◊ 11K (3n + 2)(3n + 5) = ◊ 2 ◊ 4 ◊ 6 K 2 n( 2 n + 2 ) un +1 5 ◊ 8 ◊ 11K (3n + 2) 3n + 5 2n + 2 5 3+ n = 2 2+ n =
[Using A.P.]
5.11
D’Alembert’s Ratio Test
5.51
5 3+ un n lim = lim 2 n Æ• un +1 n Æ• 2+ n 3 = >1 2 Hence, by D’Alembert’s ratio test, the series is convergent.
example 16 Test the convergence of the series Solution Let
un =
2 2 ◊ 5 2 ◊ 5 ◊ 8 2 ◊ 5 ◊ 8 ◊ 11 + L. + + + 1 1 ◊ 5 1 ◊ 5 ◊ 9 1 ◊ 5 ◊ 9 ◊ 13
2 ◊ 5 ◊ 8 ◊ 11K (3n - 1) 1 ◊ 5 ◊ 9 ◊ 13K (4 n - 3)
[Using A.P.]
2 ◊ 5 ◊ 8 ◊ 11K (3n - 1)(3n + 2) 1 ◊ 5 ◊ 9 ◊ 13K(4 n - 3)(4n + 1) 2 ◊ 5 ◊ 8 ◊ 11K(3n - 1) un 1 ◊ 5 ◊ 9 ◊ 13K(4 n - 3) lim = lim n Æ• un +1 n Æ• 2 ◊ 5 ◊ 8 ◊ 11K((3n - 1)(3n + 2) 1 ◊ 5 ◊ 9 ◊ 13K (4 n - 3)(4 n + 1) 4n + 1 = lim n Æ• 3n + 2 1 4+ n = lim 2 n Æ• 3+ n 4 = >1 3 Hence, by D’Alembert’s ratio test, the series is convergent. un+1 =
example 17
2
2
2
Ê 1ˆ Ê 1◊ 2 ˆ Ê 1◊ 2 ◊ 3 ˆ +Á + K •. Test the convergence of the series Á ˜ + Á ˜ Ë 3¯ Ë 3 ◊ 5¯ Ë 3 ◊ 5 ◊ 7 ˜¯ Solution Let
È 1 ◊ 2 ◊ 3K n ˘ un = Í ˙ Î 3 ◊ 5 ◊ 7 K (2 n + 1) ˚
2
È ˘ 1 ◊ 2 ◊ 3 K n(n + 1) un +1 = Í ˙ ( 2 n + 3 ) 3 5 7 K ( 2 1 )( ◊ ◊ n + Î ˚
[Using A.P.] 2
5.52
Chapter 5
Sequences and Series
lim
n Æ•
È 1 ◊ 2 ◊ 3K n ˘ Í 3 ◊ 5 ◊ 7 K(2 n + 1) ˙ Î ˚
2
un = lim 2 un +1 nÆ• È ˘ 1 ◊ 2 ◊ 3K n(n + 1) Í 3 ◊ 5 ◊ 7 K(2 n + 1)(2 n + 3) ˙ Î ˚ È 2n + 3 ˘ = lim Í n Æ• Î n + 1 ˙ ˚ 3˘ È Í2 + n ˙ = lim Í ˙ n Æ• Í1+ 1 ˙ ÍÎ n ˙˚ = 4 >1
2
2
Hence, by D’Alembert’s ratio test, the series is convergent.
example 18 Test the convergence of the series 2 +
3 4 5 x + x 2 + x 3 + L. 2 3 4 [Summer 2014]
Solution Let
n + 1 n -1 x n n+2 n = x n +1
un = un +1
un (n + 1) x n -1 n +1 = ◊ un +1 n ( n + 2) x n 1 1 1+ n◊ n ◊1 2 x 1 1+ n
1+ =
1 1 1+ 1+ un n◊ n ◊1 lim = lim 2 x n Æ• un +1 n Æ• 1 1+ n 1 = x
5.11
D’Alembert’s Ratio Test
By D’Alembert’s ratio test, the series is 1 (i) convergent if > 1 or x < 1 x 1 (ii) divergent if < 1 or x > 1 x 1 The test fails if = 1, i.e., x = 1. x n +1 1 = 1+ n n Ê 1ˆ lim un = lim Á 1 + ˜ n Æ• n Æ• Ë n¯
Then
un =
=1
[finite and non-zero]
Sun is divergent for x = 1. Hence, the series is convergent for x < 1 and is divergent for x ≥ 1.
example 19 •
Test the convergence of the series
Â
(n + 1)
n =1
Solution Let
un =
n +1 n
un +1 =
2
n
2
xn.
xn
n+2 (n + 1)
2
x n +1
un (n + 1) x n (n + 1)2 = ◊ 2 un +1 (n + 2) x n +1 n =
(n + 1)3 2
n ( n + 2) x 3
Ê 1ˆ ÁË 1 + n ˜¯ 1 ◊ = Ê 2ˆ x ÁË 1 + n ˜¯ 3
lim n Æ•
un un +1
Ê 1ˆ ÁË 1 + n ˜¯ 1 ◊ = lim n Æ• Ê 2ˆ x 1 + ÁË n ˜¯ =
1 x
5.53
5.54
Chapter 5
Sequences and Series
By D’Alembert’s ratio test, the series is (i) convergent if
1 > 1 or x < 1 x
(ii) divergent if
1 < 1 or x > 1 x
The test fails if
1 = 1, i.e., x = 1. x
Then
un =
vn =
lim
n →∞
n2
1 Ê 1ˆ 1+ ˜ n ÁË n¯
= Let
n +1
1 n
un 1 = lim 1 + vn n→∞ n
= 1 [finite and non-zero] 1 and Svn = Â is divergent as p = 1. n By comparison test, Sun is also divergent for x = 1. Hence, the series is convergent for x < 1 and is divergent for x ≥ 1.
example 20 Ê xn ˆ Test the convergence of the series  Á 2 ˜ , for x > 0. n =1 Ë n + 1¯ Solution •
Let
xn un =
un +1 =
2
n +1
x n +1 (n + 1)2 + 1
un x n (n + 1)2 + 1 = 2 ◊ un +1 n + 1 x n +1 2
1 Ê 1ˆ ÁË 1 + n ˜¯ + 2 1 n = ◊ 1 x 1+ 2 n
5.11
D’Alembert’s Ratio Test
2
un n Æ• un +1 lim
1 Ê 1ˆ ÁË 1 + n ˜¯ + 2 1 n = lim ◊ 1 n Æ• x 1+ 2 n 1 = x
By D’Alembert’s ratio test, the series is (i) convergent if
1 > 1 or x < 1 x
1 < 1 or x > 1 x The test fails if x = 1.
(ii) divergent if
Then
n2 + 1 1 = 1ˆ Ê n2 Á 1 + 2 ˜ Ë n ¯ 1 vn = 2 n un 1 lim = lim n Æ• vn n Æ• Ê 1ˆ ÁË 1 + 2 ˜¯ n = 1 [finite and non-zero]
Let
and Svn = Â
1
un =
1 n2
is convergent as p = 2 > 1.
By comparison test, Sun is also convergent if x = 1. Hence, the series is convergent for x £ 1 and is divergent for x > 1.
example 21 Test the convergence of the series
Â
Solution Let
un =
2n 4
n +1 2n 4
n +1
x n , x > 0.
xn
5.55
5.56
Chapter 5
Sequences and Series
2 n +1
un +1 = un
(n + 1) + 1 4
x n +1
2 n x n (n + 1)4 + 1 ◊ n 4 + 1 2 n +1 x n +1
=
un +1
4
1 Ê 1ˆ ÁË 1 + n ˜¯ + 4 n = 1ˆ Ê ÁË 1 + 4 ˜¯ 2 x n 4
lim
n Æ•
un un +1
1 Ê 1ˆ ÁË 1 + n ˜¯ + 4 n = lim n Æ• Ê 1ˆ ÁË 1 + 4 ˜¯ 2 x n 1 = 2x
By D’Alembert’s ratio test, the series is (i) convergent if 1 > 1 or x < 1 2x 2 (ii) divergent if 1 < 1 or x > 1 2x 2 The test fails if 1 = 1 or x = 1 . 2x 2 Then
un = =
2n
1 ◊ n + 1 2n 1 4
n4 + 1 1 = 1ˆ Ê n4 Á1 + 4 ˜ Ë n ¯
Let
vn = lim
n Æ•
1 n4
un 1 = lim n Æ• 1ˆ vn Ê ÁË 1 + 4 ˜¯ n
=1 and Σvn = ∑
1 is convergent as p = 4 > 1. n4
[finite and non-zero]
5.11
1 . 2
By comparison test, Sun is also convergent if x = Hence, the series is convergent for x £
D’Alembert’s Ratio Test
1 1 and is divergent for x > . 2 2
example 22
∑
Test the convergence of the series
n
xn .
2
n +1
Solution n
Let
un =
un +1 = un = un +1 =
n2 + 1
◊ xn
(n + 1) (n + 1)2 + 1 n n2 + 1
◊ xn
◊ x n +1 (n + 1)2 + 1 1 ◊ n +1 n +1 x
n (n2 + 2 n + 2) 1 ◊ ◊ (n + 1) x (n2 + 1)
Ê 2 2ˆ ÁË 1 + n + 2 ˜¯ un 1 n lim = lim ◊ n Æ• un +1 n Æ• Ê 1ˆ Ê 1ˆ x ÁË 1 + n ˜¯ ÁË 1 + 2 ˜¯ n
1 x By D’Alembert’s ratio test, the series is 1 (i) convergent if > 1, or x < 1 x =
(ii) divergent if
1 < 1, or x >1 x
The test fails for x = 1. Then
n un =
2
n +1 1
n2 = n 1+
1 n2
5.57
5.58
Chapter 5
Sequences and Series
1
=
Let
1 n2
1+
1 n2
1
vn =
1
n2 lim n Æ•
and Svn = Â
1 1 n2
un = lim vn nƕ
1
1 n2 = 1 [finite and non-zero]
is divergent for p =
1+
1 1 or x < 1 x
1 < 1 or x > 1 x
1 = 1 or x = 1. x 1 1+ n 1 = 1 n + 1 n
Then
un =
Let
vn =
1 n
un 1 = lim n Æ• vn n Æ• Ê 1 ˆ ÁË n + 1˜¯ = 1 [finite and non-zero] lim
and Σvn = ∑
1 is divergent as p = 1. n
By comparison test, Sun is also divergent for x = 1. Hence, the series is convergent for x < 1 and is divergent for x ≥ 1.
example 24 Test the convergence of the series Solution Let
x x2 x3 x4 + + + +L. 1◊ 2 3 ◊ 4 5 ◊ 6 7 ◊ 8
un =
un +1 =
xn (2 n - 1)2 n
x n +1 (2 n + 1)(2 n + 2)
un (2 n + 1)(2 n + 2) xn = ◊ un +1 (2 n - 1)2 n x n +1 1ˆ Ê ÁË 2 + n ˜¯ Ê 1ˆ 1 1+ = ◊ 1 ˆ ÁË n ˜¯ x Ê 2 ÁË n ˜¯
5.59
5.60
Chapter 5
Sequences and Series
lim
n Æ•
un un +1
1ˆ Ê ÁË 2 + n ˜¯ = lim n Æ• Ê 1ˆ ÁË 2 - n ˜¯ =
Ê 1ˆ 1 ÁË 1 + n ˜¯ ◊ x
1 x
By D’Alembert’s ratio test, the series is (i) convergent if (ii) divergent if The test fails if
1 > 1 or x < 1 x
1 < 1 or x > 1 x
1 = 1 or x = 1. x
Then
1 (2n − 1)2n
un =
1 1 2n 2 − n
=
2
Let
vn = lim
n →∞
1 n2
un 1 = lim n →∞ 1 vn 22 − n =
1 [finite and non-zero] 4
1 is convergent as p = 2 > 1. n2 By comparison test, Sun is also convergent.
and Σvn = ∑
Hence, the series is convergent for x £ 1 and is divergent for x > 1.
example 25 Test the convergence of the series
x x2 1 + + +L. 1◊ 2 ◊ 3 4 ◊ 5 ◊ 6 7 ◊ 8 ◊ 9 [Summer 2017]
5.11
D’Alembert’s Ratio Test
Solution Let
un =
xn - 1 (3n - 2)(3n - 1) 3n xn (3n + 1) (3n + 2)(3n + 3)
un + 1 =
un (3n + 1) (3n + 2) (3n + 3) xn - 1 = ◊ un + 1 (3n - 2) (3n - 1) (3n) xn Ê ÁË 3 +
1ˆ Ê 2ˆ Ê 3ˆ 3 + ˜ Á3 + ˜ ˜ Á ¯ Ë ¯ Ë n n n¯ 1 ◊ 2ˆ Ê 1ˆ x Ê ÁË 3 - n ˜¯ ÁË 3 - n ˜¯ 3
=
1ˆ Ê 2ˆ Ê 3ˆ Ê ÁË 3 + n ˜¯ ÁË 3 + n ˜¯ ÁË 3 + n ˜¯ 1 un lim = lim ◊ n Æ • un + 1 nÆ• 2ˆ Ê 1ˆ x Ê ÁË 3 - n ¯˜ ÁË 3 - n ˜¯ 3 1 x By D¢ Alembert’s ratio text, the series is 1 (i) convergent if > 1 or x < 1 x 1 (ii) divergent if < 1 or x > 1 x =
The test fails if
1 = 1 or x = 1 . x
Then
un = =
Let
vn = lim n Æ•
1 (3n - 2)(3n - 1)(3n) 1 2ˆ Ê 1ˆ Ê n3 Á 3 - ˜ Á 3 - ˜ 3 Ë n¯ Ë n¯ 1 n3
un 1 = lim n Æ• 2ˆ Ê 1ˆ vn Ê ÁË 3 - n ˜¯ ÁË 3 - n ˜¯ 3 =
1 27
[finite and non-zero]
5.61
5.62
Chapter 5
and
 vn =  n3
Sequences and Series
1
is convergent as p = 3 > 1.
 un
By comparison test,
is also convergent for x = 1.
Hence, the series is convergent for x £ 1 and is divergent for x > 1.
example 26 3 4 (n + 1) Test the convergence of the series 2 x + x 2 + x 3 + L + 3 x n + L. 8 27 n Solution Let
un +1
n +1
◊ xn n3 n+2 = ◊ x n +1 (n + 1)3
un =
un (n + 1) x n (n + 1)3 = ◊ un +1 (n + 2) x n +1 n3 4
Ê 1ˆ ÁË 1 + n ˜¯ 1 ◊ = Ê 2ˆ x 1 + ÁË n ˜¯ 4
un n Æ• un +1 lim
Ê 1ˆ ÁË 1 + n ˜¯ 1 ◊ = lim n Æ• Ê 2ˆ x 1 + ÁË n ˜¯ =
1 x
By D’Alembert’s ratio test, the series is (i) convergent if (ii) divergent if The test fails if Then
1 > 1 or x < 1 x
1 < 1 or x > 1 x
1 = 1 or x = 1 . x un =
n +1
n3 1 Ê 1ˆ = 2 Á1 + ˜ n¯ n Ë
5.11
Let
vn =
D’Alembert’s Ratio Test
5.63
1 n2
un Ê 1ˆ = lim Á 1 + ˜ n Æ• vn n Æ• Ë n¯ lim
=1 and Svn = Â
[finite and non-zero]
1
is convergent as p = 2 > 1. n2 By comparison test, Sun is also convergent for x = 1. Hence, the series is convergent for x £ 1 and is divergent for x > 1.
example 27 Test the convergence of the series
1 2 1
+
x2 x4 x6 + + + L. 3 2 4 3 5 4 [Winter 2013]
Solution Let
x2n− 2 un =
(n + 1) n
x2n u n +1 =
( n + 2) n + 1
un ( n + 2) n + 1 x2n− 2 = ⋅ un +1 (n + 1) n x2n =
( n + 2) n + 1 1 ⋅ (n + 1) n x2
Ê 2ˆ ÁË 1 + n ˜¯ 1 1 = 1+ ◊ 2 n x Ê 1ˆ ÁË 1 + n ˜¯ Ê 2ˆ ÁË 1 + n ˜¯ un 1 1 = lim lim 1+ ◊ 2 n Æ• un +1 n Æ• Ê 1ˆ n x ÁË 1 + n ˜¯ =
By D’Alembert’s ratio test, the series is 1 (i) convergent if 2 > 1 or x2 < 1 x 1 2 (ii) divergent if 2 < 1 or x > 1 x
1 x2
5.64
Chapter 5
Sequences and Series
1
The test fails if
x2
= 1 or x 2 = 1.
Then
un = =
1 (n + 1) n 1 3 n2
Ê 1ˆ ÁË 1 + n ˜¯
1
Let
vn =
3
n2
un 1 = lim n Æ• vn Ê 1ˆ ÁË 1 + n ˜¯ = 1 [finite and non-zero] 3 1 and Svn = 3 is convergent as p = > 1. 2 n2 By comparison test, Sun is also convergent for x2 = 1. lim
n Æ•
Hence, the series is convergent for x2 £ 1 and is divergent for x2 > 1.
example 28 Test the convergence of the series 1 +
3 5 7 3 9 4 x + x2 + x + x + L. 2 9 28 65
Solution un =
Let
lim m n Æ•
un un +1
xn n3 + 1 2n + 3
x n +1 (n + 1)3 + 1 (2 n + 1) x n [(n + 1)3 + 1] = ◊ n3 + 1 (2 n + 3) x n +1
un +1 = un un +1
2n + 1
3 1 ˆ ÈÊ 1 ˆ 1˘ Ê ÁË 2 + n ˜¯ ÍÁË 1 + n ˜¯ + 3 ˙ n ˙˚ ÍÎ = 3ˆ Ê 1ˆ Ê ÁË 2 + n ˜¯ ÁË 1 + 3 ˜¯ x n 3 1 ˆ ÈÊ 1 ˆ 1˘ Ê 2 1 + + + 3˙ Í ÁË ˜ Á ˜ n ¯ ÍË n¯ n ˙˚ Î = lim n Æ• 1ˆ 3ˆ Ê Ê ÁË 2 + n ˜¯ ÁË 1 + 3 ˜¯ x n
[Neglecting the first term]
5.11
D’Alembert’s Ratio Test
5.65
1 x By D’Alembert’s ratio test, the series is =
(i) convergent if
1 > 1 or x < 1 x
1 < 1 or x > 1 x The test fails if x = 1. Then
(ii) divergent if
un =
2n + 1 n3 + 1
1 n = 1 n 2 1 + 3 n 2+
Let
vn =
1 n2
1 2+ un n lim = lim n →∞ n →∞ v 1 n 1 + 3 n
=2 and Svn = Â
[finite and non-zero]
1
is convergent as p = 2 > 1. n2 By comparison test, Sun is also convergent if x = 1. Hence, the series is convergent for x £ 1 and is divergent for x > 1.
exercISe 5.3 Test the convergence of the following series: 1. 1 +
2 2 32 4 2 + + + K∞ 2! 3! 4! [ans.: Convergent]
∞
4 ⋅ 7 ⋅ 10 K(3n + 1) 2. ∑ 1⋅ 2 ⋅ 3 ⋅ 4 K n n =1
1 2 3 + + + K∞ 3. 2 1+ 5 1+ 5 1 + 53 4. 1 +
2! 3! 4! + + + K∞ 22 33 4 4
[ans.: Convergent]
[ans.: Convergent] [ans.: Convergent]
5.66
5.
Chapter 5
2 3 4 + + +L 3! 4 ! 5!
∞
[ans.: Convergent]
n2
[ans.: Convergent]
∑3 n =1 ∞
8.
n
2 n −1 n +1
∑3 n =1 ∞
9.
[ans.: Convergent]
3 32 33 34 + + + +L 2 ! 3! 4 ! 5!
6. 1 + 7.
Sequences and Series
[ans.: Convergent]
1
∑ n!
[ans.: Convergent]
n =1
n 2 (n + 1)2 n! n =1 ∞
10.
∑
[ans.: Convergent]
∞
3n + 4 n 11. ∑ n n n =1 4 + 5
[ans.: Divergent]
∞
xn 12. ∑ n 2 , x > 0 n =1 3 ⋅ n ∞
13.
3 − 2 n −1 ⋅ x , x>0 n +1 n =1
∑3 ∞
14.
[ans.: Convergent for x < 1, divergent for x > 3]
xn
∑ (2 )! n =1 ∞
15.
[ans.: Convergent for x < 3, divergent for x > 3]
n
∑ n =1
[ans.: Convergent]
n
n +1 n ⋅x , x > 0 n3 + 1
[ans.: Convergent for x < 1, divergent for x > 1]
16. x + 2 x 2 + 3x 3 + 4 x 4 + K ∞ [ans.: Convergent for x < 1, divergent for x > 1] 2
17. 1 +
3
x x x xn + + +K+ 2 + K∞ 2 5 10 n +1 [ans.: Convergent for x < 1, divergent for x > 1]
2 3 18. x + x + x + K ∞ 1⋅ 3 3 ⋅ 5 5 ⋅ 7
[ans.: Convergent for x < 1, divergent for x > 1]
5.12
19. x +
20.
Raabe’s Test
5.67
3 2 8 3 15 4 n2 − 1 n x + x + x +K + 2 x + K∞ 5 10 17 n +1 [ans.: Convergent for x < 1, divergent for x > 1]
x 2 3
+
x2 3 4
+
x3 4 5
+ K∞ [ans.: Convergent for x < 1, divergent for x > 1]
5.12
raaBe’S teSt
u n n − 1 = l , then If Sun is a positive term series and nlim →∞ un +1 (i) Sun is convergent if l > 1 (ii) Sun is divergent if l < 1 (iii) Test fails if l = 1 proof: 1 (i) Consider a number p such that p > 1. The series Σun = Σ p is convergent n if p > 1. By comparison test, Sun will be convergent if from and after some term un v (n + 1) p Ê 1 ˆ > n = = Á1 + ˜ Ë n¯ un +1 vn +1 np
p
p
un p p( p - 1) Ê 1ˆ > 1+ = 1+ + +K un +1 ÁË n ˜¯ n 2! n2 Ê u ˆ È p p( p - 1) ˘ n Á n - 1˜ > n Í + + K˙ 2 n Ë un +1 ¯ 2 n Î ˚ Ê u ˆ p( p - 1) È ˘ lim n Á n - 1˜ > lim Í p + + K˙ Æ• n 2n Ë un +1 ¯ Î ˚
n Æ•
l > p >1 Hence, Sun is convergent if l > 1. (ii) Consider a number p such that p < 1. The series Svn = S
1
is divergent if p < 1. np By comparison test, Sun will be divergent if from and after some term un v < n un +1 vn +1
5.68
Chapter 5
Sequences and Series
Proceeding as in case (i), we get u p ( p − 1) lim n n − 1 < lim p + + K →∞ n 2n un +1 l < p 1 n Æ• 2 3 3+ n = lim
n Æ•
Hence, by Raabe’s test, the series is convergent.
example 2 Test the convergence of the series Â
4 ◊ 7 ºº (3n + 1) x n . n!
5.12
Raabe’s Test
Solution un = un +1 = un un +1
lim n Æ•
4 ◊ 7 K (3n + 1) x n n! 4 ◊ 7 K (3n + 1)(3n + 4) x n +1 (n + 1)!
(n + 1)! 4 ◊ 7 K (3n + 1) x n ◊ n! 4 ◊ 7 K (3n + 1)(3n + 4) x n +1 n +1 = (3n + 4) x =
un un +1
1 n = lim n Æ• Ê 4ˆ ÁË 3 + n ˜¯ x 1+
=
1 3x
By ratio test, the series is 1 1 > 1 or x < (i) Convergent if 3 3x 1 1 < 1 or x > (ii) Divergent if 3 3x (iii) Test fails if x =
1 3 un
Then
un+1
=
n +1 (3n + 4)
=
1 3
3n + 3 3n + 4
Applying Raabe’s test, Ê u ˆ Ê 3n + 3 ˆ lim n Á n - 1˜ = lim n Á - 1˜ n Æ• Ë un +1 ¯ nÆ• Ë 3n + 4 ¯ Ê -n ˆ = lim Á ˜ n Æ• Ë 3n + 4 ¯ Ê ˆ Á -1 ˜ = lim Á 4˜ n Æ• Á 3+ ˜ Ë n¯ 1 = - 1 or x 2 < 1
x2
2
< 1 or x 2 > 1
x (iii) Test fails if x = 1 2
Then
un (2 n + 2)(2 n + 3) = un +1 (2 n + 1)2
5.12
Raabe’s Test
Applying Raabe’s test,
È (2 n + 2)(2 n + 3) ˘ Ê u ˆ - 1˙ lim n Á n - 1˜ = lim n Í 2 n Æ• u Ë n +1 ¯ Î (2 n + 1) ˚
n Æ•
= lim
n Æ•
= lim
n Æ•
=
n(6 n + 5) (2 n + 1)2 5ˆ Ê ÁË 6 + n ˜¯ 1ˆ Ê ÁË 2 + n ˜¯
2
3 >1 2
By Raabe’s test, the series is convergent if x 2 = 1 . Hence, the series is convergent if x 2 £ 1 and is divergent if x 2 > 1.
example 4 Test the convergence of the series a(a + 1)(a + 2)ºº (a + n - 1) ◊ b(b + 1)(b + 2)ºº (b + n - 1) x n .  1 ◊ 2 ◊ 3ºº n ◊ c(c + 1)(c + 2)ºº (c + n - 1)
Solution un =
a(a + 1)(a + 2)KK(a + n - 1) ◊ b(b + 1)(b + 2)KK(b + n - 1) x n 1 ◊ 2 ◊ 3KK n ◊ c(c + 1)(c + 2)KK(c + n - 1)
un a(a + 1)KK (a + n - 1) ◊ b(b + 1)KK (b + n - 1) x n = ◊ 1 ◊ 2 ◊ 3KK n ◊ c(c + 1)KK (c + n - 1) un +1 1 ◊ 2 KK n(n + 1) ◊ c(c + 1)...(c + n - 1)(c + n) a(a + 1)KK (a + n - 1)(a + n) ◊ b(b + 1)KK(b + n - 1)(b + n) x n +1 (n + 1)(c + n) = (a + n)(b + n) x
lim
n Æ•
un un +1
Ê 1ˆ Ê c ˆ ÁË 1 + n ˜¯ ËÁ n + 1˜¯ = lim n Æ• Ê a ˆÊb ˆ ÁË n + 1˜¯ ÁË n + 1˜¯ x =
1 x
5.71
5.72
Chapter 5
Sequences and Series
By ratio test, the series is (i) Convergent if
1 > 1 or x < 1 x
1 < 1 or x >1 x (iii) Test fails if x = 1 (ii) Divergent if
un (n + 1)(c + n) = un +1 (a + n)(b + n)
Then
Applying Raabe’s test, Ê u ˆ È (n + 1)(c + n) ˘ lim n Á n - 1˜ = lim n Í - 1˙ n Æ• u Ë n +1 ¯ Î (a + n)(b + n) ˚
n Æ•
È (c - ab) + n(1 + c - a - b) ˘ = lim n Í ˙ n Æ• Î (a + n)(b + n) ˚ È (c - ab) ˘ + (1 + c - a - b) ˙ Í ˙ = lim Í n n Æ• Í ˙ Êa ˆÊb ˆ ÁË n + 1˜¯ ÁË n + 1˜¯ Í ˙ Î ˚ = 1+ c - a - b By Raabe’s test, the series is (i) convergent if 1 + c - a - b > 1 or c > a + b , and (ii) divergent if 1 + c - a - b < 1 or c < a + b. Hence, the series is convergent if x < 1 and divergent if x > 1. For x = 1, the series is convergent if c > a + b and divergent if c < a + b.
exercISe 5.4 Test the convergence of the following series: 1. 1 + 1 + 1 + 1 + ... . 2 2◊4 2◊4 ◊6 2.
Â
[ans.: Divergent]
12 ◊ 52 ◊ 92 KK(4 n - 3)2 . 4 2 ◊ 82 ◊ 122 KK(4 n)2
[ans.: Convergent] 3. (i) 1 +
22 22 ◊ 4 2 22 ◊ 4 2 ◊ 6 2 + + +K. 3◊ 4 3◊ 4 ◊5◊6 3◊ 4 ◊5◊6 ◊7 ◊8
[ans.: Convergent]
5.13
(ii) 1 +
Cauchy’s Root Test
5.73
(1!)2 (2!)2 2 (3!)2 3 x+ x + x + K. 2! 4! 6! [ans.: Convergent for x < 4 and divergent for x ≥ 4]
(iii) 1 + 3 x + 3 ◊ 6 x 2 + 3 ◊ 6 ◊ 9 x 3 + K . 7 7 ◊ 10 7 ◊ 10 ◊ 13 [ans.: Convergent for x £ 1 and divergent for x > 1] 4. 1 +
22 22 ◊ 4 2 22 ◊ 4 2 ◊ 6 2 + + + K. 32 32 ◊ 52 32 ◊ 52 ◊ 7 2
[ans. : Divergent] 5.
a(a + 1) (a + 1)(a + 2) (a + 2)(a + 3) + + + K. 2! 3! 4! [ans.: Convergent for a £ 0]
6.
(n !)2
 (2n)!x
2n
. [ans.: Convergent for x < 4 and divergent for x2 ≥ 4]
5.13
cauchy’S root teSt 1
If Sun is a positive term series and if lim (un ) n = l then n Æ•
(i) Sun is convergent if l < 1. (ii) Sun is divergent if l > 1. Proof 1
Case I If lim (un ) n = l < 1. n Æ•
1
Consider a number l < r < 1 such that (un ) n < r for all n > m un < r n for all n > m The geometric series,
Sr n = r + r 2 + r 3 + K∞
... (1)
5.74
Chapter 5
Sequences and Series
Sn = r + r 2 + r 3 + K + r n r (1 - r n ) 1- r r (1 - r n ) lim Sn = lim n Æ• n Æ• 1 - r r = , which is finite 1- r =
˘ ÈQr < 1 ˙ Í n r = 0˙ ÍÎ\ xlim Æ• ˚
Hence, the series Sr n is convergent. un < r n for all n > m
From Eq. (1),
Sun < Sr n Since Sr n is convergent, Sun is also convergent. 1
Case II: If lim (un ) n = l > 1. n Æ•
1
(un ) n > 1 for all n > m
… (2)
Neglecting the first m terms, 1
1
1
1
S (un ) n = (um +1 ) m +1 + (um + 2 ) m + 2 + (um + 3 ) m + 3 + K ∞
[Using Eq. (2)]
> 1 + 1 + 1K ∞ Sn = (um +1
1 m ) +1
+ (um + 2
1 m ) +2
+ (um + 3
1 m ) +3
+ K + (um + n
1 m ) +n
> 1 + 1 + 1K n terms = n
lim Sn > lim n Æ •
n Æ•
n Æ•
•
Â
The series
un is divergent. The nature of a series remains unchanged if we
n = m +1
neglect a finite number of terms in the beginning. Hence, the series
•
 un
is divergent.
n= 1 1
Note: If lim(un ) n = 1, the root test fails, i.e., no conclusion can be drawn about the n→∞
convergence or divergence of the series.
5.13
Cauchy’s Root Test
example 1 •
Test the convergence of the series
1
Â
n=1 (log n)
n
.
Solution Let
un =
1 (log n)n
1
1 log n
(un ) n = 1
1 n Æ• log n = 0 1 and is divergent if p £ 1. 1 2 3 n=1 n [Summer 2015]
5.86
Chapter 5
Sequences and Series
Solution Let
un = f ( x) = •
Ú1
1
= f ( n)
np 1 xp
f ( x )dx = Ú
•
1
dx
xp
1
x - p +1 = lim m Æ• - p + 1
m
1
Êm 1 ˆ = lim Á m Æ• Ë 1 - p 1 - p ˜¯ 1- p
1 , 1- p = •, =-
If p = 1,
•
Ú1
f ( x )dx = Ú
• 1
p >1 p 1 and is infinite if p £ 1.
Hence, by Cauchy’s integral test, the series is convergent if p > 1 and is divergent if p £ 1.
exercISe 5.5 Test the convergence of the following series: ∞ 1 1. ∑ n n =1 ∞
2.
∑n n =1
2
1 +1
∞
3.
∑ ne
[ans.: Divergent] [ans.: Convergent]
− n2
[ans.: Convergent]
n =1 ∞
4.
1
∑ n(log n)
2
n =1
[ans.: Convergent]
5.15 Alternating Series
5.87
5.15 aLternatIng SerIeS An infinite series with alternate positive and negative terms is called an alternating series.
Leibnitz’s Test for Alternating Series •
An alternating series
•
 (-1)n -1vn =  un n =1
is convergent if
n =1
(i) each term is numerically less than its preceding term, i.e. un +1 < un or un > un +1 (ii) lim un = 0 n Æ•
example 1 Test the convergence of the series 1 −
1 1 1 + − +… . 2 3 4
Solution un = ( -1)n -1
Let
1 n
1 un = n The given series is an alternating series.
(i)
un - un +1 = =
1
1 -
n
n +1
n +1 - n n n +1
>0
for all n ΠN
\ un > un +1 (ii)
lim un = lim n Æ•
n Æ•
1 n
=0 Hence, by Leibnitz’s test, the series is convergent.
example 2 Test the convergence of the series 1 -
1 2 2
1 +
Solution Let
( -1)n -1 un = n n
3 3
1 -
4 4
+L.
5.88
Chapter 5
Sequences and Series
1
un =
n n
The given series is an alternating series. 1 1 (i) un - un +1 = n n (n + 1) n + 1 (n + 1) n + 1 - n n > 0 for all n Œ N ( n n ) ÈÎ(n + 1) ( n + 1) ˘˚ \ un > un +1 =
(ii)
lim un = lim n Æ•
1
n Æ•
n n =0 Hence, by Leibnitz’s test, the series is convergent.
example 3 1 Test the convergence of the series
1
2
1
-
2
1 +
2
Solution un = ( -1)n -1 ◊
Let
un =
1 -
2
3
42
+L.
1 n2
1 n2
The given series is an alternating series. 1 1 n2 (n + 1)2 2n + 1 = 2 > 0 for all n ΠN n (n + 1)2
(i) un - un +1 =
\
un > un +1
(ii) lim un = lim n Æ•
n Æ•
1 n2
=0 Hence, by Leibnitz’s test, the series is convergent.
example 4 1 Test the convergence of the series
1 -
1
p
1 +
2
p
1 -
3
p
4p
+ º.
5.15 Alternating Series
Solution Let
1
un = ( -1)n
np
1 un =
np
The given series is an alternating series. Case I: If p > 0, 1 1 (i) un - un +1 = p (n + 1) p n =
(n + 1) p - n p n p (n + 1) p
>0
[Q p > 0]
\ un > un +1 (ii) lim un = lim n Æ•
n Æ•
1 np
=0
[Q p > 0]
Hence, by Leibnitz’s test, the series is convergent if p > 0. Case II: If
p 0 for all n ΠN
1 2
n -1
5.89
5.90
Chapter 5
\
Sequences and Series
un > un +1
(ii) lim un = lim n Æ•
n Æ•
1 2
n -1
=0
Hence, by Leibnitz’s test, the series is convergent.
example 6 Test the convergence of the series
1 2 3 4 - + - +L. 2 5 10 17
Solution un = ( -1)n -1
Let
n 2
n +1
n un =
n2 + 1
The given series is an alternating series. n
(i) un - un +1 = = =
2
n +1
-
n +1 (n + 1)2 + 1
n(n2 + 2n + 2) - (n + 1)(n2 + 1) (n2 + 1)(n2 + 2n + 2) n2 + n - 1 (n2 + 1)(n2 + 2n + 2)
> 0 for all n ΠN
\ un > un +1
(ii) lim un = lim
n 2
n +1 1 = lim 1 n Æ• n+ n =0 Hence, by Leibnitz’s test, the series is convergent. n Æ•
n Æ•
example 7 ( -1)n+1 . Â n£1 log( n + 1) •
Test the convergence of the series Solution Let
un =
( -1)n +1 log( n + 1)
[Summer 2014]
5.15 Alternating Series
1 log(n + 1)
un =
The given series is an alternating series. (i) un - un +1 = =
1 1 log(n + 1) log (n + 2)
log(n + 2) - log(n + 1) >0 log(n + 1) ◊ log (n + 2)
\ un > un +1
(ii) lim un = lim n Æ•
n Æ•
1 log(n + 1)
=0 Hence, by Leibnitz’s test, the series is convergent.
example 8 •
Test the convergence of the series
Â
( -1)n -1
n =1 n
2
(n + 1)
.
Solution Let
( -1)n -1
un =
n2 (n + 1)
un =
n2 (n + 1)
1
The given series is an alternating series. (i) un - un +1 = =
1 n2 (n + 1)
-
1 (n + 1)2 (n + 2)
(n + 1)(n + 2) - n2
=
n2 (n + 1)2 (n + 2)
3n + 2 2
n (n + 1)2 (n + 2)
> 0 for all n ΠN
\ un > un +1
(ii) lim un = lim
1 2
n (n + 1) =0 Hence, by Leibnitz’s test, the series is convergent. n Æ•
n Æ•
5.91
5.92
Chapter 5
Sequences and Series
example 9 Test the convergence of the series x -
x2 x3 x 4 + + º(if x < 1) . 2 3 4
Solution un = ( -1)n -1
Let
un =
xn n
xn n
The given series is an alternating series. (i) un - un +1 =
\ (ii)
x n x n +1 n n +1
=
x n [(n + 1) - n x ] n(n + 1)
=
x n [1 + (1 - x )n] >0 n(n + 1)
[Q n ≥ 1 and 0 < x < 1]
un > un +1
lim un = lim n Æ•
n Æ•
xn n
ÈQ lim x n = 0 if x < 1˘ ÎÍ nÆ• ˚˙
=0 Hence, by Leibnitz’s test, the series is convergent.
example 10 Test the convergence of the series Ê 1ˆ Ê 2ˆ Ê 3ˆ Ê 4ˆ log Á ˜ - log Á ˜ + log Á ˜ - log Á ˜ + L Ë 2¯ Ë 4¯ Ë 3¯ Ë 5¯
Solution Ê 1ˆ Ê 2ˆ Ê 3ˆ Ê 4ˆ log Á ˜ - log Á ˜ + log Á ˜ - log Á ˜ + L Ë 2¯ Ë 4¯ Ë 3¯ Ë 5¯ Ê 2ˆ Ê 3ˆ Ê 4ˆ Ê 5ˆ = - log Á ˜ + log Á ˜ - log Á ˜ + log Á ˜ - L Ë 1¯ Ë 2¯ Ë 4¯ Ë 3¯
Let
Ê n + 1ˆ un = ( -1)n log Á Ë n ˜¯ Ê n + 1ˆ un = log Á Ë n ˜¯
5.15 Alternating Series
5.93
The given series is an alternating series. (i) un - un +1 = log
n +1 n+2 È n +1 n + 2 ˘ for all n Œ N ˙ - log > 0 ÍQ > n n +1 n n +1 Î ˚
\ un > un +1
Ê n + 1ˆ (ii) lim un = lim log Á Ë n ˜¯ n Æ• n Æ•
Ê 1ˆ = lim log Á 1 + ˜ Ë n Æ• n¯ = log 1 =0 Hence, by Leibnitz’s test, the series is convergent.
example 11 Test the convergence of the series
1 2 3 4 - + - +L. 2 3 4 5
Solution un = ( −1) n −1 ⋅
Let
un =
n n +1
n n +1
The given series is an alternating series. n n +1 (i) un - un +1 = n +1 n + 2 n2 + 2n - n2 - 2n - 1 (n + 1)(n + 2) 1 =1 8
•
By D’Alembert’s ratio test, gent and hence convergent.
 un
is convergent. Thus, the series is absolutely conver-
n=1
example 2 Test the series for absolute or conditional convergence 2 3 4 1- + 2 + 3 +L. 3 3 3 Solution Let
un = ( -1)n -1 ◊ •
 un
= 1+
n =1
un = un +1 = lim n Æ•
un un +1
n 3
n -1
2 3 4 + 2 + 3 +K 3 3 3
n 3n -1 n +1 3n n
= lim n Æ•
3
3
= lim n Æ•
◊ n -1
1+
= 3>1
1 n
3n n +1
5.96
Chapter 5
Sequences and Series
•
By D’Alembert’s ratio test, convergent.
 un
is convergent and hence, the series is absolutely
n =1
example 3 •
Test the series for absolute or conditional convergence
Â
(-1)n
. n + n +1 [Winter 2016]
n =1
Solution Let
(-1)n n + n +1
un =
1
un =
n + 1+n
The given series is an alternating series. 1 1 (i) un - un +1 = n + 1+ n n +1 + n + 2 =
= \ (ii)
n+2 - n ( n + 1 + n ) ( 1 + n + n + 2)
un > un +1
lim un = lim n Æ•
By Leibnitz’s test, un = =
Let
n +1 + n + 2 - n - n +1 ( n + 1 + n ) ( 1 + n + n + 2)
vn =
1
=0
n Æ•
n + 1+ n
 un
is convergent. 1
n + 1+ n 1 Ê 1ˆ n Á1 + 1 + ˜ n¯ Ë 1 n
>0
for all n ΠN
5.16 Absolute and Conditional Convergent of a Series
lim
n Æ•
un
= lim
n Æ•
vn
1 Ê 1ˆ Á1 + 1 + n ˜ Ë ¯ =6
and
 vn = Â
1 1 n2
 un
[finite and non-zero]
1 is divergent as p = . 2
By comparison test, The series
5.97
 un
is also divergent.
is convergent and
 un
is divergent.
Hence, the series is conditionally convergent.
example 4 Determine absolute or conditional convergence of the series •
 (-1) ◊ n
n =1
n2
[Winter 2013; Summer 2017]
n3 + 1
Solution un = ( -1)n ◊
Let
n2 n3 + 1
n2
un =
n3 + 1 1 = 1ˆ Ê n Á1 + 3 ˜ Ë n ¯
Let
vn =
lim n Æ•
un vn
1 n
= lim
1
1 n3 = 1 [finite and non-zero] n Æ•
1+
1 is divergent as p = 1. n By comparison test, Â un is also divergent.
and Svn = Â
Hence, Sun is not absolutely convergent. To check the conditional convergence, applying Leibnitz’s test, (i) un - un +1 =
n2 n3 + 1
-
(n + 1)2 (n + 1)3 + 1
5.98
Chapter 5
Sequences and Series
n2 (n3 + 3n2 + 3n + 2) - (n3 + 1)(n2 + 2 n + 1)
=
(n3 + 1)[(n + 1)3 + 1] n 4 + 2 n3 + n 2 - 2 n - 1
=
= =
(n3 + 1)[(n + 1)3 + 1]
n 4 + n2 (2 n + 1) - 1(2n + 1) (n3 + 1)[(n + 1)3 + 1] n 4 + (2n + 1)(n2 - 1) (n3 + 1)[[(n + 1)3 + 1]
> 0 for all n ΠN
un > un +1
(ii) lim un = lim n Æ•
n Æ•
= lim
n Æ•
=0 By Leibnitz’s test,
n2 3
n +1
1 1ˆ Ê n Á1 + 3 ˜ Ë n ¯
Su
n
is convergent.
The series Sun is convergent and
 un
is divergent.
Hence, the series is conditionally convergent.
example 5 Test the series for absolute or conditional convergence 2 3 1 4 1 5 1 - ◊ + ◊ - ◊ +º 3 4 2 5 3 6 4 Solution Ê n +1 1ˆ un = ( -1)n -1 Á ◊ Ë n + 2 n ˜¯
Let •
Â
un =
2 3 1 4 1 5 1 + ◊ + ◊ + ◊ + ◊◊◊. 3 4 2 5 3 6 4
| un | =
n +1 1 ⋅ n+2 n
n=1
Let
vn = lim
n Æ•
un vn
1 n n +1 n Æ• n + 2
= lim
5.16 Absolute and Conditional Convergent of a Series
5.99
1 n = lim 2 n Æ• 1+ n = 1 [finite and non-zero] 1+
1 is divergent as p = 1. n By comparison test, Σ un is also divergent. Hence, the series is not absolutely convergent. and Σvn = ∑
To check the conditional convergence, applying Leibnitz’s test, (i) un - un +1 = =
n +1 n+2 n(n + 2) (n + 1)(n + 3)
n2 + 3n + 3 >0 n(n + 1)(n + 2)(n + 3)
for all n ΠN
u n > u n +1 n +1 n(n + 2) 1 1+ n = lim n Æ• Ê 2ˆ n Á1 + ˜ Ë n¯
(ii) lim un = lim n Æ•
n Æ•
=0 By Leibnitz’s test, Sun is convergent. The series Sun is convergent and Σ un is divergent. Hence, the series is conditionally convergent.
example 6 Test the convergence of the series x −
x3 x5 + − L, x > 0. 3 5 [Summer 2016]
Solution Let
un = ( -1)n -1 un =
x 2 n -1 2n - 1
un +1 =
x 2 n +1 2n + 1
x 2 n -1 2n - 1
5.100
Chapter 5
Sequences and Series
un un +1
=
x 2 n -1 2 n + 1 ◊ 2 n - 1 x 2 n +1
1 n◊ 1 = 1 x2 2n 1ˆ Ê 2+ Á n˜◊ 1 = lim Á 1 ˜ x2 n Æ• Á2- ˜ Ë n¯ 1 = 2 x 2+
lim
n Æ•
un un +1
1 > 1 or x2 < 1 or x < 1 [Q x > 0] x2 Thus, the given series is absolutely convergent and hence, is convergent for x < 1. If x2 = 1 or x = 1 [Q x > 0] ( -1)n -1 un = 2n - 1 1 un = 2n - 1 The given series is an alternating series.
By D’Alembert’s ratio test,
 un
is convergent if
1 1 2n - 1 2n + 1 2 = 2 > 0 for all n ΠN 4n - 1
(i) un - un +1 =
(ii) lim un = lim n Æ•
n Æ•
1 2n - 1
=0 By Leibnitz’s test, the series is convergent for x = 1 Hence, the series is convergent for x £ 1.
exercISe 5.7 Test the following series for absolute or conditional convergence: 1. 1 −
1 1 1 1 1 + − + − +K 2 3 4 5 6
[ans.: Conditionally convergent] 1 1 1 1 1 2. 1 + 2 − 2 − 2 + 2 + 2 − K 2 3 4 5 6
[ans.: Absolutely convergent]
5.17
3. 1 −
1 2
1 +
3
Power Series
5.101
1 −
4
+K
[ans.: Conditionally convergent] sin x sin2 x sin3x 4. − + −K 13 23 33
5.17
[ans.: Absolutely convergent]
power SerIeS ∞
A power series is an infinite series of the form
∑a x n
n
= a0 + a1 ( x − c) + ..
n =1
a2 ( x − c) 2 + a3 ( x − c)3 +L, where an represents the coefficient of the nth term, c is a constant and x varies around c. When c = 0, the series becomes ∞
∑a x n
n
= a0 + a1 x + a2 x 2 + a3 x 3 + L
n=0
5.17.1
Interval and radius of convergence
A power series will converge only for certain values of x. An interval (−R, R) in which a power series converges is called the interval of convergence. The number R is called the radius of convergence, e.g., if a power series converges for all the values of x, then interval of convergence will be (−•, •) and the radius of convergence will be •.
5.17.2 test for convergence Since a power series may be positive, alternating or mixed series, the concept of absolute convergence is used to test the convergence of a power series. Applying D’Alembert’s ratio test, un = an x n un +1 = an +1 x n +1 lim
n Æ•
un an x n = lim un +1 nƕ an +1 x n +1 =
If
lim n Æ•
lim n Æ•
a 1 lim n x nƕ an +1
an = l, an +1 un 1 l = ◊l = un +1 x x
5.102
Chapter 5
Sequences and Series
By D’ Alembert’s ratio test, the series is absolutely convergent and hence is convergent l > 1, i.e., | x |< l, - l < x < l. If x Here, interval of convergence of the series is (−l, l ) and the radius of convergence is l.
example 1 ∞
Obtain the range of convergence of
xn , x > 0. ∑ n n =1 2
Solution Let
un =
xn 2n x n +1
un +1 =
2 n +1
un x n 2 n +1 = lim n ◊ n +1 n Æ• un +1 n Æ• 2 x 2 = x By D’Alembert’s ratio test, the series is 2 (i) convergent if > 1 or x < 2 x 2 (ii) divergent if < 1 or x > 2 x lim
The test fails if
2 = 1, or x = 2. x
Then
un =
2n 2n
=1
•
 un = 1 + 1 + 1 + L • n=1
which is a divergent series. Hence, the series is convergent for 0 < x < 2 and the range of convergence is 0 < x < 2.
example 2 ∞
Determine the interval of convergence for the series their behaviour at each end point.
2n x n and also, ∑ n = 0 n!
5.17
Power Series
5.103
Solution Let
un = un +1 =
2n x n n! 2 n +1 x n +1 (n + 1)!
un 2 n x n (n + 1)! = ◊ un +1 n ! 2 n +1 x n +1 n +1 = 2x
lim
n Æ•
un n +1 = lim un +1 nƕ 2 x
= • >1 Hence, By D’Alembert’s ratio test, the series is convergent for all values of x i.e. – • < x < • and interval of convergence is (– •, •).
example 3 ∞
Obtain the range of convergence of
xn , x > 0, a > 0. ∑ n n =1 a +
Solution xn
Let
un = a+ n
x n +1 un +1 =
a + n +1
un xn a + n +1 = lim ◊ n Æ• un +1 n Æ• a + n x n +1 lim
a = lim
n Æ•
+ 1+
n a n
= By D’Alembert’s ratio test, the series is 1 (i) convergent if > 1 or x < 1 x 1 (ii) divergent if < 1 or x > 1 x The test fails if x = 1.
1 x
+1
1 n 1 ◊ x
5.104
Chapter 5
Sequences and Series
Then
un =
=
Let
vn =
1 a+ n
1 1 n a + 1 n
1 n
un 1 = lim vn nƕ a +1 n = 1 [finite and non-zero]
lim n Æ•
and Σvn = ∑
1 1 2
is divergent as p =
1 < 1. 2
n By comparison test, Σun is also divergent for x = 1. Hence, the series is convergent for 0 < x < 1 and the range of convergence is 0 < x < 1.
example 4
∞
Obtain the range of convergence of Solution Let
un = un +1 = un un +1
( x + 1) n . ∑ n n =1 3 ⋅ n ( x + 1)n 3n ◊ n ( x + 1)n +1
3n +1 (n + 1) ( x + 1)n 3n +1 (n + 1) = n ◊ 3 ◊ n ( x + 1)n +1 3(n + 1) = ( x + 1)n Ê 1ˆ 3 Á1 + ˜ Ë n¯ = x +1
1 3 1 + n un lim = lim n →∞ u n →∞ x +1 n +1 =
3 x +1
5.17
Power Series
5.105
The series is convergent if 3 >1 x +1 3 > x +1
x +1 < 3 −3 < ( x + 1) < 3 −4 < x < 2
At x = 2, un = ∞
∞
∑u
∴
1 n
n
1 n =1 n
=∑
n =1
is divergent as p = 1. ( −1) n n 1 un = n
At x = – 4,
un =
The given series is an alternating series. 1 1 − n n +1 1 = >0 n(n + 1)
(i) un − un +1 =
for all n ΠN
∴ u n > u n +1
(ii) lim un = lim n →∞
n →∞
1 n
=0 By Leibnitz’s test, the series is convergent at x = – 4. Hence, the series is convergent for – 4 £ x < 2 and the range of convergence is – 4 £ x < 2.
example 5
∞
Obtain the range of convergence of
∑ n+ n =1
Solution Let
xn
.
1 + n2
xn un =
n + 1 + n2
x n +1 u n +1 =
(n + 1) + 1 + (n + 1) 2
5.106
Chapter 5
Sequences and Series
(n + 1) + 1 + (n + 1) 2 un xn = ⋅ u n +1 n + 1 + n 2 x n +1 1 1 1 + 1 + 1 + + n n2 n = 1 + 1 x 1 + n2
lim
n →∞
un u n +1
1 1 1 + 1 + 1 + + n n2 n = lim n →∞ 1 + 1 x 1 + 2 n =
The series is convergent if
2
1 x
1 >1 x x 0
for all n ŒN
∴ u n > u n +1
(ii) lim un = lim n →∞
n →∞
1 n + 1 + n2
=0
Thus, by Leibnitz’s test, the series is convergent if x = –1. Hence, the series is convergent for – 1 £ x < 1 and the range of convergence is – 1 £ x < 1.
example 6 •
Â
For the series
(-1)n - 1
n =1
convergence.
x2n - 1 , find the radius and interval of 2n - 1 [Winter 2016]
Solution Let
un =
(-1)n -1 x 2 n - 1 2n - 1
un + 1 =
(-1)n x 2 n + 2 - 1 2n + 2 - 1
=
(-1)n x 2 n + 1 2n + 1
un (-1)n - 1 x 2 n -1 2n + 1 = ◊ 2n - 1 un + 1 (-1)n x 2 n + 1 =-
(2 n + 1) 1 ◊ (2 n - 1) x 2
5.108
Chapter 5
lim n Æ•
Sequences and Series
un (2 n + 1) 1 = lim ◊ n Æ• (2 n - 1) x 2 un + 1 =
1 x2
By D’Alembert’s ratio test, the series is convergent if i.e.,
1 2 2 > 1 or 1 > x or x < 1 x2
– 1 < x < 1 and divergent for x > 1.
At x = 1, (-1)n - 1 (1)2 n - 1 2n - 1 1 1 1 =1- + - +L 3 5 7
un =
=
(-1)n - 1 2n - 1 1 2n - 1
un =
The given series is an alternating series. (i)
un - un + 1 =
1 1 2n - 1 2n + 1
=
2n + 1 - 2n + 1 (2 n - 1) (2 n + 1) =
2 >0 (2 n - 1) (2 n + 1)
un > un +1 (ii)
lim un = lim n Æ•
n Æ•
By Leibnitz’s test,
 un
1 =0 2n - 1 is convergent.
At x = –1, un =
=
(-1)n - 1 (-1)2 n - 1 2n - 1 (-1)3n - 2 2n - 1
for all n ΠN
5.17
un =
Power Series
5.109
1 2n - 1
The given series is an alternating series. Hence, by Leibnitz’s test,
 un
is convergent.
Thus, for the interval –1 £ x £ 1, given series is convergent. Since condition for convergence is | x2 | < 1, radius of convergence = 1. •
Since
 un
is convergent, the series is absolutely convergent for –1 £ x £ 1.
n =1
example 7 •
Find the interval of convergence of the series
Â
( -3)n x n
. n +1 [Winter 2013; Summer 2016] n= 0
Solution un = un +1 =
(-3)n x n n +1 (-3)n +1 x n +1 n+2
un n+2 (-3)n x n = ◊ un +1 n + 1 (-3)n +1 x n +1 =
n+2 (-3) x n + 1 1+
=
2 n
(-3) x 1 +
u lim n = lim n Æ• un +1 n Æ• 1 -3 x 1 = 3x =
1 n
1+
2 n
( -3) x 1 +
1 n
5.110
Chapter 5
Sequences and Series
1 1 > 1 or 3 x < 1 or x < or By D’Alembert’s ratio test, the series is convergent if 3x 3 1 1 - 1 x +1 2 > x +1
x +1 < 2 −2 < ( x + 1) < 2 −3 < x < 1
The series is convergent in the interval (– 3, 1).
5.112
Chapter 5
Sequences and Series
At x = –3, un = ( −1) n ∞
n( −3 + 1) n 2n
∞
∑u
n
= ∑n
n =1
n =1
which is a divergent series. At x = 1, n(1 + 1) n 2n = ( −1) n n
un = ( −1) n
un = n lim un ≠ 0
n →∞
By Leibnitz’s test, the series is not convergent at x = 1. Hence, the series is not convergent at each end point and the interval of convergence is (– 3, 1).
example 9 •
For the series
( -1)n ( x + 2)n , find the radius and interval of n n =1
Â
convergence. For what values of x does the series converge absolutely, conditionally? [Winter 2015] Solution un =
Let
un +1 =
( -1)n ( x + 2)n n ( -1)n +1 ( x + 2)n +1 n +1
un (n + 1) ( -1)n ( x + 2)n = ◊ n +1 un +1 n ( -1) ( x + 2)n +1 1 Ê 1ˆ = - Á1 + ˜ ◊ Ë n ¯ ( x + 2) lim
n Æ•
un = lim un +1 nƕ =
1 Ê 1ˆ ÁË 1 + ˜¯ ◊ n ( x + 2)
1 x+2
5.17
Power Series
The series is convergent if 1 >1 x+2 1> x+2
x+2 1, u lim n = lim n →∞ u n →∞ n +1
1 x2n 1 x 1 + 2 n x x2 +
= x > 1 Q lim x 2 n → ∞ n→∞
Thus, the series is convergent for x > 1, i.e. x > 1 and x < – 1 At x = 1, 1 2 ∞ 1 1 1 un = + + + L ∞ ∑ 2 2 2 n =1 un =
which is a divergent series. At x = – 1,
( −1) n 2 1 un = 2 1 lim un = ≠ 0 n →∞ 2 un =
Thus, by Leibnitz’s test, the series is not convergent at x = – 1. Hence, the series is convergent for x > 1 and range of convergence is x > 1.
5.17
Points to Remember
5.115
exercISe 5.9 Obtain the range of convergence of the following series: 1. 1 + x + 2 x 2 + 3x 3 + L + nx n + L 2
3
[ans.: −1< x < 1]
n
2.
1 x x x x + + + +L+ +L 2 3 4 5 n+2
3.
∑
[ans.: −1 < x < 1]
∞
(−1)n x n n =1 (n + 1) ∞
4.
∑ n=0
(x + 2)
[ans.: −3 ≤ x ≤ −1]
n +1
∞
5.
[ans.: |x| ≤ 1]
∑ (−1)
n
n=0
xn log(n + 1)
[ans.: |x| < 1]
∞
6.
1 3 Ans.: 2 < x < 2
∑ (−2) (n + 1)(x − 1) n
n
n=0 ∞
7.
∑ n !(x − 1)
n
[ans.: x = 1]
n =1
(n !)2 n x 8. ∑ n =1 (2n)! ∞
[ans.: |x| < 4]
∞
9.
(−2)n (2 x + 1)n n2 n =1
∑ ∞
10.
∑ n =1
(−1)n x 2 n
3 1 Ans.: − 4 ≤ x ≤ − 4
3
[ans.: −1 ≤ x ≤ 1]
(2n) 2
points to remember Sequence A sequence {un} is said to be convergent, divergent or oscillatory according as lim un is finite, infinite or not unique respectively. n Æ•
The infinite series Sun is said to be convergent, divergent or oscillatory according as lim Sn is finite, infinite or not unique respectively. n Æ•
If a positive term series Sun is convergent then lim un = 0 but converse is not true, n →∞ i.e., if lim un = 0, the series may converge or diverge. If lim un π 0, the series is n Æ•
not convergent.
n Æ•
5.116
Chapter 5
Sequences and Series
Comparison Test un = l (finite and non-zero) n Æ• vn
If Sun and Svn are series of positive terms such that lim then both series converge or diverge together.
D’Alembert’s Ratio Test un = l then n Æ• un +1
If Sun is a positive term series and lim (i) Sun is convergent (ii) Sun is divergent (iii) The test fails
if l > 1. if l < 1. if l = 1.
Raabe’s Test
u n n − 1 = l , then If Sun is a positive term series and nlim →∞ u n +1 (i) Sun is convergent if l > 1 (ii) Sun is divergent if l < 1 (iii) Test fails if l = 1
Cauchy’s Root Test
1
If Sun is a positive term series and if lim (un ) n = l then n Æ•
(i) Sun is convergent if l < 1. (ii) Sun is divergent if l > 1. This test is preferred when un contains nth powers of itself.
Cauchy’s Integral Test If Sun = S f (n) is a positive term series where f (n) decreases as n increases and let
Ú
•
1
f (x) dx = I then (i) Sun is convergent if I is finite. (ii) Sun is divergent if I is infinite.
This test is preferred when evaluation of the integral of f (x) is easy.
Leibnitz’s Test An alternating series
•
 (- 1)n -1 un is convergent if n =1
(i) each term is numerically less than its preceding term, i.e, un +1 < un or un > un +1 (ii)
lim un = 0
n Æ•
5.117
Multiple Choice Questions
Absolute Convergence •
The series
 un with both positive and
negative terms (not necessarily alternative) •
n=1
is called absolutely convergent if the corresponding series
 |un | with all positive n=1
terms is convergent.
Conditional Convergence •
If the series
•
•
n=1
n=1
 un is convergent and  | un | is divergent then the series  un is n=1
called conditionally convergent.
Multiple choice questions Select the most appropriate response out of the various alternatives given in each of the following questions: 1 is x
1. The value of lim x sin x Æ0
(a) 1
(b) p •
2. The sum of the series
 n =1
(a) 0
(b)
[Summer 2014] (c) 0
(d) •
1 is 2n
3 4
[Winter 2015] (c) 1
(d) 2
•
3. The series
( -1)n is n =1 n
Â
(a) divergent (c) conditionally convergent
[Winter 2015] (b) absolutely convergent (d) nothing can be said
1
 n(5 n - 1) is (a) convergent (c) oscillates finitely
4. The series
•
5. The series
 n=2
(b) divergent (d) oscillates infinitely
1 is n(log n)
(a) convergent (c) oscillates finitely
(b) divergent (d) oscillates infinitely
5.118
Chapter 5
6. The series
Sequences and Series
1
 np
is divergent if (b) p £ 1
(a) p > 1
(c) p = 1
(d) p = 0
•
un +1 > 1, then  un is x Æ 0 un n =1 (a) convergent (b) divergent (c) may or may not be convergent (d) oscillatory 2 3 8. The series a + ar + ar + ar + L oscillates finitely if (a) |r| < 1 (b) r > 1 (c) r = 1 7. If lim
3 9 27 81 + + + + L is 4 8 16 32 (a) convergent (c) oscillates finitely
(d) r £ –1
9. The series
•
10. The series
Â
(b) divergent (d) oscillates infinitely
1
is n 5n (a) convergent (b) divergent (c) oscillates finitely (d) oscillates infinitely 11. The series a + ar + ar 2 + ar 3 + L diverges if (a) |r| < 1 (b) r ≥ 1 (c) r £ –1 (d) r = 1 1 12. The series  p converges if n (a) p > 1 (b) p < 1 (c) p = 0 (d) p = 1 1 13. The series  1 is n =1
n4 (a) convergent (c) oscillates finitely • 32 n 14. The series  is n =1 4 n (a) convergent (c) oscillates finitely •
15. The series
1
 sin n
(b) divergent (d) oscillates infinitely
(b) divergent (d) oscillates infinitely
is
n =1
(a) convergent (c) oscillatory 16.
lim
nÆ a
(b) divergent (d) may or may not be convergent
un = 1 and  vn diverges then  un is vn
(a) divergent (c) oscillatory
(b) convergent (d) oscillates infinitely
Multiple Choice Questions
17.
2
3
5.119
4
+ L • converges if 1 2 3p (a) p < 2 (b) p > 2 (c) p > 1 (d) p ≥ 1 18. If Sun is a series of positive terms such that lim un = 0 then Sun is p
+
+
p
n Æ•
(a) convergent (c) may or may not be convergent 19.
1
 n(log n) p
(b) divergent (d) oscillatory
is convergent
(a) for p > 1 (c) for all real values of p
(b) for p < 1 (d) for no value of p
•
20.
 (2 x)n
is divergent if
n=0
(a) -1 £ x £ 1 (b) -
1 1
•
26. The series
 n =1
1 3
(c) -
1 1 2) to x = 4 Limits of y : y = 0 to y = 2 (iii) In subregion RSQ, strip EF starts from the parabola y2 = 4x and terminates on the line x = 4. y2 Limits of x : x = to x = 4 4 Limits of y : y = 2 to y = 4 Hence, given integral after change of order is 4
∫∫ 0
4x 4 x− x
2
f ( x, y )dy dx = ∫
2
0
∫
+∫
2− 4− y2 y2 4 2
0
∫
f ( x, y )dx dy
4
2+ 4− y
2
f ( x , y ) dx dy + ∫
4
2
∫
4 y2 4
f ( x, y )dx dy
Example 12 Change the order of integration of
2
( 4− x )2
0
4− x
∫∫
f ( x, y )dy dx.
Solution 1. The function is integrated first w.r.t. y and then w.r.t. x. 2. Limits of y : y = 4 − x to y = (4 − x) 2 . Limits of x : x = 0 to x = 2 3. The region is enclosed by the parabolas y2 = 4 – x, y = (4 – x)2, the lines x = 0 and x = 2. 4. (i) The points of intersection of x = 2 and y2 = 4 – x are obtained as
9.3
y2 = (4 – 2)
y
y = ± 2. The points of intersection are Q (2, 2 ) and Q ′ (2, − 2 ) . (ii) The point of intersection of x = 2 and y = (4 – x)2 is obtained as y = (4 – 2)2 = 4. The point of intersection is S (2, 4). (iii) The points of intersection of x = 0 and y2 = 4 – x are obtained as y2 = 4 y = ±2. The points of intersection are P (0, 2) and P¢ (0, –2). (iv) The point of intersection of x = 0 and y = (4 – x)2 is obtained as y = 16. The point of intersection is U (0, 16). 5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into three subregions PQR, PRST and STU. Draw a horizontal strip in each subregion. (i) In subregion PQR, strip AB starts from the parabola y2 = 4 – x and terminates on the line x = 2. Limits of x : x = 4 – y2 to x = 2 Limits of y : y= 2 to y = 2 (ii) In subregion PRST, strip CD starts from y-axis and terminates on the line x = 2. Limits of x : x = 0 to x = 2 Limits of y : y = 2 to y = 4
9.41
Change of Order of Integration
U (0, 16) y = (4 − x)2 B′ x=2 S (2, 4)
P (0, 2)
A′
Q (2, 2)
O
V (4, 0)
x
y 2= 4 − x Q ′(2, − 2) P′ (0, −2)
Fig. 9.45 y U (0, 16) y = (4 − x)2
E
F S (2, 4)
T C P (0, 2)
D A
R (2, 2) B
Q (2, 2)
x=2 O
V (4, 0) y2 Q ′(2, − 2) P′(0, −2)
Fig. 9.46
= (4 − x)
x
9.42
Chapter 9
Multiple Integrals
(iii) In subregion STU, strip EF starts from y-axis and terminates on the parabola y = (4 – x)2. Limits of x : x = 0 to x = 4 − y (Part of the parabola where x < 4) Limits of y : y = 4 to y = 16 Hence, the given integral after change of order is 2
( 4 − x )2
0
4− x
∫∫
f ( x, y )dy dx = ∫
2 2
∫
+∫
2 4 − y2
16 4
∫
f ( x, y )dx dy + ∫
4− y
0
4 2
∫
2
0
f ( x, y )dx dy
f ( x, y )dx dy
type II Evaluation of Double Integrals by Changing the Order of Integration
Example 1 Change the order of integration and evaluate
∫∫ a
a
0
x
( x 2 + y 2 )dy dx.
Solution
0
x
( x 2 + y 2 ) dy d x = ∫
a
=∫
a
0
∫
y
0
=
x
Q (a, a)
A′ O
x
Fig. 9.47 y y=a
P
Q (a,a)
( x 2 + y 2 )dx dy y
0
y=a
x3 + y 2 x dy 3 0
3 a y = ∫ + y 3 dy 0 3
x
a
B′
=
a
P
y
∫∫
y
y
1. Since inner limits depend on x, the function is integrated first w.r.t. y. 2. Limits of y : y = x to y = a, along vertical strip Limits of x : x = 0 to x = a 3. The region is bounded by the lines y = x, y = a and x = 0 4. The point of intersection of y = x and y = a is Q (a, a). 5. To change the order of integration, i.e. to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = a Hence, the given integral after change of order is
A
B
x
O
Fig. 9.48
9.3
=∫
9.43
4 3 y dy 3
a
0
=
Change of Order of Integration
4 y4 3 4
a
0
4
=
a 3
Example 2 Change the order of integration and evaluate
π
π
0
x
∫ ∫
sin y dy dx. y
Solution y
1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = x to y = p Limits of x : x = 0 to x = p 3. The region is bounded by the line y = x, y = p and x = 0. 4. The point of intersection of the line y = x and the line y = p is P (p, p). 5. To change the order of integration, i.e., O to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = p Hence, the given integral after change of order is
x
y
=
x
A¢
x
Fig. 9.49
y
y=p
Q
P (p, p )
A
B
π
= ∫ sin y dy
x
0
∫ ∫
π sin y y sin y dy dx = ∫ dx d y 0 y y ∫0 π sin y y =∫ x 0 dy 0 y π sin y =∫ ⋅ y dy 0 y
P (p, p )
=
π
B¢
0
y
π
y=p
π
= − cos y 0
= − cos π + cos 0 =2
O
x
Fig. 9.50
9.44
Chapter 9
Multiple Integrals
Example 3 Evaluate the iterated integral
1 1
Ú0 Úx sin y
2
dydx.
[Winter 2013]
Solution 1. Since the inner limits depend on x, the function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = x to y = 1, along vertical strip Limits of x : x = 0 to x = 1 3. The region is bounded by the lines y = x, y = 1 and x = 0. 4. The point of intersection of the line y = x and the line y = 1 is P (1, 1). 5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 1
Fig. 9.51
Hence, the given integral after change of order is 1 1
1 y
Ú0 Úx sin y dydx = Ú0 Ú0 sin y 2
1
2
dx dy y
= Ú sin y 2 x 0 dy
Fig. 9.52
0
1
= Ú sin y 2 ◊ y dy 0
1 1 sin y 2 ◊ 2 y dy 2 Ú0 1 1 = - cos y 2 dy 0 2 1 = [ - cos 1 + cos 0 ] 2 1 = [1 - cos 1] 2 =
ÈQ sin f ( y) ◊ f ¢( y) dy = - cos f ( y)˘ Î Ú ˚
9.3
9.45
Change of Order of Integration
Example 4 Change the order of integration and evaluate
∞
∞
0
x
∫ ∫
−y ∞ ye e− y dy d x = ∫ ∫ dx d y 0 0 y y
=∫
∞
0
{∫ dx} ey dy 0
∞
=∫ x0
y
0
∞
= ∫ y⋅ 0
−y
y
e− y dy y
e− y dy y
∞
= ∫ e − y dy 0
= −e − y
∞ 0
= −(e −∞ − e0 ) =1
0
x
∫ ∫
e− y dy dx. y [Winter 2015]
B′ y Æ •
= y
A′ x
O
Fig. 9.53
=
x
y
y
1. Since inner limits depend on x, the function is integrated first w.r.t. y but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = x to y Æ •, along vertical strip Limits of x : x = 0 to x Æ • 3. The region is bounded by the lines y = x and x = 0. 4. Here, the only point of intersection is origin. 5. To change the order of integration, i.e. to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = • Hence, the given integral after change of order is
∞
x
y
Solution
∞
A
B
x
O
Fig. 9.54
9.46
Chapter 9
Multiple Integrals
Example 5 Evaluate the integral
2 1 x2
Ú0 Ú y e
dx dy by changing the order of integration.
2
[Summer 2017, 2015]
Solution 1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration. y 2. Limits of x : x = to x = 1 2 along horizontal strip A¢B¢ Limits of y : y = 0 to y = 2 3. The region is bounded by the lines y = 2x, x = 1, y = 2, and y = 0. 4. The point of intersection of y = 2x and x = 1 is x = 1, y = 2. The point of intersection is Q (1, 2). 5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the line y = 2x. Limits of y : y = 0 to y = 2x Limits of x : x = 0 to x = 1 Hence, the given integral after change of order is 2 1 x2
Ú0 Ú y e 2
dx dy = Ú
1 2 x x2
Ú
e dy dx
0 0
=Ú
1
0
{Ú } 2x
0
1
Fig. 9.55
Fig. 9.56
2
dy e x dx 2
= Ú 2 x ◊ e x dx 0
= ex
2 1
0
= e1 - e0 = e -1
ÈQ e f ( x ) f ¢( x ) dx = e f ( x ) ˘ Î Ú ˚
9.3
Change of Order of Integration
9.47
Example 6 ∞
∫ ∫
Change the order of integration and evaluate
0
−
x
xe
0
x2 y
dy dx.
Solution
∞
∫ ∫ 0
x
0
−
xe
x2 y
dy dx = ∫
∞
0
∫
∞ y
−
xe
x2 y
x =
x
A′
O
Fig. 9.57
dx dy x2
1 ∞ − x2 y y 2 ∫0 e Q e ∫
B′
y
∞ y ∞ − = ∫ − ∫ e y 0 2 y
=−
y
y
1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = 0 to y = x. Limits of x : x = 0 to x Æ •. 3. The region is the part of the first quadrant bounded between the lines y = x and y = 0. 4. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip parallel to x-axis which starts from the line y = x and extends up to infinity. Limits of x : x = y to x Æ • Limits of y : y = 0 to y Æ • Hence, the given integral after change of order is
2x − y dx dy
∞
A
dy y
f ( x)
y
f ′ ( x ) dx = e
f ( x)
=
B
x
x
O
1 ∞ = − ∫ y (0 − e − y )dy 2 0 1 ∞ = − ye − y − e − y 0 2 1 = 2
Fig. 9.58
Example 7 Change the order of integration and evaluate
∫∫ a
a
0
y
x2 x2 + y 2
dx dy.
9.48
Chapter 9
Multiple Integrals
Solution
a
a
x2
0
y
x +y
∫∫
2
2
dx dy = ∫
a
0
B
x
x2
0
x +y
∫
2
dx
2
a 1 x d y x 2 dx = ∫ ∫ 2 2 0 0 x +y
( = ∫ log ( x +
= ∫ log y + y 2 + x 2 a
0
)
)
0
(
)
x 2 dx 0
x 1+ 2 2 a x dx = ∫ log 0 x = log 1 + 2
(
)∫
(
)
(
) a3
= log 1 + 2 = log 1 + 2
a
0
x 2 dx
x3 3 3
a
0
A
Fig. 9.60 x
2 x 2 − log x x 2 dx
a
O
x=a
x=a
y
=
x
1. Since inner limits depend on y, the function is integrated first w.r.t. x but evaluation becomes easier by changing the order of intey gration. 2. Limits of x : x = y to x = a, along horizontal Q (a, a) strip A¢B¢ Limits of y : y = 0 to y = a 3. The region is bounded by the lines y = x, x = a and y = 0. A′ B′ 4. The point of intersection of y = x and x = a is x = a, y = a. P x O The point of intersection is Q (a, a). 5. To change the order of integration, i.e. to integrate first w.r.t. y, draw a vertical strip AB Fig. 9.59 parallel to y-axis which starts from x-axis and y terminates on the line y = x. Q (a, a) Limits of y : y = 0 to y = x Limits of x : x = 0 to x = a x = Hence, the given integral after change of order is y
P
x
9.3
Change of Order of Integration
9.49
Example 8 [Winter 2016]
Change the order of integration and evaluate 1
∫∫
1− x
2
e
y
(e y + 1) 1 − x 2 − y 2
0 0
y
dy dx.
Q (0, 1) B′
Solution
P (1, 0)
1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = 0
to
to
x = 1 − y2
Limits of y : y = 0
to
y=1
Fig. 9.61 y Q (0, 1) x2 + y2 = 1
Hence, the given integral after change of order is
∫∫ 0
0
ey (e + 1) 1 − x − y y
2
x
y = 1 − x2
Limits of x : x = 0
1− x 2
A′
O
Limits of x : x = 0 to x = 1 3. Since given limits of x and y are positive, the region is the part of circle x2 + y2 = 1 in the first quadrant. 4. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y axis and terminates on the circle x2 + y2 = 1.
1
x2 + y2 = 1
2
1− y 2 ey y ∫ 0 e +1 0
dy dx = ∫
A
B
O
P(1, 0)
Fig. 9.62
1
1
(1 − y 2 ) − x 2
x e y sin −1 =∫ y 0 e +1 1 − y2
dx dy
1− y 2
1
dy 0
ey (sin −1 1 − sin −1 0)dy 0 ey +1 y 1 e π ⋅ dy =∫ y 0 e +1 2 f ′( y) π 1 Q∫ dy = log f ( y ) = log(e y + 1) 0 f y ( ) 2 =∫
1
π [log(e + 1) − log 2] 2 π e + 1 = log 2 2
=
x
9.50
Chapter 9
Multiple Integrals
Example 9 a
a+ a2 − y2
0
a− a2 − y2
∫∫
Change the order of integration and evaluate
dx dy.
Solution 1. Since inner limits depend on y, the function is integrated first w.r.t. x.
y P (a, a)
2. Limits of x : x = a − a 2 − y 2 to
A′
B′
x = a + a 2 − y 2 , along horizontal strip A¢B¢
Limits of y : y = 0 to y = a 3. The region is bounded by the circle (x – a)2 + y2 = a2 and the line y = 0. Since limits of y are positive, the region is the part of the circle (x – a)2 + y2 = a2 above x-axis. 4. The points of intersection of the circle with x-axis are O (0, 0) and Q (2a, 0). 5. To change the order of integration i.e. to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the circle
Q (2a, 0) O
x
(a, 0)
Fig. 9.63 y B
(x – a)2 + y2 = a2 or x2 + y2 – 2ax = 0
Q (2a, 0)
y = 2ax − x 2
Limits of y : y = 0 to
O
A
x
(a, 0)
Limits of x : x = 0 to x = 2a Hence, the given integral after change of order is
∫∫ a
0
a + a2 − y2 2
a− a − y
2
dx dy = ∫
2a
0
=∫
2a
=∫
2a
=∫
2a
0
0
0
2 ax − x 2
∫
0
Fig. 9.64
dy dx
2 ax − x 2
y
dx
0
2ax − x 2 dx a 2 − ( x − a)2 2a
=
( x − a) 2 a2 x − a a − ( x − a ) 2 + sin −1 a 2 2
=
a a (0 − a ) a 0 + sin −1 1 − 0 − sin −1 ( −1) 2 2 2 2 2
2
0
9.3
Change of Order of Integration
9.51
= a 2 sin -1 1 [Q sin -1 ( -1) = - sin -1 (1)] p 2 2 pa = 2 = a2
Example 10 1 2- x2
Evaluate
Ú Ú 0
x
x x 2 + y2
dy dx by changing the order of integration. [Summer 2016]
Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y. 2. Limits of y : y = x to y = 2 - x 2 along vertical setup A¢B¢ Limits of x : x = 0 to x = 1 3. The region is bounded by y-axis, the line y = x and the circle x2 + y2 = 2. 4. The point of intersection of the circle y = 2 - x 2 and y = x is obtained as x2 = 2 - x2 2 x2 = 2
\
x2 = 1 x = ±1 y=1
Fig. 9.65
Hence, P(1, 1) is the point of intersection. 5. To change the order of integration, i.e. to integrate first w.r.t. x, divide the region into two subregions OPR and PQR. Draw a horizontal strip parallel to x-axis in each subregion. (i) In the subregion OPR, strip AB starts from y-axis and terminates on the line y = x. Limit of x: x = 0
to
x=y
Limit of y: y = 0
to
y=1
Fig. 9.66
9.52
Chapter 9
Multiple Integrals
(ii) In the subregion PQR, strip CD starts from y-axis and terminates on the circle y = 2 - x2 . Limit of x: x = 0 to x = 2 - y 2 Limit of y: y = 1 to y = 2 1 2 - x2
Ú Ú 0
x
x 2
x +y
2
dy dx =
x
ÚÚ
2
x +y
OPR 1 y
x 2 + y2
00
2
x + y2
PQR
dy dx +
1
= Ú ÈÎ 2 y - y ˘˚ dy + 0
y2 2
1
2
Ú ÈÎ 1
x 2 + y2
0
y
˘ ˙ ˙ dy + ˙ ˚˙ 0
2
Ú 1
2 - y ˘˚ dy
È y2 ˘ + Í 2y - ˙ 2 ˙˚ Í 1 0 Î
2 1 3 - + - 2 2 2 2
= 1- 2 + = 1-
1 2
1 2
dy dx
1 È 1 Í ( x 2 + y2 ) 2 Í 1 2Í 2 ÎÍ
2
1ˆ Ê1 ˆ Ê = ( 2 - 1) Á - 0˜ + Á 2 - 1 - 2 + ˜ Ë2 ¯ Ë 2¯ =
dydx
x
Ú Ú 1
1 È 1 1 Í ( x 2 + y2 ) 2 =Ú Í 1 2Í 0 2 ÎÍ
= ( 2 - 1)
x
ÚÚ
2 2 2- y
x
= ÚÚ
dy dx +
2
˘ ˙ ˙ ˙ ˚˙ 0
2 - y2
dy
9.3
Change of Order of Integration
9.53
Example 11 Sketch the region of integration, reverse the order of integration and 2y 2 4 - x 2 xe dy dx. evaluate Ú0 Ú0 [Summer 2014] 4-y Solution 1. Since inner limit depends on x, the function is integrated first w.r.t. y. 2. Limits of y : y = 0 to y = 4 – x2 along vertical strip A¢B¢ Limits of x : x = 0 to x = 2 3. The region is bounded by the parabola x2 = 4 – y, x-axis and y-axis.
Fig. 9.67
4. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip parallel to x-axis which starts from y-axis and terminates on the parabola x2 = 4 – y. Limits of x : x = 0
to
x = 4− y
Limits of y : y = 0
to
y=4 Fig. 9.68
Hence, the given integral after change of order is 2
4 - x2
Ú0 Ú0
2y 4 4 - y xe xe2 y dydx = Ú Ú dx dy 0 0 4-y 4-y
=Ú
4
=Ú
4
0
0
=Ú
4
0
e2 y È 4 - y ˘ x dx ˙ dy 4 - y ÍÎ Ú0 ˚ e2 y x 2 4-y 2
4- y
dy 0
e2 y È 1 ˘ (4 - y)˙ dy 4 - y ÍÎ 2 ˚
9.54
Chapter 9
Multiple Integrals
=Ú
4
0
e2 y dy 2
1 2y 4 e 0 4 1 = (e8 - 1) 4 =
Example 12 Change the order of integration and evaluate
1
2− x
0
x2
∫∫
xy dx dy.
Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y.
P ′(−2, 4) y
The correct form of integral =∫
1
0
∫
2− x x2
xy dy dx
2. Limits of y : y = x2 to y = 2 – x, along vertical strip A¢B¢ Limits of x : x = 0 to x = 1 3. The region is bounded by y-axis, the line x + y = 2 and the parabola x2 = y. Since given limits of x and y are positive, the region lies in the first quadrant. 4. The points of intersection of x + y = 2 and x2 = y are obtained as x2 = 2 – x x +x–2=0 (x – 1) (x + 2) = 0 x = 1, –2 y = 1, 4
Q (0, 2)
x2 = y
B′ A′
P (1, 1) x +y=2
O
x
Fig. 9.69 P ′(−2, 4) y
2
Q (0, 2) C R A
D B
x2 = y
P (1, 1) x +y=2
O x The points of intersection are P (1, 1) and P' (–2, 4). Fig. 9.70 5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into two subregions OPR and RPQ. Draw a horizontal strip parallel to x-axis in each subregion. (i) In subregion OPR, strip AB starts from y-axis and terminates on the parabola x2 = y. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 1
9.3
Change of Order of Integration
9.55
(ii) In subregion RPQ, strip CD starts from y-axis and terminates on the line x + y = 2. Limits of x : x = 0 to x = 2 – y Limits of y : y = 1 to y = 2 Hence, the given integral after change of order is 1
2− x
0
x2
∫∫
xy dy dx = ∫
1
=∫
1
0
0
∫
y
0
x2 2
xy dx dy + ∫
2
1
y
y dy + ∫
2
1
0
∫
2− y
0
x2 2
xy dx dy
2− y
y dy 0
1 1 1 2 ( y ) y dy + ∫ ( 2 − y ) 2 y dy ∫ 2 0 2 1 1 1 2 1 2 = ∫ y d y + ∫ ( 4 y − 4 y 2 + y 3 ) dy 2 0 2 1 =
1
= = = = =
2
y3 y 4 1 y3 1 y2 + 4 −4 + 2 3 0 2 2 3 4 1 1 1 1 32 4 1 + 8 − + 4 − 2 + − 2 3 2 3 3 4 1 5 + 6 24 9 24 3 8
Example 13 Change the order of integration and evaluate
4a
2 ax
0
x2 4a
∫ ∫
Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y. x2 2. Limits of y : y = to y = 2 ax , 4a along vertical strip A¢B¢ 3. The region is bounded by the parabolas x2 = 4ay and y2 = 4ax. 4. The points of intersection of x2 = 4ay and y2 = 4ax are obtained as x4 = 16a2y2 =16a2 (4ax)
dy dx. [Winter 2014]
y y 2 = 4ax
P(4a, 4a)
B′ x 2 = 4ay A′ O
Fig. 9.71
x
9.56
Chapter 9
Multiple Integrals
x(x3 – 64a3) = 0 x = 0, x = 4a \ y = 0, y = 4a The points of intersection are O (0, 0) and P (4a, 4a). 5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from the parabola y2 = 4ax and terminates on the parabola x2 = 4ay. y
y2 Limits of x : x = to x = 2 ay 4a Limits of y : y = 0 to y = 4a Hence, the given integral after change of order is 4a
2 ax
Ú0 Ú x
2
4a
xy dy dx = Ú
0
4a
2 ay
Úy
2
0
=Ú
x
P(4a, 4a) x 2 = 4ay A
dx dy
4a 4 a 2 ay
=Ú
y 2 = 4ax
y2 4a
O
y2 ˆ 2 ay dy Á 4 a ˜¯ Ë
=2 a
3 y2
3 2
x
dy
4a Ê
0
B
Fig. 9.72
4a
1 y3 4a 3
4a
0
0
3
3
4 1 a ( 4) 2 a 2 (64 a 3 ) 3 12 a 32 2 16 2 = a - a 3 3 16 2 = a 3 =
Example 14 Change the order of integration and evaluate a
a− a2 − y2
0
0
∫∫
xy log( x + a ) dx dy. ( x − a)2
Solution 1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration.
9.3
y
2. Limits of x : x = 0 to x = a − a 2 − y 2
y=a
Limits of y : y = 0 to y = a 3. The region is bounded by the circle (x – a)2 + y2 = a2, the lines y = a and x = 0.
Q A′
4. The point of intersection of (x – a)2 + y 2 = a2 and y = a is obtained as (x – a)2 + a2 = a2 x=a The point of intersection is P (a, a). 5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from the circle (x – a)2 + y2 = a2 and terminates on the line y = a. Limits of y : y = 2ax − x 2
to
Limits of x : x = 0
to x = a
P (a, a)
B′
(x − a)2 + y 2 = a 2 x
O
Fig. 9.73 y y=a Q
B
P(a, a) (x − a)2 + y 2 = a 2
A
y=a x
O
Hence, the given integral after change of order is a
a − a2 − y2
0
0
∫∫
9.57
Change of Order of Integration
Fig. 9.74
a a xy log( x + a ) x log( x + a ) y dy d x dx dy = ∫ ∫ 2 2 ax − x 0 2 ( x − a) ( x − a)2
=∫
a
=∫
a
0
0
x log( x + a ) y 2 2 ( x − a)2
a
dx 2 ax − x 2
x log( x + a ) a 2 − 2ax + x 2 dx 2 ( x − a ) 2
1 a x log( x + a )dx 2 ∫0 a 2 a x 1 x2 1 dx = log( x + a ) − ∫ ⋅ 0 2 x+a 2 2 0 =
=
1 a 1 a2 a2 dx log 2a − ∫0 ( x − a ) + 2 2 2 x + a
a 1 a2 1 x2 2 = log 2a − − ax + a log( x + a ) 2 2 2 2 0 2 1 a = a 2 log 2a − + a 2 − a 2 log 2a + a 2 log a 4 2
9.58
Chapter 9
Multiple Integrals
=
1 a2 + a 2 log a 4 2
=
a2 (1 + 2 log a ) 8
Example 15 Change the order of integration and evaluate 1 2
∫ ∫ 0
1 + x2
1− 4 y 2
0
1 − x2 1 − x2 − y 2
dx dy. y
Solution Q 0,
1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration.
O
1 2 3. Since the limits of x and y are positive, the region is the part of the ellipse in the first quadrant. 4. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the ellipse x2 + 4y2 = 1. Limits of y : y = 0
to
x 2+ 4y 2 = 1 B′ x P (1, 0)
A′
x = 1 − 4 y2
2. Limits of x : x = 0 to
1 2
y=
Fig. 9.75 y Q 0,
1 2
B x 2+ 4y 2 = 1
O
Limits of y : y = 0
to
Limits of x : x = 0
to
Ú Ú
1 + x2 1- x
2
P (1, 0)
x=1
Hence, the given integral after change of order is 1 1- 4 y2 2 0 0
A
1 y= 1 − x2 2
2
1- x - y
2
Fig. 9.76
dxdy = Ú
1
0
=Ú
1
0
1 + x2 1- x 1+ x
2
2
1- x
2
1 1- x 2 2 0
Ú
sin -1
1 (1 - x 2 ) - y 2 y 1 - x2
dydx
1 1- x 2 2
dx 0
x
9.3
=Ú
1
0
=Ú
Change of Order of Integration
9.59
1 + x 2 Ê -1 1 -1 ˆ Á sin 2 - sin 0˜¯ dx 2 Ë 1- x x2 ) p ◊ dx 6 1 - x2
1 2 - (1 -
0 1Ê
ˆ - 1 - x 2 ˜ dx ¯ 1 - x2 2
=
p 6
=
1 p x 2 sin -1 x 1 - x 2 - sin -1 x 6 2 2 0
Ú0 ÁË
1
È x 2 a2 2 2 a - x2 + sin -1 ÍQ Ú a - x dx = 2 2 ÍÎ
=
x˘ ˙ a ˙˚
p Ê 3 -1 ˆ sin 1˜ ¯ 6 ÁË 2
p p ◊ 4 2 p2 = 8 =
Example 16 Change the order of integration and evaluate
∫∫ a
0
x dy dx
y
0
(a 2 − x 2 )(a − y )( y − x)
.
Solution y
y=a
=
x
P (a, a)
y
1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration. 2. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = a 3. The region is bounded by the line y = x, y = a and x = 0. 4. The point of intersection of y = a and y = x is P (a, a). 5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from the line y = x and terminates on the line y = a.
A′
B′
x
O
Fig. 9.77
9.60
Chapter 9
Multiple Integrals
Limits of y : y = x to y = a Limits of x : x = 0 to x = a Hence, the given integral after change of order is a y x dy dx
∫∫
B y=a P (a, a) x
(a 2 − x 2 )(a − y )( y − x)
0
=
0
y
a
=∫
a
0
∫
x d y dx
a
y
=∫
(a − x )(a − y )( y − x) 2
x
2
A
dy a ∫x (a − y )( y − x) dx 2 2 a −x x
0
x
O
Putting y – x = t2, dy = 2t dt When y = x, t = 0 When a
∫∫ 0
y = a,
t = a−x
x dy dx
y
(a − x )(a − y )( y − x) 2
0
2
Fig. 9.78
a
x
0
a −x
=∫
2
2
= 2∫
a
x
0
a −x
= 2∫
a
x
2
a2 − x2
0
= 2∫ = 2⋅
x
a
0
a − x2 2
dx
(a − x − t 2 )t 2
0
2
2t dt
a−x
∫
∫
dt
a−x
(a − x) − t 2
0
sin
a−x
t
−1
a−x
dx
dx 0
(sin −1 1 − sin −1 0)dx
1 π a 1 2 − 2 2 a x − ( − ) ( −2 x) dx ∫ 2 0 2
1
a
p = - 2( a 2 - x 2 ) 2 2
0
È [ f ( x )]n +1 ˘ n ÍQ Ú [ f ( x )] f ¢( x )dx = ˙ n + 1 ˙˚ ÍÎ
= pa
Example 17 Change the order of integration and evaluate 1
1 x
0
x
∫∫
y dy dx. (1 + xy ) (1 + y 2 ) 2
Solution 1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration.
9.3
y
1 x
xy � 1
3. The region is bounded by the rectangular hyperbola xy = 1, the line y = x and y-axis in the first quadrant. 4. The point of intersection of xy = 1 and y = x in the first quadrant is obtained as x2 = 1 x=1 \ y=1 The point of intersection is P (1, 1). 5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into two subregions OPQ and QPR. Draw a horizontal strip parallel to x-axis in each subregion. (i) In subregion OPQ, strip AB starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 1
B′ P (1, 1) A′ O
x
Fig. 9.79 y
C
R
D
x
x=1
=
to
y
Limits of x : x = 0
x
y=
�
to
y
2. Limits of y : y = x
9.61
Change of Order of Integration
xy = 1 P (1, 1)
Q
(ii) In subregion QPR, strip CD starts from y-axis and terminates on the rectangular hyperbola xy = 1. 1 Limits of x : x = 0 to x = y Limits of y : y = 1 to y → ∞.
A
B
O
x
Fig. 9.80
Hence, the given integral after change of order is 1
1 x
0
x
∫∫
1 ∞ 1 y y y y 1 1 y x x y d d = d d + ∫0 ∫0 1 + y 2 (1 + xy)2 ∫1 ∫0y 1 + y 2 (1 + xy)2 dx dy (1 + xy ) 2 (1 + y 2 ) y
1
y ∞ y y 1 1 dy + ∫ dy =∫ − − 2 2 0 1+ y 1 1+ y y (1 + xy ) 0 y (1 + xy ) 0 1
= −∫
1
0
∞ 1 1 1 1 − 1 dy − ∫ − 1 dy 1 1+ y2 2 1 + y 2 1 + y 2
1 1 1 1 ∞ 1 = −∫ − dy + ∫ dy 2 2 2 0 (1 + y ) 1+ y 2 1 1+ y2
9.62
Chapter 9
Multiple Integrals
Putting y = tan q in the first term of first integral, dy = sec2q dq, When y = 0, q = 0 When y = 1, θ = 1
1 x
0
x
∫∫
π 4
π 1 ∞ y sec 2 θ dθ 1 4 y x d d = − + tan −1 y 0 + tan −1 y 1 4 2 2 ∫ 0 2 (1 + xy ) (1 + y ) sec θ π
= −∫ 4 0
(1 + cos 2θ ) 1 dθ + (tan −1 1 − tan −1 0) + (tan −1 ∞ − tan −1 1) 2 2
sin 2θ 1 =− θ+ 2 2
π 4
+
0
π 1π π + − 4 2 2 4
π 1 π 3π − sin + 8 4 2 8 π −1 = 4 =−
Example 18 Change the order of integration and evaluate 1
∫∫ 0
1− y 2
0
cos −1 x 1 − x2 1 − x2 − y 2
dx dy.
Solution y
1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration. 2. Limits of x : x = 0 to x = 1 − y 2 Limits of y : y = 0 to y = 1 3. Since given limits of x and y are positive, the region is the part of the circle x2 + y 2 = 1 in the first quadrant. 4. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis in the region. AB starts from x-axis and terminates on the circle x2 + y 2 = 1. Limits of y : y = 0 to y = 1 − x 2 Limits of x : x = 0 to x = 1.
Q (0, 1) A′ O
Fig. 9.81
B′
x2 + y2 = 1
P (1, 0) x
9.3
Change of Order of Integration
Hence, the given integral after change of order is 1− y
1
∫∫ 0
9.63
y
−1
cos x
2
1− x
0
2
1 − x2 − y 2
1− x 2
=∫
1
=∫
1
cos −1 x
0
1 − x2
∫
0
0
=∫
1
cos −1 x
dx d y
Q (0, 1) B
cos −1 x
1
1 − x2
(1 − x 2 ) − y 2
O
1− x 2
sin
−1
y 1 − x2
A
P (1, 0) x
dx 0
(sin −1 1 − sin −1 0)dx
1− x 1 π 1 −1 = − ∫ cos x − dx 0 2 1 − x2 2
0
x2 + y2 = 1
dy dx
π (cos −1 x) 2 =− 2 2
1
0
Fig. 9.82
[ f ( x)]2 Q ∫ f ( x) f ′ ( x)d x = 2
π (cos −1 1) 2 − (cos −1 0) 2 4 2 π π = − 0 − 4 2
=−
=
π3 16
ExERCIsE 9.3 Change the order of integration of the following integrals: 1.
6
2+ x
0
2− x
∫∫
f (x, y )dy dx Ans. : 2 6 f (x, y )dy dx + 8 6 f (x, y )dy dx ∫− 4 ∫2−y ∫2 ∫y −2
2.
1
∫∫
2x
0 x
f (x, y )dy dx 1 y 2 1 Ans. : ∫0 ∫y f (x, y )dx dy + ∫1 ∫ y f (x, y )dx dy 2 2
3.
1
∫∫
y
0 − y
f (x, y )dx dy Ans. : 1 1 f (x, y )dx dy ∫−1 ∫x 2
9.64
4.
Chapter 9
y2 a
a
∫ ∫
−a 0
Multiple Integrals
f (x, y )dx dy a − ax a a Ans. : ∫0 ∫− a f (x, y )dy dx + ∫0 ∫ ax f (x, y )dy dx
5.
a
2a −x
0
x2 a
∫∫
f (x, y )dy dx a ay 2a 2a−y Ans. : ∫0 ∫0 f (x, y )dx dy + ∫a ∫0 f (x, y )dx dy
6.
3
∫ ∫
y
−2 y 2 − 6
f (x, y )dx dy −2 x +6 3 x +6 Ans. : ∫−6 ∫− x +6 f (x, y )dy dx + ∫−2 ∫x f (x, y )dy dx
7.
1
∫∫
2(1+ 1− y )
0 2y
f ( x , y ) dx dy 4 x −x x 2 4 Ans. : ∫0 ∫02 f (x, y )dy dx + ∫2 ∫0 4 f (x, y )dy dx 2
8.
2
∫∫ 0
6−x 2 x2 +4 4
f ( x , y ) dy dx 2 2 Ans. : ∫1 ∫0
9.
2
x +6 a
0
4−x2
∫∫
y −1
f (x, y )dx dy + ∫
3
2
∫
6 −2 y
0
f (x, y )dx dy
f (x, y )dy dx
6a 2 6a+2 2 Ans. : 2 2 ∫0 ∫ 4 −y 2 f(x, y)dx dy + ∫2 ∫0 f(x, y)dx dy + ∫6 a ∫y −6 a f(x, y)dx dy a
10.
∫ ∫
y +a a2 − y 2
0
f (x , y )dxdy 2a a Ans. : a a ∫0 ∫ a2 − x 2 f(x, y)dy dx + ∫a ∫x −a f(x, y)dy dx
11.
2a
2 ax
0
2 ax − x 2
∫ ∫
f (x, y )dx dy
Ans. : a 2a a a− f (x, y )dx dy + ∫ ∫ y 2 2 2 ∫ ∫ 0 0 a + a − y 2a
a2 − y 2
f (x, y )dx dy + ∫
2a
a
∫
2a y2 2a
f (x, y )dx dy
9.3
12.
a
a2 − x 2
0
a2 − x 2 4
∫∫
a2 x
a
∫∫ 0
9.65
f (x, y )dy dx a 2 Ans. : ∫0 ∫
13.
Change of Order of Integration
a2 − y 2 a2 − 4 y
a
f (x, y )dx dy + ∫a ∫ 2 2
a2 − y 2
0
f (x, y )dx dy
f (x, y )dy dx
x
a a y ∞ Ans. : ∫0 ∫0 f (x, y )dx dy + ∫a ∫0 y f (x, y )dx dy 2
14.
b
mx
a
k x
∫ ∫
f (x, y )dy dx k b mb b ma b a Ans. : ∫bk ∫yk f(x, y)dx dy + ∫ak ∫a f(x, y)dx dy + ∫ma ∫my f(x, y) dx dy
15.
1
∫∫
ex
f (x, y )dy dx
0 1
Ans. : e 1 f (x, y )dx dy ∫1 ∫log y 16.
17.
2
x3
0
0
∫∫ 1 2
8 2 Ans. : ∫0 ∫ 1 f (x, y )dx dy y3
1+ x 2
1− 4 y 2
∫ ∫ 0
f (x, y )dy dx
1− x2 − y 2
0
dx dy 1− x 2 1 1+ x 2 2π Ans. : ∫ ∫ 2 dy dx = 0 0 3 1− x2 − y 2
x2 2
2
18.
∫∫ 0
x x + y2 +1 2
0
dy dx 2 2 Ans. : ∫ ∫ 0 2y
19.
a
a
0
y
∫∫
x dy dx x + y2
x x2 + y2 + 1
dx dy =
1 (5 log 5 − 4) 4
2
πa Ans. : 4
9.66
Chapter 9
9.4
DOublE IntEGRAls In POlAR COORDInAtEs θ2
The integral
∫∫
θ1
r2 r1
Multiple Integrals
f ( r , θ )drdθ represents the polar form of the double integration. This
integral is first integrated w.r.t. r keeping q constant and then the resulting expression is integrated w.r.t. q. R
limits of Integration If the limits of integration are not given then these limits are obtained from the equations of the given curves. Let the region be bounded by the curves r = r1(q ), r = r2 (q ) and the lines q = q1, q = q2.
B q = q2
r = r2(q ) Q
S A
q = q1
dq
The region of integration is PQRS. Draw an elementary radius vector AB from origin which enters in the region from the curve r = r1(q ) and leaves at the curve r = r2(q ). Therefore, limits for r are r1(q ) to r2(q ).
q2
P q1
r = r1(q )
O
q =0
Fig. 9.83
To cover the entire region PQRS, rotate elementary radius vector AB from PQ to RS. Therefore, q varies from q1 to q 2. θ2
r2 (θ )
θ1
r1 (θ )
∫∫ f (r , θ ) dr dθ = ∫ ∫
f (r , θ ) dr dθ
type I Evaluation of Double Integrals in Polar Coordinates
Example 1 Evaluate
π 1 4 0 0
∫ ∫ r dr dθ.
Solution π 4 0
π
1 r d r dθ = ∫ 4 ∫ r d r d θ 0 0 0
∫ ∫
1
π 4 0
=∫
1
r2 dθ 2 0
π 1 = ∫ 4 dθ 0 2 1 π = θ 04 2
9.4
Double Integrals in Polar Coordinates
9.67
1 π ⋅ 2 4 π = 8 =
Another method: Since both the limits are constant and integrand (function) is explicit in r and q, the integral can be written as π 4 0
∫ ∫
1
0
π
1
r dr dθ = ∫ 4 dθ ⋅ ∫ r dr 0
0
2 1
π 4 0
=θ ⋅
r 2
0
π 1 ⋅ 4 2 π = 8 =
Example 2 Evaluate
π
sin θ
0
0
∫ ∫
r dr dθ.
Solution π
sin θ
0
0
∫ ∫
π sin θ r dr dθ = ∫ ∫ r dr dθ 0 0
=∫
π
0
r2 2
sin θ
dθ 0
1 π 2 sin θ dθ 2 ∫0 1 π 1 − cos 2θ = ∫ dθ 2 0 2
=
=
1 sin 2θ θ− 4 2
π
0
1 sin 2π π − 4 2 π = 4 =
9.68
Chapter 9
Multiple Integrals
Example 3 Evaluate the integral
p 2 1 - sin q r 2 0 0
Ú Ú
cos q dr dq .
[Summer 2014]
Solution p 1 - sin q 2 2 r 0 0
Ú Ú
cos q dr dq =
=
p 2 0
È
r dr ˘˙ cos q dq ˚
1 - sin q 2
Ú ÍÎÚ0 p 2 0
1 - sin q
r3 3
Ú
cos q dq 0
=
p 2 0
1 3
Ú
(1 - sin q )3 cos q dq
1 (1 - sin q )4 = 3 4 =
p 2 0
1È 1˘ 0- ˙ Í 3Î 4˚
= -
1 12
Example 4 π 2 a cos θ 2 0 0
∫ ∫
Evaluate
r 2 sin θ dθ dr.
Solution Since inner limits depend on q, the function is integrated first w.r.t. r. π
2 a cos θ
The correct form of the integral = ∫02 ∫0 π 2 a cos θ 2 0 0
∫ ∫
π
2 a cos θ
r 2 sin θ dr dθ = ∫ 2 ∫ 0 0 π
=∫2 0
r 2 sin θ dr dθ
r3 3
r 2 dr sin θ dθ
2 a cos θ
sin θ dθ 0
π
=
8a 3 2 cos3 θ sin θ dθ 3 ∫0 π
=−
8a 3 2 cos3 θ ( − sin θ ) dθ 3 ∫0
9.4
8a 3 cos 4 q =3 4
Double Integrals in Polar Coordinates
p 2
9.69
È ˘ [ f (q )]n +1 n , n π - 1˙ ÍQ Ú [ f (q )] f ¢ (q )dq = n +1 ÍÎ ˙˚
0
3
==
8a Ê 4 p ˆ - cos 4 0˜ Á cos ¯ 12 Ë 2
2 3 a 3
Example 5 Evaluate
π a (1+ sin θ ) 2 0 0
∫ ∫
r 2 cos θ dθ dr.
Solution Since inner limits depend on q, the function is integrated first w.r.t. r. π
a (1+ sin θ )
The correct form of the integral = ∫02 ∫0 π 2 0
∫ ∫
a (1+ sin θ )
0
π
a (1+ sin θ )
r 2 cos θ dr dθ = ∫ 2 ∫ 0 0 π 2 0
=∫
r3 3
r 2 cos θ dr dθ
r 2 dr cos θ dθ
a (1+ sin θ )
cos θ dθ 0
p
=
1 2 3 a (1 + sin q )3 cos q dq 3 Ú0
a 3 (1 + sin q ) 4 = 3 4 =
a3 4 2 − 1 12 5 = a3 4
Example 6 Evaluate
∫ ∫
0
cos 2θ
0
È [ f (q )]n +1 n [ f ( q )] f ( q ) d q , = ¢ ÍQ n +1 Í Ú ÍÎ n π -1
4 π a 3 4 1 + sin − (1 + sin 0) 12 2
=
π 4 0
p 2
r dr dθ. (1 + r 2 ) 2
˘ ˙ ˙ ˙˚
9.70
Chapter 9
Multiple Integrals
Solution p
Ú04 Ú0
cos 2q
1 2r ◊ dr dq 2 (1 + r 2 )2 p
1 4 È cos 2q ˘ (1 + r 2 )-2 ◊ 2r dr ˙ dq Ú Ú Í 0 0 Î ˚ 2 È p [ f (r )]n+1 ˘˙ cos 2q 1 4 ÍQ Ú [ f (r )]n f ¢(r ) dr = 2 -1 dq Í = Ú -(1 + r ) n +1 ˙ 0 2 0 Í ˙ π 1 n Î ˚ =
p
=-
1 4Ê 1 ˆ - 1˜ dq Á Ú 0 Ë 2 1 + cos 2q ¯
=-
1 4Ê 1 ˆ - 1˜ dq Á Ú 2 0 2 Ë 2 cos q ¯
=-
1 4Ê1 ˆ sec 2 q - 1˜ dq Á Ú 0 ¯ 2 Ë2
p
p
p 4 1 1 =tan q - q 2 2 0
1Ê1 p pˆ = - Á tan - ˜ 2Ë2 4 4¯ =
1 (p - 2) 8
Example 7 Evaluate
p p 2 2 0 0
Ú Ú
sin(q + f )dq df.
Solution p p 2 2 0 0
Ú Ú
p È p ˘ sin(q + f ) dq df = Ú 2 Í Ú 2 sin(q + f ) dq ˙ df 0 ÍÎ 0 ˙˚ p 2
p 2 0
=Ú
- cos(q + f ) df 0
p 2 0
˘ È Êp ˆ = - Ú Ícos Á + f ˜ - cos f ˙ df Ë ¯ 2 ˚ Î p
= - Ú 2 ( - sin f - cos f ) df 0
9.4
Double Integrals in Polar Coordinates
= - cos f - sin f
9.71
p 2 0
p p Ê ˆ = - Á cos - sin - cos 0 + sin 0˜ Ë ¯ 2 2 =2
type II Evaluation of Double Integrals Over a Given Region in Polar Coordinates
Example 1 Evaluate
ÚÚr
3
sin 2q dr dq over the area bounded in the first quadrant
R
between the circle r = 2 and r = 4.
[Summer 2016]
Solution 1. The region of integration is the interior of the circle between r = 2 to r = 4. 2. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = 2 and leaves at the circle r = 4. 3. Limits of r: r = 2 to r = 4 Limits of q: q = 0 to q =
p 2
I = ÚÚ r 3 sin 2q dr dq p 2 4
=
Ú Úr
3
sin 2q dr dq
0 2
r4 = 4
4
cos 2q 2 2
p 2 0
=
1È 4 Ê 1ˆ (4) - (2)4 ˘˚ Á ˜ (cos p - cos 0) Î Ë 2¯ 2
=
1 Ê 1ˆ [256 - 16] Á - ˜ (-2) Ë 2¯ 2
1 [240] 2 = 120 =
Fig. 9.84
9.72
Chapter 9
Multiple Integrals
Example 2 a 2 - r 2 dr dq
ÚÚ r
Evaluate
over the upper half of the circle
r = a cos q.
[Summer 2017]
Solution
q= p 2
1. The region of integration is the upper half of the circle r = a cos q. 2. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = a cosq. 3. Limits of r : r = 0
to
r = a cosq
Limits of q : q = 0
to
θ=
A r = a cos q
O
π 2
q =0
I = ÚÚ r a 2 - r 2 dr dq p
=Ú2Ú 0
=-
1
a cosq Ê
1ˆ 2 2 2 ÁË - 2 ˜¯ (a - r ) (-2r ) dr dq
0
p 2 0
1 2Ú
2( a
2
3 - r2 )2
3
Fig. 9.85
a cosq
È [ f (r )]n +1 ˘ n [ f ( r )] f ( r ) r = Q d ¢ Í Ú ˙ n +1 ˚ Î
dq 0
p 2 0
=-
1 (a 3 sin 3 q - a 3 ) dq 3Ú
=-
a3 3
p 2 0
Ú
Ê 3 sin q - sin 3q ˆ - 1˜ dq ÁË ¯ 4 p
2 a3 1 Ê cos 3q ˆ -3 cos q + =-q Á ˜ 3 4Ë 3 ¯ 0
=-
a3 3
p 1 3p p 3 1 Ê 3 ˆ - + cos 0 - cos 0˜ ÁË - cos + cos ¯ 12 4 2 12 2 2 4
=-
a3 3
Ê p 3 1ˆ ÁË - 2 + 4 - 12 ˜¯
=-
a3 Ê 2 p ˆ 3 ÁË 3 2 ˜¯
9.4
Double Integrals in Polar Coordinates
9.73
Example 3
ÚÚ r
Evaluate
4
cos3 q dr dq over the interior of the circle r = 2a cos q.
Solution 1. The region of integration is the interior of the circle r = 2a cos θ. 2. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 2a cos θ. 3. Limits of r : r = 0 to r = 2a cos θ Limits of θ : θ = −
π π to θ = 2 2
q= p 2 A r = 2a cos q
O
q=0
I = Ú Ú r 4 cos3 q dr dq p
= Ú 2p cos3q Ú -
=Ú
2 a cosq 4
r dr d q
0
2
p 2 p 2
5
cos3 q
Fig. 9.86
2 a cosq
r 5
dq 0
p
=
1 2 cos3q (2 a cos q )5 dq 5 Ú- p 2
5
p
=
32 a ◊ 2 Ú 2 cos8 q dq 0 5
=
32 a 5 Ê 9 ◊ BÁ , Ë2 5
1ˆ 2 ˜¯
p È Ê p + 1 q + 1ˆ ˘ 2 sin p q cosq q dq ˙ 2 , = ÍQB Á ˜ Ú 0 2 ¯ ÍÎ Ë 2 ˙˚
9 1 32 a 5 2 2 = ◊ 5 5 7 5 3 1 1 1 ◊ ◊ ◊ ◊ 32 a 5 2 2 2 2 2 2 = ◊ 24 5 7p 5 a = 4
Example 4 Evaluate
2 ÚÚ r sinq dr dq over the cardioid r = a (1 + cos q ) above the
initial line.
9.74
Chapter 9
Multiple Integrals
Solution 1. The region of integration is the part of the cardioid r = a (1 + cos q ) above the initial line (q = 0). 2. Draw an elementary radius vector OA which starts from the origin and terminates on the cardioid r = a (1 + cos q ). 3. Limits of r : r = 0 to r = a (1 + cos q ) Limits of q : q = 0
to q = p
q=p 2
I = ÚÚ r 2 sin q dr dq =Ú
p
=Ú
p
a (1+ cosq ) 2 r sinq 0
Ú
0
r3 3
0
=
A r = a (1 + cos q )
dr dq O
a (1+ cosq )
q =0
sin q dq 0
1 p 3 a (1 + cos q )3 sin q ◊ dq 3 Ú0
=-
a3 3
p
Ú0
(1 + cos q )3 (- sin q )dq p
a 3 (1 + cos q )4 =3 4
0
Fig. 9.87
È [ f (q )]n +1 ˘ n ÍQ Ú [ f (q )] f ¢(q )dq = ˙ n + 1 ˙˚ ÍÎ
3
=-
a È 4 4 Î(1 + cos p ) - (1 + cos 0) ˘˚ 12
=-
a3 (0 - 16) 12
=
4 2 a 3
Example 5 Evaluate
ÚÚ
r dr dq 2
r +a
2
over one loop of the lemniscate r2 = a2 cos 2q.
Solution 1. The region of integration is one loop of the lemniscate r2 = a2 cos 2q bounded π π between the lines θ = − and θ = . 4 4
9.4
Double Integrals in Polar Coordinates
2. Draw an elementary radius vector OA which starts from the origin and terminates on the lemniscate r 2 = a2 cos 2q.
q= p 2 q= p 4 r 2 = a 2 cos 2q A q=0
3. Limits of r: r = 0 to r = a cos 2θ Limits of θ : θ = −
=Ú
2
r +a
q=− p 4
2
p a cos 2q 4 p 0 4 p a cos 2q 4 p 0 4
Ú
r 2 + a2 1
1 2 (r + a 2 ) 2 (2r )dr dq 2 1 a cos 2q
p
4
1 2Ú
p 4 p 4
p
= a Ú 4p -
=a
Fig. 9.88
r d r dq
Ú
1 = Ú 4p 2(r 2 + a 2 ) 2 2 =
O
r dr dq
I = ÚÚ =Ú
π π to θ = 4 4
9.75
dq 0
È ˘ [ f (r )]n +1 n , n π - 1˙ ÍQ Ú [ f (r )] f ¢ (r )dr = 1 n + ÍÎ ˙˚
1 È ˘ 2a Í(cos 2q + 1) 2 - 1˙ dq ÍÎ ˙˚
(
)
2 cos q - 1 dq
4
2 sin q - q
p 4 -
p 4
È p p Ê p ˆ Ê p ˆ˘ = a Í 2 sin - - 2 sin Á - ˜ + Á - ˜ ˙ Ë 4¯ Ë 4¯˚ 4 4 Î pˆ Ê = aÁ2- ˜ Ë 2¯ =
a (4 - p ) 2
Example 6 Evaluate
2 ÚÚ r dr dq over the area between the circles r = a sin q and
r = 2a sin q. Solution 1. The region of integration is the area bounded between the circle r = a sin q and r = 2a sin q.
9.76
Chapter 9
Multiple Integrals
2. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = a sin q and leaves at the circle r = 2a sin q. 3. Limits of r : r = a sin q to r = 2a sin q Limits of q : q = 0 to q = p
q= p 2
B
I = ∫∫ r 2 dr dθ =∫
π
=∫
π
0
0
∫
2 a sin θ a sin θ
r3 3
r = 2a sin q A
r 2 d r dθ
r = a sin q
2 a sin θ
dθ O
a sin θ
1 π = ∫ (8a 3 sin 3 θ − a 3 sin 3 θ )dθ 3 0 7a3 π 3 = sin θ dθ 3 ∫0 7 a 3 π 3 sin θ − sin 3θ = dθ 3 ∫0 4
=
cos 3θ 7a3 −3 cos θ + 12 3
q =0
Fig. 9.89
π
0
7a3 = 12
1 −3(cos π − cos 0) + 3 (cos 3π − cos 0) 7 a 3 16 = 12 3 28 3 = a 9
ExERCIsE 9.4 Evaluate the following integrals: −r2
1.
∫∫ re
a2
cos q sin q dr dq over the upper half of the circle r = 2a cos q. a2 1 Ans. : 3+ 4 16 e
2.
∫∫ r dr dq 3
over the region between the circles r = 2 sin q and r = 4 sin q. 45π Ans. : 2
9.5
3.
∫∫ r sinq dA
Multiple Integrals by Substitution
9.77
over the cardioid r = a (1+ cos q ) above the initial line. 4 3 Ans. : 3 a
4.
∫∫
r r +4 2
dr dq over one loop of the lemniscate r2 = 4 cos 2q. Ans. :(4 − π )
9.5
MultIPlE IntEGRAls by substItutIOn
9.5.1
Change of Variables from Cartesian to Polar Coordinates
The double integral can be changed from Cartesian coordinates (x, y) to polar coordinates (r, q ) by putting x = r cos q, y = r sin q. Then ∫∫ f ( x, y )dy dx =
∫∫ f (r cos θ , r sin θ ) J dr dθ where
J is the Jacobian (functional determinant)
defined as ∂x ∂x ∂ ( x , y ) ∂r ∂θ = J= ∂ ( r , θ ) ∂y ∂y ∂r ∂θ cos θ − r sin θ = sin θ r cos θ = r (cos 2 θ + sin 2 θ ) =r
Hence,
∫∫ f ( x, y)dy dx = ∫∫ f (r cos θ , r sin θ ) r dr dθ = ∫∫ f (r cos θ , r sin θ )r dr dθ
Example 1 Evaluate
∫∫
1 − x2 − y 2 dx dy over the first quadrant of the circle 1 + x2 + y 2
x2 + y 2 = 1. Solution 1. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y2 = 1 is obtained as r = 1. 2. The region of integration is the part of the circle r = 1 in the first quadrant.
9.78
Chapter 9
Multiple Integrals
3. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 1. 4. Limits of r : r = 0 to r = 1 π Limits of q : q = 0 to θ = 2 Hence, the polar form of the given integral is
q= p 2
A r=1 O
q =0
1− x − y dx d y 1 + x2 + y 2 2
I = ∫∫ π
= ∫2∫ 0
2
1− r2 r dr dθ 1+ r2
1
0
2
Putting r = cos2t, 2r dr = – 2 sin 2t dt π When r = 0, t = 4 When r = 1, t = 0 π
0
I = ∫ 2 ∫π 0
π 2 0
=∫
4
1 − cos 2t ( − sin 2t dt )dθ 1 + cos 2t
π 4 0
2 sin 2 t sin 2t dt dθ 2 cos 2 t
∫
π
π
= ∫2∫4 0
0
π
Fig. 9.90
sin t ⋅ 2 sin t cos t dt dθ cos t π
= ∫ 2 dθ ∫ 4 (1 − cos 2t )dt 0
=θ
0
π 2 0
sin 2t t− 2
π 4
dθ
0
π π 1 − 2 4 2 π = (π − 2) 8 =
Example 2 Evaluate
4 xy
ÚÚ x 2 + y2 e
- x 2 - y2
dx dy over the region bounded by the circle
x2 + y 2 – x = 0 in the first quadrant. Solution 1. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y2 – x = 0 is r 2 – r cos q = 0, r = cos q.
9.5
Multiple Integrals by Substitution
2. The region of integration is the part of the circle r = cos q in the first quadrant. 3. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = cos q. 4. Limits of r : r = 0 to r = cos q
π 2 Hence, the polar form of the given integral is 4 xy − x − y I = ∫∫ 2 e dx dy x + y2 Limits of q : q = 0
2
π
= ∫2∫ 0
cos θ
0
θ=
to
q= p 2 A r = cos q O
q =0
2
4r 2 cos θ sin θ − r 2 e r dr dθ r2
π
cos θ
= −2∫ 2 cos θ sin θ ∫ 0 0 π
= −2∫ 2 cos θ sin θ e − r
2
0
π
9.79
(
e − r ( −2r ) dr dθ
cos θ 0
Fig. 9.91
2
Q e f ( r ) f ′ (r )dr = e f ( r ) ∫
dθ
)
= −2∫ 2 cos θ sin θ e − cos θ − 1 dθ 0
π 2 0
2
= − ∫ e − cos θ (2 cos θ sin θ ) − sin 2θ dθ
=−e
− cos 2 θ
2
cos 2θ + 2
π 2 0
cos π cos 0 = − e0 + − e −1 − 2 2 1 = e
Example 3 Evaluate
x 2 y2
ÚÚ ( x 2 + y2 ) dx dy
over the region bounded by the circles
x2 + y2 = a2 and x2 + y2 = b2 (a > b). Solution 1. Putting x = r cos q, y = r sin q, polar form of (i) the circle x2 + y2 = a2 is r2 = a2, r = a. (ii) the circle x2 + y2 = b2 is r 2 = b2, r = b.
9.80
Chapter 9
Multiple Integrals
2. The region of integration is the part bounded between the circles r = a and r = b. 3. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = b and terminates on the circle r = a. 4. Limits of r : r = b to r = a Limits of q : q = 0 to q = 2p Hence, the polar form of the given integral is I = ÚÚ
x2 y 2 ( x2 + y 2 )
=Ú
2p
=Ú
2p
0
a
Úb
B r=a A O
r=b q=0
dx dy
r 4 cos 2 q sin 2 q r2
◊ r dr dq Fig. 9.92
a
0
q= p 2
r4 cos 2 q sin 2 q dq 4 2
4
b 4
sin 2q (a - b ) ◊ dq 4 4 a 4 - b 4 2p (1 - cos 4q ) = dq 16 Ú0 2 =Ú
2p
0
Ê a 4 - b4 ˆ sin 4q =Á ˜q- 4 32 Ë ¯
2p 0
Ê a -b ˆ =Á ˜ (2p ) Ë 32 ¯ 4
=
4
p 4 (a - b4 ) 16
Example 4 Evaluate
ÚÚ
( x 2 + y 2 )2
dx dy over the region common to the circles x 2 y2 x2 + y 2 = ax and x2 + y2 = by (a, b > 0). Solution 1. Putting x = r cos q, y = r sin q, polar form of (i) the circle x2 + y 2 = ax is r 2 = ar cos q, r = a cos q. (ii) the circle x2 + y 2 = by is r 2 = br sin q, r = b sin q. 2. The region of integration is the common part of the circles r = a cos q and r = b sin q.
9.5
Multiple Integrals by Substitution
9.81
q= p 2
3. The point of intersection of the circle r = a cos q and r = b sin q, is obtained as b sin θ = a cos θ a tan θ = b a θ = tan −1 b
r = b sin q q = tan−1 a b P
B
r = a cos q
A O
q=0
a Hence, θ = tan at P. b 4. Divide the region into two subregions OAP and OBP. Draw an elementary Fig. 9.93 radius vector OA and OB in each subregion. (i) In subregion OAP, elementary radius vector OA starts from the origin and terminates on the circle r = b sin q. Limits of r : r = 0 to r = b sin q a Limits of q : q = 0 to θ = tan −1 b (ii) In subregion OBP, elementary radius vector OB starts from the origin and terminates on the circle r = a cos q. Limits of r : r = 0 to r = a cos q a π Limits of θ : θ = tan −1 to θ = b 2 Hence, the polar form of the given integral is −1
I = ÚÚ =Ú
( x 2 + y 2 )2 x 2 y2
tan -1
a b
0
=Ú
tan -1
a b
0
dx dy
Ú0
p
r4
b sin q
r 4 sin 2 q cos2 q
r2 sin 2 q cos2 q 2
◊ r dr dq + Ú 2 -1 a Ú tan
b sin q
1
p
dq + Ú 2 -1 a tan
0
b
r2 sin 2 q cos2 q 2
◊ r dr dq
a cosq
dq 0
1 tan -1 b 1 1 1 ◊ b2 sin 2 q dq + Ú 2 -1 a ◊ a 2 cos2 q dq 2 2 2 2 2 Ú0 2 tan sin q cos q sin q cos q b
2
=
r 4 sin 2 q cos2 q
1
p
a
=
0
b
r4
a cosq
b 2
tan
Ú0
-1 a
b sec 2 q
2
dq +
a 2
p 2
Útan-1 a cosec q dq 2
b
9.82
Chapter 9
tan b2 = tan q 0 2
-1 a
b
a2 + - cot q 2
p 2 tan -1
˘ a2 È -1 Ê a ˆ tan tan tan 0 ˙Í ÁË b ˜¯ ˚ 2 Î
b2 2
=
Multiple Integrals
b2 È a ˘ a2 0 ˙- 2 2 ÍÎ b ˚ ab ab = + 2 2 = ab =
a b
È p Ê -1 a ˆ ˘ ˙ Ícot - cot ÁË tan b ˜¯ ˚ 2 Î
b˘ È Í0 - a ˙ Î ˚
Example 5 Evaluate
∞
∞
0
0
∫ ∫
2
2
e − ( x + y )dx dy.
Solution
y
1. Limits of x : x = 0 to x → ∞ Limits of y : y = 0 to y → ∞ 2. The region of integration is the first quadrant. 3. Putting x = r cos q, y = r sin q, the integral changes to polar form. 4. Draw an elementary radius vector which starts from the origin and extends up to infinity. Limits of r : r = 0 to r → ∞
π 2 Hence, the polar form of the given integral is Limits of q : q = 0 to
I=∫
∞
0
∫
∞
0
p
e−( x
•
2
+ y2 )
x
O
Fig. 9.94
θ=
q=
p 2
r→∞
dx dy
2
= Ú 2 Ú e - r r dr dq 0
0
p
=-
1 2 • -r 2 e ( -2r )dr dq 2 Ú0 Ú0
=-
1 2 -r 2 • dq e 2 Ú0 0
=-
1 2 (0 - e0 )dq 2 Ú0
p
p
ÈQ e f (r ) f ¢(r )dr = e f (r ) ˘ Î Ú ˚ O
q=0
Fig. 9.95
9.5
==
Multiple Integrals by Substitution
9.83
p 2 0
1 -q 2
p 4
Example 6 ∞
∫ ∫
Evaluate
dx dy
∞
−∞ −∞
3 2 2
.
(1 + x + y ) 2
Solution 1. Limits of x : x → −∞
to
x→∞
Limits of y : y → −∞
to
y→∞
2. The region of integration is the entire coordinate plane. 3. Putting x = r cos q, y = r sin q, integral changes to polar form. 4. Draw an elementary radius vector which starts from origin and extends up to •. Limits of r : r = 0 to r → ∞ Limits of q : q = 0 to q = 2p Hence, the polar form of the given integral is ∞ ∞ dx d y I=∫ ∫ 3 −∞ −∞ (1 + x 2 + y 2 ) 2 2π ∞ r dr dθ =∫ ∫ 3 0 0 (1 + r 2 ) 2
y
x
O
Fig. 9.96
3
=
− 1 2π ∞ 2 2 1 ( + r ) (2r )dr dθ 2 ∫0 ∫0
q= p 2
∞
1 − 1 2π = ∫ −2(1 + r 2 ) 2 dθ 2 0 0
[ f (r )]n +1 n Q ∫ [ f (r )] f ′ (r )dr = n +1 = −∫
2π
0
1 1+ r2
∞
r→∞
O
dθ 0
2π
= − ∫ (0 − 1)dθ 0
2π
= θ0 = 2π
Fig. 9.97
q=0
9.84
Chapter 9
Multiple Integrals
Example 7 Evaluate ∫0
∫
a
a
x dx dy by transforming into polar coordinates. x + y2 [Winter 2013] 2
y
Solution
a
0
∫
π
=∫4∫ 0
a y
π 0
y =
P
O
x
Fig. 9.98
x dx dy x + y2
a sec θ
a sec θ
r cos θ ⋅ r d r dθ r2
y q=p 2
cos θ dr dθ
0
π 4 0
x=a
2
0
= ∫ 4∫
Q (a, a)
Q a sec θ
=∫ r0
cos θ dθ = p 4
I=∫
y
x
1. Limits of x : x = y to x = a Limits of y : y = 0 to y = a 2. The region of integration is bounded by the lines y = x, x = a and y = 0. 3. Putting x = r cos q, y = r sin q, polar form of (i) the line y = x is r sin q = r cos q, tan π q = 1, θ = . 4 (ii) the line x = a is r cos q = a, r = a sec q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the line r = a sec q. Limits of r : r = 0 to r = a sec q π Limits of q : q = 0 to θ = 4 Hence, the polar form of the given integral is
π
q
= ∫ 4 a sec θ cos θ dθ
A
0
r = a secq
π
= a ∫ 4 dθ 0
π
= a θ 04
π 4 πa = 4
P q=0
O
= a⋅
Fig. 9.99
9.5
Multiple Integrals by Substitution
9.85
Example 8 2
∫∫
Evaluate
0
2 x − x2
x dy dx by transforming into polar coordinates. x + y2 2
0
Solution 1. Limits of y : y = 0 to y = 2 x − x 2 Limits of x : x = 0 to x = 2 2. The region of integration is bounded by the circle x 2 + y 2 – 2x = 0 and the lines y = 0, x = 0. Since the limits of x and y are positive, the region of integration is the part of the circle in the first quadrant. 3. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y 2 – 2x = 0 is
y
x 2+y 2−2x = 0
(2, 0) x
O
r2 – 2r cos q = 0 r = 2 cos q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 2 cos q. Limits of r : r = 0 to r = 2 cos q
Fig. 9.100
π 2 Hence, the polar form of the given integral is Limits of q : q = 0
2 x - x2
2
I=Ú
0 Ú0 p
0
x
dy dx
r cos q r2
0
q=
p 2 A r = 2 cosq
◊ r dr dq
(2, 0) O
p
=Ú2Ú
2 cosq
0
0
p 2 0
r0
=Ú
θ=
x 2 + y2 2 cosq
=Ú2Ú
to
q=0
cos q dr dq
2 cosq
cos q dq
p
= Ú 2 2 cos2 q dq 0
p 2 0
= Ú (1 + cos 2q )dq sin 2q = q+ 2
p 2 0
Fig. 9.101
9.86
Chapter 9
Multiple Integrals
p 1 1 + sin p - sin 0 2 2 2 p = 2 =
Example 9 2 x − x2
1
∫∫
Evaluate
0
( x 2 + y 2 )dx dy.
x
y
Solution y=x
1. Limits of y : y = x to y = 2 x − x 2 Limits of x : x = 0 to x = 1 2. The region of integration is bounded by the line y = x and the circle x2 + y 2 – 2x = 0. O 3. Putting x = r cos q, y = r sin q, polar form of (i) the line y = x is π r sin θ = r cos θ , tan θ = 1, θ = . 4 2 2 (ii) the circle x + y – 2x = 0 is r2 – 2r cos q = 0 r = 2 cos q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 2 cos q. Limits of r : r = 0 to r = 2 cos q π π to θ = 4 2 Hence, the polar form of the given integral is
P (1, 1)
x 2 + y 2 − 2x = 0
(1, 0)
x
Fig. 9.102 q= p 2 A
P
q= p 4 r = 2 cos q
Limits of θ : θ =
I=Ú
1
0 Úx
2 x - x2
O
q=0
( x 2 + y 2 )dx dy
p
= Ú p2 Ú
2 cosq 2
r ◊ r d r dq
0
4 p
= Ú p2 4
r4 4
p 2 p 4
2 cosq
dq 0
= 4Ú cos4q dq
Fig. 9.103
9.5
p
Multiple Integrals by Substitution
9.87
2
Ê 1 + cos 2q ˆ = 4Ú p2 Á ˜¯ dq Ë 2 4
p 2 p 4 p 2 p 4
= Ú (1 + 2 cos 2q + cos2 2q ) dq 1 + cos 4q ˆ Ê = Ú Á 1 + 2 cos 2q + ˜¯ dq Ë 2 3 2 sin 2q sin 4q = q+ + 2 2 8
p 2 p 4
=
3Êp pˆ Ê pˆ 1 - ˜ - Á sin p - sin ˜ + (sin 2p - sin p ) Á ¯ Ë Ë 2 2 4 2¯ 8
=
3p +1 8
Example 10 Evaluate the integral
a
Ú0 Ú0
a2 - y2
y 2 x 2 + y 2 dy dx by changing into [Summer 2014]
polar coordinates. Solution 1. Limits of x : x = 0 to x =
a2 - y2
Limits of y : y = 0 to y = a 2. The region of integration is bounded by x = 0, x=
a 2 - y 2 , y = 0 and y = a.
3. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y2 = a2 is (i) r2 = a2 \r=a 4. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = a. Limits of r: r = 0 to r = a p Limit of q: q = 0 to q = 2
Fig. 9.104
Fig. 9.105
9.88
Chapter 9
Multiple Integrals
Hence, the polar form of the given integral is I= =
=
p a 2 2 r 0 0
Ú Ú
sin 2 q ◊ r ◊ r dr dq
p 2 0
a sin 2 q ÈÍ Ú r 4 dr ˘˙ dq Î 0 ˚
p 2 0
sin 2 q
Ú Ú
r5 5
a
dq 0
5
p 2 0
=
a 5
=
a5 1 p ◊ ◊ 5 2 2
=
a 5p 20
Ú
sin 2 q dq
Example 11 2
Evaluate
1
Ú0 Ú
1- x 2 x - x2
2
xye - ( x + y ) dx dy. x 2 + y2
Solution 1. Limits of y : y = x − x 2
to
y = 1 − x2
Limits of x : x = 0 to x = 1 2. The region of integration is the part of the first quadrant bounded by the circles x 2 + y 2 – x = 0 and x 2 + y 2 = 1. 3. Putting x = r cos q, y = r sin q, polar form of (i) the circle x 2 + y 2 – x = 0 is r2 – r cos q = 0, r = cos q. (ii) the circle x2 + y2 = 1 is r2 = 1, r = 1. 4. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = cos q and terminates on the circle r = 1. Limits of r : r = cos q to r = 1 p Limits of q : q = 0 to q = 2
y
x2 + y2 = 1 x2 + y2 – x = 0 P (1, 0) O
x
Fig. 9.106
9.5
Hence, the polar form of the given integral is I=Ú
1
0
1- x 2
Ú
x - x2
p
=Ú2Ú 0
xye - ( x
2
+y )
x 2 + y2
dx dy
r2
cosq
q= p 2
2
r 2 sin q cos q e - r
1
9.89
Multiple Integrals by Substitution
r=1 B
2
◊ r d r dq
A
r = cos q
p
=-
2 1 1 2 sin q cos q Ú e - r (-2r ) dr dq Ú cos 0 q 2
=-
2 1 2 sin q cos q e - r 2 Ú0
p
O
q =0
1
dq cosq
ÈQ e f (r ) f ¢(r )dr = e f (r ) ˘ Î Ú ˚ p
=-
2 1 2 sin q cos q (e -1 - e - cos q )dq 2 Ú0
=-
2 1 2 1Ê1 ˆ sin 2q - e - cos q ◊ 2 sin q cos q ˜ dq Á Ú 0 ¯ 2 2Ëe
Fig. 9.107
p
1 1 Ê cos 2q ˆ - cos2 q =-e 4 e ÁË 2 ˜¯
p 2 0
ÈQ e f (q ) f ¢(q )dq = e f (q ) ˘ Î Ú ˚
pˆ Ê È ˘ - Á cos2 ˜ 2 1Í 1 Ë 2¯ + e - cos 0 ˙ = - - (cos p - cos 0) - e ˙ 4 Í 2e Î ˚ 1È 1 ˘ = - Í- (-2) - e0 + e -1 ˙ 4 Î 2e ˚
==
1 È1 1˘ -1+ ˙ 4 ÍÎ e e˚
1 È 2˘ 14 ÍÎ e ˙˚
Example 12 Evaluate
4a
y
Ú0 Ú y
2
4a
x2 - y 2 dx dy. x2 + y 2
Solution y2 to x = y 4a Limits of y : y = 0 to y = 4a.
1. Limits of x : x =
[Winter 2013]
9.90
Chapter 9
Multiple Integrals
2. The region of integration is bounded by the line y = x and the parabola y 2 = 4ax. 3. Putting x = r cos q, y = r sin q, polar form of
y y 2 = 4ax
r 2 sin2 q = 4ar cos q, r = 4a cot q cosec q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the parabola r = 4a cotq cosecq. Limits of r : r = 0 to Limits of q : q =
p to 4
x
=
y
(i) the line y = x is r sin q = r cos q, p tan q = 1, q = . 4 (ii) the parabola y2 = 4ax is O
x
r = 4a cot q cosec q q=
p 2
Fig. 9.108
Hence, the polar form of the given integral is I=Ú
4a
0
2
x -y
y
Úy
2
4a
dx dy
r = 4a cot q cosec q P
p 4 a cot q cosec q r 2 (cos2 q - sin 2 q ) 2 ◊ r dr dq p 0 2 r 4 4 a cot q cosec q p 2 2 (1 - 2 sin 2 q ) r dq p 2 0 4 p 1 2 (1 - 2 sin 2 q )(4 a )2 cot 2 q cosec 2q dq 2 p 4 p 8a 2 p2 (cot 2 q cosec 2q - 2 cot 2 q )dq 4
=Ú
q= p 2
2
x 2 + y2
P(4a, 4a)
A
Ú
q= p 4
=Ú = =
O
q=0
Ú
= 8a
Ú
2
Ú
p 2 p 4
= 8a 2 -
{
}
È -(cot 2 q )(-cosec 2q ) - 2cosec 2q + 2 ˘ dq Î ˚ cot 3 q + 2 cot q + 2q 3
p 2 p 4
Fig. 9.109
È [ f (q )]n +1 ˘ n ˙ ÍQ Ú [ f (q )] f ¢(q )dq = n + 1 ˙˚ ÍÎ
È 1Ê p pˆ p pˆ Ê Ê p p ˆ˘ = 8a 2 Í- Á cot 3 - cot 3 ˜ + 2 Á cot - cot ˜ + 2 Á - ˜ ˙ Ë ¯ Ë ¯ Ë 2 4¯˚ 3 2 4 2 4 Î
9.5
Multiple Integrals by Substitution
9.91
p˘ È 1 = 8a 2 Í- (-1) + 2(-1) + 2 ◊ ˙ 4˚ Î 3 È 5 p˘ = 8a 2 Í - + ˙ Î 3 2˚ Èp 5˘ = 8a 2 Í - ˙ Î 2 3˚
Example 13 Evaluate
a
Ú0 Ú2
5 ax - x 2
x2 + y 2 y2
ax
dx dy.
Solution 1. Limits of y : y = 2 ax to
y y 2 = 4ax
2
y = 5ax − x Limits of x : x = 0 to x = a 2. Since the limits of x and y are positive, the region of integration is the part of the first quadrant bounded by the parabola y 2 = 4ax and the circle x2 + y 2 – 5ax = 0 3. Putting x = r cos q, y = r sin q, polar form of (i) the parabola y2 = 4ax is r2 sin2 q = 4ar cos q, r = 4a cot q cosecq. (ii) the circle x2 + y2 – 5ax = 0 is r2 – 5ar cos q = 0, r = 5a cos q. 4. The points of intersection of r = 4a cotq cosecq and r = 5a cos q are obtained as 4a cot q cosec q = 5a cos q 4 sin 2 q = 5 2 q = ± sin -1 5 Hence, q = sin
-1
2 5
P
x 2 + y 2 – 5ax = 0
O
x
Fig. 9.110 q= p 2 B
r = 4a cotq cosecq P r = 5a cosq
A
O
q=0
Fig. 9.111
at P.
9.92
Chapter 9
Multiple Integrals
5. Draw an elementary radius vector OAB from the origin which enters in the region from the parabola r = 4a cotq cosecq and terminates on the circle r = 5a cos q. Limits of r : r = 4a cotq cosecq to r = 5a cos q p 2 Limits of q : q = sin -1 to q = 2 5 Hence, the polar form of the given integral is I=Ú
a
0
=Ú
Ú2
5 ax - x 2
2
p 2
sin
d x dy r
5 a cosq
-1
5
=Ú
y2
ax
p 2
sin
x 2 + y2
Ú4 a cot q cosecq r 2 sin2 q r dr dq 5 a cosq
-1
2
cosec 2q r 4 a cot q cosecq dq
5
=Ú
p 2
sin -1
2
cosec 2q (5a cos q - 4 a cot q cosecq )dq
5
=Ú
p 2
sin -1
2
È5a cot q cosecq + 4 a cosec 2q (-cosec q cot q )˘ dq Î ˚
5
cosec3q = -5acosecq + 4 a 3
p 2 sin -1
2 5
È [ f (q )]n +1 ˘ n ÍQ Ú [ f (q )] f ¢(q )dq = ˙ n + 1 ˙˚ ÍÎ
È Ê Ê 2 ˆ˘ p p 4a 2 ˆ 4a cosec3 cosec3 Á sin -1 = Í-5acosec + 5acosec Á sin -1 + ˙ ˜ 2 2 3 Ë Ë 5 ˜¯ ˚˙ 5¯ 3 ÍÎ 3˘ È 5 4a 4a Ê 5 ˆ ˙ Í = -5a + 5a + Í 2 3 3 ÁË 2 ˜¯ ˙ Î ˚
=
a (5 5 - 11) 3
È Ê Ê 2 ˆ 5ˆ˘ = cosec Á cosec -1 ÍQ cosec Á sin -1 ˙ ˜ 2 ˜¯ ˙ Ë 5¯ Ë Í Í ˙ 5 Í ˙ = ÍÎ ˙˚ 2
Example 14 n
Ê x2 y 2 ˆ 2 Evaluate ÚÚ xy Á + ˜ dx dy over the first quadrant of the ellipse Ë a 2 b2 ¯ x2 y 2 + = 1. a 2 b2
9.5
Multiple Integrals by Substitution
9.93
Solution Let
x = ar cos q, y = br cos q ∂x ∂x ∂( x, y) ∂r ∂q J= = ∂(r , q ) ∂y ∂y ∂r ∂q a cos q - ar sin q = = abr b sin q br cos q
y
x
O
dx dy = J dr dq = abr dr dq
Under the transformation x = ar cos q, x2 y 2 y = br sin q, the ellipse 2 + 2 = 1 in the a b xy-plane gets transformed to r 2 = 1 or r = 1, circle with centre (0, 0) and radius 1 in the rq -plane. The region of integration is the part of the circle r = 1 in first quadrant in the rq -plane. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 1. Limits of r : r = 0 to r = 1 p Limits of q : q = 0 to q = 2 Hence, the polar form of the given integral is
Fig. 9.112 q=
p 2 A r=1
O
n
Ê x 2 y2 ˆ 2 I = ÚÚ xy Á 2 + 2 ˜ dx dy b ¯ Ëa p
Fig. 9.113 n
1
= Ú 2 Ú abr 2 cos q sin q (r 2 ) 2 abr dr dq 0
0
p
= a 2 b2 Ú 2 0
sin 2q 1 n + 3 (r ) dr 2 Ú0
a 2 b2 cos 2q = 2 2 2 2
p 2 0
r n+4 n+4
1
0
=
a b 1 (- cos p + cos 0) ◊ 4 n+4
=
a 2 b2 2(n + 4)
q=0
9.94
Chapter 9
Multiple Integrals
ExERCIsE 9.5 Change to polar coordinates and evaluate the following integrals: 1
∫∫
1.
dx dy over the region bounded by the semicircle x 2 + y 2 – x = 0,
xy y ≥ 0.
∫∫ y
2.
2
p Ans. : 2
dx dy over the area outside the circle x 2 + y 2 – ax = 0 and inside the
circle x 2 + y 2 – 2ax = 0.
3.
∫∫ sin(x
15p a 4 Ans. : 64
+ y 2 )dx dy over the circle x2 + y2 = a2.
2
Ans. : p(1 − cos a 2 ) 4.
∫∫ xy(x
3
+ y 2 ) 2 dx dy over the first quadrant of the circle x2 + y2 = a2.
2
a7 Ans. : 14 3
5.
6.
3x
∫∫ 0
0
a
x
0
0
Ú Ú
dy dx x2 + y2
3 Ans. : 2 log 3
x 3 dx dy x2 + y2
a4 Ans. : log 1 + 2 4
(
7.
a
∫ ∫ 0
a2 − x 2
0
)
p sin 2 (a 2 − x 2 − y 2 ) dx dy a a2 Ans. : 2
8.
a
∫ ∫ 0
0
a2 − x 2
e−x
2
−y2
dx dy
(
)
p −a 2 Ans. : 4 1 − e
9.5
9.
∫ ∫ 0
x - x2
ÚÚ 0
12.
È 3p a 4 ˘ Í Ans. : ˙ 4 ˚ Î
2 2 4 xy e -( x + y ) dx dy 2 x +y
Ú Ú
a2 - y 2
2
0
1˘ È Í Ans. : e ˙ Î ˚
log e(x 2 + y 2 )dx dy
y
È p 2Ê 1ˆ ˘ Í Ans. : a ÁË log a - ˜¯ ˙ 4 2 ˚ Î
a
a + a2 - y 2
0
y
Ú Ú
9.95
2
0
a
11.
(x 2 + y 2 )dx dy
0
1
10.
2 ax − x 2
2a
Multiple Integrals by Substitution
dx dy (4 a 2 + x 2 + y 2 )3 È 1 Êp 1 1 ˆ˘ tan-1 Í Ans. : 2 Á ˜˙ 4 Ë 8a 2 2¯˚ Î
13.
a
a2 - x 2
0
ax - x 2
Ú Ú
dx dy a - x2 - y2 2
ÈÎAns. : a ˘˚ 14.
a
a2 - x 2
0
ax - x 2
Ú Ú
2 2 xy e -( x + y ) dx dy 2 x +y
2
1 È È - a2 ˘ ˘ 2 Í Ans. : 4 a 2 Î1 - (1 + a )e ˚ ˙ Î ˚ 15.
1
x
0
x2
ÚÚ
dx dy x2 + y2 È Ans. : 2 - 1˘ Î ˚
9.5.2
Change of Variables from Cartesian to Other Coordinates
In some cases, evaluation of double integral becomes easier by changing the variables. Let the variables x, y be replaced by new variables u, v by the transformation x = f1 (u, v), y = f2 (u, v), then
ÚÚ f ( x, y)dx dy = ÚÚ f ( f1 , f2 ) J dudv where
∂x ∂( x, y) ∂u Jacobian J = = ∂(u, v) ∂y ∂u
∂x ∂v ∂y ∂v
Using Eq. (1), the double integral can be transformed to new variables.
... (1)
9.96
Chapter 9
Multiple Integrals
Example 1 Ê x - yˆ dxdy Using the transformation x - y = u, x + y = v, evaluate ÚÚ cos Á Ë x + y ˜¯ over the region bounded by the lines x = 0, y = 0, x + y = 1. Solution
x - y = u, x + y = v u+v v-u x= ,y = 2 2 ∂x ∂( x, y) ∂u J= = ∂(u, v) ∂y ∂u
∂x 1 ∂v 2 = ∂y 1 ∂v 2 1 dxdy = J dudv = dudv 2
1 1 1 1 2 = + = 1 4 4 2 2
The region bounded by the lines x = 0, y = 0 and x + y = 1 in xy-plane is a triangle OPQ. u+v v-u and y = Under the transformation x = , 2 2 (i) the line x = 0 gets transformed to the line u = –v y v Q
v=1
C Q' x+y=1
x=0
P' A
B
u = −v
u=v
P O
x
O
u
y=0
Fig. 9.114
(ii) the line y = 0 gets transformed to the line u = v (iii) the line x + y = 1 gets transformed to the line v = 1 Thus, triangle OPQ in xy-plane gets transformed to triangle OP'Q' in uv-plane bounded by the lines u = v, u = –v and v = 1. In the region, draw a horizontal strip AB parallel to u-axis which starts from the line u = –v and terminates on the line u = v. Limits of u : u = –v to u = v Limits of v : v = 0 to v = 1
9.5
Multiple Integrals by Substitution
9.97
Ê x - yˆ I = ÚÚ cos Á dx dy Ë x + y ˜¯ =Ú
Ê uˆ 1
1 v
cos Á ˜ du dv 0 Ú- v Ë v¯ 2
1 1 Ê uˆ = Ú v sin Á ˜ Ë v¯ 2 0
=
v
dv -v
1 1 v [sin 1 - sin(-1)] dv 2 Ú0
1 v2 = ◊ 2 sin 1 2 2
1
0
1 = sin 1 2
Example 2
2 2 Using the transformation x 2 - y 2 = u , 2 xy = v, find ÚÚ ( x + y ) dx dy over the region in the first quadrant bounded by x2 – y2 = 1, x2 – y2 = 2, xy = 4, xy = 2.
Solution x2 – y2 = u, 2xy = v It is difficult to express x and y in terms of u and v, therefore we write Jacobian of u, v in terms of x and y.
∂u ∂ (u, v) ∂x J= = ∂ ( x , y ) ∂v ∂x
∂u ∂y 2x − 2y = = 4 ( x 2 + y2 ) ∂v 2y 2x ∂y
(
)
dudv = J dxdy = 4 x 2 + y 2 dxdy dxdy =
1
(
2
4 x + y2
)
dudv
The region in xy-plane bounded by the curves x2 – y2 = 1, x2 – y2 = 2, xy = 4, xy = 2 is transformed to a square in uv-plane bounded by the lines u = 1, u = 2, v = 4, v = 8. In the region, draw a vertical strip AB parallel to the v-axis which starts from the line v = 4 and terminates on the line v = 8. I = ÚÚ ( x 2 + y 2 ) dx dy
9.98
Chapter 9
Multiple Integrals
=Ú
2 8
Ú4 ( x
1
2
+ y2 )
1 2
4 ( x + y2 )
du dv
1 2 8 u v 4 1 4 =1 =
v
y xy = 4 xy = 2
v=8 R'
B
Q'
S'
A
P'
x2 − y2 = 1
R Q
x2 − y2 = 2
S P
u=2
u=1 x
O
O
v=4 u
Fig. 9.115
Example 3 Evaluate
Ú Ú (x
2
+ y 2 ) dA by changing the variables, where R is
R
the region lying in the first quadrant and bounded by the hyperbola x2 – y2 = 1, x2 – y2 = 9, xy = 2 and xy = 4. [Summer 2014] Solution Let
u = x2 – y2
J=
v = xy ∂u ∂x
∂(u, v) = ∂v ∂( x, y) ∂x
∂u ∂y ∂v ∂y
=
2 x -2 y y x
= 2x2 + 2y2 = 2(x2 + y2) du dv = |J| dx dy = 2(x2 + y2) dx dy 1 dx dy = ◊ du dv 2 2( x + y 2 ) The region in the xy-plane bounded by the curves x2 – y2 – 1, x2 – y2 = 9, xy = 2 and xy = 4 is transformed to a square in the uv-plane bounded by the lines u = 1, u = 9, v = 2, v = 4.
9.5
Multiple Integrals by Substitution
9.99
v
y xy = 4 xy = 2
v=8 R'
B
Q'
S'
A
P'
x2 − y2 = 1
R Q
x2 − y2 = 9
S P
u=2
u=1 x
O
O
v=4 u
Fig. 9.116
I=
ÚÚ ( x
2
+ y 2 ) dx dy
2
+ y2 ) ◊
R
=
ÚÚ ( x R
=
1 2
=
1 2
1 2
2( x + y 2 )
du dv
ÚÚ du dv 49
Ú Ú du dv 21
1 4 9 v u1 2 2 1 = (4 - 2) (9 - 1) 2 1 = (2)(8) 2 =8 =
Example 4 Using the transformation x + y = u, y = uv, show that 1 1- x
Ú0 Ú0
e
y x+ y
dy dx =
1 (e - 1). 2
[Winter 2014]
9.100
Chapter 9
Multiple Integrals
Solution x + y = u, y = uv x = u(1 - v), y = uv ∂x ∂x ∂( x, y) ∂u ∂v 1 - v - u J= = = = (1 - v)u + uv = u v u ∂(u, v) ∂y ∂y ∂u ∂v dxdy = J du udv = udu dv Limits of y : y = 0 to y = 1 – x Limits of x : x = 0 to x = 1. The region in xy-plane is the triangle OPQ bounded by the lines x = 0, y = 0 and x + y = 1. Under the transformation x = u (1 – v) and y = uv, (i) the line x = 0 gets transformed to the line u = 0 or v = 1 (ii) the line y = 0 gets transformed to the line u = 0 or v = 0 (iii) the line x + y = 1 gets transformed to the line u = 1 v y
v=1 R'
Q
B
Q' (1, 1)
u=0
u=1
P x
O
O
A
P'
u
v=0
Fig. 9.117
Thus, the triangle OPQ in the xy-plane gets transformed to the square OP'Q'R' in uv-plane bounded by the lines u = 0, v = 0, u = 1 and v = 1. In the region, draw a vertical strip AB parallel to the v-axis which starts from the u-axis and terminates on the line v = 1. Limits of v : v = 0 to v = 1 Limits of u : u = 0 to u = 1 y I=Ú
1 1- x
Ú
0 0
e x + y dx dy
1 1 v e udu dv 0 0
=Ú
= ev
Ú
1 0
u2 2
1
0
= (e1 - e0 ) ◊ =
1 (ee -1) 2
1 2
9.5
Multiple Integrals by Substitution
9.101
Example 5 Evaluate
u=
4
Ú0 Ú
y +1 2 2x y 2 2
y
dx dy by applying the transformations
y 2x - y , v = . Draw both regions. 2 2
[Winter 2015]
Solution 1. The function is integrated first w.r.t. x. y y to x = + 1 2. Limits of x : x = 2 2 Limits of y : y = 0 to y = 4. 3. The region is the parallelogram bounded y y , x = + 1, y = 0 and by the lines x = 2 2 y = 4 in xy-plane. Applying the transformations 2x - y y , v= 2 2 y = x2 y the line x - = 0 mapped to the line u = 0 2 y the line x - = 1 mapped to the line u = 1 2 the line y = 0 mapped to the line v = 0 the line y = 4 mapped to the line v = 2 u=
(i) (ii) (iii) (iv)
Fig. 9.118
Hence, the parallelogram OABC in the xy-plane mapped to the rectangle O¢A¢B¢C¢ in uv-plane, bounded by the lines u = 0, u = 1, v = 0 and v = 2. Fig. 9.119 In the region, draw a vertical strip AB parallel to the v-axis which starts from the u-axis and terminates on the line v = 2. Limits of u : u = 0 to u = 1 Limits of v : v = 0 to v = 2 dxdy = J du dv where Jacobian, J = J* =
∂( x, y) ∂(u, v) 1 ∂(u, v) = J ∂( x, y)
9.102
Chapter 9
Multiple Integrals
∂u ∂x = ∂v ∂x
∂u ∂y ∂v ∂y
1 = 0
1 2 1 2
1 2 1 \ dxdy = dudv 2 Hence, the new form of the integral is 4 y +1 2 x - y I = Ú Úy dxdy 0 2 2 =
=Ú
2
Ú
1
v=0 u=0
1 u ◊ du dv 2 1
1 2 u2 = Ú dv 2 0 2 0
1 2 = v0 4 1 = (2 ) 4 =2
Example 6 Using the transformation x = u (1 + v), y = v(1 + u ), u ≥ 0, v ≥ 0, evaluate
2
y
Ú0 Ú0 ÈÎ( x - y)
2
-
1 2
+ 2( x + y ) + 1˘˚ dydx.
Solution x = u(1 + v), y = v(1 + u)
∂x ∂x ∂( x, y) ∂u ∂v 1 + v u J= = = = 1+ u + v v 1+ u ∂(u, v) ∂y ∂y ∂u ∂v dxdy = J dudv = (1 + u + v) dudv
9.5
Multiple Integrals by Substitution
9.103
v y
y=x
y=2
v (1 + u) = 2
Q P
Q'
x=0
u = −1 u=0
P'(1, 1) A
x
O
v=u
B
u
O
Fig. 9.120
Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 2. The region in the xy-plane is the DOPQ bounded by the lines x = 0, y = 2 and y = x. Under the transformation x = u (1 + v), y = v (1 + u), u ≥ 0, v ≥ 0 (i) the line x = 0 gets transformed to the line u = 0 (ii) the line y = 2 gets transformed to the curve v (1 + u) = 2 (iii) the line y = x gets transformed to the line u = v Thus, the triangle OPQ in the xy-plane gets transformed to the region OP'Q' in uv plane bounded by the lines u = 0, u = v and the curve v (1 + u) = 2. The point of intersection of u = v and v (1 + u) = 2 is obtained as u2 + u – 2 = 0, u = 1, –2 and v = 1, –2. The point of intersection is P' (1,1). In the region, draw a vertical strip AB parallel to the v-axis which starts from the line u = v and terminates on the curve v (1 + u) = 2. 2 Limits of v : v = u to v = 1+ u Limits of u : u = 0 to u = 1 I=Ú
2
Ú
y
0 0
È( x - y)2 + 2( x + y) + 1˘ Î ˚
2 1+ u 0 u
=Ú
1
Ú
-
1 2
dy dx -
È(u - v)2 + 2(u + v + 2uv) + 1˘ Î ˚
2 1+ u (1 + u + v ) -1 (1 + u + v ) dv du 0 u
=Ú
1
Ú
2 1+ u dv du u 0 u
=Ú
1
Ú
1
2 1+ u
0
u
=Ú v
du
1Ê 2 ˆ =Ú Á - u˜ du 0 Ë 1+ u ¯
1 2
(1 + u + v)du dv
9.104
Chapter 9
Multiple Integrals
= 2 log(1 + u) -
u2 2
1
0
1 = 2 log 2 2
Example 7 Evaluate ÚÚ xy dxdy by changing the variables over the region in the first quadrant bounded by the hyperbolas x2 – y2 = a2, x2 – y2 = b2 and the circles x2 + y2 = c2, x2 + y2 = d2 with 0 < a < b < c < d. Solution Let
x 2 - y 2 = u, x 2 + y 2 = v u+v 2 v-u x2 = ,y = 2 2 1 1 ∂x ∂x 1 ∂( x, y) ∂u ∂v 4x 4x = = J= = 1 1 8 xy ∂(u, v) ∂y ∂y 4y 4y ∂u ∂v dxdy = J dudv = xydxdy =
1 dudv 8 xy
dudv 8
The region bounded by the hyperbolas x2 – y2 = a2, x2 – y2 = b2 and the circles x2 + y2 = c2, x2 + y2 = d 2 in xy-plane is the curvilinear rectangle PQRS. v y S' x2 R
−
y2
P
v = d2
R'
a2
= x2 − y2 = b2
Q
S
B
x2 + y 2 = d2 x x2 + y2 = c2
u = b2
u = a2
Fig. 9.121
P' O
A
v = c2
Q' u
9.5
9.105
Multiple Integrals by Substitution
Under the transformation x2 – y2 = u and x2 + y2 = v, (i) the hyperbolas x2 – y2 = a2, x2 – y2 = b2 get transformed to the lines u = a2, u = b2 respectively. (ii) the circles x2 + y2 = c2, x2 + y2 = d 2 get transformed to the lines v = c2, v = d 2 respectively. Thus, the curvilinear rectangle PQRS in the xy-plane gets transformed to the rectangle P'Q'R'S' in uv-plane bounded by the lines u = a2, u = b2, v = c2 and v = d 2. In the region, draw a vertical strip AB parallel to v-axis which starts from the line v = c2 and terminates on the line v = d 2. Limits of v : v = c2 to v = d 2 Limits of u : u = a2 to u = b2 I = ÚÚ xydxdy =Ú
b2
u=a
d2
2
Úv = c
2
1 dudv 8
1 b2 d 2 u 2 v 2 8 a c 1 = (b 2 - a 2 )(d 2 - c 2 ) 8 =
Example 8 2 Evaluate ∫∫ ( x + y ) dx dy, by changing the variables parallelogram with vertices (1, 0), (3, 1), (2, 2), (0, 1).
over
the
Solution The region of integration in xy-plane is the parallelogram PQRS. Equations of the sides of the parallelogram are obtained as (i) PQ : y − 0 =
1− 0 ( x − 1) 3 −1
y
2 y = x −1
R (2, 2)
x − 2y =1 2 −1 ( x − 0) 2−0 2y − 2 = x x − 2 y = −2
(ii) RS : y − 1 =
(iii) PS : y − 0 = 1 − 0 ( x − 1) 0 −1 x + y =1
S (0, 1)
O
Q (3, 1)
P (1, 0)
Fig. 9.122
x
9.106
Chapter 9
Multiple Integrals
2 −1 ( x − 3) 2−3 y −1 = −x + 3 x+ y = 4
(iv) QR : y − 1 =
Let x − 2 y = u, x + y = v x=
u + 2v v−u ,y= 3 3
∂x ∂( x, y) ∂u = J= ∂(u, v) ∂y ∂u
∂x 1 2 ∂v 3 3 1 = = ∂y 1 1 3 ∂v 3 3 1 dxdy = J dudv = du udv 3
Under the transformation x – 2y = u, and x + y = v (i) the lines x – 2y = 1, x – 2y = –2 get transformed to the lines u = 1, u = –2 respectively. (ii) the lines x + y = 1, x + y = 4 get transformed to the lines v = 1, v = 4 respectively Thus, the parallelogram PQRS in the xyv v=4 plane gets transformed to a square P'Q'R'S' in uv-plane bounded by the lines u = 1, u = B R' Q' –2, v = 1 and v = 4. In the region, draw a vertical strip AB parallel to v-axis which starts from the line v = 1 and terminates on the line v = 4. Limits of v : v = 1 to v = 4 u=1 u = −2 Limits of u : u = –2 to u = 1 I = ÚÚ ( x + y)2 dxdy =Ú
1
4
v u = 2 Úv =1
1 1 v3 = u -2 3 3
2
1 dudv 3
A
S'
P'
v=1
4
u
O 1
Fig. 9.123
= 21
Example 9 n
Ê x2 y 2 ˆ 2 Evaluate ÚÚ xy Á 2 + 2 ˜ over the first quadrant of the ellipse b ¯ Ëa x2 y 2 + = 1. a 2 b2
9.5
Multiple Integrals by Substitution
9.107
Solution Let x = ar cos q, y = br sin q ∂ ( x, y ) ∂(r , q )
J=
a cos q
=
b sin q
y
∂x ∂x ∂r ∂ q = ∂y ∂y ∂r ∂ q − ar sin q = abr br cos q
x
O
dx dy = J dr dq = abr dr dq Under the transformation x = ar cos q, y = br sin q, the ellipse
x2
y2
= 1. in the a 2 b2 xy-plane gets transformed to r2 = 1 or r = 1, circle with centre (0, 0) and radius 1 in the rq -plane. +
Fig. 9.124
The region of integration is the part of the circle r = 1 in first quadrant in the rq-plane. In the region, draw an elementary radius vector OA from the pole which terminates on the circle r = 1. Limits of r : r = 0 to r = 1 Limits of q : q = 0 to q =
p 2
q=p 2
n
A
Ê x 2 y2 ˆ 2 I = ÚÚ xy Á 2 + 2 ˜ dxdy b ¯ Ëa p
r=1 n
1
= Ú 2 Ú abr 2 cos q sin q (r 2 ) 2 abr dr dq 0
O
0
p
= a 2 b2 Ú 2 0
sin 2q 2
a 2 b2 cos 2q = 2 2 2 2
1
Ú0 (r ) p 2 0
n +3
r n+4 n+4
dr 1
0
=
a b 1 (- cos p + cos 0) ◊ 4 n+4
=
a 2 b2 2(n + 4)
Fig. 9.125
q=0
9.108
Chapter 9
Multiple Integrals
ExERCIsE 9.6 x -y
1. Using the transformation x + y = u, x – y = v, evaluate
x +y ÚÚ e dxdy over
the region bounded by x = 0, y = 0 and x + y = 1. È 1Ê 1ˆ ˘ Í Ans.: ÁË e - ˜¯ ˙ 4 e ˚ Î 2 2 2. Using the transformation x2 – y2 = u, 2xy = v, evaluate ÚÚ (x - y )dxdy over the region bounded by the hyperbolas x2 – y2 = 1, x2 – y2 = 9, xy = 2 and xy = 4.
[Ans.: 4] 3. Using the transformation x + y = u, y = uv, evaluate •
•
0
0
Ú Ú
e -( x + y ) x p -1y q -1dx dy . È Ans.: p q ˘ Î ˚ 1
4. Using the transformation x = u, y = uv, evaluate
∫∫
x
0 0
x 2 + y 2 dx dy .
È ˘˘ 1È 2 1 + log 1 + 2 ˙ ˙ Í Ans.: Í 3 ÍÎ 2 2 ÍÎ ˚˙ ˙˚
(
5. Evaluate
)
∫∫ (x + y) dx dy by changing the variables over the region bounded 2
by the parallelogram with sides x + y = 0, x + y = 2, 3x – 2y = 0 and 3x – 2y = 3. 8 Ans. : 5
6. Evaluate ∫∫ (x − y )4 e x + y dx dy , by changing the variables over the region bounded by the square with vertices at (1, 0), (2, 1), (1, 2), (0, 1). e3 − e Ans. : 5 1
7. Evaluate ∫∫ [ xy (1 − x − y )]2 dx dy , by changing the variables over the region bounded by the triangle with sides x = 0, y = 0, x + y = 1. 2p ˘ È Í Ans.: 105 ˙ Î ˚
9.6
Triple Integrals
9.109
9.6 tRIPlE IntEGRAls Let f (x, y, z) be a continuous function defined in a closed and bounded region V in 3-dimensional space. Divide the region V into small elementary parallelopipeds by drawing planes parallel to the coordinate planes. Let the total number of complete parallelopipeds which lie inside the region V be n. Let d Vr be the volume of the rth parallelopiped and (xr, yr, zr) be any point in this parallelopiped. Consider the sum n
S = Â f ( xr , yr , zr )d Vr
…(1)
r =1
d Vr = d xr ◊ d yr ◊ d zr
where,
If we increase the number of elementary parallelopipeds, n, then the volume of each parallelopiped decreases. Hence as n Æ •, d Vr Æ 0. The limit of the sum given by Eq. (1), if it exists is called the triple integral of f (x, y, z) over the region V and is denoted by
ÚÚÚ f ( x, y, z)dV V
Hence,
ÚÚÚ V
where
n
 f ( xr , yr , zr )d Vr n ƕ
f ( x, y, z )dV = lim
d Vr Æ 0 r =1
dV = dx dy dz
9.6.1 triple Integrals in Cartesian Coordinates Triple integral of a continuous function f (x, y, z) over a region V can be evaluated by three successive integrations. Let the region V be bounded below by a surface z = z1 (x, y) and above by a surface z = z2 (x, y). Let the projection of region V in xy-plane be R which be bounded by the curves y = y1(x), y = y2(x) and x = a, x = b. Then the triple integral is defined as b È y2 ( x ) I = Ú ÍÚ a Î y1 ( x )
{Ú
z2 ( x , y ) z1 ( x , y )
}
˘ f ( x, y, z ) dz dy ˙ dx ˚
Note: The order of variables in dx dy dz indicates the order of integration. In some cases this order is not maintained. Therefore, it is advisable to identify the order of integration with the help of the limits.
9.6.2 triple Integrals in Cylindrical Coordinates Cylindrical coordinates r, q, z are used to evaluate the integral in the regions which are bounded by cylinders along z-axis, planes through z-axis, planes perpendicular to the z-axis.
9.110
Chapter 9
Multiple Integrals
Relations between Cartesian (rectangular) coordinates (x, y, z) and cylindrical coordinates (r, q, f ) are given as x = r cos q y = r sin q z=z z
Then
ÚÚÚ
f ( x, y, z )dx dy dz = ÚÚÚ f (r cos q , r sin q , z ) J dz dr dq P (r, q, z)
∂x ∂z ∂y ∂z ∂z ∂z
z=z y = r sin q
N
os
q
O rc
y
90°
q
=
r
x
where,
∂x ∂x ∂r ∂q ∂( x, y, z ) ∂y ∂y J= = ∂(r , q , z ) ∂r ∂q ∂z ∂z ∂r ∂q cos q -r sin q 0 = sin q r cos q 0 0 0 1
90° M
Q
x
Fig. 9.126
= cos q (r cos q ) + r sin q (sin q ) =r
Hence,
ÚÚÚ f ( x, y, z)dx dy dz = ÚÚÚ f (r cos q , r sin q , z)r dz dr dq
9.6.3 triple Integrals in spherical Coordinates Spherical coordinates (r, q, f ) are used to evaluate the integral in the regions which are bounded by the sphere with centre at the origin.
z
P (x, y, z)
Relations between cartesian (rectangular) coordinates (x, y, z) and spherical coordinates (r, q, f ) are given as x = r sin q cos f
r q
z = r cos q N
y = r sin q sin f
O f
y
in
q
co
s
y = r sin q sin f = x
Then
90° rs
in
rs
z = r cos q
f
q
90°
M
x ÚÚÚ f ( x, y, z)dx dy dz = ÚÚÚ f (r sin q cos f , r sin q sin f , r cos q ) J dr dq df
Q
Fig. 9.127
9.6
where
J=
∂x ∂r
∂x ∂q
∂x ∂f
∂( x, y, z ) ∂y = ∂(r , q , f ) ∂r
∂y ∂q
∂y ∂f
∂z ∂r
∂z ∂q
∂z ∂f
sin q cos f r cos q cos f = sin q sin f cos q
r cos q sin f -r sin q
Triple Integrals
9.111
-r sin q sin f r sin q cos f 0
= sin q cos f (r 2 sin 2 q cos f ) - r cos q cos f (-r sin q cos f cos q ) - r sin q sin f (-r sin 2 q sin f - r cos2 q sin f ) = r 2 sin q cos2 f (sin 2 q + cos2 q ) + r 2 sin q sin 2 f = r 2 sin q
Hence,
ÚÚÚ f ( x, y, z)dx dy dz = ÚÚÚ f (r sin q cos f, r sin q sin f, r cos q )r
2
sin q dr dq df.
Note: If the region of integration is a sphere x2 + y2 + z2 = a2 with centre at (0, 0, 0) and radius a, then limits of r, q, f are (i) For positive octant of the sphere, r : r = 0 to r = a p q : q = 0 to q = 2 p f : f = 0 to f = 2 (ii) For hemisphere, r : r = 0 to r = a
p 2 f : f = 0 to f = 2p
q : q = 0 to q =
(iii) For complete sphere, r : r = 0 to r = a q : q = 0 to q = p f : f = 0 to f = 2p
9.6.4
Change of Variables
In some cases, evaluation of a triple integral becomes easier by changing the variables. Let the variables x, y, z be replaced by new variables u, v, w by the transformation x = f1 (u, v, w), y = f2 (u, v, w), z = f3 (u, v, w).
9.112
Chapter 9
Multiple Integrals
ÚÚÚ f ( x, y, z)dx dy dz = ÚÚÚ f ( f1 , f2 , f3 ) J du dv dw
Then
∂x ∂u ∂( x, y, z ) ∂y J= = ∂(u, v, w) ∂u ∂z ∂u
where,
∂x ∂v ∂y ∂v ∂z ∂v
∂x ∂w ∂y ∂w ∂z ∂w
9.6.5 Working Rule for Evaluation of triple Integrals 1. 2. 3. 4.
Draw all the planes and surfaces and identify the region of integration. Draw an elementary volume parallel to z ( y or x) axis. Find the variation of z ( y or x) along the elementary volume. Lower and upper limits of z ( y or x) are obtained from the equation of the surface (or plane) where elementary volume starts and terminates respectively. 5. Find the projection of the region on xy (zx or yz) plane. 6. Draw the region of projection in xy (zx or yz) plane. 7. Follow the steps of double integration to find the limits of x and y (z and x or y and z).
Note: (1) If the region is bounded by the cylinders along the z-axis, planes through z-axis, the planes perpendicular to the z-axis, then the cylindrical coordinates are used. (2) If the region is bounded by the sphere, then the spherical coordinates are used.
type I Evaluation of triple Integrals when limits are Given
Example 1 Evaluate
1 2 e
Ú0 Ú0 Ú0 dy dx dz.
Solution 1 2
1 2È e
e
˘
Ú0 Ú0 Ú0 dy dx dz = Ú0 Ú0 ÍÎ Ú0 dy ˙˚dx dz =Ú
e
1 2
0 Ú0
y dx dz 0
1È 2
= Ú Í Ú e dx ˘˙ dz 0Î 0 ˚ 1
2
0
0
= e Ú x dz 1
= e Ú 2 dz 0
Triple Integrals
9.6
9.113
1
= 2e z 0 = 2e
Another method Since all the limits are constant and integrand (function) is explicit in x, y and z, the integral can be written as 1 2
1
e
2
e
Ú0 Ú0 Ú0 dy dx dz = Ú0 dz ◊Ú0 dx ◊ Ú0 dy 1
2
= z0◊ x0◊ y0 e
= 1◊ 2 ◊ e = 2e.
Example 2 Evaluate
2 3 2
Ú0 Ú1 Ú1
xy 2 z dz dy dx.
Solution Since all the limits are constant and integrand (function) is explicit in x, y and z, the integral can be written as 2
3 2
Ú0 Ú1 Ú1 xy
2
2
3
2
0
1
1
z dz dy dx = Ú x dx ◊Ú y 2 dy ◊ Ú z dz =
x2 2
2
◊
3
2
1
1
y3 z2 ◊ 3 2
0
26 3 = 2◊ ◊ 3 2 = 26
Example 3 Evaluate
1 p
p
Ú0 Ú0 Ú0
[Winter 2013]
y sin z dx dy dz.
Solution Since all the limits are constant and integrand (function) is explicit in x, y, and z, the integral can be written as 1 p
p
1
p
p
Ú0 Ú0 Ú0 y sin z dx dy dz = Ú0 sin z dzÚ0 y dyÚ0 dx 1
= - cos z 0 ◊
y2 2
p p
◊ x0 0
9.114
Chapter 9
Multiple Integrals
Ê p2 ˆ = ( - cos 1 + cos 0) Á ˜ (p ) Ë 2 ¯ =
p3 (1 - cos 1) 2
Example 4 1
z 2p
Ú0 Ú0 Ú0
Evaluate
(r 2 cos2 q + z 2 )rdq drdz.
[Winter 2016]
Solution 1
z
2p
Ú0 Ú0 Ú0
(r 2 cos2 q + z 2 )rdq drdz = Ú
1
z
{
2p
=Ú
1
z
ÏÔ
2p
Ú0 0 Ú0
Ì Ú0 0 Ú0 Ô
=Ú
1
Ó
z
Ú
0 0
= 2p Ú
1
= 2p Ú
1
È 2 Ê 1 + cos 2q ˆ ˘ ¸Ô + z 2 ˙ dq ˝ r drdz Ír ÁË ˜ ¯ 2 Î ˚ ˛Ô
2p ¸ ÏÔ È 1 Ê sin 2q ˆ Ô 2 2 ˘ + z q ˙ ˝ d r dz Ìr Í r Á q + ˜ Ë ¯ 2 ˚ 0 Ô˛ ÔÓ Î 2
Ú
z
0 0
0
}
[r 2 cos2 q + z 2 ] dq r drdz
È1 3 2 ˘ Í 2 r + z r ˙ drdz Î ˚ 2
r4 r2 + z2 8 2
dz 0
= 2p Ú
1
0
z2 z3 + dz 8 2 1
È z3 z 4 ˘ = 2p Í + ˙ ÍÎ 24 8 ˙˚ 0 È 1 1˘ = 2p Í + ˙ Î 24 8 ˚ È 3 + 1˘ = 2p Í ˙ Î 24 ˚
9.6
= 2p ◊ =
Triple Integrals
4 24
p 3
Example 5 Evaluate
1 1- x
x+ y
Ú0 Ú0 Ú0
dx dy dz.
Solution 1 1- x
x+ y
Ú0 Ú0 Ú0
dx dy dz = Ú
1 1- x È x + y
Ú
ÍÎ Ú0
0 0
=Ú
x+ y
1 1- x
0 Ú0
=Ú
1 1- x
Ú
0 0 1
dy dx
z 0
( x + y)dy dx
= Ú xy + 0
dz ˘˙ dy dx ˚
y2 2
1- x
dx 0
1È (1 - x )2 ˘ = Ú Í x(1 - x ) + ˙ dx 0 2 ˙˚ ÍÎ 1È (1 - x )2 ˘ = Ú Í x - x2 + ˙ dx 0Î 2 ˚
x 2 x 3 1 (1 - x )3 = + ◊ 2 3 2 (-3)
1
0
1 1 1 - + 2 3 6 1 = 3 =
Example 6 Evaluate
1 1- x
x+ y z
Ú0 Ú0 Ú0
e dx dy dz.
Solution 1 1- x
x+ y z
Ú0 Ú0 Ú0
1 1- x È x + y z ˘ e dz ˙ dy dx 0 0 Í Î 0 ˚
e dx dy dz = Ú
Ú
Ú
9.115
9.116
Chapter 9
Multiple Integrals
=Ú
1 1- x
0 Ú0
ez
x+ y 0
dy dx
1 1- x = Ú ÈÍ Ú (e x + y - e0 )dy ˘˙ dx 0Î 0 ˚
1
1- x
0 1
0
= Ú ex+ y - y
dx
= Ú ÈÎ(e - e x ) - (1 - x )˘˚ dx 0
x2 = (e - 1) x + - ex 2
1
0
1 = (e - 1) + - e + e0 2 1 = 2
Example 7 Evaluate
x+ y
x+ y+ z dz dy dx. Ú0 Ú0 Ú0 e a
x
Solution a
x
x+ y x+ y+ z
Ú0 Ú0 Ú0
e
a x x+ y dz dy dx = Ú e x Ú e y ÈÍ Ú e z dz ˘˙ dy dx 0 0 0 Î ˚
a
x
= Ú e x Ú e y ez
x+ y
0 0 0 a x x y x+ y e e e 0 0 a x È x x 2y e Í e e 0 Î 0
=Ú =Ú
(
Ú
Ú
(
dy d x
) - e ) dy ˘˙ dx ˚ - e 0 dy dx y
x
a x e 0
=Ú
e2 y - e y dx e ◊ 2 x
0
ÈÊ ˘ a 1ˆ e = Ú e x ÍÁ e x ◊ - e x ◊ ˜ - (e x - e0 )˙ dx Ë ¯ 0 2 2 Î ˚ a Ê 1 4x 3 2x ˆ = Ú Á e - e + e x ˜ dx 0 Ë2 ¯ 2 2x
=
1 e 4 x 3 e2 x ◊ - ◊ + ex 2 4 2 2
a
0
1 3 = (e 4 a - e0 ) - (e2 a - e0 ) + (e a - e0 ) 8 4 1 4a 3 2a 3 = e - e + ea 8 4 8
9.6
Triple Integrals
9.117
Example 8 Evaluate the integral
2 2 yz
Ú0 Ú1 Ú0
xyz dx dy dz.
Solution 2
2
yz
Ú0 Ú1 Ú0
xyz dx dy dz = =
2
2
È
2
2
x2 yz 2
x dx ˘˙ dy dz ˚
yz
Ú0 Ú1 yz ÍÎÚ0 Ú0 Ú1
yz
dy dz 0
2
=
2
1
Ú0 Ú1 2 y
1 y4 = 2 4
3 3
z dy dz
2
z4 4
1
2
0
=
1 Ê 1ˆ 1 [16 - 1] [16 - 0] Á ˜ 2 Ë 4¯ 4
=
15 2
Example 9 1 xy x 1 1 0 3
Ú Ú Ú
Evaluate
xy dz dy dx.
Solution 1 xy x 1 1 0 3
Ú Ú Ú
1 x 1 1
xy dz dy dx = Ú
3
1 x 1 1
=Ú
3
È
Ú ÍÎÚ0 Ú
xy
˘ dz ˙ xy dy dx ˚
xy
z 0 dy dx
1 ˘ 3È = Ú Í Ú x xy dy ˙ dx 1 Î 1 ˚
=Ú
1 3 x 2y 2
3
1
x
3
dx
1
Ê ˆ 2 3 1 Á = Ú x 3 - 1˜ dx Á ˜ 3 1 Ë x2 ¯
[Summer 2014]
9.118
Chapter 9
Multiple Integrals
=
=
1ˆ 2 3Ê 1 2 dx x Á ˜¯ 3 Ú1 Ë x 3 2x 2
2 log x 3 3
3
1
(
)
3 ˘ 2È 2 Í( log 3 - log 1) - 3 2 - 1 ˙ 3Î 3 ˚ 2È 2˘ = Ílog 3 - 2 3 + ˙ 3Î 3˚
=
Example 10 Evaluate
2 z
yz
Ú0 Ú1 Ú0 xyz dx dy dz.
[Summer 2017]
Solution The innermost limits depend on y and z. Hence, integrating first w.r.t. x, 2
z
yz
Ú0 Ú1 Ú0
xyz dx dy dz = Ú
2
Ú
z
0 1
x2 2
yz
yz dy dz 0
1 2 z 2 2 ( y z ) yz dy dz 2 Ú0 Ú1 z 1 2 = Ú z 3 ÈÍ Ú y3 dy ˘˙ dz ˚ 2 0 Î1 =
z
=
1 2 3 y4 z dzz 2 Ú0 4 1
1 2 = Ú z 3 ( z 4 - 1)dz 8 0 =
1 z8 z 4 8 8 4
1 = (32 - 4) 8 7 = 2
2
0
9.6
Triple Integrals
9.119
Example 11 Evaluate
1 1- x 1- x - y
Ú0 Ú0 Ú0
1 dx dy dz. ( x + y + z + 1)3
Solution The innermost limits depend on x and y. Hence, integrating first w.r.t. z, 1 1- x 1- x - y
Ú0 Ú0 Ú0
1 ( x + y + z + 1)3
=Ú
1 1- x È 1- x - y
Ú
0 0
=Ú
dx dy dz
Í Ú0 Î
1 ( x + y + z + 1)3
1 1- x
Ú
0 0
˘ dz ˙ dy dx ˚
1- x - y
1 -2( x + y + z + 1)2
dy dx 0
=-
˘ 1 1 1- x È 1 1 Í ˙ dy dx Ú Ú 2 2 2 0 0 ÍÎ {x + y + (1 - x - y) + 1} ( x + y + 1) ˙˚
=-
1 1 È 1- x ÔÏ 1 1 Ô¸ ˘ ÍÚ Ì ˝ dy ˙ dx Ú 2 0 0 2 ÍÎ ÔÓ 4 ( x + y + 1) Ô˛ ˙˚
=-
1 1 y 1 + dx Ú 0 2 4 x + y +1 0
1- x
1 1 È1 - x 1 1 ˘ + dx Í Ú 2 0Î 4 x + (1 - x ) + 1 x + 1 ˙˚ 1 ˆ 1 1Ê 1 - x 1 + =- Ú Á ˜ dx 0 Ë 4 2 x + 1¯ 2 =-
=-
1 x x2 x + - log( x + 1) 2 4 8 2
1
0
1Ê5 ˆ = - Á - log 2˜ Ë ¯ 2 8
Example 12 Evaluate
1
Ú0 Ú0
1- x 2
Ú0
1- x 2 - y2
[Winter 2014]
xyz dz dy dx.
Solution 1
Ú0 Ú0
1- x 2
È 1- x2 - y2 ˘ xy Í Ú z dz ˙ dx dy = Î 0 ˚
1
Ú0 Ú0
1- x 2
z2 xy z
1- x 2 - y2
dx dy 0
9.120
Chapter 9
Multiple Integrals
=
1 1 1- x2 xy(1 - x 2 - y 2 ) dx dy 2 Ú0 Ú0
=
˘ 1 1 È 1- x2 È 2 3˘ (1 ) x ÍÚ x y y Ú Î ˚ ˙˚ dy dx 2 0 Î 0
1 1 y2 y 4 = Ú x (1 - x 2 ) 2 0 2 4
1- x 2
dx 0
=
2 Ê 1 - x 2 )2 ˆ ˘ 1 1 È 2 (1 - x ) x (1 x ) Í ÁË ˜ ˙ dx 2 Ú0 Î 2 4 ¯˚
=
1 1 È (1 - x 2 )2 ˘ xÍ ˙ dx 2 Ú0 Î 4 ˚
=
1 1 x(1 - x 2 )2 dx 8 Ú0
=
1 1 {x(1 - 2 x 2 + x 4 )} dx 8 Ú0
=
1 1 ( x - 2 x 3 + x 5 ) dx 8 Ú0
1 x2 2 x 4 x6 + = 8 2 4 6 =
1 Ê 1 1 1ˆ Á - + ˜ 8 Ë 2 2 6¯
=
1 48
1
0
Example 13 Evaluate
e log y e x
Ú1 Ú1
Ú1 log z dx dy dz.
[Summer 2016]
Solution The inner most limit depends on x and middle limit depends on y. Hence, integrating first w.r.t. z, e
log y
ex
Ú1 Ú1 Ú1
log z dx dy dz = Ú
e
log y
ex
Ú1 1 Ú1
log z dz dx dy
9.6
=Ú
e
=Ú
e
Ú
Triple Integrals
9.121
log y Ê
1 1 log y
1 Ú1
ex 1 ˆ ex z log z ÁË Ú1 z ◊ z dz˜¯ dx dy 1
(e log e x
x
ex
)
- log 1 - z 1 dx dy
e log y x (e x - e x + 1) dx ˘˙ dy = Ú ÈÍ Ú 1 Î 1 ˚ e
log y
1
1
= Ú xe x - e x - e x + x
dy
= Ú ÈÎe log y (log y - 2) + log y - e(1 - 2) - 1˘˚ dy 1 e
= Ú [ y(log y - 2) + log y + e - 1] dy e
1 e
= Ú [( y + 1) log y - 2 y + e - 1] dy 1
e
2 Ê y2 ˆ ˆ e e1Ê y e = log y Á + y˜ - Ú Á + y˜ dy - y 2 + (e - 1) y 1 1 1 yË 2 Ë 2 ¯1 ¯ e
Ê e2 ˆ Ê 1 ˆ y2 = log e Á + e˜ - log 1Á + 1˜ + y - (e2 - 1) + [(e - 1)(e - 1)] Ë ¯ 2 2 4 Ë ¯ 1
=
e2 È1 ˘ + e - Í (e2 - 1) + (e - 1)˙ - e2 + 1 + e2 - 2e + 1 2 Î4 ˚
=
13 e2 - 2e + 4 4
Example 14 • • •
Evaluate
Ú0 Ú0 Ú0
dx dy dz . (1 + x 2 + y 2 + z 2 )2
Solution
z
1. It is difficult to integrate this integral in cartesian form. Putting x = r sin q cos f, y = r sinq sin f, z = r cosq integral changes to spherical form. 2. Limits of x : x = 0 to x Æ •
r →∞
O
Limits of y : y = 0 to y Æ • Limits of z : z = 0
y
to z Æ •
The region of integration is the positive octant of the plane. Limits of r : r = 0 to r Æ •
x
Fig. 9.128
9.122
Chapter 9
Multiple Integrals
Limits of q = 0
to q =
p 2
p 2 Hence, the spherical form of the given integral is Limits of f : f = 0 to f =
I=Ú
•
0
=Ú
•
0
=Ú
•
dxdydz (1 + x + y 2 + z 2 )2
p p 2 2 0 0
Ú Ú
2
r 2 sin q dr dq df (1 + r 2 )2
r 2 dr
•
0
•
Ú0 Ú0
(1 + r 2 )2
p 2 0
Ú
p
sinq dq Ú 2 df 0
Putting r = tan t , dr = sec 2 t dt When r = 0, t = 0 When r Æ •, t =
p 2 p
p
tan 2 t
0
sec 4 t
I = Ú 2 sin q dq Ú 2 0
= - cos q
p 2 0
p
sec 2 t dt Ú 2 df
Ê p 2 ˆ ÁË Ú 2 sin t dt ˜¯ f 0
0
p 2 0
ˆ p p Ê ˆ Ê p = Á - cos + cos 0˜ ◊ Á 2 1 - cos 2t dt ˜ ◊ Ë ¯ Ë Ú0 2 ¯ 2 2 p
1 sin 2t 2 p = t2 2 0 2 1 Èp 1 ˘p = Í - (sin p - sin 0 )˙ 2 Î2 2 ˚2 p2 = 8
Example 15 Evaluate
a
Ú0 Ú0
a2 - x2
Ú0
a2 - x 2 - y2
xyz dx dy dz.
Solution 1. It is difficult to integrate this integral in cartesian form. Putting x = r sin q cos f , y = r sin q sin f , z = r cos q integral changes to spherical form.
9.6
Triple Integrals
2 2 2 2. Limits of z : z = 0 to z = a - x - y
9.123
z
Limits of y : y = 0 to y = a 2 - x 2 Limits of x : x = 0 to x = a The region of integration is the positive octant of the sphere x 2 + y 2 + z 2 = a 2 .
r q
Limits of r : r = 0 to r = a
I=Ú
a
0
Ú0
a -x
p
2
p
=Ú2
Ú0 a
3
p
p
0
2
a -x -y
2 r f = 0 Úq = 0 Úr = 0
=Ú2
φ
x
Fig. 9.129
2
xyz dx dy dz
sin 2 q cos q ◊ cos f sin f ◊ r 2 sin q dr dq df
a sin 2f df Ú 2 sin 3q cos q dq Ú r 5 dr 0 0 2
1 - cos 2f = 2 2 =
2
y
O
p Limits of q : q = 0 to q = 2 p Limits of f : f = 0 to f = 2 Hence, the spherical form of the given integral is 2
P
p 2 0
sin 4 q ◊ 4
p 2 0
r6 ◊ 6
a
0
È [ f (q )]n +1 ˘˙ ÍQ Ú [ f (q )]n f ¢(q )dq = n +1 ˙ Í Î ˚
1Ê p 1 1 ˆ a6 ◊ (- cos p + cos 0) ◊ Á sin - sin 0˜ ◊ ¯ 6 4Ë 2 2 2
1 2 1 a6 ◊ ◊ ◊ 2 2 4 6 a6 = 48 =
Example 16 1
Evaluate
Ú0 Ú0
1- x 2
Ú
dz dy dx
1 x 2 + y2
x 2 + y2 + z2
polar coordinates. Solution 2 2 1. Limits of z : z = x + y to
Limits of y : y = 0 Limits of x : x = 0
to to
z=1 y = 1 − x2 x=1
by transforming into spherical
9.124
Chapter 9
Multiple Integrals
2. The region of integration is the part of the cone z2 = x2 + y 2 bounded above by the plane z = 1 in the positive octant (since all three limits are positive). 3. Putting x = r sinq cos f, y = r sinq sin f, z = r cosq, spherical polar form of (i) the cone z2 = x2 + y 2 is r2 cos2q = r2 sin2q (cos2f + sin2f ) = r2 sin2q cosq = sin q tanq = 1 p q= 4 (ii) the plane z = 1 is r cos q = 1 r = secq z
r = sec q A y P
q=
f O
O x
p 4
f
y P
x
Fig. 9.130
4. Draw an elementary radius vector OA which starts from the origin and terminates on the plane r = secq. Limits of r : r = 0 to r = secq p Limits of q : q = 0 to q = 4 p (in positive octant) 2 Hence, the spherical form of the given integral is
Limits of f : f = 0 to f =
I=Ú
1
Ú
1- x 2
0 0
Ú
dz dy dx
1 x 2 + y2
p sec q 4 f =0 q =0 r =0 p
=Ú2
Ú Ú
p p 2 4 0 0
=Ú
È
sec q
Ú ÍÎÚ0
x 2 + y2 + z2
r 2 sin q dr dq df r
r dr ˘˙ sin q dq df ˚
9.6
p
=Ú2
p 4 0
Triple Integrals
9.125
sec q
r2 2
0
Ú
p 2 0
È p sec 2 q ˘ sin q dq ˙ df ÍÚ 4 2 Î 0 ˚
sin q dq df 0
=Ú
p
=
1 2 df 2 Ú0
p 4 0
Ú
p
tan q sec q dq p
1 2 f sec q 04 2 0 1 pÊ p ˆ = . Á sec - sec 0˜ ¯ 2 2Ë 4 =
=
p 4
(
)
2 -1
type II Evaluation of triple Integrals Over the Given Region
Example 1 Evaluate ÚÚÚ x 2 yz dx dy dz over the region bounded by the planes x = 0, y = 0, z = 0 and x + y + z = 1. Solution 1. Draw an elementary volume AB parallel to z-axis in the region. AB starts from xy-plane and terminates on the plane x + y + z = 1. z
y
R (0, 0, 1)
Q (0, 1) x+y+z=1 B′ x
B Q (0, 1, 0)
O A x+
y=
y
1
P (1, 0, 0) x
Fig. 9.131
O
A′
+
y
=
1
P (1, 0) x
9.126
Chapter 9
Multiple Integrals
Limits of z : z = 0 to z = 1 – x – y 2. Projection of the plane x + y + z = 1 in xy-plane is DOPQ. Putting z = 0 in x + y + z = 1, the equation of the line PQ is obtained as x + y = 1. 3. Draw a vertical strip A¢B¢ in the region OPQ. A¢B¢ starts from the x-axis and terminates on the line x + y = 1. Limits of y : y = 0 to y = 1 – x Limits of x : x = 0 to x = 1 I=Ú
1 1- x 1- x - y 2 x yz dz dy dx 0
Ú Ú
0 0
=Ú
1 1- x
Ú
0 0
x2 y
z2 2
1- x - y
dy dx 0
{
} }
1- x 1 1 = Ú x 2 ÈÍ Ú y (1 - x )2 + y 2 - 2 y(1 - x ) dy ˘˙ dx 0 0 Î ˚ 2 1- x 1 1 y(1 - x )2 + y3 - 2 y 2 (1 - x ) dy ˘˙ dx = Ú x 2 ÈÍ Ú ˚ 2 0 Î 0
{
1 1 y2 y 4 y3 = Ú x 2 (1 - x )2 + - 2(1 - x ) 2 0 2 4 3
1- x
dx 0
=
2 1 1 2È (1 - x ) 4 (1 - x )3 ˘ 2 (1 - x ) 1 x ( x ) ◊ + - 2(1 - x ) ◊ Í ˙ dx Ú 2 0 Î 2 4 3 ˚
=
1 1 x2 (1 - x )4 dx 2 Ú0 12
(1 - x )7 1 (1 - x )5 2 (1 - x )6 = ◊x ◊ 2x + ◊2 -210 24 -5 30 =
1 Ê 1 ˆ 0+ 24 ÁË 105 ˜¯
=
1 2520
1
0
Example 2 Evaluate
ÚÚÚ 2x dV ,
where E is the region under the plane
E
2x + 3y + z = 6 that lies in the first octant.
[Winter 2015]
Solution 1. Draw an elementary volume AB parallel to z-axis in the region. AB starts from xy-plane and terminates on the plane 2x + 3y + z = 6. Limits of z : z = 0 to z = 6 – 2x – 3y
9.6
Triple Integrals
9.127
2. Projection of the plane 2x + 3y + z = 6 in xy-plane is DOPQ. Putting z = 0 in 2x + 3y + z = 6, the equation of the line PQ is obtained as 2x + 3y = 6. z
y
R (0, 0, 6)
Q (0, 2) B′ B Q (0, 2, 0)
O
y
A x+
y=
O
A′
P (3, 0) x
1
P (3, 0, 0) x
Fig. 9.132
3. Draw a vertical strip A¢B¢ in the region OPQ. A¢B¢ starts from the x-axis and terminates on the line 2x + 3y = 6. 6 - 2x 3
Limits of y : y = 0
to
y=
Limits of x : x = 0
to
x=3
I = ÚÚÚ 2 x dV
E 6 -2 x 3 6 -2 x -3 y 3 2 x dz dy dx 0 0 0
=Ú
Ú
Ú
6 -2 x 3 3 0 0
=Ú
Ú
6 -2 x 3 0 0
=2Ú
3
=2Ú
3
0
Ú
6 -2 x -3 y
2x z 0
dy dx
x(6 - 2 x - 3 y) dy dx
3 y2 x (6 - 2 x ) y 2
6 -2 x 3
dx 0
2 3 È (6 - 2 x ) 3 Ê 6 - 2 x ˆ ˘ ˙ dx = 2Ú x Í(6 - 2 x ) - Á 0 3 2 Ë 3 ˜¯ ˙ ÍÎ ˚
9.128
Chapter 9
Multiple Integrals
3
= 2Ú x 0
(6 - 2 x )2 dx 6
4 3 = Ú x (9 + x 2 - 6 x )dx 3 0 4 3 = Ú (9 x + x 3 - 6 x 2 ) dx 3 0 x3 4 x2 x4 + -6 = 9 4 3 3 2
3
0
=9
Example 3 Evaluate ÚÚÚ xyz dx dy dz over the positive octant of the sphere x2 + y2 + z2 = 4. Solution Putting x = r sin q cos f, y = r sin q sin f, z = r cos q, the equation of the sphere x2 + y2 + z2 = 4 reduces to r2 sin2q cos2f + r2 sin2q sin2f + r2 cos2 q = 4, r2 = 4, r = 2.
z
The region is the positive octant of the sphere r = 2. Limits of r : r = 0 to r = 2 Limits of
q=0
to
θ=
Limits of
f=0
to
φ=
r
P
q y
O φ
π 2 π
x
2
Fig. 9.133
Hence, the spherical form of the given integral is I = ÚÚÚ xyz dx dy dz p
p
0
0
2
= Ú 2 Ú 2 Ú (r 3 sin 2 q cos q cos f sin f )r 2 sin q dr dq df 0
p
p
0
0
= Ú 2 sin 3 q cos q dq Ú 2 sin 4 q = 4
p 2 0
- cos 2f 4
p 2 0
2 sin 2f df Ú r 5 dr 0 2
r6 6
2
0
˘ È [ f (q )]n +1 n , n π 1˙ ÍQ Ú [ f (q )] f ¢(q )dq = n 1 + ˚ Î
9.6
=
1Ê 4 p ˆÈ 1 ˘ Ê 26 ˆ sin - sin 0˜ Í - (cos p - cos 0)˙ Á ˜ Á ¯Î 4 4Ë 2 ˚Ë 6 ¯
=
4 3
Triple Integrals
9.129
Example 4 Evaluate
dx dy dz
ÚÚÚ
over the region bounded by the sphere
a - x 2 - y2 - z2 x2 + y2 + z2 = a2. 2
Solution 1. Putting x = r sin q cos f, y = r sin q sin f, z = r cos q, the equation of the sphere x2 + y2 + z2 = a2 reduces to r = a. 2. For the complete sphere, limits of r : r = 0 to r = a limits of q : q = 0 to q = p limits of f : f = 0 to f = 2p Hence, the spherical form of the given integral is dx dy dz
I = ÚÚÚ =Ú
2p
=Ú
2p
2
a - x 2 - y2 - z2 p
0
0
a r2
Ú0 Ú0
sin q dr dq df a2 - r 2
p
a
0
0
df Ú sin q dq Ú
r 2 + a2 - a2 a2 - r 2
dr
ˆ aÊ a2 2p p = f 0 ◊ - cos q 0 ◊ Ú Á - a 2 - r 2 ˜ dr 0 ¯ Ë a2 - r 2 2
= (2p )(- cos p + cos 0) a sin
-1
Ê ˆ a sin -1 1˜ = 2p (2) Á a 2 sin -1 1 Ë ¯ 2 2
Ê a2 ˆ = 4p Á sin -1 1˜ Ë 2 ¯ = 4p ◊
a2 p ◊ 2 2
= p 2 a2
r r a2 r a2 - r 2 sin -1 a 2 a 2
a
0
9.130
Chapter 9
Multiple Integrals
Example 5 Evaluate
dx dy dz
∫∫∫
1 over the region bounded by the spheres ( x2 + y 2 + z 2 ) 2 x2 + y 2 + z 2 = a2 and x2 + y 2 + z 2 = b2, a > b > 0.
z
Solution 1. Putting x = r sinq cos f, y = r sinq sin f, z = r cosq, equations of the spheres x2 + y2 + z2 = a2 and x 2 + y 2 + z 2 = b2 reduce to r = a and r = b respectively. 2. Draw an elementary radius vector OAB from the origin in the region. This radius vector enters in the region from the sphere r = b and terminates on the sphere r = a. 3. Limits of r : r = b to r = a. For the complete sphere, limits of q : q = 0 to q = p limits of f : f = 0 to f = 2p Hence, the spherical form of the given integral is dx dy dz I = ÚÚÚ
B A O
r=a r=b y
x
Fig. 9.134 1
( x 2 + y2 + z2 ) 2 =Ú
2p
=Ú
2p
0
0
p
a r2
Ú0 Úb
sinq dr dq df r
p
a
0
b
df Ú sin q dq Ú r dr
2p
p
= f 0 ◊ - cos q 0 ◊
r2 2
= 2p (- cos p + cos 0)
a
b
(a 2 - b2 ) 2
= 2p (aa 2 - b2 ).
Example 6 Evaluate
ÚÚÚ z
2
dx dy dz over the region common to the sphere
x2 + y2 + z2 = 4 and the cylinder x2 + y2 = 2x.
9.6
Triple Integrals
9.131
Solution 1. Putting x = r cos q, y = r sin q, z = z, the equation of (i) the sphere x 2 + y 2 + z 2 = 4 reduces to r2 + z2 = 4 z 2 = 4 – r 2. (ii) the cylinder x2 + y 2 = 2x reduces to r 2 = 2r cosq, r = 2 cosq. 2. Draw an elementary volume parallel to z-axis in the region. This elementary volume starts from the part of the sphere z2 = 4 – r 2, below xyplane and terminates on the part of the sphere z2 = 4 – r 2, above xy-plane.
q=p 2
A r = 2cos q O
q=0
Fig. 9.135
Limits of r : z = - 4 - r 2 to z = 4 - r 2 3. Projection of the region in rq -plane is the circle r = 2 cosq. 4. Draw an elementary radius vector OA in the region (r = 2 cosq ) which starts from the origin and terminates on the circle r = 2 cosq Limits of r : r = 0 to r = 2 cos q p p Limits of q : q = to q = 2 2 Hence, the cylindrical form of the given integral is I = ÚÚÚ z 2 dx dy dz =Ú
p 2 cosq 4 -r 2 2 -p 0 - 4 -r 2 2
=Ú
p 2 cosq 2 p 0 2
Ú
Ú
Ú
z3 3
z 2 r dz dr dq
4 -r 2
r dr dq - 4 -r 2
p
=
3
1 2 2 cosq 2(4 - r 2 ) 2 r dr dq 3 Ú- p Ú0 2
=
1 3Ú
=-
=-
p 2 cosq 2 p 0 2
1 3Ú
Ú
p 2 p 2
2 15 Ú
p 2 p 2
3 È ˘ ÍÎ -(4 - r 2 ) 2 ( -2r )dr ˙˚ dq
5 2 2 2( 4 - r )
5
2 cosq
dq 0
È [ f (r )]n +1 ˘ n ÍQ Ú [ f (r )] f ¢(r )dr = ˙ n +1 ˚ Î
5 5˘ È dq ÍÎ(4 - 4 cos2 q ) 2 - (4) 2 ˙˚
9.132
Chapter 9
Multiple Integrals
p
2 = - Ú 2p (25 sin 5 q - 25 ) dq 15 2
ÈQ a f (q )dq = 0, if f (-q ) = - f (q )˘ Í Ú- a ˙ Í ˙ 5 5 ÎHere sin (-q ) = - sin q ˚
p ˘ 2 È = - Í0 - 2 5 q 2 p ˙ 15 Í - ˙ 2˚ Î
26 p 15 64p = 15 =
Example 7 Evaluate ÚÚÚ xyz dx dy dz over the region bounded by the planes x = 0, y = 0, z = 0, z = 1 and the cylinder x2 + y 2 = 1. Solution 1. Putting x = r cosq, y = r sinq, z = z, equation of the cylinder x2 + y2 = 1 reduces to r 2 = 1, r = 1. z
B
q= p 2
z=1
A′ r=1
O
y
O
q=0
A
x
Fig. 9.136
2. Draw an elementary volume AB parallel to z-axis in the region. This elementary volume AB starts from xy-plane and terminates on the plane z = 1. Limits of z : z = 0 to z = 1
9.6
Triple Integrals
9.133
3. Projection of the region in rq -plane is the part of the circle r = 1 in the first quadrant. 4. Draw an elementary radius vector OA¢ in the region in the rq -plane which starts from the origin and terminates on the circle r = 1. Limits of r : r = 0
to
r=1 p Limits of q : q = 0 to q = 2 Hence, the cylindrical form of the given integral is I = ÚÚÚ xyz dx dy dz =Ú
p 1 2 r2 z=0 q =0 r =0 1
Ú Ú p
1
= Ú z dz Ú 2 0
z2 = 2
0
=
1 sin 2q dq Ú r 3 d r 0 2
1
cos 2q 4 0
cos q sin q ◊ zr dz dr dq
p 2 0
r4 4
1
0
1 16
Example 8 Evaluate
∫∫∫
x 2 + y 2 dx dy dz over the region bounded by the right
circular cone x2 + y2 = z2, z > 0 and the planes z = 0 and z = 1. Solution 1. Putting x = r cosq, y = r sinq, z = z, the equation of the cone x 2 + y 2 = z 2 reduces to r 2 = z 2, r = z. 2. Draw an elementary volume AB parallel to z-axis in the region, which starts from the cone r = z and terminates on the plane z = 1. Limits of z : z = r to z = 1. 3. Projection of the region in rq -plane is the curve of intersection of the cone r = z and the plane z = 1 which is obtained as r = 1, a circle with centre at the origin and radius 1. 4. Draw an elementary radius vector OA¢ in the region which starts from the origin and terminates on the circle r = 1. Limits of r : r = 0 to r = 1 Limits of q : q = 0 to q = 2p
9.134
Chapter 9
Multiple Integrals
z
z=1
B
q= p 2
r=z A′
A
r=1 y
O
O
q=0
x
Fig. 9.137
Hence, the cylindrical form of the given integral is I = ÚÚÚ x 2 + y 2 dx dy dz =Ú
2p
=Ú
2p
=Ú
2p
1
Ú Ú
1
q = 0 r =0 z =r
0
0
1 2
Ú0 r
r ◊ r dz dr dq
1
z r dr dq
1
dq Ú r 2 (1 - r )dr 0
2π
= θ0
r3 r4 − 3 4
1
0
1 = 2π ⋅ 12 π = 6
Example 9 Evaluate ÚÚÚ ( x 2 + y 2 )dx dy dz over the region bounded by the paraboloid x2 + y2 = 3z and the plane z = 3. Solution 1. Putting x = r cos q, y = r sin q, z = z, the equation of the paraboloid x2 + y2 = 3z reduces to r 2 = 3z. 2. Draw an elementary volume AB parallel to z-axis in the region which starts from the paraboloid r 2 = 3z and terminates on the plane z = 3.
9.6
Triple Integrals
9.135
r2 to z = 3 3 3. Projection of the region in rq -plane is the curve of intersection of the paraboloid r 2 = 3z and the plane z = 3 which is obtained as r 2 = 9, r = 3, a circle with centre at the origin and radius 1. Limits of z : z =
z
B
q= p 2
z=3
A′
r 2 = 3z A
q y
O
O
r=3 q=0
x
Fig. 9.138
4. Draw an elementary radius vector OA¢ in the region (circle r = 3) which starts from origin and terminates on the circle r = 3. Limits of r : r = 0 to r = 3 Limits of q : q = 0 to q = 2p Hence, the cylindrical form of the given integral is I = ÚÚÚ ( x 2 + y 2 )dx dy dz =Ú
2p
=Ú
2p
=Ú
2p
0
0
0
=q
3 3
Ú0 Ú r
2
3 3 3 r 0
Ú
r 2 ◊ r d z d r dq z
3 r2 3
d r dq
Ê 3 r2 ˆ dq ◊ Ú r 3 Á 3 - ˜ dr 0 3¯ Ë
2p 0
3r 4 r 6 4 18
3
0
Ê 35 36 ˆ = 2p Á - ˜ Ë 4 18 ¯ 81p = 2
9.136
Chapter 9
Multiple Integrals
Example 10 Evaluate
ÚÚÚ V
x 2 y2 z2 1 - 2 - 2 - 2 dx dy dz, where V is the volume of the a b c
x 2 y2 z2 ellipsoid 2 + 2 + 2 = 1. a b c Solution It is difficult to integrate this integral in cartesian form. Therefore, transforming the ellipsoid into a sphere using following change of variables. x 2 y2 z2 y z x = u, = v, = w, equation of the ellipsoid 2 + 2 + 2 = 1 reduces to a b c a b c 2 2 2 u + v + w = 1, which is a sphere of radius 1 and centre at the origin,
Putting
dx dy dz = |J| du dv dw
where,
∂x ∂u ∂( x, y, z ) ∂y J= = ∂(u, v, w) ∂u ∂z ∂u
∂x ∂v ∂y ∂v ∂z ∂v
∂x ∂w ∂y ∂w ∂z ∂w
a 0 0 = 0 b 0 = abc 0 0 c Therefore, dx dy dz = abc du dv dw New form of the integral is I = Ú Ú Ú 1-
x2 a2
-
y2 b2
-
z2 c2
dx dy dz
= Ú Ú Ú 1 - u2 - v 2 - w 2 ◊ abc du dv dw Since in the new coordinate system u, v, w, the region of integration is a sphere, therefore using spherical coordinates u = r sinq cos f, v = r sin q sin f, w = r cosq and du dv dw = r2 sin q dr dq df, the equation of the sphere u2 + v2 + w2 = 1 reduce to r2 =1, r = 1. For complete sphere limits of r : r = 0 to r = 1 (radius of sphere) limits of θ : θ = 0 to θ = π limits of φ : φ = 0 to φ = 2π
9.6
Triple Integrals
9.137
Hence, the spherical form of the given integral is I=Ú
2p
0
p
1
2p
df Ú sin q dq Ú r 2 1 - r 2 dr
Ú0 Ú0
= abc Ú
0
Putting r = sin t,
1 - r 2 abc ◊ r 2 sin q dr dq df p
1
0
0
dr = cos t dt
When r = 0,
t=0
When r = 1,
t=
p 2 2p
p
\ I = abc f 0 - cos q 0
p 2 0
Ú
sin 2 t ◊ cos t ◊ cos t dt
1 Ê 3 3ˆ = abc (2p )(2) ◊ B Á , ˜ 2 Ë 2 2¯ 3 3 2 2 = 2p abc 3 Ê 1 1ˆ Á 2 2˜ Ë ¯ = 2p abc 2 2 p abc = 4
2
Example 11 Evaluate ÚÚÚ x 2 y 2 z 2 dx dy dz over the region bounded by the surfaces xy = 4, xy = 9, yz = 1, yz = 4, zx = 25, zx = 49. Solution Evaluation of integral becomes easier by changing the variables. Under the transformation xy = u, yz = v, zx = w, the surfaces get transformed to u = 4, u = 9, v = 1, v = 4, w = 25, w = 49. These equations represent the planes parallel to v w, w u and uv planes in the new coordinate system. It is easier to find partial derivatives of u, v, w w.r.t. x, y and z. ∂u ∂x ∂(u, v, w) ∂v = ∂( x, y, z ) ∂x ∂w ∂x
∂u ∂y ∂v ∂y ∂w ∂y
∂u ∂z ∂v ∂z ∂w ∂z
9.138
Chapter 9
Multiple Integrals
y = 0 z
x 0 z y 0
x
= y( zx - 0) - x(0 - yz ) = 2 xyz du dv dw = J dx dy dz = 2 xyz dx dy dz
dx dy dz =
1 du dv dw 2 xyz
1 du dv dw 2 uvw Limits of u : u = 4 to u = 9 Limits of v : v = 1 to v = 4 Limits of w : w = 25 to w = 49 Hence, the new form of the integral is
[Q x 2 y 2 z 2 = uvw]
=
I = ÚÚÚ x 2 y 2 z 2 dx dy dz =Ú
49
4
9
uvw ◊ w = 25 Úv =1 Úu = 4 2 1
=
1
1
du dv dw
uvw 1
4 9 1 49 2 w dw Ú v 2 dv Ú u 2 du Ú 1 4 2 25 3 2w 2
49
3 2v 2
4
3 2u 2
9
1 2 3 25 3 1 3 4 4 = (343 - 125)(8 - 1)(27 - 8) 27 115976 = 27 =
ExERCIsE 9.7 (I) Evaluate the following integrals: 1.
1
2
0
0
∫ dx ∫
2
dy ∫ x 2 yz dz 1
Ans. :1 2.
log 2
x
x +y
0
0
0
∫ ∫ ∫
e x + y + z d z dy d x 5 Ans. : 8
9.6
π 2 0
3.
∫ ∫
4.
∫ ∫
5.
6.
7.
a cos θ
0
a2 −r 2
∫
π
a(1+ cos θ )
0
0
4
2 z
0
0
∫
r 2 1 − r dz dr d θ 1 + a ( cos θ )
4 z−x2
0
a sin θ
0
a −r a
∫
2
y
x +y
0
0
x −y
a3 π 2 Ans. : − 3 2 3
π a 2h Ans. : 2
dy dx dz
2
r dz dr dθ
0
∫∫ ∫
9.139
Ans. : 8π 2
∫ ∫
h
0
∫∫ ∫ π 2 0
r dz dr dθ
0
Triple Integrals
5a 3 Ans. : 64
(x + y + z)dz dx dy Ans. :16
8.
a
∫∫ 0
0
a2 − x 2
∫
a2 − x 2 − y 2
0
xyz dz dy dx.
a6 Ans. : 48
(II) Evaluate the following integrals over the given region of integration: 1.
∫∫∫ (x + y + z)dx dy dz
over the tetrahedron bounded by the planes
x = 0, y = 0, z = 0 and x + y + z = 1. 1 Ans. : 8 2.
dx dy dz
∫∫∫ (1 + x + y + z)
3
over the tetrahedron bounded by the planes x = 0,
y = 0, z = 0 and x + y + z = 1.
3.
∫∫∫ xyz dx dy dz
1 5 Ans. : log 2 − 2 8
over the positive octant of the sphere x2 + y2 + z2 = a2. a6 Ans. : 48
4.
∫∫∫ xyz(x
2
+ y 2 + z 2 )dx dy dz over the positive octant of the sphere
x 2 + y 2 + z 2 = a2.
a8 Ans. : 64
9.140
5.
Chapter 9
∫∫∫ (y
2
Multiple Integrals
z 2 + z 2 x 2 + x 2 y 2 ) dx dy dz over the sphere of radius a and centre
at the origin.
6.
7.
∫∫∫ x
2
z2 dx dy dz over the sphere x 2 + y 2 + z 2 = 2. + y 2 + z2
dx dy dz
∫∫∫
3 2 2
4π a 7 Ans. : 35 8π 2 Ans. : 9
over the region bounded by the spheres x2 + y2 + z2 = a2
(x + y + z ) and x2 + y 2 + z2 = b2, a > b > 0. 2
2
a Ans. : 4π log b 8.
∫∫∫ z dx dy dz 2
over the region common to the spheres x 2 + y 2 + z 2 = a 2
and cylinder x 2 + y 2 = ax.
9.
∫∫∫ (x
2
+ y 2 )dx dy dz over the region bounded by the paraboloid x2 + y2 = 2z
and the plane z = 2.
∫∫∫ x
2π a 5 Ans. : 15
16π Ans. : 3
2
10.
y z dx dy dz over the tetrahedron bounded by the planes x = 0, x y z y = 0, z = 0 and + + = 1. a b c a 3 b 2c 2 Ans. : 2520
11.
∫∫∫ xyz dx dy dz 2
2
over the positive octant of the ellipsoid
2
x y z + 2 + 2 ≤ 1. 2 a b c
12.
a 2 b 2c 2 Ans. : 48
x 2 y 2 z2 ∫∫∫ 1 + 4 + 9 dx dy dz over the region bounded by the ellipsoid x 2 y 2 z2 + + = 1. 1 4 9 Ans. : 8π
9.7 Area by Double Integrals
9.141
9.7 AREA by DOublE IntEGRAls 9.7.1 Area in Cartesian Coordinates (i) The area A bounded by the curves y = y1(x) and y = y2(x) intersecting at the points P (a, b) and Q (c, d ) is c a
∫
y2 ( x ) y1 ( x )
d y dx
y=
y2
(x
)
d
∫
b
x2 ( y ) x1 ( y )
y=
y
1
(ii) If equation of the curves are represented as x = x1(y) and x = x2(y) then A=∫
Q (c, d )
(x )
A=∫
y
P (a, b)
dx dy
Note: Consider the symmetricity of the region while calculating area.
x
O
Fig. 9.139
Example 1 Find the area bounded by the ellipse
x2 y2 + = 1, above x-axis. a2 b2
Solution 1. The region is symmetric about y-axis. Total area = 2 (area bounded by the ellipse in the first quadrant) 2. Draw a vertical strip AB in the region which lies in the first quadrant. AB starts from the x-axis and terminates on the x2 y 2 ellipse 2 + 2 = 1. a b x2 Limits of y : y = 0 to y = b 1 − 2 a Limits of x : x = 0 to x = a A = 2∫
a
0 a
∫
b 1−
0 b 1−
0
= 2∫ b 1 − 0
=
Q (0, b)
O
B
A
x2 a2
= 2∫ y 0 a
y
dy dx
x2 a2
dx
x2 dx a2
2b a 2 a − x 2 dx a ∫0
Fig. 9.140
2 x2 + y =1 2 b2 a
x P (a, 0)
9.142
Chapter 9
Multiple Integrals
a
2b x 2 a2 x a − x 2 + sin −1 a 2 a0 2 2 2b a −1 = sin 1 a 2 =
2b a 2 π ⋅ a 2 2 π ab = 2
=
Example 2 Find the area bounded by the parabola y 2 = 4x and the line 2x –3y + 4 = 0. Solution 1. The points of intersection of the parabola y2 = 4x and the line 2x – 3y + 4 = 0 are obtained as 2 2x + 4 = 4x 3 ( x + 2) 2 = 9 x
y
x2 − 5x + 4 = 0 x = 1, 4
B
∴ y = 2, 4
A
The points of intersection are P (1, 2) and Q (4, 4). 2. Draw a vertical strip AB which starts from the line 2x – 3y + 4 = 0 and terminates on the parabola y2 = 4x. 2x + 4 Limits of y : y = to y = 2 x 3 Limits of x : x = 1 to x = 4 A = 2∫
4
1
4
∫
2 x 2 x+4 3
2x – 3y + 4 = 0
3
1
2x + 4 2 x − dx 3 3 2
x
O y 2 = 4x
Fig. 9.141
2 x
1
=∫
P (1, 2)
dy dx
= ∫ y 2 x + 4 dx 4
Q (4, 4)
4 2
= 2 ⋅ 2 ⋅ x − x − 4x 3 3 3 1 4 1 4 = (8 − 1) − (16 − 1) − (4 − 1) 3 3 3 1 = 3
9.7 Area by Double Integrals
9.143
Example 3 Find the area enclosed by the curves y = x2 and y = x. Solution y
1. The points of intersection of the parabola y = x2 and the line y = x are obtained as x = x2 x = 0, 1 \ y = 0, 1 The points of intersection are O (0, 0) and P (1, 1). 2. Draw a vertical strip AB which starts from the parabola y = x2 and terminates on the line y = x. Limits of y : y = x2 to y = x Limits of x : x = 0 to x = 1 A=∫
1
0
∫
x x2
y =x
P(1,1) y = x2
B A O
x
Fig. 9.142
d y dx
1
x
0
x
= ∫ y 2 dx 1
= ∫ ( x − x 2 ) dx 0
=
x 2 x3 − 2 3
1
0
1 1 = − 2 3 1 = 6
Example 4 Find the area enclosed by the parabola y2 = 4ax and the lines x + y = 3a, y = 0 in the first quadrant. Solution 1. The points of intersection of the parabola y2 = 4ax and the line x + y = 3a are obtained as y2 = 4a(3a – y) y2 + 4ay – 12a2 = 0 y = 2a, – 6a \ x = a, 9a The point of intersection is Q (a, 2a) which lies in the first quadrant. 2. Area enclosed in the first quadrant is OPQ.
9.144
Chapter 9
Multiple Integrals
Draw a horizontal strip AB which starts from the parabola y2 = 4ax and terminates on the line x + y = 3a. y y2 4a Limits of y : y = 0 Limits of x : x = A=∫
2a
∫
3a − y
to x = 3a – y
0
x
2
y 4a
y 2 = 4ax
Q(a, 2a)
d x dy
y2 0 4a 3a − y 2a
=∫
x + y = 3a
to y = 2a
A
dy
B
P(3, 0) x
O
2a y2 = ∫ 3a − y − dy 0 4a
y 2 1 y3 = 3ay − − ⋅ 2 4a 3
2a
0
Fig. 9.143
1 8a 3 = 6a 2 − 2a 2 − ⋅ 4a 3 10 2 = a 3
Note: In case of vertical strip, two vertical strips are required to cover the entire region. Therefore one horizontal strip is preferred over vertical strip.
Example 5 Find the area bounded by the parabolas y2 = 4ax and x2 = 4ay. Solution 1. The points of intersection of the parabolas y2 = 4ax and x2 = 4ay are obtained as 2
x2 = 4ax 4a x4 = 16a2 (4ax) 3 x(x – 64a3) = 0 x = 0, x = 4a \ y = 0, y = 4a The points of intersection are O (0, 0) and P (4a, 4a). 2. Draw a vertical strip AB which starts from the parabola x2 = 4ay and terminates on the parabola y2 = 4ax. x2 Limits of y : y = to y = 2 ax 4a Limits of x : x = 0 to x = 4a
y y 2 = 4ax B
P(4a,4a)
x 2 = 4ay A O
Fig. 9.144
x
9.7 Area by Double Integrals
A=∫
4a
=∫
4a
0
0
∫
2 ax x2 4a
dy dx
2 ax
y
x2 4a
9.145
dx
4a x2 = ∫ 2 ax − dx 0 4a 4a
2 3 1 x3 = 2 a ⋅ x2 − ⋅ 3 4a 3
0
3 4 1 = a ( 4a ) 2 − ( 4a )3 3 12a 32 2 16 2 = a − a 3 3 16 2 = a 3
Example 6 Find the area enclosed by the curves y = 2 – x and y2 = 2(2 – x). Solution 1. The points of intersection of the line y = 2 – x and the parabola y2 = 2(2 – x) are obtained as y (2 – x)2 = 2(2 – x) (2 – x)(2 – x – 2) = 0 (2 – x)(–x) = 0 x = 2, 0 y = 0, 2
Q (0, 2) y =2−x
B y 2 = 2 (2 −x) A
The points of intersection are P (2, 0) and Q (0, 2). 2. Draw a vertical strip AB which starts from the line y = 2 – x and terminates on the parabola y2 = 2 (2 – x). Limits of y : y = 2 – x to y = 2(2 − x) Limits of x : x = 0 to x = 2 A=∫
2
0 2
∫
=∫ y 0
2( 2− x )
2− x
d y dx
2( 2− x ) 2− x
dx
1 2 = ∫ 2 ( 2 − x ) 2 − ( 2 − x ) dx 0
O
Fig. 9.145
P (2, 0) x
9.146
Chapter 9
Multiple Integrals
=
2⋅
2
3 2
x 2( 2 − x ) − 2x + −3 2
2
0
8 1 = 0 + − 2( 2 − 0 ) + ( 4 − 0 ) 3 2 =
2 3
Example 7 Find the area bounded between the parabolas x2 = 4ay and x2 = – 4a (y – 2a). Solution 1. The parabola x2 = 4ay has vertex (0, 0) and the parabola x2 = – 4a(y – 2a) has vertex (0, 2a). Both the parabolas are y symmetric about the y-axis. 2. The points of intersection of Q (0, 2a) x2 = 4ay and x2 = – 4a(y – 2a) are B x 2 = 4ay obtained as 4ay = −4a ( y − 2a ) R P (2a, a) (−2a, a)
8ay = 8a 2 y=a
x 2 = −4a (y − 2a) O
A
x
∴ x = ± 2a The points of intersection are P (2a, a) and R (–2a, a). 3. The region is symmetric about y-axis.
Fig. 9.146
Total area = 2 (Area in the first quadrant) 4. Draw a vertical strip AB in the region which lies in the first quadrant. AB starts from the parabola x2 = 4ay and terminates on the parabola x2 = – 4a(y – 2a). x2 x2 Limit of y : y = to y = 2a − 4a 4a Limits of x : x = 0 to x = 2a A = 2∫
2a
= 2∫
2a
0
0
∫
2a −
x2 4a
x2 4a
y
2a − x2 4a
d y dx
x2 4a
dx
2a x2 x2 = 2∫ 2 a − − dx 0 4a 4a
9.7 Area by Double Integrals
9.147
2a
x3 6a 0 4 = 2 4a 2 − a 2 3 16 = a2 3 = 2 2ax −
Example 8 Find smaller of the area enclosed by the curves y = 2 – x and x2 + y2 = 4. Solution 1. The points of intersection of the line y = 2 – x and the circle x2 + y2 = 4 are obtained as x2 + (2 – x)2 = 4 x + 4 – 4x + x2 = 4 2x2 = 4x x = 2, 0 \ y = 0, 2
y
2
Q (0, 2) B y
x2 + y2 = 4
=
The points of intersection are P (2, 0) and Q (0, 2). 2. Draw a vertical strip AB which starts from the line y = 2 – x and terminates on the circle x2 + y2 = 4.
2 − x
A O
Limits of y : y = 2 – x to y = 4 − x 2 Limits of x : x = 0
to x = 2
A=∫
2
0 2
∫
4 − x2 2− x
=∫ y 0
dy dx
Fig. 9.147
4− x2 2− x
dx
= ∫ 4 − x 2 − ( 2 − x ) dx 0 2
=
x 4 x x2 4 − x 2 + sin −1 − 2 x + 2 2 2 2
2
0
= 2 sin −1 1 − 2 =π −2
Example 9 Find the area of the loop of the curve x(x2 + y2) = a (x2 – y2).
P (2, 0) x
9.148
Chapter 9
Multiple Integrals
Solution
a − x The equation of the curve can be rewritten as y 2 = x 2 a + x 1. The points of intersection of the curve with x-axis ( y = 0) are obtained as x2(a – x) = 0 x = 0, x = a. y The loop of the curve lies between the points O (0, 0) and P (a, 0). 2. The region is symmetric about x-axis Total area = 2 (Area above x-axis) 3. Draw a vertical strip AB in the region above x-axis. AB starts from x-axis and a − x terminates on the curve y 2 = x 2 . a + x Limits of y : y = 0 to
y=x
Limits of x : x = 0
x=a
to
A = 2∫
a
0
∫
a− x a+ x
x
0 x
= 2∫ y 0 a
0
= 2∫ x a
0
Putting When When
B P (a, 0) O x=−a
a−x a+ x
dy dx
a−x a+ x
dx
a−x dx a+ x
x = a cosq, dx = – a sin q dq π x = 0, θ = 2 x = a, q = 0 0
A = 2 Ú p a cos q 2
a - a cos q (- a sin q )dq a + a cos q
q 2 dq = Ú cos q sin q ◊ 2 q 2 cos 2 q p sin q q 2 dq = 2 a 2 Ú 2 cos q ◊ 2 sin cos ◊ 0 q 2 2 cos 2 p 2a 2 2 0
2 sin 2
p
= 2 a 2 Ú 2 cos q (1 - cos q ) dq 0
A
Fig. 9.148
x
9.149
9.7 Area by Double Integrals
p
1 + cos 2q ˆ Ê = 2 a 2 Ú 2 Á cos q ˜¯ dq 0 Ë 2 q sin 2q = 2 a sin q - 2 4
p 2
2
0
ÈÊ p ˘ ˆ 1Êp ˆ 1 = 2 a 2 ÍÁ sin - sin 0˜ - Á - 0˜ - (sin p - sin 0 )˙ Ë ¯ Ë ¯ 2 2 2 4 Î ˚ Ê pˆ = 2a2 Á 1 - ˜ Ë 4¯
Example 10 Find the area included between the curve y2(2a – x) = x3 and its asymptote. [Summer 2017] Solution The equation of the curve can be rewritten as x3 2a - x 1. The point of intersection of the curve with x-axis (y = 0) is x = 0. 2. The region is symmetric about x-axis. Total area = 2 (Area above x-axis) 3. Draw a vertical strip AB in the region above x-axis. AB starts from x-axis and terminates on the curve x3 . y2 = 2a - x
y
y2 =
Limits of y : y = 0 to y = x
x 2a - x
Limits of x : x = 0 to x = 2 a A = 2Ú
2a
= 2Ú
2a
= 2Ú
2a
0
x
Ú0
x
y
0
0
x 2a - x
x
0
dy dx
x 2a - x
dx
x dx 2a - x
B
O
A
Fig. 9.149
Asymptote
x x = 2a
9.150
Chapter 9
Multiple Integrals
x = 2a sin2q, dx = 2a (2 sinq cosq dq) When x = 0, q = 0 p When x = 2 a, q = 2 Putting
p
2 a sin 2 q
A = 2 Ú 2 2 a sin 2 q
2 a cos2 q
0
p
= 2 Ú 2 2 a sin 2 q ◊ 0
◊ 4 a sin q cos q dq
sin q ◊ 4 a sin q cos q dq cos q
p
= 16 a 2 Ú 2 sin 4 q dq 0
= 16 a 2 ◊
3 1 p ◊ ◊ 4 4 4
= 3p a 2
Example 11 Find the area between the rectangular hyperbola 3xy = 2 and the line 12x + y = 6. Solution 1. The points of intersection of the rectangular hyperbola 3xy = 2 and the line 12x + y = 6 are obtained as y 3 x (6 − 12 x) = 2 Q
18 x 2 − 9 x + 1 = 0
1 , 4 6
1 1 x= , 3 6 ∴ y = 2, 4 The points of intersection are 1 1 P , 2 and Q , 4 . 3 6 2. Draw a vertical strip AB in the region which starts from the rectangular hyperbola 3xy = 2 and terminates on the line 12x + y = 6. 2 Limits of y : y = to y = 6 – 12x 3x 1 1 Limits of x : x = to x = 6 3
12x + y = 6 B 3xy = 2
A
P 1, 2 3 O
x
Fig. 9.150
9.151
9.7 Area by Double Integrals
1
6 −12 x
A = ∫ 13 ∫ 2 6 1
= ∫ 13 y 6
d y dx
3x 6 −12 x 2 3x
dx
1
2 = ∫ 13 6 − 12 x − dx 3x 6 1 3 2 = 6 x − 6 x − log x 1 3 2
6
1 1 1 1 2 = (2 − 1) − 6 − − log − log 9 36 3 3 6 1 2 = − log 2 2 3
Example 12
2
2
3 3 Find the area bounded by the hypocycloid x + y = 1. a b Solution
1. The hypocycloid is symmetric in all the quadrants. Total area = 4 (area in the first quadrant) 2. Draw a vertical strip AB parallel to y-axis in the region which lies in the first quadrant. AB starts from x-axis and terminates 2 3
2 3
y Q (0, b) x a B
O
x y on the curve + = 1. a b
2 3
A
3 2 2 x 3 to y = b 1 − a to x = a
Limits of y : y = 0 Limits of x : x = 0
3
A = 4∫
a
0
a
∫
2 2 x 3 b 1− a
0
= 4∫ y 0
d y dx
3 2 2 x 3
b 1− a 0
dx 3
2 2 a x 3 = 4∫ b 1 − dx 0 a
Fig. 9.151
+
y b
2 3
= 1
P (a, 0)
x
9.152
Chapter 9
Multiple Integrals
Putting
x = a cos3 t, dx = 3a cos2 t (–sin t) dt
When
π 2 x = a, t = 0
When
x = 0, t =
3
0
A = 4∫π b(1 − cos 2 t ) 2 ( −3a cos 2 t sin t )dt 2
π
= 12ab∫ 2 sin 4 t cos 2 t dt 0
1 Ê 5 3ˆ = 12 ab B Á , ˜ 2 Ë 2 2¯ 5 3 = 6 ab 2 2 4 3 1 1 1 1 ◊ ◊ 2 2 2 2 2 = 6 ab 3! 3 = p ab 8
ExERCIsE 9.8 1. Find the area bounded by y-axis, the line y = 2x and the line y = 4. [Ans. : 4] 2. Find the area bounded by the lines y = 2 + x, y = 2 – x and x = 5. [Ans. : 25] 2
3. Find the area bounded by the parabola y + x = 0, and the line y = x + 2. 9 Ans. : 2 4. Find the area bounded by the parabola x = y – y 2 and the line x + y = 0. 4 Ans. : 3 5. Find the area bounded by the curves y 2 = 4x and 2x – 3y + 4 = 0. 1 Ans. : 3
9.7 Area by Double Integrals
9.153
6. Find the area bounded by the parabola y = x2 – 3x and the line y = 2x. 125 Ans. : 6 7. Find the area bounded by the parabolas y2 = x, x2 = –8y. 8 Ans. : 3 8. Find the area bounded by the parabolas y = ax2 and y = 1 − a > 0.
x2 , where a
a 4 Ans. : 2 3 a + 1 a + x 9. Find the area of the loop of the curve y 2 = x 2 a − x 2 π Ans. : 2a − 1 4 10. Find the area of one of the loops of x4 + y4 = 2a2xy.
π a2 Ans. : 4
11. Find the area enclosed by the curve 9xy = 4 and the line 2x + y = 2. 1 4 Ans. : 3 − 9 log 2 12. Find the area of the smaller region bounded by the circle x2 + y2 = 9 and a straight line x = 3 – y. 2π 3 − Ans. : 4 2 3 13. Find the area bounded by the x-axis, circle x2 + y2 = 16 and the line y = x. [Ans. : 2p ] 14. Find the area bounded between the curves y = 3x2 – x – 3 and y = – 2x2 + 4x + 7. 45 Ans. : 2 2
2
2
15. Find the area bounded by the asteroid (x) 3 + (y ) 3 = (a) 3 . 3 2 Ans. : 8 π a
9.154
Chapter 9
Multiple Integrals
9.7.2 Area in Polar Coordinates
p 2
q =
The area A bounded by the curves r = r1 (q ), r = r2 (q ) and the lines q = q1 and q = q2 is A=∫
θ2
θ1
∫
r2 (θ )
r1 (θ )
q = q2 R q = q1
S
r dr dθ
q2
Note: Consider the symmetricity of the region while calculating the area.
q1 P
Q r = r2(q ) r = r1(q )
O
q =0
Fig. 9.152
Example 1 Find the area between the circles r = 2 sinq and r = 4 sinq. Solution 1. The region is symmetric about the line θ =
π . 2
Total area = 2 (area in the first quadrant) 2. Draw an elementary radius vector OAB from the origin in the region which lies in the first quadrant. OAB enters in the region from the circle r = 2sinq and terminates on at the circle r = 4 sinq. Limits of r : r = 2sinq to r = 4sinq π Limits of θ : θ = 0 to θ = 2 π
A = 2∫ 2 ∫ 0
π
= 2∫ 2 0
4 sin θ 2 sin θ
r2 2
P B r = 4 sin q Q
A
r dr dθ
r = 2 sinq q
4 sin θ
O
dθ 2 sin θ
π 2 0
= ∫ (16 sin 2 θ − 4 sin 2 θ ) dθ π
= ∫ 2 12 sin 2 θ dθ 0
π
= ∫ 2 6(1 − cos 2θ ) dθ 0
sin 2θ =6θ− 2
π 2
dθ
0
π sin π − sin 0 = 6 − 2 2 = 3π
q= p 2
Fig. 9.153
q =0
9.7 Area by Double Integrals
9.155
Example 2 Use double integral in polar form to find the area enclosed by the three petalled rose r = sin 3q. [Winter 2015] Solution 1. This curve consists of three similar loops. Total area = 3 (area of the loop in the first quadrant) 2. When r = 0, sin 3q = 0 3q = 0, p, 2p, 3p, ... p 2p , p , ... q = 0, , 3 3 Since, in the first quadrant, p r = 0 at q = 0, Fig. 9.154 3 p loop exists between q = 0 and q = . 3 3. Draw an elementary radius vector OA from the origin in the loop which lies in the first quadrant. OA starts from the origin and terminates on the curve = sin q. Limits of r : r = 0 to r = sin 3q Limits of q : q = 0 to q =
p 3 p
A = 3Ú 3 Ú
sin 3q
0
0
p 3 0
r2 2
= 3Ú
r dr dq sin 3q
dq 0
p
=
3 3 2 sin 3q dq 2 Ú0
=
3 3 Ê 1 - coss 6q ˆ ˜¯ dq 2 Ú0 ÁË 2
p
sin 6q 3 = q4 6
p 3 0
3Êp 1 ˆ = Á - sin 2p ˜ ¯ 4Ë 3 6 =
3Êpˆ 4 ÁË 3 ˜¯
=
p 4
9.156
Chapter 9
Multiple Integrals
Example 3 Find the area of the crescent bounded by the circles r = 3 and r = 2cosq. Solution p 2 r = √3
q=
1. The points of intersection of r = 3 and r = 2cosq are obtained as 3 = 2 cos θ
P
3 cos θ = 2 π θ=± 6 Hence, θ =
p 6
q=
A
r = 2 cosq B
q q =0
O
π at P. 6
q =−
p 6
2. The region is symmetric about the Fig. 9.155 initial line, q = 0. Area of the crescent = 2 (area above the initial line, q = 0) 3. Draw an elementary radius vector OAB from the origin in the region above the initial line. OAB enters in the region from the circle r = 3 and terminates on at the circle r = 2 cosq. Limits of r : r = 3 to r = 2 cos θ π Limits of θ : θ = 0 to θ = 6 π
A = 2∫ 6 ∫
2 cos θ
0
3
π
r2 2
= 2∫ 6 0
r dr dθ
2 cos θ
dθ 3
π
= ∫ 6 (4 cos 2 θ − 3)dθ 0
π
= ∫ 6 [2(1 + cos 2θ ) − 3]dθ 0
π
6 sin 2θ −θ = 2 2 0
π π − 3 6 3 π = − 2 6 = sin
9.157
9.7 Area by Double Integrals
Example 4 Find the area which lies inside the circle r = 3a cosq and outside the cardioid r = a (1 + cosq ). q=
Solution
p 2
1. The points of intersection of the circle r = 3a cos q and the cardioid r = a (1 + cos q ) are obtained as
q=
p 3 r = 3a cosq
R
B A
3a cos θ = a (1 + cos θ ) 1 cos θ = 2 π θ=± 3
Q
q O
P r = a (1 + cosq )
q =0
π p q =− at R. 3 3 2. The region is symmetric about the iniFig. 9.156 tial line q = 0. Total area = 2 (area above the initial line) 3. Draw an elementary radius vector OAB from the origin in the region above the initial line. OAB enters in the region from the cardioid r = a (1 + cosq ) and terminates on the circle r = 3a cosq. Limits of r : r = a (1 + cosq ) to r = 3a cosq π Limits of θ : θ = 0 to θ = 3 Hence, θ =
p
A = 2Ú 3 Ú
3 a cosq
0
a (1+ cosq )
p 3 0
r2 2
= 2Ú
r dr dq
3 a cosq
dq a (1+ cosq )
p
= Ú 3 ÈÎ9a 2 cos2 q - a 2 (1 + cos q )2 ˘˚ dq 0
p
= Ú 3 ÈÎ8a 2 cos2 q - a 2 - 2 a 2 cos q ˘˚ dq 0
p
= a 2 Ú 3 [ 4(1 + cos 2q ) - 1 - 2 cos q ] dq 0
p 3 4 sin 2q = a 3q + - 2 sin q 2 0 2
9.158
Chapter 9
Multiple Integrals
2p pˆ Ê p = a 2 Á 3 + 2 sin - 2 sin ˜ Ë 3 3 3¯ = p a2
Example 5 Find the area lying inside the circle r = a sin q and outside the cardioid r = a (1 – cos q). [Summer 2016] Solution 1. The points of intersection of circle r = a sinq and the cardioid r = a (1 – cosq) are obtained as a sinq = a(1 – cosq) 2 sin
q q q cos = 2 sin 2 2 2 2
sin
q q = 0, tan = 1 2 2 q q p = 0, = 2 2 4 p q = 0, q = 2
Fig. 9.157
p at P. 2 2. Draw an elementary radius vector OAB from origin in the region. OAB enters in the region from the cardioid r = a(1 – cos q) and terminates on the circle r = a sinq. Hence, q = 0 at origin and q =
Limit of r: r = a(1 – cos q) to r = a sin q Limit of q: q = 0 to q = p
A=Ú2Ú
a sin q
0
a (1- cosq )
p
r2 2
=Ú2 0
p 2
r dr dq
a sin q
dq a (1- cosq )
p 2 0
=
1 Èa 2 sin 2 q - a 2 (1 - cos q )2 ˘ dq ˚ 2Ú Î
=
a2 2
p 2 0
Ú
Èsin 2 q - (1 - 2 cos q + cos2 q )˘ dq Î ˚
9.7 Area by Double Integrals
=
a2 2
p 2 0
Ú
9.159
Èsin 2 q - cos2 q + 2 cos q - 1˘ dq Î ˚
p p p ˘ a2 È 2 Í Ú (- cos 2q )dq + 2 Ú 2 cos q dq - Ú 2 dq ˙ 0 0 ˚ 2 Î 0 p È p p ˘ a 2 Í sin 2q 2 2 = + 2 sin q 0 - q 02 ˙˙ Í 2 Î 2 0 ˚
=
=
a2 2
p p˘ È 1 ÍÎ- 2 (sin p - sin q ) + 2(sin 2 - sin 0) - 2 ˙˚
a2 È p ˘ 2- ˙ 2 ÍÎ 2˚ Ê pˆ = a2 Á1 - ˜ Ë 4¯
=
Example 6 Find the area common to the cardioids r = a(1 + cosq ) and r = a(1 – cos q ). Solution 1. The points of intersection of the cardioids r = a (1 + cos θ ) and r = a (1 − cos θ ) are obtained as a (1 + cosθ ) = a (1 − cosθ ) cosθ = 0 π θ=± 2 Hence, θ =
q= r = a (1 − cosq )
p 2 r = a (1 + cosq )
P A
O
q =0
π at P. 2
2. The region is symmetric in all the quadrants Total area = 4 (area in the first quadrant)
Fig. 9.158
3. Draw an elementary radius vector OA from the origin in the region which lies in the first quadrant. OA starts from the origin and terminates on the cardioid r = a(1 – cos q ).
9.160
Chapter 9
Multiple Integrals
Limits of r : r = 0 to r = a (1 − cos θ ) Limits of θ : θ = 0 to θ =
π 2
p
a (1- cosq )
A = 4Ú 2 Ú 0
0
p
r2 2
= 4Ú 2 0
r dr dq
a (1- cosq )
dq 0
p 2 0
= 2 Ú a 2 (1 - cos q )2 dq p
= 2 a 2 Ú 2 (1 - 2 cos q + cos2 q )dq 0
p
1 + cos 2q ˆ Ê = 2 a 2 Ú 2 Á 1 - 2 cos q + ˜¯ dq 0 Ë 2 3 sin 2θ = 2a θ − 2 sin θ + 2 4 2
π 2 0
3π = 2a 2 − 2 4
Example 7 Find the area inside the cardioid r = 3(1 + cos θ ) and outside the 3 parabola r = . 1 + cos θ Solution 1. The points of intersection of the cardioid r = 3(1 + cos θ ) and the 3 parabola r = are ob1 + cos θ tained as
P
3 3(1 + cos θ ) = (1 + cos θ )
O
(1 + cos θ ) 2 = 1 cos θ = 0
π θ=± 2 π Hence, θ = at P. 2
q=
p 2
B A r = 3 (1 + cosq ) q =0
r=
3 1 + cosq
Fig. 9.159
9.7 Area by Double Integrals
9.161
2. The region is symmetric about the initial line q = 0. Total area = 2 (area above the initial line) 3. Draw an elementary radius vector OAB from the origin in the region above the 3 initial line q = 0. OAB enters in the region from the parabola r = and 1 + cos θ terminates on the cardioid r = 3(1 + cos θ ). 3 Limits of r : r = to r = 3(1 + cos q ) 1 + cos q Limits of θ : θ = 0 p
A = 2Ú 2 Ú 0
p 2 0
= 2Ú
3(1+ cosq ) r 3 1+ cosq
r2 2
to θ =
π 2
dr dq
3(1+ cosq )
dq 3 1+ cosq
p È ˘ 1 = Ú 2 9 Í(1 + cos q )2 ˙ dq 2 0 (1 + cos q ) ˚ Î p
1 + cos 2q 1 È ˘ = 9 Ú 2 Í1 + 2 cos q + ˙ dq 2 0 2 Ê 2 qˆ ˙ Í ÁË 2 cos 2 ˜¯ ˙ ÍÎ ˚ p È3 q˘ cos 2q 1 Ê qˆ = 9 Ú 2 Í + 2 cos q + - Á 1 + tan 2 ˜ sec 2 ˙ dq 0 2 ¯ Ë 2 4 2 2˚ Î
p È3 cos 2q 1 q 1 q Ê1 q ˆ˘ = 9 Ú 2 Í + 2 cos q + - sec 2 - ◊ tan 2 Á sec 2 ˜ ˙ dq 0 2 Ë 2 4 2 2 2 2 2¯˚ Î
q tan 3q sin 2q 1 q 1 2 =9 + 2 sin q + - ◊ 2 tan 2 4 4 2 2 3 3
p 2
0
È [ f (q )]n+1 ˘˙ ÍQ Ú [ f (q )]n f ¢(q )dq = n +1 ˙ Í Î ˚ p sin p 1 p 1 3 pˆ Ê 3p = 9Á + 2 sin + - tan - tan ˜ Ë 4 2 4 2 4 6 4¯ Ê 3p 4 ˆ = 9Á + Ë 4 3 ˜¯
9.162
Chapter 9
Multiple Integrals
Example 8 Find the area common to both the circles r = cos q and r = sin q. [Winter 2013] Solution q=
1. So the point of intersection of the circles r = cos q and r = sin q is obtained as
p 2 r = sin q
cos q = sin q
P
q=
p 4
B
tan q = 1 p q = 4
A O
r = cos q q =0
p Hence, q = is the point of intersection 4 2. The point of intersection divides the region into two subregions OAP and OBP. Fig. 9.160 3. Draw an elementary radius in each subregion. (i) In subregion OAP, radius vector OA starts from the origin and terminates on the circle r = sin q. Limit of r: r = 0 to r = sin q p Limit of q: q = 0 to q = 4 (ii) In the subregion OBP, the radius vector OB starts from the origin and terminates on the circle r = cos q. Limit of r: r = 0 to r = cos q p p Limit of q: q = to q = 4 2 A=
p 4
Ú Ú 0
=
Ú
p 4
0
sin q
0
r2 2
r dr dq +
sin q
dq + 0
1È = Í 2Í Î =
1È Í 2Í Î
Ú
p 4
0
Ú
p 4
0
2
Ú
p 2
Ú Ú 0
p 2 p 4
sin q dq +
Ú
cosq
r dr dq
0
r2 2
cosq
dq 0
p 2 p 4
˘ cos2 q dq ˙ ˙ ˚
Ê 1 - cos 2q ˆ ÁË ˜¯ dq + 2
Ú
p 2 p 4
Ê 1 + cos 2q ˆ ˘ ÁË ˜¯ dq ˙ 2 ˙ ˚
9.7 Area by Double Integrals
=
1È Í 4Í Î
Ú
p 4
(1 - cos 2q ) dq +
0
È 1Í sin 2q = Íq 4 2 ÎÍ
p 4 0
Ú
p 2 p 4
9.163
˘ (1 + cos 2q ) dq ˙ ˙ ˚
sin 2q +q+ 2
=
1 ÈÊ p 1 ˆ Ê p p 1 ˆ ˘ + - - ˙ Í 4 ÎÁË 4 2 ˜¯ ÁË 2 4 2 ˜¯ ˚
=
1Êp ˆ -1 4 ÁË 2 ˜¯
=
p 1 8 4
˘ ˙ p ˙ ˙ 4˚ p 2
Example 9 Find the area common to the circles r = cos θ and r = 3 sin θ . Solution 1. The point of intersection of the circles r = cos θ and r = 3 sin θ is obtained as 3 sin θ = cos θ q=
1 tan θ = 3 π θ= 6
p 2
q=
π Hence, θ = at P. 6
B
2. Divide the region OAPBO into two subregions OAP and OBP. Draw an elementary radius vector in each subregion. (i) In subregion OAP, radius vector OA starts from the origin and terminates on the circle r = 3 sin θ. Limits of r : r = 0 to r = 3 sin θ Limits of θ : θ = 0 to θ =
π 6
p 6
P r = √3 sin q A
O
r = cos q q =0
Fig. 9.161
9.164
Chapter 9
Multiple Integrals
(ii) In the subregion OBP, the radius vector OB starts from the origin and terminates on the circle r = cos q. Limits of r : r = 0 to r = cos q Limits of q : q =
p 6
to q =
p 2 p
p
A=Ú6Ú 0
3 sin q
0
r dr dq + Ú p2 Ú
cosq
r dr dq
0
6 p 6 0
=Ú
3 sin q
r2 2
dq + Ú 0
p 2 p 6
r2 2
cosq
dq 0
p
p
1 1 = Ú 6 3 sin 2 q dq + Ú p2 cos2 q dq 0 2 2 6 p
p
3 Ê 1 - cos 2q ˆ 1 Ê 1 + cos 2q ˆ = Ú6Á dq + Ú p2 Á ˜ ˜¯ dq 0 ¯ 2 Ë 2 2 Ë 2 6
sin 2q 3 = q4 2
p 6 0
sin 2q 1 + q+ 4 2
p 2 p 6
=
3Êp 1 pˆ 1Êp p 1 1 pˆ - sin ˜ + Á - + sin p - sin ˜ Á Ë ¯ Ë 4 6 2 3 4 2 6 2 2 3¯
=
5p 3 24 4
Example 10 Find the area common to the circle r = a and the cardioid r = a(1 + cos q ). Solution 1. The points of intersection of the circle r = a and the cardioid r = a (1 + cos q ) are obtained as a = a(1 + cos q ) cos q = 0 p q=± 2 π Hence, θ = at Q. 2
9.7 Area by Double Integrals
9.165
2. The region is symmetric about the initial line q = 0. Total area = 2 (area above the initial line) 3. Divide the region OPQO above the initial line into two subregions OPQ and OBQ. Draw an elementary radius vector in each subregion. (i) In the subregion OPQ the radius vector OA starts from the origin and terminates on the circle r = a. Limits of r : r = 0 to r = a q=
p 2
Q A
B
r=a
r = a (1 + cos q )
O
P
q =0
Fig. 9.162
π 2 (ii) In the subregion OBQ, radius vector OB starts from the origin and terminates on the cardioid r = a (1 + cos q ). Limits of r : r = 0 to r = a (1 + cos θ ) π Limits of θ : θ = to θ = π 2 Limits of θ : θ = 0 to θ =
Ê p a ˆ p a (1+ cosq ) A = 2 Á Ú 2 Ú r dr dq + Ú p Ú r dr dq ˜ 0 Ë 0 0 ¯ 2 Ê p 2 a p r2 r = 2Á Ú 2 dq + Ú p 2 ÁË 0 2 0 2 p 2 0
p
a (1+ cosq )
0
= Ú a 2 dq + Ú p a 2 (1 + coss q )2 dq 2
ˆ dq ˜ ˜¯
9.166
Chapter 9
Multiple Integrals
p
pÊ 1 + cos 2q ˆ = a 2 q 02 + a 2 Ú p Á 1 + 2 cos q + ˜¯ dq Ë 2 2
= a2 .
=
p 3q sin 2q + a2 + 2 sin q + 4 2 2
p a 2 3a 2 + 2 2
p p 2
pˆ p ˆ a2 Ê 2Ê p 2 sin p sin (sin 2p - sin p ) + a + ÁË ÁË 2 ˜¯ 2 ˜¯ 4
Ê 5p ˆ = a2 Á - 2˜ Ë 4 ¯
Example 11 Find the area between the curve r = a (secq + cosq ) and its asymptote r = a sec q.
q = p 2
r = a (sec q + cos q )
Solution 1. The region is symmetric about the initial line q = 0 Total area = 2(area above the initial line) 2. Draw an elementary radius vector OAB in the region above the initial line. OAB enters in the region from the line r = a sec q and terminates on the curve r = a (secq + cosq ). Limits of r : r = a sec q to r = a (sec q + cos q ) π Limits of q : q = 0 to θ = 2 p
A = 2Ú 2 Ú 0
p
a (sec q + cosq ) a sec q
r2 2
= 2Ú 2 0
r dr dq
a (sec q + cosq )
dq a sec q
p 2 [ a 2 (sec q 0
=Ú
+ cos q )2 - a 2 sec 2 q ]dq
p
= a 2 Ú 2 (cos2 q + 2)dq 0
B r = a sec q
A O
q =0
Fig. 9.163
9.7 Area by Double Integrals
È1 Ê 3 1ˆ = a 2 Í B Á , ˜ + 2q ÍÎ 2 Ë 2 2 ¯
p 2 0
9.167
˘ ˙ ˙˚
È 3 1 ˘ Í1 2p ˙ ˙ = a2 Í 2 2 + 2 ˙ Í2 2 ÍÎ ˙˚ 5p 2 a = 4
Example 12 Find the area of the loop of the curve x 4 + y 4 = 8 xy. Solution 1. The equation of the curve in polar form is r 4 (cos 4 θ + sin 4 θ ) = 8r 2 cos θ sin θ r2 =
8 cos θ sin θ cos 4 θ + sin 4 θ
2. Draw an elementary radius vector OA from the origin in the region which lies in the first quadrant. OA starts from the origin and terminates 8 cos θ sin θ on the curve r 2 = . cos 4 θ + sin 4 θ Limits of 8 cos θ sin θ r : r = 0 to r = cos 4 θ + sin 4 θ π Limits of θ : θ = 0 to θ = 2 8 cosq sin q p 4 cos q + sin 4 q 2 0 0
A=Ú
Ú
r dr dq
8 cosq sin q p 2 0
=Ú
r2 2
cos4 q + sin 4 q
dq
0
=
p 2 0
1 Ê 8 cos q sin q ˆ dq 2 Ú ÁË cos4 q + sin 4 q ˜¯
π tan θ sec 2 θ = 4∫ 2 dθ 0 1 + tan 4 θ
q=
p 2
q =
A
r2= O
Fig. 9.164
p 4
8 cos q sin q cos4 q + sin4 q
q =0
9.168
Chapter 9
Multiple Integrals
Putting tan 2 θ = t , 2 tan θ sec 2 θ dθ = dt When
θ = 0, t = 0
When
θ=
π ,t → ∞ 2
A = 2Ú
•
0
dt 1 + t2
= 2 tan -1 t
• 0
Êpˆ = 2Á ˜ Ë 2¯ =p
ExERCIsE 9.9 1. Find the area common to the circles r = a and r = 2acos q. 3 2 2π − Ans. : a 2 3 2. Find the area of the crescent bounded by the circles r = 2 and r = 2 cos θ. Ans. :1 3. Find the area which lies inside the cardioid r = 2a (1 + cos q ) and 2a outside the parabola r = . 1 + cos θ 16a 2 2 Ans. : 3π a + 3 4. Find the area bounded between the circles r = 2a sin q, r = 2b sin q (b > a). Ans. : π(b 2 − a 2 ) 5. Find the area outside the circle r = a and inside the cardioid r = a(1 + cos q ). a2 Ans. : (π + 8) 4
Points to Remember
9.169
Points to Remember Double Integrals The double integral of a function f (x, y) over the region R is denoted by ÚÚ f ( x, y)dx dy. R
Double Integrals over Rectangles and General Regions Method-I: When the region R is bounded by the curves y = y1(x), y = y2(x) and x = a, x = b, b È y2 ( x )
ÚÚ f ( x, y)dx dy = Úa ÍÎÚy ( x ) 1
R
f ( x, y)dy ˘˙ dx ˚
Method-II: When the region R is bounded by the curves x = x1(y), x = x2(y) and y = c, y = d, d È x2 ( y )
ÚÚ f ( x, y)dx dy = Úc ÍÎÚx ( y) 1
R
f ( x, y)dx ˘˙ dy ˚
If all the four limits are constant and f (x, y) is explicit, then the f (x, y) can be integrated w.r.t. any variable first and also can be written as product of two single integrals.
Change of Order of Integration Sometimes evaluation of double integral becomes easier by changing the order of integration. To change the order of integration, 1. Draw the region of integration with the help of the given limits. 2. Draw vertical or horizontal strip as per the required order of integration 3. Find the limits of integration b È y2 ( x )
Úa ÍÎÚy ( x ) 1
d x2 ( y ) f ( x, y)dy ˘˙ dx = Ú ÈÍ Ú f ( x, y)dx ˘˙ dy c Î x1 ( y ) ˚ ˚
Double Integrals in Polar Coordinates Putting x = r cos q, y = r sin q,
ÚÚ f ( x, y)dy dx = ÚÚ f (r cos q , r sin q ) J dr dq where Hence,
Jacobian, J = r
ÚÚ f ( x, y)dy dx = ÚÚ f (r cos q , r sin q )r dr dq
Triple Integrals The triple integral of a continuous function f (x, y, z) over a region V is denoted by ÚÚÚ f ( x, y, z)dx dy dz. V
9.170
Chapter 9
Multiple Integrals
Triple Integrals in Cartesian Coordinates If the region V is bounded below by a surface z = z1 (x, y) and above by a surface z = z2 (x, y) and if the projection of region V in xy-plane is R which is bounded by the curves y = y1 (x), y = y2 (x) and x = a, x = b then b È y2 ( x )
{
z2 ( x , y )
ÚÚÚ f ( x, y, z)dx dy dz = Úa ÍÎÚy ( x ) Úz ( x, y) 1
V
1
}
˘ f ( x, y, z )dz dy ˙ dx ˚
Triple Integrals in Cylindrical Coordinates Putting x = r cos q, y = r sin q, z = z,
ÚÚÚ f ( x, y, z)d x dy dz = ÚÚÚ f (r cos q , r sin q , z) J dz dr dq where
Jacobian, J = r
Hence, ÚÚÚ f ( x, y, z )dx dy dz = ÚÚÚ f (r cos q , r sin q , z )r dz dr dq
Triple Integrals in Spherical Coordinates Putting x = r sin q cos f, y = r sin q sin f, z = r cos q,
ÚÚÚ f ( x, y, z)dx dy dz = ÚÚÚ f (r sin q cos f, r sin q sin f, r cos q ) J dr dq df where Hence,
Jacobian,J = r 2 sin q
ÚÚÚ f ( x, y, z)dx dy dz = ÚÚÚ f (r sin q cos f , r sin q sin f , r cos q ) ◊ r 2
sin q dr dq df
Area by Double Integrals Area in Cartesian Coordinates (i) The area bounded by the curves y = y1(x) and y = y2(x) intersecting at the points P (a, b) and Q (c, d ) is A=Ú
c a
y2 ( x )
Úy ( x ) dy dx 1
(ii) The area bounded by the curves x = x1(y) and x = x2(y) and intersecting at the points P (a, b) and Q (c, d ) is A=Ú
c
y2 ( x )
dy dx a Úy ( x ) 1
Area in Polar Coordinates The area bounded by the curves r = r1(q ), r = r2(q ) and the lines q = q1 and q = q2 is A=Ú
q2
q1
r2 (q )
Úr (q ) r dr dq 1
Multiple Choice Questions
9.171
Multiple Choice Questions Select the most appropriate response out of the various alternatives given in each of the following questions: •
1. To evaluate
Ú0 Ú0 xe x
-
x2 y
dx dy by change of order of integration, the lower limit
for the variable x is equal to (a) y2 (b) 0 2.
(c) •
(d) y
4 3 2
Ú0 Ú0 Ú0 dx dy dz = (a) 9
(b) 24
(c) 1
(d) 0 2 ex
Ú0 Ú1
3. By changing the order of integration, the integral the double integral _______. e 2
Ú1 Úlog y dx dy
(a)
1
log y
Úe Ú2
(c)
2
(b) (d)
dx dy
e2
2
e2
log y
dy dx is equivalent to
Ú1 Úlog y dx dy Ú1 Ú2
4. By changing to spherical polar co-ordinates,
dx dy
ÚÚÚ dy dx dz, where R is the region R
of hemisphere x 2 + y 2 + z 2 = a 2 is equivalent to triple integral ______ 2p
p p a 2 2 2 r 0 0 0
Ú Ú Ú
(c) 5.
p a 2 2 r 0 0
Ú0 Ú Ú
(a)
x
(a)
sin q dr dq df
(d)
Ú Ú Ú
(c)
a4 48
y
a6 24
p
(b)
Ú0 Ú0 Ú0 xyz dz dy dx a
2p
Ú0 Ú0 Ú0 r
sin q dr dq df
(b)
a 2
p p a 2 2 2 r 0 0 0
sin q dr dq df cos q dr dq df
= a6 48
(d)
a4 24
6. In evaluating Ú Ú xy ( x + y) dx dy over the region between y = x2 and y = x, the limits are (a) x = 0 to 1, xy = 0 to 1 (b) x = 0 to 1, y = 0 to x (c) x = 0 to 1, y = 0 to x2 (d) x = 0 to 1, y = x2 to x 7.
Ú0 Ú0 a
(a)
a2 - y2
p a4 8
( a 2 - x 2 - y 2 ) dx dy = (b)
p a4 4
(c)
p a4 2
(d) pa4
9.172
Chapter 9
Multiple Integrals
8. After transforming to polar co-ordinates p 1 -r 2 2 e dr 0 0
Ú Ú
(a)
9.
p • -r 2 2 e r 0 0
(c)
Ú Ú
p
a cosq
Ú0 Ú0 (a)
Ú0 Ú0 xy (a)
11.
2
Ú Ú
Ú Ú
(b)
dr dq
(d)
Ú Ú
d r dq
p • - y2 2 e r 0 0
dr dq
(b)
a2 3
(c)
a2 2
(d)
a2 6
(b)
1 3
(c)
2 3
(d)
4 3
(c)
p 2
(d) 2
dy dx =
5 3
p p 2 2 0 0
p 1 -r 2 2 e r 0 0
dq
dx dy =
r sin q dr dq =
a2 4
1 2
10.
• • - ( x 2 + y2 )
Ú0 Ú0 e
sin ( x + y) dx dy is
(a) 0
(b) p
12. The value of the integral Ú Ú xy dx dy over the region bounded by the x-axis, ordinate at x = 2a and the parabola x2 = 4ay is (a)
a4 3
(b)
13. The triple integral
a4 5
(c)
ÚÚÚ dx dy dz
a4 7
(d)
a4 9
gives
R
(a) volume
(b) area 1 1
14. The value of (a)
4 35
1- x
Ú0 Úy Ú0 2
(b)
p 3
(b)
3 35
p 2
(b)
(c) 2p
p
p 3
1 2
Ú0 Ú03 Ú0 r
p 6
16. The value of the integral (a)
(d) density
x dz dx dy is
15. The value of the integral (a)
(c) surface area
8 35
(d)
6 35
(d)
p 4
(d)
p 6
sin q dr dq df is (c)
• • - x 2 (1 + y2 )
Ú0 Ú0 e
(c)
2p 3 x dx dy is p 4
Multiple Choice Questions
9.173
• 2
17. Changing the order of integration in the double integral to
s q
(b) 16y2
(c) x
18. The limits of integration of y = x2 and y2 = x are (a) x = 0 to 1, y = x2 to x (c) x = y2 to y , y = 0 to 1
ÚÚ
xy 1- y
ÚÚr
3
Ú Ú (x
2
(d) 8
2
+ y ) dx dy over the domain bounded by (b) x = 0 to 1, y = 0 to 1 (d) x = 0 to y, y = x to x2
dx dy over the positive quadrant of the circle x2 + y2 = 1 is
1 6
(a) 20.
2
2 3
(b)
5 6
(c)
(d)
5 3
dr dq over the region included between the circles r = 2 sinq and
r = 4 sinq is p
(a)
(c)
p
4sin q 3 r 2sin q
Ú0 Ú
(b)
dr dq
p
4sin q 3 r - p 2sin q
Ú Ú
(d)
dr dq
21. On converting into polar co-ordinates (a) (c)
leads
4
Úr Ú p f ( x, y) dx dy , then q is
(a) 4y
19.
Ú0 Ú x f ( x, y) dy dx
p 2 r2 0 0 p a 4 r3 0 0
p 4sin q 3 2 r 0 sin q
Ú Ú
Ú0 Ú0 a
4sin q 3
Ú02 Ú2sinq r
a2 - x 2
dr dq
dr dq
( x 2 + y 2 ) dy dx =
p
Ú Ú
dr d q
(b)
Ú0 Ú02 r
Ú Ú
dr dq
(d)
Ú0 Ú
a
a
a
3
p 4 r2 0
dr dq dr d q
22. In spherical co-ordinates, dx dy dz is equal to (a) r dq df dr (d) r sinq dq df dr (c) r2 sinq dq df dr (d) r2 dq df dr 23. The value of the integral (a) 1 24. The value of (a) 4pa2
(b)
Ú0 dx Ú0 a
(b)
1 1 1
Ú0 Ú0 Ú0 ( x
1 3 a2 - x 2
+ y 2 + z 2 ) dz dy dx is (c)
dy Ú
0
p a3 6
2
a2 - x 2 - y2
2 3
(d) 3
dz is
(c) 4pa3
(d)
p 2 a 3
9.174
Chapter 9
Multiple Integrals
25. The transformations x + y = u, y = uv transform the area element dy dx into J du dv, where J is equal to (a) 1
(b) u
26. The value of
ÚÚ x
3
(d) u2
(c) –1
y dx dy, where R is region enclosed by the ellipse
R
x2 a2
+
y2 b2
=1
in the first quadrant is b2 a 4 24
(a)
b3 a 4 24
(b)
(c)
y
(c)
Ú0 Ú1
2p
28.
0
Ú1 Ú1
Ú0
1 y
(d)
Ú0 Ú0
f ( x, y)dy dx =
f ( x, y)dx dy
(b)
Ú1 Úy f ( x, y)dx dy
f ( x, y)dx dy
(d)
Ú0 Úy f ( x, y)dx dy
27. By changing the order of integration, (a)
1 x
ba 4 24
b2 a 2 24
0 1
1 1
1
dq Ú e2r dr is equal to 0
(a) e2 – 1 29. The value of
(b)
ÚÚ x
p 2 (e - 1) 2
(c) p(e2 – 1)
(d) 2p(e2 – 1)
2 3
y dx dy, where R is the region bounded by the rectangle
R
0 £ x £ 1 and 0 £ y £ 3 is 27 4
(a)
(b)
27 8
(c)
29 4
(d)
29 8
30. The value of Ú Ú 3 y dx dy over the triangle with vertices (–1, 1), (0, 0) and (1, 1) is [Winter 2015] (a) 0 (b) 1 (c) 2 (d) 3 31. The area of the curve y = x2 + 1 bounded by the x-axis and the line x = 1 and x = 2 is [Summer 2014] 3 10
(a)
(b)
10 3
(c) 6
(d)
1 6
32. The equation of a cylindrical surface x2 + y2 = 9 becomes ____ when converted to cylindrical polar coordinates. [Summer 2016] 2 (a) r = 9 (b) r = 9 (c) r = ± 3 (d) r = 3 33.
2
x2
Ú0 Ú0
y
[Summer 2016]
e x dydx is equal to
(a) e2 – 1
(b) e2
(c) e2 + 1
(d) e–2
Multiple Choice Questions
1 dxdy = xy (a) 0 (b) (log 2)2
9.175
2 2
34.
Ú1 Ú1
[Winter 2016] (c) 1
(d) log 2
4 6
35. The region of
Ú1 Ú2 dx dy
[Winter 2016]
represents
(a) rectangle (b) square
(c) circle
(d) triangle
2 2
36. The region
Ú1 Ú1 dx dy
(a) rectangle (b) square 37. The value of (a) 0
1 1
Ú0 Ú0 (3 x
2
[Summer 2017]
represents (c) circle
(d) triangle
- 2 y 2 ) dx dy is
(b) 1
[Summer 2017]
(c) –1
(d)
1 3
Answers 1. (d) 10. (d) 19. (a) 28. (c) 37. (d)
2. (b) 11. (d) 20. (a) 29. (a)
3. (b) 12. (a) 21. (b) 30. (c)
4. (a) 13. (a) 22. (c) 31. (b)
5. (b) 14. (a) 23. (a) 32. (d)
6. (d) 15. (a) 24. (b) 33. (a)
7. (a) 16. (c) 25. (b) 34. (b)
8. (c) 17. (a) 26. (a) 35. (a)
9. (b) 18. (a) 27. (d) 36. (a)
UNIT-6 Chapter 10. Matrices
CHAPTER
10
Matrices chapter outline 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13 10.14 10.15 10.16
10.1
Introduction Matrix Some Definitions Associated with Matrices Elementary Row Operations in Matrix Row Echelon and Reduced Row Echelon Form of a Matrix Rank of a Matrix Inverse of a Matrix by Gauss–Jordan Method System of Non-homogeneous Linear Equations System of Homogeneous Linear Equations Eigenvalues and Eigenvectors Properties of Eigenvalues Linear Dependence and Independence of Eigenvectors Properties of Eigenvectors Cayley-Hamilton Theorem Similarity Transformation Diagonalization of a Matrix
IntroductIon
A matrix is a rectangular table of elements which may be numbers or any abstract quantities that can be added and multiplied. Matrices are used to describe linear equations, keep track of the coefficients of linear transformation and record data that depend on multiple parameters. There are many applications of matrices in mathematics, viz. graph theory, probability theory, statistics, computer graphics, geometrical optics, etc.
10.2
10.2
Chapter 10
Matrices
MAtrIX
A set of mn elements (real or complex) arranged in a rectangular array of m rows and n columns is called a matrix of order m by n, written as m × n. An m × n matrix is usually written as a11 a12 a 21 a22 A = … … … … am1 am 2
a1n a2 n … … … amn m × n … … … …
The matrix can also be expressed in the form A = [aij ]m× n , where aij is the element in the i th row and j th column, written as (i, j)th element of the matrix.
10.3
SoMe defInItIonS ASSocIAted wIth MAtrIceS
1. row Matrix A matrix having only one row and any number of columns, is called a row matrix or row vector, e.g. [2 5 -3 4] 2. column Matrix A matrix, having only one column and any number of rows, is called a column matrix or column vector, e.g. 1 −3 4 3. Zero or null Matrix A matrix, whose all the elements are zero, is called zero or null matrix and is denoted by 0, e.g. 0 0 0 0 0 , 0 0 0 0 0 0 0 0 4. Square Matrix A matrix, in which the number of rows is equal to the number of columns, is called a square matrix, e.g.
10.3 Some Definitions Associated with Matrices 10.3
1 3 2 2 3 1 4 , −1 4 −5 2 6 8 5. diagonal Matrix A square matrix, whose all non-diagonal elements are zero and at least one diagonal element is non-zero, is called a diagonal matrix. e.g. 1 0 0 2 0 0 4 , 0 4 0 0 0 8 6. unit or Identity Matrix A diagonal matrix, whose all diagonal elements are unity, is called a unit or identity matrix and is denoted by I, e.g. 1 0 0 1 0 0 1 , 0 1 0 0 0 1 7. Scalar Matrix A square matrix, whose all diagonal elements are equal, is called a scalar matrix, e.g. 2 0 0 3 0 0 3 , 0 2 0 0 0 2 8. upper triangular Matrix A square matrix, in which all the elements below the diagonal are zero, is called an upper triangular matrix, e.g. 2 1 3 0 4 −5 0 0 8 9. Lower triangular Matrix A square matrix, in which all the elements lower triangular matrix, e.g. 1 0 −1 4 2 6
above the diagonal are zero, is called a 0 0 8
10.4
Chapter 10
Matrices
10. trace of a Matrix The sum of all the diagonal elements of a square matrix is called the trace of a matrix, e.g.
0 2 −1 A = 4 6 − 2 −1 0 3 Trace of A = 2 + 6 + 3 = 11
11. transpose of a Matrix A matrix obtained by interchanging rows and columns of a matrix is called transpose of a matrix and is denoted by A′ or AT, e.g. If
1 −1 3 A = 0 2 6 then −4 1 5
1 0 −4 A = −1 2 1 3 6 5 T
i.e.,if A = [aij]m×n , then AT = [aji]n×m 12. Symmetric Matrix A square matrix A = [aij]m×m is called symmetric if aij = aji for all i and j, i.e., A = AT, e.g. 1 2 4 4 3 , i −3 i
i −3 i −2 4 4 3
13. Skew-Symmetric Matrix A square matrix A = [aij]m×m is called skew-symmetric if aij = -aji for all i and j, i.e., A = -AT. Thus, the diagonal elements of a skew-symmetric matrix are all zero, e.g. 0 −3i −4 3i 0 8 4 −8 0 14. conjugate of a Matrix A matrix obtained from any given matrix A, on replacing its elements by the corresponding conjugate complex numbers is called the conjugate of A and is denoted by A , e.g. 8 1 + 3i 2 + 5i A= , − i 6 9 − i
8 1 − 3i 2 − 5i A= i 6 9 + i
10.3 Some Definitions Associated with Matrices 10.5
15. transposed conjugate of a Matrix The conjugate of the transpose of a matrix A is called the conjugate transpose or transposed conjugate of A and is denoted by Aθ , e.g. Aq = ( A)T = ( AT ) 1 − 2i 2 + 3i 3 + 4i For example, If A = 4 − 5i 5 + 6i 6 − 7i , 8 7 + 8i 7 Then,
8 1 − 2i 4 − 5i A = 2 + 3i 5 + 6i 7 + 8i 3 + 4i 6 − 7i 7 T
8 1 + 2i 4 + 5i Aθ = 2 − 3i 5 − 6i 7 − 8i 3 − 4i 6 + 7 i 7
16. hermitian Matrix ___
A square matrix A = [aij] is called Hermitian if aij = a ji for all i and j, i.e., A = Aθ , e.g. 2 + 3i 3 − 4i 1 2 − 3i 0 2 − 7i i i 3 + 4 2 + 7 2 17. Skew-hermitian Matrix ___
A square matrix A = [aij] is called skew-Hermitian if aij = - a ji for all i and j, i.e., A = - Aq . Hence, diagonal elements of a skew-Hermitian matrix must be either purely
imaginary or zero, e.g. 2 + 3i i −2 + 3i 0 18. orthogonal Matrix A square matrix A is called orthogonal if AAT = I. 19. unitary Matrix A square matrix A is called unitary if AAq = I . 20. determinant of a Matrix If A is a square matrix, determinant of A is represented as | A |.
If
2 3 1 2 3 1 A = 1 2 3 , then | A | = 1 2 3 1 1 0 1 1 0
10.6
Chapter 10
Matrices
21. Singular and non-Singular Matrices A square matrix A is called singular if | A | = 0 and non-singular if | A | ≠ 0. 22. Minor of an element of a determinant a11
a12
a13
If | A |= a21 a31
a22 a32
a23 , minor of an element of a determinant is a determinant a33
obtained by leaving the row and column passing through that element, e.g., minor of element a11 = minor of element a13 =
a22
a23
a32
a33
a21
a22
a31
a32
, minor of element a12 =
a21
a23
a31
a33
,
23. cofactor of an element of a determinant Cofactor of an element aij of a determinant is the minor multiplied by (-1)i+j, e.g., Cofactor of the element a11 = (−1)1+1
a22
a23
a32
a33
Cofactor of the element a12 = (−1)1+ 2
a21
a23
a31
a33
Cofactor of the element a13 = (−1)1+ 3
a21
a22
a31
a32
24. Inverse of a Matrix If A is a square matrix and A ≠ 0 , AA−1 = I = A−1 A
where, A-1 is called inverse of the matrix A.
10.4
eLeMentAry row operAtIonS In MAtrIX
Elementary transformation is any one of the following operations on a matrix. (i) The interchange of any two rows (or columns) (ii) The multiplication of the elements of any row (or column) by any non-zero number (iii) The addition or subtraction of k items the elements of a row (or column) to the corresponding elements of another row (or column), where k π 0 Symbols to be used for elementary transformation: (i) Rij :
Interchange of i th and j th row
10.5 Row Echelon and Reduced Row Echelon Forms of a Matrix 10.7
(ii) kRi : Multiplication of i th row by a non zero number k (iii) Ri + kRj : Addition of k times the j th row to the i th row The corresponding column transformations are denoted by Cij, kCi and Ci + kCj respectively.
10.4.1
elementary Matrices
A matrix obtained from a unit matrix by subjecting it to any row or column transformation is called an elementary matrix.
10.4.2
equivalence of Matrices
If B be an m × n matrix obtained from an m × n matrix by elementary transformation of A, then A is called the equivalent to B. Symbolically, we can write A ~ B.
10.5
row echeLon And reduced row echeLon forMS of A MAtrIX
A matrix A is said to be in row echelon form if it satisfies the following properties: (i) Every zero row of the matrix A occurs below a non-zero row. (ii) The first non-zero number from the left of a non-zero row is a 1. This is called a leading 1. (iii) For each non-zero row, the leading 1 appears to the right and below any leading 1 in the preceding rows. The following matrices are in row echelon form. 1 1 0 1 2 −1 3 0 1 3 5 0 0 1 0 , 0 1 5 6 , 0 0 1 −1 0 0 0 0 0 0 1 4 0 0 0 0 1 A matrix A is said to be in reduced row echelon form if each column that contains a leading 1 in row echelon form of the matrix A has zeros everywhere else in that column. The following matrices are in reduced row echelon form. 0 0 0 0 ,
1 0 0 1 0 0 2 0 1 0 , 0 1 0 5 , 0 0 1 0 0 1 −1
0 0 0 0
1 − 4 0 1 0 0 1 4 0 0 0 0 0 0 0 0
example 1 In each part determine whether the matrix is in row echelon form, reduced row echelon form, both or neither.
10.8
(i)
(iii)
Chapter 10
È1 Í0 Í Í0 Í Î0
Matrices
2 0 3 0˘ 0 1 1 0 ˙˙ 0 0 0 1˙ ˙ 0 0 0 0˚
(ii)
È 1 -6 4 3˘ Í0 1 3 2 ˙˚ Î
(iv)
0 0 5˘ 0 1 2 ˙˙ 1 0 7˙˚ 0 0 0 0 1 2 0 0 1 0 0 0
È1 Í0 Í ÍÎ0 È0 Í0 Í Í0 Í Î0
0˘ -3˙˙ 0˙ ˙ 0˚
Solution (i) The given matrix is in reduced row echelon form and row echelon form since it satisfies properties (i), (ii), (iii) and columns containing leading 1 have zero everywhere else. (ii) The given matrix is neither in row echelon form nor in reduced row echelon form since it does not satisfy the property (iii). (iii) The given matrix is in row echelon form since it satisfies properties (i), (ii) and (iii). (iv) The given matrix is neither in row echelon form nor in reduced row echelon form since it does not satisfy the property (i).
example 2 Find a row echelon form of the following matrix: È0 -1 2 Í2 3 4 Í Í 1 3 -1 Í Î3 2 4
3˘ 5˙˙ 2˙ ˙ 1˚
Solution 0 −1 2 2 3 4 1 3 −1 3 2 4
3 5 2 1
R13 1 3 −1 2 3 4 ~ 0 −1 2 3 2 4
2 5 3 1
10.5 Row Echelon and Reduced Row Echelon Forms of a Matrix 10.9
R2 − 2 R1 , R4 − 3R1 1 3 −1 2 0 −3 6 1 ~ 0 −1 2 3 0 −7 7 −5 R23 1 3 −1 2 0 −1 2 3 ~ 0 −3 6 1 0 −7 7 −5 ( −1) R2 1 3 −1 2 0 1 −2 −3 ~ 0 −3 6 1 0 −7 7 −5 R3 + 3R2 , R4 + 7 R2 1 0 ~ 0 0
3 −1 1 −2 0 0
2 −3 −8 0 −7 −26
R34 1 0 ~ 0 0
3 −1 2 1 −2 −3 0 −7 −26 0 0 −8
1 − R3 7 1 0 ~ 0 0
3 −1 1 −2 0 0
2 −3 26 1 7 0 −8
10.10
Chapter 10
Matrices
1 − R4 8 1 0 ~ 0 0
3 −1 1 −2 0 0
2 −3 26 1 7 0 1
example 3 Find a row echelon form of the following matrix: È 1 Í-1 Í Í 0 Í Î 2 Solution 1 −1 0 2
2 -3 0 3
1˘ 4 ˙˙ 2 -1˙ ˙ 0 -3˚
1 3
2 −3 0 3 1 3
1 4 2 −1 0 −3
R2 + R1 , R4 − 2 R1 1 1 2 −3 0 2 0 5 ~ 0 1 2 −1 0 −1 6 −5 R23 1 1 2 −3 0 1 2 −1 ~ 0 2 0 5 0 −1 6 −5 R3 − 2 R2 , R4 + R2 1 0 ~ 0 0
−3
1 1 2 −1 0 −4 7 0 8 −6
2
10.5 Row Echelon and Reduced Row Echelon Forms of a Matrix 10.11
1 − R3 4 1 0 ~ 0 0
2 −3 1 2 0 0
1 −1 7 1 − 4 8 −6
R4 − 8 R3 1 0 ~ 0 0
2 −3 1 2 0 0
1 −1 7 1 − 4 0 8
1 R4 8 1 0 ~ 0 0
2 −3 1 2 0 0
1 −1 7 1 − 4 0 1
example 4 Find the reduced row echelon form of the following matrix: È0 -1 2 Í2 3 4 Í Í 1 3 -1 Í Î3 2 4
3˘ 5˙˙ 2˙ ˙ 1˚
10.12
Chapter 10
Matrices
Solution The row echelon form of the matrix is 1 0 0 0
3 −1
2 1 −2 −3 26 0 1 7 0 0 1
Beginning with the last non-zero row and working upward, we add suitable multiples of each row to the rows above to introduce zeros above the leading 1’s. 26 R4 , R2 + 3R4 , R1 − 2 R4 7 1 3 −1 0 0 1 −2 0 ~ 0 0 1 0 0 1 0 0 R3 −
R2 + 2 R3 , R1 + R3 1 0 ~ 0 0
3 0 0 1 0 0 0 1 0 0 0 1
R1 − 3R2 1 0 ~ 0 0
0 0 0 1 0 0 0 1 0 0 0 1
example 5 Find the reduced row echelon form of the following matrix: È 1 Í-1 Í Í 0 Í Î 2
2 -3 0 1 3
1˘ 3 4 ˙˙ 2 -1˙ ˙ 0 -3˚
10.6 Rank of a Matrix 10.13
Solution The row echelon form of the matrix is 1 0 0 0
2 −3 1 0 0
1 −1 7 1 − 4 0 1 2
Beginning with the last non-zero row and working upward, we add suitable multiples of each row to the rows above to introduce zeros above the leading 1’s. R3 + 1 0 ~ 0 0
7 R4 , R2 + R4 , R1 − R4 4 2 −3 0 1 2 0 0 1 0 0 0 1
R2 − 2 R3 , R1 + 3R3 1 0 ~ 0 0
2 0 0 1 0 0 0 1 0 0 0 1
R1 − 2 R2 1 0 ~ 0 0
10.6
0 0 0 1 0 0 0 1 0 0 0 1
rAnk of A MAtrIX
The positive integer r is said to be the rank of a matrix A if it possesses the following properties: (i) There is at least one minor of order r which is nonzero. (ii) Every minor of order greater than r is zero. Rank of matrix A is denoted by r (A). Note (1): The rank of a matrix remains unchanged by elementary transformations.
10.14
Chapter 10
Matrices
Note (2): The rank of the transpose of a matrix is same as that of the original matrix. Note (3): The rank of the product of two matrices cannot exceed the rank of either matrix. r (AB) ≤ r (A)
10.6.1
or
r (AB) ≤ r (B)
rank of a Matrix by echelon form
The rank of a matrix in echelon form is equal to the number of nonzero rows of the matrix, e.g., 1 3 −1 A = 0 1 4 0 0 0 The matrix A is in echelon form and the number of nonzero rows is two. Hence, rank of the matrix is two, i.e., ρ ( A) = 2
example 1 Find the rank of the following matrix by reducing it to echelon form:
Solution Let
5 3 14 4 0 1 2 1 1 −1 2 0 5 3 14 4 A = 0 1 2 1 1 −1 2 0
Reducing the matrix A to echelon form, R13 1 −1 2 0 ~ 0 1 2 1 5 3 14 4 R3 − 5 R1 1 −1 2 0 ~ 0 1 2 1 0 8 4 4
10.6 Rank of a Matrix 10.15
R3 − 8 R2 2 0 1 −1 ~ 0 1 2 1 0 0 − 12 − 4 1 − R3 12 1 −1 2 0 1 2 1 ~ 0 0 0 1 1 3 The equivalent matrix is in echelon form. Number of nonzero rows = 3 r (A) = 3
example 2 Find the rank of the following matrix by reducing it to echelon form: 3 −1 1 2 −2 −1 −3 −1 1 0 1 1 1 1 −1 0 Solution Let
1 2 3 −1 −2 −1 −3 −1 A= 1 0 1 1 0 1 1 −1
Reducing the matrix A to echelon form, R2 + 2 R1 , R3 − R1 3 −1 1 2 0 3 3 −3 ~ 0 − 2 − 2 2 1 1 −1 0
10.16
Chapter 10
Matrices
R24 3 −1 1 2 0 1 1 −1 ~ 0 −2 −2 2 3 3 −3 0 R3 + 2 R2 , R4 − 3R2 1 0 ~ 0 0
2 1 0 0
3 −1 1 −1 0 0 0 0
The equivalent matrix is in echelon form. Number of nonzero rows = 2
ρ ( A) = 2
example 3 Find the rank of the following matrix by reducing it to echelon form:
Solution
Let
3 −2 0 −1 0 2 2 1 1 −2 −3 2 1 2 1 0
3 −2 0 −1 0 2 2 1 A= 1 −2 −3 2 1 2 1 0
Reducing the matrix A to echelon form, R13 1 −2 −3 2 0 2 2 1 ~ 3 − 2 0 − 1 1 2 1 0 R3 − 3R1 , R24 1 −2 −3 2 0 1 2 1 ~ 4 9 −7 0 2 2 1 0
10.6 Rank of a Matrix 10.17
R3 − 4 R2 , R4 − 2 R2 2 1 −2 −3 0 1 2 1 ~ 0 1 −11 0 0 −2 −1 0 R4 + 2 R3 2 1 −2 −3 0 1 2 1 ~ 0 1 −11 0 0 0 −23 0 1 − R4 23 2 1 −2 −3 0 1 2 1 ~ 0 0 1 11 − 0 0 1 0 The equivalent matrix is in echelon form. Number of nonzero rows = 4 r(A) = 4
eXercISe 10.2 1. Find the ranks of the following matrices by reducing them to echelon forms: 1 1 −1 1 (i) 1 −1 2 −1 3 1 0 1
(iii)
3 1 2 −2 2 5 −4 6 −1 −3 2 −2 2 4 −1 6
3 −2 0 −1 −7 0 2 2 1 −5 (v) 1 −2 −3 −2 1 1 2 1 6 0
3 4 2 (ii) 8 4 6 −2 −1 −1.5 0 0 (iv) 3 1
1 −3 −1 0 1 1 1 0 2 1 2 0
1 2 −1 4 (vi) 2 4 3 5 −1 −2 6 7 Ans. : (i) 2 (ii) 1 (iii) 4 (iv) 2 (v) 4 (vi) 2
10.18
10.7
Chapter 10
Matrices
InverSe of A MAtrIX by GAuSS–JordAn Method
Let A be any non-singular matrix. Then A = IA. Applying suitable elementary row transformation to A on the L.H.S and to I on the R.H.S, of a matrix A reduces to I and I reduces to any matrix B. Hence, I = BA B = A−1
example 1 Find the inverse of the following matrix by elementary transformations (Gauss–Jordan method): 2 4 1
3 4 3 1 2 4
Solution 2 3 4 A = 4 3 1 1 2 4 A = I3 A
Let
2 3 4 1 0 0 4 3 1 = 0 1 0 A 1 2 4 0 0 1 Reducing the matrix A to reduced row echelon form, R13 1 2 4 0 0 1 4 3 1 = 0 1 0 A 2 3 4 1 0 0 R2 − 4 R1 , R3 − 2 R1 4 0 0 1 1 2 0 −5 −15 = 0 1 − 4 A 0 −1 − 4 1 0 −2
10.7 Inverse of a Matrix by Gauss–Jordan Method 10.19
1 − R2 5 0 1 0 4 1 2 1 4 = 0 − 0 A 1 3 5 5 0 −1 − 4 0 −2 1 R3 + R2 0 1 0 1 2 4 4 0 1 3 = 0 − 1 A 5 5 0 0 −1 1 6 1 − − 5 5 ( −1) R3 0 0 1 2 4 1 0 1 3 = 0 − 5 0 0 1 1 −1 5
1 4 A 5 6 5
R2 − 3R3 , R1 − 4 R3 4 19 4 −5 − 5 1 2 0 0 1 0 = 3 − 4 − 14 A 5 5 0 0 1 1 6 −1 5 5 R1 − 2 R2 4 9 −2 5 5 1 0 0 0 1 0 = 3 − 4 − 14 A 5 5 0 0 1 1 6 −1 5 5 −1 I3 = A A 4 −2 5 4 −1 ∴A = 3 − 5 1 −1 5
9 5 4 9 −10 14 1 − = 15 −4 −14 5 5 −5 1 6 6 5
10.20
Chapter 10
Matrices
example 2 Find the inverse of the following matrix by elementary transformations (Gauss–Jordan method): 2 1 −1 0 1 1 −1 0 2 1 2 1 6 3 −2 1 Solution 1 −1 0 1 A= 2 1 3 −2
Let
0 2 1 −1 2 1 1 6
A = I4 A 1 −1 0 1 2 1 3 −2
2 1 1 −1 0 = 2 1 0 1 6 0
0 0 0 1 0 0 A 0 1 0 0 0 1 Reducing the matrix A to reduced row echelon form, 0
R3 − 2 R1 , R4 − 3R1 1 −1 0 1 0 3 1 0
0 2 1 1 −1 0 = 2 −3 −2 1 0 −3
0 1 0 0
0 0 1 0
0 0 A 0 1
R3 − 3R2 , R4 − R2 1 −1 0 2 1 0 0 1 1 −1 0 1 = 0 0 −1 0 −2 −3 1 −3 −1 0 0 0
0 0 0 0 A 1 0 0 1
( −1) R3 1 −1 0 1 0 0 0 0
0 2 1 0 0 1 −1 0 1 0 = 1 0 2 3 −1 0 1 −3 −1 0
0 0 A 0 1
10.7 Inverse of a Matrix by Gauss–Jordan Method 10.21
R2 + R4 , R1 − 2 R4 1 −1 0 1 0 0 0 0
0 0 7 2 0 −2 1 0 −3 0 0 1 A = 1 0 2 3 −1 0 1 0 1 −3 −1 0
R2 − R3 1 −1 0 1 0 0 0 0
0 0 7 2 0 −2 0 0 −5 −3 1 1 A = 1 0 2 3 −1 0 0 1 −3 −1 0 1
R1 + R2 1 0 0 0
0 0 0 2 −1 1 −1 1 0 0 −5 −3 1 1 A = 0 1 0 2 3 −1 0 0 0 1 −3 −1 0 1 I 4 = A−1 A 2 −1 1 −1 −5 −3 1 1 ∴ A−1 = 2 3 −1 0 1 −3 −1 0
eXercISe 10.3 1. Using elementary row transformations, find the inverses of the following matrices: 5 5 -1 8 4 3 3 -3 4 0 (i) 2 1 1 (ii) 0 2 (iii) 2 -3 4 -5 3 -15 1 2 1 0 -1 1
(iv)
1 1 -1 2 2 -3 1 - 4 9
(v)
1 2 3 2 3 0 0 1 2
(vi)
2 3 2 4
2 6 5 2 5 2 -3 5 14 14
4
3
10.22
Chapter 10
Matrices
(vii)
0 1 2 2
2 3 3 3
(ix)
2 -6 5 -13 -1 4 1 0
1 1 2 3
Ans.: 1 -2 1 (i) 1 -5 3 3 12 6 1 ( iv ) -21 17 -10 3 -3 3 -4 (vii) -3 4 2 -2 5 0 0 1 1 0 (x) 0 0 2 1 0 0
10.8
2 2 2 3
(viii)
-2 -4 1 0
-1 2 0
5 8 3
-3 -7 2 1
(x)
-1 -3 3 -1 1 1 -1 0 2 -5 2 -3 1 -1 1 0 2 2 1 -2
9 -10 4 1 15 -4 -14 (ii) 5 1 6 -5
1 6 1 5 ( v) - 4 4 4 2
-3 2 4 -2 -5 3 3 -2 2 2 1 1
0 1 (viii) 1 -1
-1 -9 2 6 -1 -1
0 6 0 0
0 -4 0 -16 3 -5 0 10
1 -1 0 (iii) -2 3 -4 -2 3 3 64 -23 29 - 5 26 10 -12 5 (vi) 6 1 -2 5 3 2 -2 5
2 1 3 -2 1 1 -1 -2 (ix) -4 2 0 1 1 2 6 -1
1 0
0 2
1 -3 0 -2
18 - 5 7 5 2 5 1 5 1 -1 1 2
SySteM of non-hoMoGeneouS LIneAr equAtIonS
A system of m non-homogeneous linear equations in n variables x1, x2, ..., xn or simply a linear system, is a set of m linear equations, each in n variables. A linear system is represented by
10.8 System of Non-Homogeneous Linear Equations 10.23
a11 x1 + a12 x2 + + a1n xn = b1 a21 x1 + a22 x2 + + a2 n xn = b2 am1 x1 + am 2 x2 + + amn xn = bm Writing these equations in matrix form, Ax = B a11 a 21 where A = am1
a12 a22 am 2
… a1n … a2 n is called coefficient matrix of order m × n, … amn
x1 x 2 x = is any vector of order n × 1. xn b1 b 2 B = is any vector of order m × 1. bm
10.8.1
Solutions of System of Linear equations: Gauss elimination and Gauss–Jordan Methods
For a system of m linear equations in n variables, there are three possibilities of the solutions to the system: (i) The system has unique solution. (ii) The system has infinite solutions. (iii) The system has no solution. When the system of linear equations has one or more solutions, the system is said to be consistent, otherwise it is inconsistent.
The matrix
a11 a A : B = [ ] 21 a m1
a12 a22
… a1n … a2 n
am 2 … amn
b1 b2 bm
is called the augmented matrix of the given system of linear equations. To solve a system of linear equations, elementary transformations are used to reduce the augmented matrix to either row echelon form or reduced row echelon form.
10.24
Chapter 10
Matrices
Reducing the augmented matrix to row echelon form is called Gauss elimination method. Reducing the augmented matrix to reduced row echelon form is called Gauss–Jordan method. The Gauss elimination method for solving the linear system is as follows: Step 1: Write the augmented matrix. Step 2: Obtain the row echelon form of the augmented matrix by using elementary row operations. Step 3: Write the corresponding linear system of equations from row echelon form. Step 4: Solve the corresponding linear system of equations by back substitution. The Gauss–Jordan method for solving the linear system is as follows: Step 1: Write the augmented matrix. Step 2: Obtain the reduced row echelon form of the augmented matrix by using elementary row operations. Step 3: For each non-zero row of the matrix, solve the corresponding system of equations for the variables associated with the leading one in that row. Note: The linear system has a unique solution if det (A) π 0
example 1 Solve the following system by Gauss elimination method: x + y + 2z = 9 2 x + 4 y - 3z = 1 3x + 6 y - 5 z = 0 Solution The matrix form of the system is 1 1 2 x 9 2 4 −3 y = 1 3 6 −5 z 0 The augmented matrix of the system is 1 1 2 9 2 4 −3 1 3 6 −5 0
10.8 System of Non-Homogeneous Linear Equations 10.25
Reducing the augmented matrix to row echelon form, R2 − 2 R1 , R3 − 3R1 2 9 1 1 ~ 0 2 −7 −17 0 3 −11 −27 1 R2 2 2 9 1 1 7 17 − ~ 0 1 − 2 2 0 3 −11 −27 R3 − 3R2 2 9 1 1 7 17 ~ 0 1 − − 2 2 1 3 0 0 − − 2 2 ( −2) R3 2 9 1 1 7 17 ~ 0 1 − − 2 2 0 0 1 3 The corresponding system of equations is x + y + 2z = 9 7 17 y− z=− 2 2 z=3 Solving these equations, x = 1, y = 2 Hence, x = 1, y = 2, z = 3 is the solution of the system.
10.26
Chapter 10
Matrices
example 2 Solve the following system by Gauss elimination method: 4x - 2 y + 6z = 8 x + y - 3 z = -1 15 x - 3 y + 9 z = 21 Solution The matrix form of the system is 4 −2 6 x 8 1 1 −3 y = −1 15 −3 9 z 21 The augmented matrix of the system is 4 −2 6 8 1 1 −3 −1 15 −3 9 21 Reducing the augmented matrix to row echelon form, R12 1 −3 −1 1 ~ 4 −2 6 8 15 −3 9 21 R2 − 4 R1 , R3 − 15 R1 1 −3 −1 1 ~ 0 −6 18 12 0 −18 54 36 1 1 − R2 , − R3 6 18 1 1 −3 −1 ~ 0 1 −3 −2 0 1 −3 −2 R3 − R2 1 1 −3 −1 ~ 0 1 −3 −2 0 0 0 0
10.8 System of Non-Homogeneous Linear Equations 10.27
The corresponding system of equations is x + y − 3 z = −1 y − 3 z = −2 The leading ones are in columns 1 and 2. Hence, the variables x and y are called leading variables whereas the variable z is called a free variable. Assigning the free variable z an arbitrary value t, y = 3t − 2 x = −1 − 3t + 2 + 3t = 1 Hence, x = 1, y = 3t – 2, z = t is the solution of the system where t is a parameter.
example 3 Solve the following system by Gauss elimination method: 3 x + y − 3 z = 13 2x − 3y + 7z = 5 2 x + 19 y − 47 z = 32 Solution The matrix form of the system is 1 −3 3 2 −3 7 2 19 −47
x 13 y = 5 z 32
The augmented matrix of the system is 1 −3 13 3 2 −3 7 5 2 19 −47 32 Reducing the augmented matrix to row echelon form, 1 R1 3 1 13 −1 1 3 3 ~ 2 −3 7 5 2 19 − 47 32
10.28
Chapter 10
Matrices
R2 − 2 R1 , R3 − 2 R1 1 13 −1 1 3 3 11 11 ~ 0 − 9 − 3 3 55 70 −45 0 3 3 3 − R2 11 1 13 −1 1 3 3 27 1 − 1 ~ 0 11 55 70 − 45 0 3 3 55 R2 3 1 13 −1 1 3 3 27 ~ 0 1 − 1 11 0 0 0 5 R3 −
From the last row of the augmented matrix, 0x + 0 y + 0z = 5 Hence, the system is inconsistent and has no solution.
example 4 Solve the following system for x, y and z. 1 3 4 − + + = 30 x y z 3 2 1 + − = 9 x y z 2 1 2 − + = 10 x y z
10.8 System of Non-Homogeneous Linear Equations 10.29
Solution The matrix form of the system is
−1 3 4 3 2 −1 2 −1 2
1 x 30 1 = 9 y 10 1 z
The augmented matrix of the system is −1 3 4 30 3 2 −1 9 2 −1 2 10 Reducing the augmented matrix to row echelon form, ( −1) R1 1 −3 −4 −30 9 ~ 3 2 −1 2 −1 2 10 R2 − 3R1 , R3 − 2 R1 1 −3 −4 −30 ~ 0 11 11 99 0 5 10 70 1 1 R2 , R3 11 5 1 −3 − 4 −30 ~ 0 1 1 9 0 1 2 14 R3 − R2 1 −3 −4 −30 ~ 0 1 1 9 0 0 1 5
10.30
Chapter 10
Matrices
The corresponding system of equations is 1 3 4 − − = −30 x y z 1 1 + =9 y z 1 =5 z Solving these equations, 1 1 1 x= , y= , z= 2 4 5 1 1 1 Hence, x = , y = , z = is the solution of the system. 2 4 5
example 5 Solve the following system of non-linear equations for the unknown angles a, b and g, where 0 £ a £ 2p, 0 £ b £ 2p and 0 £ g < p. 2 sin a - cos b + 3 tan g = 3 4 sin a + 2 cos b - 2 tan g = 2 6 sin a - 3 cos b + tan g = 9 Solution The matrix form of the system is 2 −1 3 sin α 3 4 2 −2 cos β = 2 6 −3 1 tan γ 9 The augmented matrix of the system is 2 −1 3 3 4 2 −2 2 6 −3 1 9 Reducing the augmented matrix to row echelon form, 1 R1 2 1 3 3 1 − 2 2 2 ~ 2 −2 2 4 6 −3 1 9
10.8 System of Non-Homogeneous Linear Equations 10.31
R2 − 4 R1 , R3 − 6 R1 1 3 3 1 − 2 2 2 ~ 0 4 −8 −4 0 0 −8 0 1 1 R2 , − R3 4 8 1 3 3 1 − 2 2 2 1 −2 −1 ~ 0 0 0 1 0 The corresponding system of equations is 1 3 3 sin α − cos β + tan γ = 2 2 2 cos β − 2 tan γ = −1 tan γ = 0 Solving these equations,
γ =0 cos β = −1 ⇒ β = π 1 3 3 cos β − tan γ + 2 2 2 1 3 3 = (−1) − (0) + = 1 2 2 2 π α= 2
sin α =
Hence, α =
π , β = π , γ = 0 is the solution of the system. 2
example 6 Investigate for what values of l and m the equations x + 2y + z = 8 2 x + 2 y + 2 z = 13 3x + 4 y + l z = m have (i) no solution, (ii) a unique solution, and (iii) many solutions.
10.32
Chapter 10
Matrices
Solution The matrix form of the system is 1 2 1 2 2 2 3 4 λ
x 8 y = 13 z µ
The augmented matrix of the system is 1 2 1 8 2 2 2 13 3 4 λ µ Reducing the augmented matrix to row echelon form, R2 − 2 R1 , R3 − 3R1 1 8 1 2 ~ 0 −2 0 −3 0 −2 λ − 3 µ − 24 1 − R2 2 1 8 1 2 3 ∼ 0 1 0 2 0 −2 λ − 3 µ − 24 R3 + 2 R2 1 8 1 2 3 ~ 0 1 0 2 0 0 λ − 3 µ − 21 (i) If l = 3 and m π 21, the system is inconsistent and has no solution. (ii) If l π 3 and m has any value, the system is consistent and has a unique solution. (iii) If l = 3 and m = 21, the system is consistent and has infinite (many) solutions.
example 7 Determine the values of l for which the following equations are consistent. Also, solve the system for these values of l.
10.8 System of Non-Homogeneous Linear Equations 10.33
x + 2y + z = 3 x+ y + z = λ 3x + y + 3z = λ 2
Solution The matrix form of the system is 1 2 1 x 3 1 1 1 y = λ 3 1 3 z λ 2 The augmented matrix of the system is 1 2 1 3 1 1 1 λ 3 1 3 λ 2 Reducing the augmented matrix to row echelon form, R2 − R1 , R3 − 3R1 3 1 2 1 ~ 0 −1 0 λ − 3 0 −5 0 λ 2 − 9 ( −1) R2 3 1 2 1 ~ 0 1 0 3 − λ 0 −5 0 λ 2 − 9 R3 + 5 R2 3 1 2 1 ~ 0 1 0 3 − λ 2 0 0 0 λ − 5λ + 6 The equations will be consistent if l 2 – 5l + 6 = 0, i.e. l = 3 or l = 2. Case I: When l = 3, x + 2y + z = 3 y =0 Assigning the free variable z any arbitrary value t, x = 3 − 2(0) − t = 3 − t Hence, x = 3 – t, y = 0, z = t is the solution of the system where t is a parameter.
10.34
Chapter 10
Matrices
Case II: When l = 2, x + 2y + z = 3 y
=1
Assigning the free variable z any arbitrary value t, x = 3 − 2(1) − t = 1 − t Hence, x = 1 – t, y = 1, z = t is the solution of the system where t is a parameter.
example 8 Show that the system of equations 3x + 4 y + 5 z = a 4x + 5 y + 6z = b 5x + 6 y + 7 z = g is consistent only if a, b and l are in arithmetic progression (A.P.). Solution The matrix form of the system is 3 4 5 x α 4 5 6 y = β 5 6 7 z γ The augmented matrix of the system is 3 4 5 α 4 5 6 β 5 6 7 γ Reducing the augmented matrix to row echelon form, R2 − R1 , R3 − R1
α 3 4 5 ~ 1 1 1 β − α 2 2 2 γ − α R12 1 1 1 β − α ~ 3 4 5 α 2 2 2 γ − α
10.8 System of Non-Homogeneous Linear Equations 10.35
R2 − 3R1 , R3 − 2 R1
β −α 1 1 1 ~ 0 1 2 4α − 3β 0 0 0 α − 2β + γ The system of equations is consistent if,
α − 2β + γ = 0 α +γ β= 2 i.e. a, b and g are in arithmetic progression (A.P.)
example 9 Show that if l π 0, the system of equations 2x + y =a x + ly - z = b y + 2z = c has a unique solution for every value of a, b, c. If l = 0, determine the relation satisfied by a, b, c such that the system is consistent. Find the solution by taking l = 0, a = 1, b = 1, c = –1. Solution The matrix form of the system is 2 1 0 x a 1 λ −1 y = b 0 1 2 z c The system has a unique solution if det (A) π 0 det( A) = 2 (2λ + 1) − 1 (2 + 0) ≠ 0 4λ ≠ 0 λ≠0 Hence, the system of equations has a unique solution if l π 0 for any value of a, b, c. If l = 0, the system is either inconsistent or has an infinite number of solutions.
10.36
Chapter 10
Matrices
For l = 0, the augmented matrix of the system is 2 1 0 a 1 0 −1 b 0 1 2 c Reducing the augmented matrix to row echelon form, R12 1 0 −1 b ~ 2 1 0 a 0 1 2 c R2 − 2 R1 b 1 0 −1 ~ 0 1 2 a − 2b 0 1 2 c R3 − R2 b 1 0 −1 ~ 0 1 2 a − 2b 0 0 0 c − a + 2b The system is consistent if c – a + 2b = 0. The corresponding system of equations is x− z = b y + 2 z = a − 2b Assigning the free variable z any arbitrary value t, y = a − 2b − 2t x = b+t Hence, x = b + t, y = a – 2b – 2t, z = t is the solution of the system where t is a parameter. When a = 1, b = 1, c = –1 x = 1+ t y = −1 − 2t z=t
10.8 System of Non-Homogeneous Linear Equations 10.37
example 10 Solve the following system by Gauss–Jordan method: x1 + x2 + 2 x3 = 8 - x1 - 2 x2 + 3 x3 = 1 3 x1 - 7 x2 + 4 x3 = 10 Solution The matrix form of the system is 1 2 x1 8 1 −1 −2 3 x = 1 2 3 −7 4 x3 10 The augmented matrix of the system is 1 2 8 1 −1 −2 3 1 3 −7 4 10 Reducing the augmented matrix to reduced row echelon form, R2 + R1 , R3 − 3R1 1 2 8 1 ~ 0 −1 5 9 0 −10 −2 −14 ( −1) R2 1 2 8 1 1 −5 −9 ~ 0 0 −10 −2 −14 R3 + 10 R2 2 8 1 1 ~ 0 1 −5 −9 0 0 −52 −104
10.38
Chapter 10
Matrices
1 − R3 52 8 1 1 2 ~ 0 1 −5 −9 0 0 1 2 R2 + 5 R3 , R1 − 2 R3 1 1 0 4 ~ 0 1 0 1 0 0 1 2 R1 − R2 1 0 0 3 ~ 0 1 0 1 0 0 1 2 The corresponding system of equations is x1 = 3 x2 = 1 x3 = 2 Hence, x1 = 3, x2 = 1, x3 = 2 is the solution of the system.
example 11 Solve the following system by Gauss–Jordan method: 2 x1 + 2 x2 + 2 x3 = 0 -2 x1 + 5 x2 + 2 x3 = 1 8 x1 + x2 + 4 x3 = -1 Solution The matrix form of the system is 2 2 2 x1 0 −2 5 2 x = 1 2 8 1 4 x3 −1
10.8 System of Non-Homogeneous Linear Equations 10.39
The augmented matrix of the system is 2 2 2 0 −2 5 2 1 8 1 4 −1 Reducing the augmented matrix to reduced row echelon form, 1 R1 2 1 1 1 0 ~ −2 5 2 1 8 1 4 −1 R2 + 2 R1 , R3 − 8 R1 1 1 0 1 ~ 0 7 4 1 0 −7 − 4 −1 R3 + R2 1 1 1 0 ~ 0 7 4 1 0 0 0 0 1 R2 7 1 1 1 0 4 1 ~ 0 1 7 7 0 0 0 0 R1 − R2 3 1 1 0 7 − 7 4 1 ~ 0 1 7 7 0 0 0 0
10.40
Chapter 10
Matrices
The corresponding system of equations is 3 1 x3 = − 7 7 4 1 x2 + x3 = 7 7 x1 +
Since leading ones are in columns 1 and 2, x1 and x2 are called leading variables, whereas x3 is a free variable. Assigning the free variable x3 any arbitrary value t, 1 3 x1 = − − t 7 7 1 4 x2 = − t 7 7 1 3 1 4 Hence, x1 = − − t , x2 = − t , x3 = t is the solution of the system where t is a 7 7 7 7 parameter.
example 12 Solve the following system by Gauss–Jordan method: x - y + 2z - w = 2 x + y - 2 z - 2w = - x + 2 y - 4z + w = 3x 3w =
-1 -2 1 -3
Solution The matrix form of the system is 1 −1 2 −1 x −1 2 1 −2 −2 y −2 = −1 2 − 4 1 z 1 0 −3 w −3 3 0 The augmented matrix of the system is 1 −1 2 −1 −1 2 1 −2 −2 −2 −1 2 − 4 1 1 0 −3 −3 3 0
10.8 System of Non-Homogeneous Linear Equations 10.41
Reducing the augmented matrix to reduced row echelon form, R2 − 2 R1 , R3 + R1 , R4 − 3R1 1 −1 2 −1 −1 0 3 −6 0 0 ~ 0 1 −2 0 0 0 3 −6 0 0 1 R2 3 1 −1 2 −1 −1 0 1 −2 0 0 ~ 0 1 −2 0 0 0 3 −6 0 0 R3 − R2 , R4 − 3R2 1 −1 2 −1 −1 0 1 −2 0 0 ~ 0 0 0 0 0 0 0 0 0 0 R1 + R2 1 0 ~ 0 0
0 −1 −1 1 −2 0 0 0 0 0 0 0 0 0 0
0
The corresponding system of equations is x− y − 2z
w = −1 =0
The leading ones are in columns 1 and 2. Hence, the variables x and y are called leading variables whereas the variables z and w are called free variables. Assigning the free variables z and w any arbitrary values t1 and t2 respectively, x = −1 + t2 and
y = 2t1
Hence, x = −1 + t2 , y = 2t1 , z = t1 , w = t2 is the solution of the system where t1 and t2 are parameters.
10.42
Chapter 10
Matrices
example 13 Solve the following system by Gauss–Jordan method: - 2 y + 3z = 1 3 x + 6 y - 3 z = -2 6 x + 6 y + 3z = 5 Solution The matrix form of the system is 0 −2 3 x 1 3 6 −3 y = −2 6 6 3 z 5 The augmented matrix of the system is 1 0 −2 3 3 6 −3 −2 6 6 3 5 Reducing the augmented matrix to reduced row echelon form, R12 3 6 −3 −2 ~ 0 −2 3 1 6 6 3 5 1 R1 3 2 1 2 −1 − 3 ~ 0 −2 3 1 6 6 3 5 R3 − 6 R1 2 1 2 −1 − 3 ~ 0 −2 3 1 0 −6 9 9
10.8 System of Non-Homogeneous Linear Equations 10.43
1 − R2 2 2 1 2 −1 − 3 3 1 1 − ~ 0 − 2 2 9 9 0 −6 R3 + 6 R2 2 1 2 −1 − 3 3 1 ~ 0 1 − − 2 2 0 0 0 6 From the last row of the augmented matrix, 0x + 0 y + 0z = 6 Hence, the system is inconsistent and has no solution.
example 14 Solve the following system by Gauss–Jordan method: x1 - 2 x2 - x3 + 3 x4 = 1 2 x1 - 4 x2 + x3
=5
x1 - 2 x2 + 2 x3 - 3 x4 = 4 Solution The matrix form of the system is x1 1 −2 −1 3 1 2 − 4 1 0 x2 = 5 x 1 −2 2 −3 3 4 x4
10.44
Chapter 10
Matrices
The augmented matrix of the system is 1 −2 −1 3 1 2 − 4 1 0 5 1 −2 2 −3 4 Reducing the augmented matrix to reduced row echelon form, R2 − 2 R1 , R3 − R1 1 −2 −1 3 1 ~ 0 0 3 −6 3 0 0 3 −6 3 R3 − R2 1 −2 −1 3 1 ~ 0 0 3 −6 3 0 0 0 0 0 1 R2 3 1 −2 −1 3 1 ~ 0 0 1 −2 1 0 0 0 0 0 R1 + R2 1 2 1 −2 0 ~ 0 0 1 −2 1 0 0 0 0 0 The corresponding system of equations is x1 − 2 x2 +
x4 = 2 x3 − 2 x4 = 1
The leading ones are in columns 1 and 3. Hence, the variables x1 and x3 are called leading variables whereas the variables x2 and x4 are called free variables. Assigning the free variables x2 and x4 any arbitrary values t1 and t2 respectively, x1 = 2 + 2t1 − t2 x3 = 1 + 2t2
10.8 System of Non-Homogeneous Linear Equations 10.45
Hence, x1 = 2 + 2t1 − t2 , x2 = t1 , x3 = 1 + 2t2 , x4 = t2 is the solution of the system where t1 and t2 are the parameters.
example 15 Solve the following system by Gauss–Jordan method: 2x 3x 4x -x +
y+ z y+ z y + 2z y- z
=9 =6 =7 =4
Solution The matrix form of the system is 9 2 −1 1 3 −1 1 x 6 y = 4 −1 2 7 z 4 −1 1 −1 The augmented matrix of the system is 2 −1 1 3 −1 1 4 −1 2 −1 1 −1
9 6 7 4
Reducing the augmented matrix to reduced row echelon form, R14 −1 1 −1 3 −1 1 ~ 4 −1 2 2 −1 1
4 6 7 9
( −1) R1 1 3 ~ 4 2
−1 −1 −1 −1
1 − 4 1 6 2 7 1 9
10.46
Chapter 10
Matrices
R2 − 3R1 , R3 − 4 R1 , R4 − 2 R1 1 −1 1 − 4 0 2 −2 18 ~ 0 3 −2 23 1 −1 17 0 R24 1 − 4 1 −1 0 1 −1 17 ~ 0 3 −2 23 0 2 −2 18 R3 − 3R2 , R4 − 2 R2 1 −1 1 − 4 0 1 −1 17 ~ 0 0 1 −28 0 0 0 −16 From the last row of the augmented matrix, 0 x + 0 y + 0 z = −16 Hence, the system is inconsistent and has no solution.
eXercISe 10.3 1. Solve the following systems of equations by Gauss elimination method: (i) 2 x - 3y - z = 3 x + 2y - z = 4 5x - 4 y - 3z = -2
(ii)
(iii) 6 x + y + z = - 4 2 x - 3y - z = 0 - x - 7 y - 2z = 7 (v)
2 x1 + x 2 + 2 x 3 +
x + 2y - z = 1 x + y + 2z = 9 2x + y - z = 2
(iv) 2 x - y - z = 2 x + 2y + z = 2 4 x - 7 y - 5z = 2 x4 = 6
6 x1 - 6 x 2 + 6 x 3 + 12 x 4 = 36 4 x1 + 3 x 2 + 3 x 3 - 3 x 4 = 1 2 x1 + 2 x 2 - x 3 + x 4 = 10
10.8 System of Non-Homogeneous Linear Equations 10.47
È Ans.: ˘ Í ˙ (ii) consistent Í (i) inconistent ˙ Í ˙ x = 2, y = 1, z = 3 Í ˙ (iv) consistent Í(iii) consistent ˙ Í ˙ 6 +t 2 - 3t x = -1, y = -2, z = - 4 x= ,y = , z = t˙ Í 5 5 Í ˙ Í (v) consistent ˙ Í ˙ x1 = 2, x 2 = 1, x 3 = -1, x 4 = 3 Î ˚ 2. Solve the following system of equations by Gauss–Jordan method: (i)
x + 2y + z = - 1 6x + y + z = -4 2 x - 3y - z = - x - 7 y - 2z = x- y =
0 7 1
(ii)
x+ y+ z= 6 x - 2y + 2z = 5 3x + y + z = 8 2 x - 2y + 3z = 7
(iii) 2 x1 + x 2 + 5x 4 = 4 3 x1 - 2 x 2 + 2 x 3 = 2 5 x1 - 8 x 2 - 4 x 3 = 1 È Ans.: Í Í (i) consistent Í x = -1, y = -2, z = 4 Í Î(iii) inconsistent
˘ ˙ (ii) consistent ˙ x = -1, y = -2, z = 3˙ ˙ ˚
3. Investigate for what values of l and m, the system of simultaneous equations x+ y+ z= 6 x + 2y + 3z = 10 x + 2y + l z = m has (i) no solution, (ii) a unique solution, and (iii) infinite number of solutions. È Ans.: Í Î (i) l = 3, m π 10
(ii) l π 3, any value of m
˘ ˙ (iii) l = 3, m = 10 ˚
10.48
Chapter 10
Matrices
4. Investigate for what values of k the equations x+y+ z= 1 2x + y + 4 z = k 4 x + y + 10 z = k 2 have infinite number of solutions. [Ans.: k = 1, 2] 5. Determine the values of l for which the following system of equations. 3x - y + l z = 0 2x + y + z = 2 x - 2y - l z = -1 will fail to have a unique solution. For this value of l, are the equations consistent? 7 È ˘ Í Ans.: l = - 2 , no solution ˙ Î ˚ 6. Find for what values l, the set of equations 2 x - 3y + 6 z - 5t = 3 y - 4z + t = 1 4 x - 5y + 8 z - 9t = l has (i) no solution, and (ii) infinite number of solutions and find the solutions of the equations when they are consistent. È Ans.: (i) Í (ii) Í Í Í ÎÍ
l π 7, ˘ ˙ l = 7, x = 3k1 + k2 + 3,˙ y = 4k1 - k2 + 1, z = k1,˙ ˙ t = k2 ˙˚
10.9 SySteM of hoMoGeneouS LIneAr equAtIonS A system of m homogeneous linear equations in n variables x1, x2, ..., xn or simply a linear system, is a set of m linear equations each in n variables. A linear system is represented by a11 x1 + a12 x2 + + a1n xn = 0 a21 x1 + a22 x2 + + a2 n xn = 0 am1 x1 + am 2 x2 + + amn xn = 0
10.9 System of Homogeneous Linear Equations 10.49
Writing these equations in matrix form, Ax = 0 where A is any matrix of order m × n, x is a vector of order n × 1 and 0 is a null vector of order m × 1. The matrix A is called coefficient matrix of the system of equations.
10.9.1
Solutions of a System of Linear equations
For a system of m linear equations in n variables, there are two possibilities of the solutions to the system. (i) The system has exactly one solution, i.e. x1 = 0, x2 = 0 ..., xn = 0. This solution is called the trivial solution. (ii) The system has infinite solutions. Note: The system of equations has a non-trivial solution if det (A) = 0.
example 1 Solve the following system of equations by the Gauss–Jordan method. 3x - y - z = 0 x + y + 2z = 0 5 x + y + 3z = 0 Solution The matrix form of the system is 3 −1 −1 x 0 1 1 2 y = 0 5 1 3 z 0 The augmented matrix of the system is 3 −1 −1 0 1 1 2 0 5 1 3 0 Reducing the augmented matrix to reduced row echelon form, R12 1 1 2 0 ~ 3 −1 −1 0 5 1 3 0
10.50
Chapter 10
Matrices
R2 − 3R1 , R3 − 5 R1 1 2 0 1 ~ 0 − 4 −7 0 0 − 4 −7 0 1 1 − R2 , − R3 4 4 1 1 2 0 7 0 ~ 0 1 4 7 0 1 0 4 R3 − R2 1 1 2 0 7 ~ 0 1 0 4 0 0 0 0 R1 − R2 1 0 ~ 0 1 0 0
1 0 4 7 0 4 0 0
The corresponding system of equations is 1 z=0 4 7 y+ z = 0 4 x+
Solving for the leading variables,
1 x=− z 4 7 y=− z 4 Assigning the free variable z an arbitrary value t, 1 x=− t 4 7 y=− t 4
10.9 System of Homogeneous Linear Equations 10.51
1 7 Hence, x = − t , y = − t is the non-trivial solution of the system where t is a 4 4 parameter.
example 2 Solve the following system of equations by the Gauss–Jordan method. x+ y- z+ w=0 x - y + 2z - w = 0 3x + y +w=0 Solution The matrix form of the system is x 1 1 −1 1 0 1 −1 2 −1 y = 0 z 3 1 0 1 0 w The augmented matrix of the system is 1 1 −1 1 0 1 −1 2 −1 0 3 1 0 1 0 Reducing the augmented matrix to the reduced row echelon form, R2 − R1 , R3 − 3R1 1 −1 1 0 1 ~ 0 −2 3 −2 0 0 −2 3 −2 0 1 1 − R2 , − R3 2 2 1 1 −1 1 0 3 1 0 ~ 0 1 − 2 3 0 1 − 1 0 2
10.52
Chapter 10
Matrices
R3 − R2 1 1 −1 1 0 3 ~ 0 1 − 1 0 2 0 0 0 0 0 R1 − R2 1 0 0 1 0 2 3 ~ 0 1 − 1 0 2 0 0 0 0 0 The corresponding system of equations is 1 x+ z =0 2 3 y− z+w =0 2 Solving for the leading variables, 1 x=− z 2 3 y= z−w 2 Assigning the free variables z and w arbitrary values t1 and t2 respectively, 1 x = − t1 2 3 y = t1 − t2 2 1 3 Hence, x = − t1 , y = t1 − t2 , z = t1 , w = t2 is the non-trivial solution of the system 2 2 where t1 and t2 are parameters.
10.9 System of Homogeneous Linear Equations 10.53
example 3 Solve the following system of equations by the Gauss–Jordan method. 2 x1 + x2 + 3 x3 = 0 x1 + 2 x2
=0
x2 + x3 = 0 Solution The matrix form of the system is 2 1 3 x1 0 1 2 0 x = 0 2 0 1 1 x3 0 The augmented matrix of the system is 2 1 3 0 1 2 0 0 0 1 1 0 Reducing the augmented matrix to reduced row echelon form, R12 1 2 0 0 ~ 2 1 3 0 0 1 1 0 R2 − 2 R1 1 2 0 0 ~ 0 −3 3 0 0 1 1 0 1 − R2 3 1 2 0 0 ~ 0 1 −1 0 0 1 1 0 R3 − R2 1 2 0 0 ~ 0 1 −1 0 0 0 2 0
10.54
Chapter 10
Matrices
1 R3 2 1 2 0 0 ~ 0 1 −1 0 0 0 1 0 R2 + R3 1 2 0 0 ~ 0 1 0 0 0 0 1 0 R1 − 2 R2 1 0 0 0 ~ 0 1 0 0 0 0 1 0 The corresponding system of equations is x=0 y=0 z=0 Hence, the system has a trivial solution, i.e. x = 0, y = 0, z = 0.
example 4 Show that the following non-linear system has 18 solutions if 0 £ a £ 2p , 0 £ b £ 2p and 0 £ g < 2p. sin a + 2 cos b + 3 tan g = 0 2 sin a + 5 cos b + 3 tan g = 0 - sin a - 5 cos b + 5 tan g = 0 Solution The matrix form of the system is 1 2 3 sin α 0 2 5 3 cos β = 0 −1 −5 5 tan γ 0
10.9 System of Homogeneous Linear Equations 10.55
The augmented matrix of the system is 1 2 3 0 2 5 3 0 −1 −5 5 0 Reducing the augmented matrix to reduced row echelon form, R2 − 2 R1 , R3 + R1 1 2 3 0 ~ 0 1 −3 0 0 −3 8 0 R3 + 3R2 3 0 1 2 ~ 0 1 −3 0 0 0 −1 0 ( −1) R3 1 2 3 0 ~ 0 1 −3 0 0 0 1 0 R2 + 3R3 , R1 − 3R3 1 2 0 0 ~ 0 1 0 0 0 0 1 0 R1 − 2 R2 1 0 0 0 ~ 0 1 0 0 0 0 1 0 The corresponding system of equations is sin α = 0 cos β = 0 tan γ = 0
10.56
Chapter 10
Matrices
From these equations,
α = 0, π , 2π π 3π β= , 2 2 γ = 0, π , 2π
[∵ α , β and γ lie between 0 and 2π ]
Hence, there are 3 · 2 · 3 = 18 possible solutions which satisfy the system of equations.
example 5 For what value of l does the following system of equations possess a non-trivial solution? Obtain the solution for real values of l. x + 2 y + 3z = l x 3x + y + 2 z = l y 2x + 3y + z = l z Solution The system of equations is (1 - l ) x + 2 y + 3 z = 0 3 x + (1 - l ) y + 2 z = 0 2 x + 3 y + (1 - l ) z = 0 The matrix form of the system is 2 3 x 0 1 − l 3 l 1 − 2 y = 0 2 3 1 − l z 0 The system will possess a non-trivial solution if det (A) = 0. 1− λ 3 2
2
3
1− λ 2 =0 3 1− λ
(1 − λ )[(1 − λ ) 2 − 6] − 2(3 − 3λ − 4) + 3(9 − 2 + 2λ ) = 0 (1 − λ )(λ 2 − 2λ − 5) + 2 + 6λ + 21 + 6λ = 0
λ 2 − 2λ − 5 − λ 3 + 2λ 2 + 5λ + 12λ + 23 = 0 − λ 3 + 3λ 2 +115λ + 18 = 0
λ = 6, λ = −1.5 ± 0.866 i
10.9 System of Homogeneous Linear Equations 10.57
For real value of l, i.e. l = 6, the augmented matrix of the system is −5 2 3 0 3 −5 2 0 2 3 −5 0 Reducing the augmented matrix to reduced row echelon form, R2 − R3 −5 2 3 0 ~ 1 −8 7 0 2 3 −5 0 R12 1 −8 7 0 ~ −5 2 3 0 2 3 −5 0 R2 + 5 R1 , R3 − 2 R1 7 0 1 −8 ~ 0 −38 38 0 0 19 −19 0 1 1 − R2 , R3 38 19 1 −8 7 0 ~ 0 1 −1 0 0 1 −1 0 R3 − R2 1 −8 7 0 ~ 0 1 −1 0 0 0 0 0 R1 + 8 R2 1 0 −1 0 ~ 0 1 −1 0 0 0 0 0
10.58
Chapter 10
Matrices
The corresponding system of equations is x −z = 0 y−z =0 Solving for the leading variables, x=z y=z Assigning the free variable z an arbitrary value t, x=t y=t Hence, x = t , y = t , z = t is the non-trivial solution of the system where t is a parameter.
example 6 If the following system has a non-trivial solution, then prove that a + b + c = 0 or a = b = c and hence find the solution in each case. ax + by + cz = 0 bx + cy + az = 0 cx + ay + bz = 0 Solution The matrix form of the system is a b c x 0 b c a y = 0 c a b z 0 The system has a non-trivial solution if det (A) = 0. a b c b c a =0 c a b a (bc − a 2 ) − b(b 2 − ac) + c(ab − c 2 ) = 0 − a 3 + b3 + c 3 − 3abc = 0 −(a + b + c)(a 2 + b 2 + c 2 − ab − bc − ca ) = 0
10.9 System of Homogeneous Linear Equations 10.59
a+b+c = 0 2
or
2
2
a + b + c − ab − bc − ca = 0 1 [( a − b) 2 + (b − c) 2 + (c − a) 2 ] = 0 2 a − b = 0, b − c = 0, c − a = 0 a = b, b = c , c = a a=b=c
Hence, the system has a non-trivial solution if a + b + c = 0 or a = b = c. The augmented matrix of the system is a b c 0 b c a 0 c a b 0 R3 + R1 + R2 b c 0 a ~ b c a 0 a + b + c a + b + c a + b + c 0 The corresponding system of equations is ax + by + cz = 0 bx + cy + az = 0 (a + b + c) x + (a + b + c) y + (a + b + c) z = 0 (i) When a + b + c = 0, we have only two equations. ax + by + cz = 0 bx + cy + az = 0 x b c c a
y =−
a c b a
z =
a b b c
=t
x y z =− 2 = =t 2 ab − c a − bc ac − b 2 Hence, x = (ab − c 2 )t , y = (bc − a 2 )t , z = (ac − b 2 )t is the solution of the system where t is a parameter. (ii) When a = b = c, we have only one equation. x+ y+ z = 0
10.60
Chapter 10
Matrices
y = t1
Let
z = t2
Then
x = −t1 − t2
Hence, x = −t1 − t2 , y = t1 , z = t2 is the solution of the system where t1 and t2 are parameters.
example 7 Discuss for all values of k, the system of equations 2x + 3ky + (3k + 4) z = 0 x + (k + 4) y + (4k + 2) z = 0 x + 2(k + 1) y + (3k + 4) z = 0 Solution The matrix form of the system is 3k 3k + 4 x 0 2 1 k + 4 4k + 2 y = 0 1 2k + 2 3k + 4 z 0 R12 1 k + 4 4k + 2 x 0 2 3k 3k + 4 y = 0 1 2k + 2 3k + 4 z 0 R2 − 2 R1 , R3 − R1 1 k + 4 4k + 2 x 0 0 k − 8 −5k y = 0 0 k − 2 − k + 2 z 0 1 k + 4 4k + 2 det ( A) = 0 k − 8 −5k 0 k − 2 −k + 2 = (k − 8)( − k + 2) + 5k (k − 2) = (k − 2)( − k + 8 + 5k ) = 4(k − 2)(k + 2)
10.9 System of Homogeneous Linear Equations 10.61
(i) When k ≠ ± 2, det ( A) ≠ 0, the system has a trivial solution, i.e. x = 0, y = 0, z = 0. (ii) When k = ±2, det ( A) = 0, the system has non-trivial solutions. Case I: When k = 2, the augmented matrix of the system is 10 0 1 6 0 −6 −10 0 0 0 0 0 1 − R2 6 1 6 10 10 ~ 0 1 6 0 0 0
0 0 0
R1 − 6 R2 1 0 0 10 ~ 0 1 6 0 0 0
0 0 0
The corresponding system of equations is x=0 10 y+ z = 0 6 Solving for the leading variables, x=0 y=−
10 z 6
Assigning the free variable z any arbitrary value t, y=−
10 5 t=− t 6 3
5 Hence, x = 0, y = − t , z = t is the solution of the system where t is a parameter. 3
10.62
Chapter 10
Matrices
Case II: When k = –2, the augmented matrix of the system is 2 −6 0 1 0 −10 10 0 0 − 4 4 0 Reducing the augmented matrix to reduced row echelon form, 1 1 − R2 , − R3 10 4 1 2 −6 0 ~ 0 1 −1 0 0 1 −1 0 R3 − R2 1 2 −6 0 ~ 0 1 −1 0 0 0 0 0 R1 − 2 R2 1 0 − 4 0 ~ 0 1 −1 0 0 0 0 0 The corresponding system of equations is x − 4z = 0 y− z =0 Solving for the leading variables, x = 4z y=z Assigning the free variable z any arbitrary value t, x = 4t y=t Hence, x = 4t , y = t , z = t is the solution of the system where t is a parameter.
10.9 System of Homogeneous Linear Equations 10.63
eXercISe 10.4 1. Solve the following equations: (i)
x- y+ z=0 x + 2y + z = 0 2 x + y + 3z = 0
(iii)
2 x - 2y + 5z + 3w = 0 4x - y + z + w = 0 3 x - 2 y + 3 z + 4w = 0 x - 3y + 7 z + 6w = 0
(v) 7 x + y - 2 z = 0 x + 5y - 4 z = 0 3x - 2y + z = 0 2 x - 7 y + 5z = 0 (vii)
(ii)
x - 2y + 3z = 0 2 x + 5y + 6 z = 0
(iv) 2 x - y + 3z = 0 3x + 2y + z = 0 x - 4 y + 5z = 0 (vi) 3x + 4 y -
z - 9w = 0
2 x + 3y + 2 z - 3w = 0 2 x + y - 14 z - 12w = 0 x + 3y + 13z + 3w = 0
x1 + 2 x 2 + 3x 3 + x 4 = 0 (viii) 2 x1 - x 2 + 3x 3 = 0 x1 + x 2 - x 3 - x 4 = 0 3 x1 + 2 x 2 + x 3 = 0 3 x1 - x 2 + 2 x 3 + 3 x 4 = 0
x1 - 4 x 2 + 5 x 3 = 0
(ii) x = -3t, y = 0, z = t ˘ È Ans.: (i) x = 0, y = 0, z = 0 Í ˙ 211 7 Í ˙ (iii) x = t, y = 4t, z = t, w = t Í ˙ 9 9 Í ˙ 3 13 Í (iv) x = -t, y = t, z = t (v) x = t, y = t, z = t ˙ 17 17 Í ˙ Í ˙ (vi) x = 11t, y = -8t, z = t, w = 0 Í ˙ 1 2 2 Í ˙ (vii) x1 = - t, x 2 = t, x 3 = - t, x 4 = t Í ˙ 3 3 3 Í ˙ ( ) x = x = x = viii t 1 2 3 Î ˚ 2. For what value of l does the following system of equations possess a nontrivial solution? Obtain the solution for real values of l. 2x2 + 3x 3 = 0 (i) 3x + y - l z = 0 (ii) (1 - l)x1 + 4 x - 2y - 3z = 0 2l x + 4 y - l z = 0
3x1 + (1 - l)x 2 + 2 x1 +
2x3 = 0
3x 2 + (1 - l)x 3 = 0
È Ans.: ˘ Í ˙ Í (i) Non-trivial solution l = 1, -9 ˙ Í For l = 1, x = -t, y = -t, z = -2t ˙ Í ˙ For l = -9, x = -3t, y = -9t, z = 2t ˙ Í Í(ii) l = 6, x = y = z = t ˙ Î ˚
10.64
Chapter 10
Matrices
3. Show that the system of equations 2 x - 2y + z = l x, 2 x - 3y + 2 z = l y , - x + 2y = l z can posses a non-trivial solution only if l = 1, l = –3. Obtain the general solution in each case. È Ans. : For l = 1, x = 2t2 - t1 y = t2 , z = t1 ˘ Í ˙ For l = - 3, x = -t, y = -2t, z = t ˚ Î
10.10 eIGenvALueS And eIGenvectorS Eigenvalues and eigenvectors are important concepts in linear algebra. They are derived from the German word ‘eigen’ which means proper or characteristic. Eigenvectors are nonzero vectors that get mapped into scalar multiples of themselves under a linear operator. Any nonzero vector x is said to be a characteristic vector or eigenvector of a matrix A if there exists a number l such that Ax = λ x x1 x 2 where A = [aij ]n× n is an n-rowed square matrix and x = is a non-zero column xn vector. Also, l is said to be characteristic root or characteristic value or eigenvalue of the matrix A. Depending on the sign and the magnitude of the eigenvalue l corresponding to x, the linear operator Ax = l x compresses or stretches the eigenvector x by a factor l. If l is negative, the direction of the eigenvector reverses. (Fig. 10.1). Now Ax = λ x = λ Ix ( A - λ I )x = 0 lx
x lx
x
x lx
(i) 0 ≤ l ≤ 1
x
(ii) l ≥ 1
(iii) −1 ≤ l ≤ 0
lx (iv) l ≤ −1
fig. 6.1
The matrix A – l I is called the characteristic matrix of A where I is the unit matrix of order n.
10.10 Eigenvalues and Eigenvectors 10.65
a11 - λ det( A - λ I ) =
a21 an1
a12
a1n a22 - λ a2 n an 2 ann - λ
which is an ordinary polynomial in l of degree n, is called the characteristic polynomial of A. The equation det(A – l I ) = 0 is called the characteristic equation of A and the roots of this equation are called the eigenvalues of the matrix A. The set of all eigenvectors is called the eigenspace of A corresponding to l. The set of all eigenvalues of A is called the spectrum of A.
Note: (1) The characteristic equation of the matrix A of order 2 can be obtained from l 2 – S1 l + S2 = 0 where S1 = sum of principal diagonal elements and S2 = determinant A (2) The characteristic equation of the matrix A of order 3 can be obtained from l 3 – S1 l2 + S2 l – S3 = 0 where S1 = sum of principal diagonal elements, S2 = sum of minors of principal diagonal elements and S3 = determinant A (3) The sum of the eigenvalues of a matrix is the sum of its principal diagonal elements. (4) The product of the eigenvalues of a matrix is the determinant of the matrix.
10.11 propertIeS of eIGenvALueS Property 1: If l is an eigenvalue of the matrix A then l is also an eigenvalue of AT. Proof: Let l be an eigenvalue of the matrix A. The characteristic equation of A is det (A – l I ) = 0. The characteristic equation of AT is det (AT – l I ) = 0. The determinant value does not change by the interchange of rows and columns. det ( A - λ I ) = det ( AT - λ I ) The characteristic equations are same for both A and AT. Hence, l is also an eigenvalue of AT. 1 Property 2: If l is an eigenvalue of the non-singular matrix A then is an –1 λ eigenvalue of A . Proof: Let l be an eigenvalue of the non-singular matrix A. Ax = λ x
…(10.1)
10.66
Chapter 10
Matrices
Premultiplying both sides of Eq. (10.1) by A–1, A-1 Ax = A-1λ x Ix = λ A-1x x = λ A-1x 1 A-1x = x λ 1 is an eigenvalue of A–1. λ Property 3: If l is an eigenvalue of the matrix A then lk is an eigenvalue of Ak. Hence,
Proof: Let l be an eigenvalue of the matrix A. Ax = λ x
…(10.2)
Premultiplying both sides of Eq. (10.2) by A, AAx = Al x A2 x = l ( Ax) = l ( l x) = l 2x
Similarly, A3x = l3x In general, Akx = lkx Hence, lk is an eigenvalue of Ak.
Property 4: If l is an eigenvalue of the matrix A then l ± k is an eigenvalue of A ± kI. Proof: Let l be an eigenvalue of the matrix A. Ax = λ x
…(10.3)
Adding kIx on both sides of Eq. (10.3), Ax + kIx = λ x + kIx ( A + kI )x = λ x + k x = (λ + k )x Similarly, ( A - kI )x = (λ - k )x In general, ( A ± kI ) x = (λ ± k ) x Hence, l ± k is an eigenvalue of A ± kI.
Property 5: If l is an eigenvalue of the matrix A then kl is an eigenvalue of kA. Proof: Let l be an eigenvalue of the matrix A. Ax = λ x Multiplying both sides of Eq. (10.4) by the scalar k,
…(10.4)
10.10 Eigenvalues and Eigenvectors 10.67
k Ax = k λ x Hence, kl is an eigenvalue of kA. Property 6: The eigenvalues of a triangular matrix are the diagonal elements of the matrix. a11 a12 a1n 0 a a2 n 22 Proof: Let A = be a triangular matrix of order n. 0 0 ann The characteristic equation is det( A - λ I ) = 0 a11 - λ 0 0
a1n a2 n =0 0 ann - λ (a11 - λ )(a22 - λ ) … (ann - λ ) = 0 a12 a22 - λ
λ = a11 , a22 , …, ann Hence, the eigenvalues of A are a11, a22, … , ann which are the diagonal elements of the matrix. Property 7: The eigenvalues of a real symmetric matrix are real. Proof: Let l be an eigenvalue of the real symmetric matrix. Ax = λ x
... (10.5)
Premultiplying both sides of Eq. (10.5) by xT , xT Ax = λ xT x
... (10.6)
Taking complex conjugate on both sides of Eq. (10.6), xT A x = λ xT x xT Ax = λ xT x (∵ A = A for real matrix)
... (10.7)
Taking transpose on both sides of Eq. (10.7), xT AT x = λ xT x T xT Ax = λ xT x (∵ A = A for symmetric matrix) ... (10.8) From Eqs (10.6) and (10.8), λxT x = λ xT x
(λ - λ ) x T x = 0
10.68
Chapter 10
Matrices
xT x is a 1 × 1 matrix, i.e., a single element which is positive,
λ -λ = 0 i.e., l is real. Hence, the eigenvalues of a real symmetric matrix are real.
example 1
2 -3 Find the sum and product of the eigenvalues of the matrix A = . 4 -2 Solution Sum of the eigenvalues of A = Sum of principal diagonal elements of A =2–2 =0 Product of the eigenvalues of A = Determinant of A 2 -3 = 4 -2 = -4 + 12 =8
example 2 1 2 5 Find the sum and product of the eigenvalues of the matrix A = 2 3 4 . 3 6 7 Solution Sum of the eigenvalues of A = Sum of principal diagonal elements of A =1+3+7 = 11 Product of the eigenvalues of A = Determinant of A 1 2 5 = 2 3 4 3 6 7 = 1(21 - 24) - 2(14 - 12) + 5(12 - 9) = -3 - 4 + 15 =8
10.10 Eigenvalues and Eigenvectors 10.69
example 3
6 -2 2 The product of two eigenvalues of the matrix A = -2 3 -1 is 16. 2 -1 3 Find the third eigenvalue. Solution Let l 1, l 2 and l 3 be the eigenvalues of the matrix A. l 1l 2 = 16 Product of the eigenvalues of A = Determinant of A 6 -2 2 λ1λ 2 λ3 = -2 3 -1 2 -1 3 16λ3 = 6(9 - 1) + 2( - 6 + 2) + 2(2 - 6) = 48 - 8 - 8 λ3 = 2
example 4
2 2 1 1 Two eigenvalues of the matrix A = 1 3 1 are equal and are times 5 1 2 2 to the third. Find the eigenvalues. Solution Let l 1, l 2 and l 3 be the eigenvalues of the matrix A. λ1 = λ 2 1 λ1 = λ3 5 1 λ 2 = λ3 5 Sum of the eigenvalues of A = Sum of principal diagonal elements of A λ1 + λ 2 + λ3 = 2 + 3 + 2 1 1 λ3 + λ3 + λ3 = 7 5 5 7 λ3 = 7 5 λ3 = 5 ∴ λ1 = λ 2 = 1 Hence, the eigenvalues of A are 1, 1, 5.
10.70
Chapter 10
Matrices
example 5
2 -2 2 If 2 is an eigenvalue of the matrix A = 1 1 1 , find the other two 1 3 -1 eigen values. Solution Let l 1, l 2 and l 3 be the eigenvalues of the matrix A. l1 = 2 Sum of the eigenvalues of A = Sum of principal diagonal elements of A 2 + l2 + l3 = 2 + 1 – 1 l2 + l3 = 0
…(1)
Product of the eigenvalues of A = Determinant of A 2 -2 2 2λ 2 λ 3 = 1 1 1 1 3 -1 = 2( -1 - 3) + 2( -1 - 1) + 2(3 - 1) = -8 - 4 + 4 = -8 l2l3 = – 4
…(2)
Solving Eqs (1) and (2), l 2 = 2, l 3 = – 2 Hence, the other two eigenvalues are 2, –2.
example 6 For a given matrix A of order 3, | A| = 32 and two of its eigenvalues are 8 and 2. Find the sum of the eigenvalues. Solution Let l 1, l 2 and l 3 be the eigenvalues of the matrix A. l1 = 8, l 2 = 2 Product of the eigenvalues of A = Determinant of A l 1l 2l 3 = A 8 × 2 × l 3 = 32 l3 = 2 Hence, the sum of the eigenvalues = 8 + 2 + 2 = 12
10.10 Eigenvalues and Eigenvectors 10.71
example 7 For a singular matrix of order three, 2 and 3 are the eigenvalues. Find its third eigenvalue. Solution Let l1, l 2 and l 3 be the eigenvalues of the matrix A. l1 = 2, l2 = 3 For a singular matrix, | A| = 0 Product of the eigenvalues of A = Determinant of A l 1l 2l 3 =A 2 × 3 × l3 = 0 l3 = 0 Hence, the third eigenvalue = 0
example 8
2 0 1 If 2, 3 are the eigenvalues of 0 2 0 , find the value of a. a 0 2 Solution Let l1, l 2 and l 3 be the eigenvalues of the matrix A. l1 = 2, l 2 = 3 Sum of the eigenvalues of A = Sum of principal diagonal elements of A 2 + 3 + l3 = 2 + 2 + 2 l3 = 1 Product of the eigenvalues of A = Determinant of A
λ1λ 2 λ3 = A 2 0 1 2 × 3 ×1 = 0 2 0 a 0 2 6 = 2(4 - 0) - 0(0 - 0) + 1(0 - 2a ) = 8 - 2a 2a = 2 a =1
10.72
Chapter 10
Matrices
example 9
2 8 -6 7 - 4 , find | A| If 3 and 15 are the two eigenvalues of A = - 6 2 - 4 3 without expanding the determinant. Solution Let l1, l 2 and l 3 be the eigenvalues of the matrix A. l1 = 3, l 2 = 15 Sum of the eigenvalues of A = Sum of principal diagonal elements of A 3 + 15 + l 3 = 8 + 7 + 3 l3 = 0 Product of the eigenvalues of A = Determinant of A 3 × 15 × 0 = A A = 0
example 10
a 4 Find a and b such that A = has 3 and –2 as eigenvalues. 1 b Solution Let l1 and l 2 be the eigenvalues of the matrix A. l1 = 3, l 2 = –2 Sum of the eigenvalues of A = Sum of principal diagonal elements of A 3–2=a+b a+b=1
…(1)
Product of the eigenvalues of A = Determinant of A
λ1λ 2 = A (3)( -2) =
a 4 1 b
= ab - 4 ab = –2 From Eq. (1),
a=1–b
…(2)
10.10 Eigenvalues and Eigenvectors 10.73
Substituting in Eq. (2), (1 - b)b = -2 b - b 2 = -2 b2 - b - 2 = 0 b = 2 or b = -1 ∴ a = -1 or a = 2
example 11
2 2 1 Two eigenvalues of the matrix A = 1 3 1 are 1 and 1. Find the eigenvalues of A–1. 1 2 2 Solution Let l1, l 2 and l 3 be the eigenvalues of the matrix A. l1 = l 2 = 1 Sum of the eigenvalues of A = Sum of principal diagonal elements of A 1 + 1 + l3 = 2 + 3 + 2 l3 = 5 Hence, the eigenvalues of A are 1, 1, 5. 1 1 1 1 The eigenvalues of A–1 are , , , i.e., 1, 1, . 1 1 5 5
example 12
3 -1 1 Two of the eigenvalues of the matrix A = -1 5 -1 are 3 and 6. –1 3 1 1 3 Find the eigenvalues of A and A . Solution Let l1, l 2 and l 3 be the eigenvalues of the matrix A. l1 = 3, l 2 = 6 Sum of the eigenvalues of A = Sum of the principal diagonal elements of A 3 + 6 + l3 = 3 + 5 + 3 l3 = 2 The eigenvalues of A are 3, 6, 2.
10.74
Chapter 10
Matrices
1 1 1 Hence, the eigenvalues of A–1 are , , , and the eigenvalues of A3 are 33, 63, 23, i.e., 3 6 2 27, 216, 8.
example 13 Form the matrix whose eigenvalues are a – 5, b – 5, g – 5 where a, b, -1 -2 -3 g are the eigenvalues of A = 4 5 -6 . 7 -8 9 Solution If l1, l 2 and l 3 are eigenvalues of the matrix A then l1 – k, l 2 – k and l 3 – k are the eigenvalues of A – kI. 1 0 0 -1 -2 -3 5 - 6 - 5 0 1 0 Required matrix = A - 5 I = 4 0 0 1 7 -8 9 - 6 -2 -3 0 - 6 = 4 7 -8 4
example 14
3 -1 If a and b are the eigenvalues of , form a matrix whose -1 5 3 3 eigenvalues are a and b . Solution If l1 and l 2 are the eigenvalues of the matrix A then λ1k and λ 2k are the eigenvalues of Ak. 3 -1 Let A= -1 5 3 -1 3 -1 10 -8 A2 = AA = = -1 5 -1 5 -8 26 10 -8 3 -1 38 -50 Required matrix x = A3 = A2 A = = -8 26 -1 5 -50 138
10.10 Eigenvalues and Eigenvectors 10.75
example 15 3 -1 1 If A = -1 5 -1 , find the eigenvalues for the following matrices: 1 -1 3 (i) A (ii) AT (iii) A–1 (iv) 4A–1 (v) A2 (vi) A2 – 2A + I (vii) A3 + 2I Solution
3 -1 1 A = -1 5 -1 1 -1 3
The characteristic equation is
det( A - λ I ) = 0 3- λ -1 -1 5 - λ -1
1
1 -1 = 0 3-λ
λ - S1λ + S 2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 3 + 5 + 3 = 11 S2 = Sum of the minors of principal diagonal elements of A =
5 -1 -1
3
+
3 1 1 3
+
3 -1 -1
5
= (15 - 1) + (9 - 1) + (15 - 1) = 14 + 8 + 14 = 36 3 -1 1 S3 = det( A) = -1 5 -1 1 -1
3
= 3(15 - 1) + 1( -3 + 1) + 1(1 - 5) = 42 - 2 - 4 = 36 Hence, the characteristic equation is l3 – 11l2 + 36l – 36 = 0 l = 2, 3, 6 (i) Eigenvalues of A = l (ii) Eigenvalues of AT = l
: :
(iii) Eigenvalues of A-1 = λ -1
:
2, 3, 6 2, 3, 6 1 1 1 , , 2 3 6
10.76
Chapter 10
Matrices
(iv) Eigenvalues of 4A-1 = 4λ -1 (v) Eigenvalues of A2 = l2 (vi) Eigenvalues of A2 – 2A + I = l2 – 2l + 1 (vii) Eigenvalues of A3 + 2I = l3 + 2
10.12
10.12.1
4 2 2, , 3 3 : 4, 9, 36 : 1, 4, 25 : 10, 29, 218
:
LIneAr dependence And Independence of eIGenvectorS Linear dependence
A set of r eigenvectors x1, x2, … xr is said to be linearly dependent if there exist r scalars (numbers) k1, k2, … kr not all zero, such that k1x1 + k2 x 2 + + kr x r = 0
10.12.2
Linear Independence
A set of r eigenvectors x1, x2, … xr is said to be linearly independent if there exist r scalars (numbers) k1, k2, … kr such that if k1x1 + k2 x 2 + + kr x r = 0 then k1 = k2 = … = kr = 0
Note: (1) If a set of eigenvectors is linearly dependent then at least one eigenvector of the set can be expressed as a linear combination of the remaining eigenvectors. (2) If a set of eigenvectors is linearly independent then no eigenvector of the set can be expressed as a linear combination of the remaining eigenvectors.
10.13
propertIeS of eIGenvectorS
Property 1: If x is an eigenvector of a matrix A corresponding to the eigenvalue l then k x is also an eigenvector of A corresponding to the same eigenvalue l , where k is a nonzero scalar. Proof: Let x be an eigenvector of a matrix A corresponding to the eigenvalue l . Ax = lx and x≠0 If k is any nonzero scalar then k x ≠ 0. Also, A(k x) = k (Ax) = k (l x) = l (k x) Hence, k x is an eigenvector of A corresponding to the eigenvalue l . Thus, corresponding to an eigenvalue l , there is more than one eigenvectors.
10.13 Properties of Eigenvectors 10.77
Property 2: If x is an eigenvector of a matrix A then x cannot correspond to more than one eigenvalue of A. Proof: Let x be an eigenvector of a matrix A corresponding to two eigenvalues l 1 and l 2. Ax = l1x and Ax = l 2x ∴ λ1x = λ 2 x
( λ1 - λ2 ) x = 0
λ1 - λ 2 = 0 (∵ x ≠ 0) λ1 = λ 2 Hence, x cannot correspond to more than one eigenvalue of A. Property 3: The eigenvectors corresponding to distinct eigenvalues of a matrix are linearly independent. Proof: Let x1, x2, …, xm be the eigenvectors of a matrix A corresponding to distinct eigenvalues l1, l 2, …, lm. Axi = λi xi , i = 1, 2, …, m Let the eigenvectors x1, x2, …, xm be linearly dependent. Choose r such that the eigenvectors x1, x2, …, xr are linearly independent but x1, x2, …, xr, xr + 1 are linearly dependent, where 1 ≤ r < m. Then
k1x1 + k2 x 2 + + kr x r + kr +1x r +1 = 0
... (10.9)
where k1, k2, …, kr, kr + 1 are scalars not all zero. A(k1x1 + k2 x 2 + + kr x r + kr +1x r +1 ) = 0 k1 ( Ax1 ) + k2 ( Ax 2 ) + + kr ( Ax r ) + kr +1 ( Ax r +1 ) = 0 k1 (λ1x1 ) + k2 (λ 2 x 2 ) + + kr (λ r x r ) + kr +1 (λ r +1x r +1 ) = 0
... (10.10)
Multiplying Eq. (10.9) by lr + 1 and subtracting from Eq. (10.10), k1(λ1 - λ r +1 )x1 + k 2 (λ 2 - λ r +1 )x 2 + + k r (λ r - λ r +1 )x r = 0
... (10.11)
Since x1, x2, …, xr are linearly independent, k1(l1 – lr+1) = 0, k2(l 2 – lr + 1) = 0, … kr(lr – lr + 1) = 0 But l 1, l 2, …, l r, l r + 1 are distinct. ∴ k1 = 0, k2 = 0, …, kr = 0 Putting in Eq. (10.9), kr +1x r +1 = 0 kr +1 = 0 [∵ x r +1 ≠ 0] Thus, for Eq. (10.9), k1 = 0, k2 = 0, kr = 0, kr + 1 = 0 which contradicts our assumption that k1, k2, …, kr , kr + 1 are not all zero. Thus, x1, x2, …, xr + 1 are linearly independent and hence x1, x2, …, xm are linearly independent.
10.78
Chapter 10
Matrices
Note: If two or more eigenvalues are equal then the corresponding eigenvectors may or may not be linearly independent. Property 4: The eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal. Proof: Let l1 and l2 be the two distinct eigenvalues of a real symmetric matrix A and x1 and x2 be the corresponding eigenvectors respectively. and
Ax1 = λ1x1
…(10.12)
Ax2 = l 2x2
…(10.13)
T 2
Premultiplying both sides of Eq. (10.12) by x , xT2 Ax1 = λ1xT2 x1 Taking the transpose on both sides, x1T Ax 2 = λ1x1T x 2
(∵ AT = A)
…(10.14)
Premultiplying both sides of Eq. (10.13) by x1T , x1T Ax 2 = λ 2 x1T x 2 From Eqs (10.14) and (10.15),
…(10.15)
λ1x1T x 2 = λ 2 x1T x 2
(λ1 - λ 2 )x1T x 2 = 0 x1T x 2 = 0 (∵ λ1 ≠ λ 2 ) Hence, the eigenvectors x1 and x2 are orthogonal. working rule for finding the eigenvalues and eigenvectors (i) Write characteristic equation det (A – l I ) = 0 for the given square matrix. (ii) Find the eigenvalues of the matrix by solving characteristic equation. (iii) Find eigenvectors corresponding to each eigenvalues from the equation [A – l I ] x = 0.
example 1
5 4 Find the eigenvalues and eigenvectors of the matrix . 1 2 Solution Let The characteristic equation is
5 4 A= 1 2
det( A - λ I ) = 0 5-λ 1
4 =0 2-λ
λ 2 - S1λ + S 2 = 0
10.13 Properties of Eigenvectors 10.79
where S1 = Sum of the principal diagonal elements of A = 5 + 2 = 7 5 4 S 2 = det( A) = 1 2 = 10 - 4 =6 Hence, the characteristic equation is l2 – 7l2 + 6 = 0 l = 6, 1 (a) For l = 6,
[A – l I ]x = 0 4 x 0 -1 1 - 4 y = 0 -x + 4y = 0
Let
y=t x = 4t
4 x 4t x = = = t = t x1 where x1 is an eigenvector corresponding to l = 6. 1 y t (b) For l = 1,
[A – lI ]x = 0 4 4 x 0 1 1 y = 0 x+ y = 0
Let
y=t x = –t
-1 x -t x = = = t = t x 2 where x2 is an eigenvector corresponding to l = 1. 1 y t
example 2
6 6 4 Find the eigenvalues and eigenvectors of the matrix 1 3 2 . -1 - 4 -3 Solution Let
6 6 4 3 2 A= 1 -1 - 4 -3
10.80
Chapter 10
Matrices
The characteristic equation is det( A - λ I ) = 0 4-λ 6 1 3-λ -1 -4
6 2 =0 -3 - λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 4 + 3 – 3 = 4 S2 = Sum of the minors of principal diagonal elements of A =
3 2 4 6 4 6 + + - 4 -3 -1 -3 1 3
= ( -9 + 8) + ( -12 + 6) + (12 - 6) = -1 - 6 + 6 = -1 4 S3 = det( A) = 1
6 3
6 2
-1 - 4 -3 = 4( -9 + 8) - 6( -3 + 2) + 6( - 4 + 3) = -4 + 6 - 6 = -4 Hence, the characteristic equation is l3 – 4l2 – l + 4 = 0 l = –1, 1, 4 (a) For l = –1,
[A – lI ]x = 0 6 6 x 0 5 1 4 2 y = 0 -1 - 4 -2 z 0 5x + 6 y + 6 z = 0 x + 4 y + 2z = 0
By Cramer’s rule, x 6 6 4 2
=-
y 5 6 1 2
=
z 5 6 1 4
=t
10.13 Properties of Eigenvectors 10.81
x y z = = =t -12 - 4 14 x y z = = =t - 6 -2 7 - 6 x - 6t x = y = -2t = t -2 = tx1 where x1 is an eigenvector corresponding to l = –1. 7 z 7t (b) For l = 1,
[A – l I ]x = 0 6 6 x 0 3 1 2 2 y = 0 -1 - 4 - 4 z 0 x + 2 y + 2z = 0 - x - 4 y - 4z = 0
By Cramer’s rule, x y z == =t 2 2 1 2 1 2 -4 -4 -1 - 4 -1 - 4 x = 0 x = 0
y z = =t 2 -2 y z = =t 1 -1
0 x 0 x = y = t = t 1 = tx 2 where x2 is an eigenvector corresponding to l = 1. -1 z -t (c) For l = 4,
[A – lI ]x = 0 6 6 x 0 0 1 -1 2 y = 0 -1 - 4 -7 z 0 0x + 6 y + 6z = 0 x - y + 2z = 0 x -4 y - 7z = 0
By Cramer’s rule, x y z == =t 6 6 0 6 0 6 -1 2 1 2 1 -1
10.82
x = 18 x = 3
Chapter 10
Matrices
y z = =t 6 -6 y z = =t 1 -1
3 x 3t x = y = t = t 1 = tx3 where x3 is an eigenvector corresponding to l = 4. -1 z -t
example 3
1 0 -1 Find the eigenvalues and eigenvectors of the matrix 1 2 1 . 2 2 3 Solution 1 0 -1 1 A = 1 2 2 2 3
Let The characteristic equation is
det( A - λ I ) = 0 1- λ 1
0 2-λ
2
2
-1 1 =0 3- λ
λ - S1λ + S 2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 1 + 2 + 3 = 6 S2 = Sum of the minors of principal diagonal elements of A =
2 1 2 3
+
1 -1 2
3
+
1 0 1 2
= (6 - 2) + (3 + 2) + (2 - 0) = 4+5+ 2 = 11 1 0 -1 S3 = det( A) = 1 2 1 2 2 3 = 1(6 - 2) - 0 - 1(2 - 4) =6
10.13 Properties of Eigenvectors 10.83
Hence, the characteristic equation is l3 – 6l2 + 11l – 6 = 0 l = 1, 2, 3 (a) For l = 1,
[A – l I ]x = 0 0 0 -1 x 0 1 1 1 y = 0 2 2 2 z 0 0x + 0 y - z = 0 x+ y + z = 0
By Cramer’s rule, x 0 -1 1 1
=-
y z = =t 0 -1 0 0 1 1 1 1 x y z = = =t 1 -1 0
1 x t x = y = -t = t -1 = tx1 where x1 is an eigenvector corresponding to l = 1. 0 z 0 (b) For l = 2,
[A – l I ]x = 0 -1 0 -1 x 0 1 0 1 y = 0 2 2 1 z 0 x + 0y + z = 0 2x + 2 y + z = 0
By Cramer’s rule, x y z == =t 0 1 1 1 1 0 2 1 2 1 2 2 x y z = = =t -2 1 2 -2 x -2t x = y = t = t 1 = tx 2 where x2 is an eigenvector corresponding to l = 2. 2 z 2t
10.84
Chapter 10
Matrices
(c) For l = 3,
[A – lI ]x = 0 -2 0 -1 x 0 1 -1 1 y = 0 2 2 0 z 0 -2 x + 0 y - z = 0 x- y+ z=0 2x + 2 y + 0z = 0
By Cramer’s rule, x y z == =t 0 -1 -2 -1 -2 0 -1 1 1 1 1 -1 x y z = = =t -1 1 2 -1 x -t x = y = t = t 1 = tx3 where x3 is an eigenvector corresponding to l = 3. 2 z 2t
example 4
2 8 -6 Find the eigenvalues and eigenvectors of the matrix -6 7 -4 . 2 -4 3 Solution Let
2 8 -6 7 - 4 A = -6 2 - 4 3
The characteristic equation is det( A - λ I ) = 0 8 - λ -6 -2 -6 7 - λ -4 = 0 2 -4 3 - λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 8 + 7 + 3 = 18 S2 = Sum of the minors of principal diagonal elements of A
10.13 Properties of Eigenvectors 10.85
=
7 -4 -4
3
+
8 2 2 3
+
8 -6 -6
7
= (21 - 16) + (24 - 4) + (56 - 36) = 5 + 20 + 20 = 45 8 -6 2 S3 = det( A) = - 6 7 -4 2 -4 3 = 8(21 - 16) + 6( -18 + 8) + 2(24 - 14) = 40 - 60 + 20 =0 Hence, the characteristic equation is l3 – 18l2 + 45l = 0 l = 0, 3, 15 (a) For l = 0,
[A – l I ]x = 0 2 x 0 8 -6 -6 7 - 4 y = 0 2 - 4 3 z 0 8x - 6 y + 2 z = 0 -6 x + 7 y - 4 z = 0 2 x - 4 y + 3z = 0
By Cramer’s rule, x -6 2 7 -4
=-
y 8 2 -6 - 4
=
z 8 -6 -6 7
=t
x y z = = =t 10 20 20 x y z = = =t 1 2 2 1 x t x = y = 2t = t 2 = tx1 where x1 is an eigenvector corresponding to l = 0. 2 z 2t
10.86
Chapter 10
Matrices
(b) For l = 3,
[A – l I ]x = 0 2 x 0 5 -6 -6 4 - 4 y = 0 2 -4 0 z 0 5x - 6 y + 2 z = 0 -6x + 4 y - 4z = 0 2x - 4 y + 0z = 0
By Cramer’s rule, x -6 2 4 -4
=-
y 5 2 -6 -4
=
z 5 -6 -6 4
=t
x y z = = =t 16 8 -16 x y z = = =t 2 1 -2 2 x 2t x = y = t = t 1 = tx 2 where x2 is an eigenvector corresponding to l = 3. -2 z -2t (c) For l = 15,
[A – lI ]x = 0 2 x 0 -7 - 6 - 6 -8 - 4 y = 0 2 - 4 -12 z 0 -7 x - 6 y + 2 z = 0 -6 x - 8 y - 4 z = 0 2 x - 4 y - 12 z = 0
By Cramer’s rule, x y z = = =t -6 2 -7 2 -7 - 6 -8 - 4 -6 - 4 - 6 -8 x y z == =t 40 40 20 x y z = = =t 2 -2 1
10.13 Properties of Eigenvectors 10.87
2 x 2t x = y = -2t = t -2 = tx3 where x3 is an eigenvector corresponding to l = 15. 1 z t Note: The eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal which can be verified with this example. 2 T x1 x 2 = [1 2 2] 1 = [0] = 0 -2 2 x x3 = [ 2 1 -2] -2 = [0] = 0 1 T 2
1 x x1 = [ 2 -2 1] 2 = [0] = 0 2 T 3
Thus, x1, x2 and x3 are orthogonal to each other.
example 5
4 2 -2 Find the eigenvalues and eigenvectors of the matrix -5 3 2 . -2 4 1 Solution 4 2 -2 A = -5 3 2 -2 4 1
Let The characteristic equation is
det( A - λ I ) = 0 4-λ 2 -5 3 - λ -2
4
-2 2 =0 1- λ
λ - S1λ + S 2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 4 + 3 + 1 = 8 S2 = Sum of the minors of principal diagonal elements of A 3 2 4 -2 4 2 = + + 4 1 -2 1 -5 3 = (3 - 8) + (4 - 4) + (12 + 10) = -5 + 0 + 22 = 17
10.88
Chapter 10
Matrices
4 2 -2 S3 = det( A) = -5 3 2 -2 4 1 = 4(3 - 8) - 2( -5 + 4) - 2( -20 + 6) = -20 + 2 + 28 = 10 Hence, the characteristic equation is
λ 3 - 8λ 2 + 17λ - 10 = 0 λ = 1, 2, 5 (a) For l = 1,
[A – l I ]x = 0 3 2 -2 x 0 -5 2 2 y = 0 -2 4 0 z 0 3x + 2 y - 2 z = 0 -5 x + 2 y + 2 z = 0 -2 x + 2 y + 0 z = 0
By Cramer’s rule, x y z == =t 2 -2 3 -2 3 2 2 2 -5 2 -5 2 x y z = = =t 8 4 16 x y z = = =t 2 1 4
2 x 2t x = y = t = t 1 = tx1 where vector x1 is an eigenvector corresponding 4 z 4t to l = 1. (b) For l = 2,
[A – l I ]x = 0 2 2 -2 x 0 -5 1 2 y = 0 -2 4 -1 z 0 2 x + 2y - 2 z = 0 -5 x + y + 2 z = 0 -2 x + 2y - z = 0
10.13 Properties of Eigenvectors 10.89
By Cramer’s rule, x y z == =t 2 -2 2 -2 2 2 1 2 -5 2 -5 1 x y z = = =t 6 6 12 x y z = = =t 1 1 2 1 x t x = y t = t 1 = tx 2 where vector x2 is an eigenvector corresponding to l = 2. 2 z 2t (c) For l = 5, [A – l I ]x = 0 -1 2 -2 x 0 -5 -2 2 y = 0 -2 4 - 4 z 0 - x + 2y - 2 z = 0 - 5x - 2 y + 2 z = 0 By Cramer’s rule, x y z == =t 2 -2 -1 -2 -1 2 -2 2 -5 2 -5 -2 x y z = = =t 0 12 12 x y z = = =t 0 1 1 0 x 0 x = y = t = t 1 = tx3 where x3 is an eigenvector corresponding to l = 5. 1 z t
example 6
2 -3 -2 2 1 -6 . Find the eigenvalues and eigenvectors of the matrix -1 -2 0 Solution Let
-2 2 -3 1 - 6 A = 2 -1 -2 0
10.90
Chapter 10
Matrices
The characteristic equation is
det( A - λ I ) = 0 -2-λ 2 2 1- λ -1 -2
-3 -6 = 0 -λ
λ 3 - S1λ 2 + S 2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = –2 + 1 + 0 = –1 S2 = Sum of the minors of principal diagonal elements of A 1 -6 -2 -3 -2 2 = + + -2 0 -1 0 2 1 = (0 - 12) + (0 - 3) + ( -2 - 4) = -12 - 3 - 6 = -21 -2 2 -3 S3 = det( A) = 2 1 -6 -1 -2
0
= ( -2)(0 - 12) - 2(0 - 6) - 3( -4 + 1) = 24 + 12 + 9 = 45 Hence, the characteristic equation is l3 + l2 – 21l – 45 = 0 l = 5, –3, –3 (a) For l = 5, [A – l I ]x = 0 -7 2 -3 x 0 2 - 4 - 6 y = 0 -1 -2 -5 z 0 -7 x + 2 y - 3 z = 0 2x - 4 y - 6z = 0 - x - 2 y - 5z = 0 By Cramer’s rule, x 2 -3 - 4 -6
=-
y -7 -3 2 -6
=
z -7 2 2 -4
=t
x y z = = =t -24 - 48 24
10.13 Properties of Eigenvectors 10.91
x y z = = =t 1 2 -1 1 x t x = y = 2t = t 2 = tx1 where x1 is an eigenvector corresponding to l = 5. -1 z -t [ A - λ I ]x = 0
(b) For l = –3,
1 2 -3 x 0 2 4 -6 y = 0 -1 -2 3 z 0 x + 2y - 3 z = 0 y = t1 and z = t2 x = –2t1 + 3t2
Let
3 -2 x -2t1 + 3t2 -2t1 3t2 t1 x = y = = t1 + 0 = t1 1 + t2 0 = t1x 2 + t2 x3 0 t2 1 0 z t2 where x2 and x3 are linearly independent eigenvectors corresponding to l = -3.
example 7
0 1 1 Find the eigenvalues and eigenvectors of the matrix 1 0 1 . 1 1 0 Solution 0 1 1 A = 1 0 1 1 1 0
Let The characteristic equation is
det (A – l I ) = 0 -λ 1 1 1 -λ 1 =0 1 1 -λ
l3 – S1l2 + S2l – S3 = 0 where S1 = Sum of the principal diagonal elements of A = 0 + 0 + 0 = 0 S2 = Sum of the minors of principal diagonal elements of A
10.92
Chapter 10
=
0 1 1 0
Matrices
+
0 1 1 0
+
0 1 1 0
= (0 - 1) + (0 - 1) + (0 - 1) = -3 0 1 1 S3 = det( A) = 1 0 1 1 1 0 = 0 - 1(0 - 1) + 1(1 - 0) =2 Hence, the characteristic equation is l3 – 3l – 2 = 0 (a) For l = 2,
l = 2, –1, –1 [A – l I ]x = 0 1 1 x 0 -2 1 -2 1 y = 0 1 1 -2 z 0 -2 x + y + z = 0 x - 2y + z = 0 x + y - 2z = 0
By Cramer’s rule, x y z == =t 1 1 -2 1 -2 1 -2 1 1 1 1 -2 x y z = = =t 3 3 3 x y z = = =t 1 1 1 1 x t x = y = t = t 1 = tx1 where x1 is an eigenvector corresponding to l = 2. 1 z t (b) For l = –1, [A – l I ]x = 0 1 1 1 x 0 1 1 1 y = 0 1 1 1 z 0 x+ y+ z = 0
10.13 Properties of Eigenvectors 10.93
y = t1 and z = t2 x = –t1 – t2 -1 -1 x -t1 - t2 -t1 -t2 x = y = t1 = t1 + 0 = t1 1 + t2 0 = t1x 2 + t2 x3 z t2 0 t2 1 0
Let
where x2 and x3 are linearly independent eigenvectors corresponding to l = –1.
example 8
-9 4 4 Find the eigenvalues and eigenvectors of the matrix -8 3 4 . -16 8 7 Solution -9 4 4 Let A = -8 3 4 -16 8 7 The characteristic equation is det( A - λ I ) = 0 -9 - λ 4 -8 3- λ -16
8
4 4
=0
7-λ
λ - S1λ + S 2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = –9 + 3 + 7 = 1 S2 = Sum of the minors of principal diagonal elements of A 3 4 -9 4 -9 4 = + + 8 7 -16 7 -8 3 = (21 - 32) + ( - 63 + 64) + ( -27 + 32) = -11 + 1 + 5 = -5 -9 4 4 S3 = det( A) = -8 3 4 -16 8 7 = -9(21 - 32) - 4( -56 + 64) + 4( - 64 + 48) = 99 - 32 - 64 =3 Hence, the characteristic equation is
λ 3 - λ 2 - 5λ - 3 = 0 λ = -1, - 1, 3
10.94
Chapter 10
Matrices
(a) For l = –1,
[A – l I ] x = 0 -8 4 4 x 0 -8 4 4 y = 0 -16 8 8 z 0 -8 x + 4 y + 4 z = 0
Let
y = t1 and z = t2 x=
1 1 t1 + t2 2 2
1 1 1 1 1 1 t1 + t2 t1 t2 2 x 2 2 2 2 2 x= y = t1 = t1 + 0 = t1 1 + t2 0 = t1x1 + t2 x 2 t2 0 1 0 t2 z where x1 and x2 are linearly independent eigenvectors corresponding to l = –1. (b) For l = 3,
[A – l I ] x = 0 -12 4 4 x 0 -8 0 4 y = 0 -16 8 4 z 0 -12 x + 4 y + 4 z = 0 -8 x + 0 y + 4 z = 0 -16 x + 8 y + 4 z = 0
By Cramer’s rule, x y z == =t 4 4 -12 4 -12 4 0 4 -8 4 -8 0 x y z = = =t 16 16 32 x y z = = =t 1 1 2 1 x t x = y = t = t 1 = tx3 where x3 is an eigenvector corresponding to l = 3. 2 z 2t
10.13 Properties of Eigenvectors 10.95
example 9
1 - 6 - 4 Find the eigenvalues and eigenvectors of the matrix 0 4 2 . 0 - 6 -3 Solution 1 - 6 - 4 4 2 A = 0 0 - 6 -3
Let
The characteristic equation is det( A - λ I ) = 0 1- λ 0
-6 4-λ
-4 2
0
-6
-3 - λ
=0
λ - S1λ + S2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 1 + 4 – 3 = 2 S2 = Sum of the minors of principal diagonal elements of A =
4
2
- 6 -3
+
1 -4 0
-3
+
1 -6 0
4
= ( -12 + 12) + ( -3 + 0) + (4 + 0) =1 1 -6 - 4 S3 = det( A) = 0 4 2 0 -6
-3
= 1( -12 + 12) - 6(0 - 0) - 4(0 - 0) =0 Hence, the characteristic equation is
λ 3 - 2λ 2 + λ = 0 λ = 0, 1, 1 (a) For l = 0,
[A – l I ] x = 0 1 - 6 - 4 x 0 0 4 2 y = 0 0 - 6 -3 z 0 x - 6 y - 4z = 0 0x + 4 y + 2z = 0 0 x - 6 y - 3z = 0
10.96
Chapter 10
Matrices
By Cramer’s rule, x -6 - 4 4 2
=-
y z = =t 1 -4 1 -6 0 2 0 4 x y = = 4 -2 x y = = 2 -1
z =t 4 z =t 2
2 x 2t x = y = -t = t -1 = tx1 where x1 is an eigenvector corresponding to l = 0. 2 z 2t (b) For l = 1,
[A – l I ] x = 0 0 - 6 - 4 x 0 0 3 2 y = 0 0 - 6 - 4 z 0 0x + 3y + 2z = 0
Let
x = t1 and z = t2 2 y = - t2 3 0 t 0 1 x 1 t1 2 2 2 x = y = - t2 = 0 + - t2 = t1 0 + t2 - = t1x 2 + t2 x3 3 3 3 0 z 1 0 t t 2 2
where x2 and x3 are linearly independent eigenvectors corresponding to l = 1.
example 10
1 2 2 Find the eigenvalues and eigenvectors of the matrix 0 2 1 . -1 2 2 Solution 1 2 2 Let
A = 0 2 1 -1 2 2
The characteristic equation is
det ( A - λ I ) = 0 1- λ 0 -1
2 2-λ 2
2 1 =0 2-λ
λ 3 - S1λ 2 + S 2 λ - S3 = 0
10.13 Properties of Eigenvectors 10.97
where S1 = Sum of the principal diagonal elements of A = 1 + 2 + 2 = 5 S2 = Sum of the minors of principal diagonal elements of A 2 1 1 2 1 2 = + + 2 2 -1 2 0 2 = (4 - 2) + (2 + 2) + (2 - 0) = 2+4+2 =8 1 2 2 S3 = det( A) = 0 2 1 -1 2 2 = 1(4 - 2) - 2(0 + 1) + 2(0 + 2) = 2-2+4 =4 Hence, the characteristic equation is λ 3 - 5λ 2 + 8λ - 4 = 0 λ = 1, 2, 2 (a) For l = 1,
[A – l I ] x = 0 0 2 2 x 0 0 1 1 y = 0 -1 2 1 z 0 0x + y + z = 0 -x + 2y + z = 0
By Cramer’s rule,
x y z == =t 1 1 0 1 0 1 2 1 -1 1 -1 2 x y z = = =t -1 -1 1 x y z = = =t 1 1 -1
1 x t x = y = t = t 1 = tx1 where x1 is an eigenvector corresponding to l = 1. -1 z -t (b) For l = 2,
[A – l I ] x = 0
10.98
Chapter 10
Matrices
-1 2 2 x 0 0 0 1 y = 0 -1 2 0 z 0 - x + 2 y + 2z = 0 0x + 0 y + z = 0 - x + 2 y + 0z = 0 By Cramer’s rule,
x y z == =t 2 2 -1 2 -1 2 0 1 0 1 0 0 x y z = = =t 2 1 0
2 x 2t x = y = t = t 1 = tx 2 where x2 is an eigenvector corresponding to l = 2. 0 z 0 Hence, there is only one eigenvector corresponding to repeated root l = 2.
example 11
1 0 0 Find the eigenvalues and eigenvectors of the matrix 0 0 1 . 1 -3 3 Solution 0 1 0 A = 0 0 1 1 -3 3
Let The characteristic equation is
det ( A - λ I ) = 0 -λ 0
1 -λ
0 1
=0
-3 3 - λ
1
λ - S1λ + S 2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 0 + 0 + 3 = 3 S2 = Sum of the minors of principal diagonal elements of A =
0 1 0 0 0 1 + + -3 3 1 3 0 0
= (0 + 3) + (0) + (0) =3
10.13 Properties of Eigenvectors 10.99
0 1 0 S3 = det( A) = 0 0 1 1 -3 3 = 0 - 1(0 - 1) + 0 =1 Hence, the characteristic equation is
For l = 1,
λ 3 - 3λ 2 + 3λ - 1 = 0 λ = 1, 1, 1 [A – l I ]x = 0 -1 1 0 x 0 0 -1 1 y = 0 1 -3 2 z 0 - x + y + 0z = 0 0x - y + z = 0 x - 3y + 2z = 0
By Cramer’s rule, x y z == =t 1 0 -1 0 -1 1 -1 1 0 1 0 -1 x y z = = =t 1 1 1 1 x t x = y = t = t 1 = tx1 where x1 is an eigenvector corresponding to l = 1. 1 z t Hence, there is only one eigenvector corresponding to repeated root l = 1.
example 12 Find the values of m which satisfy the equation A100 x = m x where
1 -1 2 A = 0 -2 -2 . 1 1 0
Solution 1 -1 2 A = 0 -2 -2 1 1 0
10.100 Chapter 10
Matrices
The characteristic equation is det( A - λ I ) = 0 2-λ 0 1
1 -2 - λ 1
-1 -2 = 0 -λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 2 – 2 + 0 = 0 S2 = Sum of the minors of principal diagonal elements of A -2 -2 2 -1 2 1 = + + 1 0 1 0 0 -2 = (0 + 2) + (0 + 1) + ( - 4 - 0) = -1 2 1 -1 S3 = det( A) = 0 -2 -2 1 1 0 = 2(0 + 2) - 1(0 + 2) - 1(0 + 2) = 4-2-2 =0 Hence, the characteristic equation is λ3 - λ = 0 λ = 0, 1, - 1 If l is an eigen value of A, it satisfies the equation Ax = lx. For equation A100 x = m x, m represents eigen values of A100. Eigenvalues of A100 = l100, i.e., 0, 1, 1. Hence, values of m are 0, 1, 1.
example 13 Find
the characteristic roots and characteristic vectors of cos θ - sin θ A= and verify that characteristic roots are of unit sin θ cos θ modulus. Solution cos θ A= sin θ
- sin θ cos θ
10.13 Properties of Eigenvectors 10.101
The characteristic equation is det( A - λ I ) = 0 cos θ - λ - sin θ =0 sin θ cos θ - λ (cos θ - λ ) 2 + sin 2 θ = 0 (cos θ - λ ) 2 = - sin 2 θ cos θ - λ = ± i sin θ λ = cos θ ± i sin θ |λ | = cos 2 θ + sin 2 θ = 1 Hence, the characteristic roots are of unit modules. (a) For l = cos q + isin q,
[A – l I ]x = 0
-i sin θ - sin θ x 0 = sin θ -i sin θ y 0 -i sin θ x - sin θ y = 0 Let
y=t x = it
i x it x = = = t = tx1 where x1 is an eigenvector corresponding to 1 y t l = cos q + isin q. (b) For l = cos q – isin q,
[A – l I ]x = 0 i sin θ - sin θ x 0 sin θ i sin θ y = 0 i sin θ x - sin θ y = 0
Let
y=t x = – it
-i x -it x = = = t = tx 2 where x2 is an eigenvector corresponding to y t 1 l = cos q – isin q.
example 14
1 2 3 Find orthogonal eigenvectors for the matrix 2 4 6 . 3 6 9
10.102 Chapter 10
Matrices
Solution
1 2 3 A = 2 4 6 3 6 9
Let
The characteristic equation is det( A - λ I ) = 0 1- λ 2 3
2 3 4-λ 6 =0 6 9-λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 1 + 4 + 9 = 14 S2 = Sum of the minors of principal diagonal elements of A =
4 6 6 9
+
1 3 3 9
+
1 2 2 4
= (36 - 36) + (9 - 9) + (4 - 4) =0 1 2 3 S3 = det( A) = 2 4 6 3 6 9 = 1(36 - 36) - 2(18 - 18) + 3(12 - 12) =0 Hence, the characteristic equation is
λ 3 - 14λ 2 = 0 λ = 0, 0, 14 (a) For l = 14,
[A – l I ]x = 0 2 3 x 0 -13 2 -10 6 y = 0 3 6 -5 z 0 - 13 x + 2 y + 3 z = 0 2 x - 10 y + 6 z = 0 3x + 6 y - 5 z = 0
10.13 Properties of Eigenvectors 10.103
By Cramer’s rule, x y z = = =t 2 3 -13 3 -13 2 -10 6 2 6 2 -10 x y z = = =t 42 84 126 x y z = = =t 1 2 3 1 x t x = y = 2t = t 2 = t x1 where x1 is an eigenvector corresponding to l = 14. 3 z 3t (b) For l = 0,
[A – l I ]x = 0 1 2 3 x 0 2 4 6 y = 0 3 6 9 z 0 x + 2 y + 3z = 0 y = t1 and z = t2 x = –2t1 – 3t2
Let .
-3 -2 x -2t1 - 3t2 -2t1 -3t2 t1 x = y = = t1 + 0 = t1 1 + t2 0 = t1x 2 + t2 x3 1 0 t2 0 z t2 where x2 and x3 are linearly independent eigenvectors corresponding to l = 0. Since x2 and x3 are not orthogonal, we must choose x3 such that x1, x2, x3 are orthogonal. l x3 = m n
Let
For x1 and x3 to be orthogonal,
x1T x3 = 0
l [1 2 3] m = 0 n l + 2m + 3n = 0
…(1)
10.104 Chapter 10
Matrices
For x2 and x3 to be orthogonal,
xT2 x3 = 0
l [ -2 1 0] m = 0 n –2l + m = 0
…(2)
Solving Eqs (1) and (2) by Cramer’s rule, l 2 3 1 0
=-
m n = =t 1 3 1 2 -2 0 -2 1 l m n = = =t -3 - 6 5 l m n = = =t 3 6 5
3 l 3t x = m = 6t = t 6 = tx3 where x3 is an eigenvector corresponding to l = 14. 5 n 5t
eXercISe 10.5 1. Find the sum and product of the eigenvalues of the following matrices: -1 1 1 (i) 1 -1 1 1 1 -1
2 -3 -2 (ii) 2 1 -6 -1 -2 0
[Ans.: (i) -3, 4 (ii) -1, 45] 1 2 1 2. The matrix A = 2 0 -2 is singular. One of its eigenvalues is 2. Find 1 2 3 the other two eigenvalues. [Ans.: 1 + 5, 1 - 5] 6 -2 2 3. If two of the eigenvalues of -2 3 -1 are 2 and 8, find the third 2 -1 3 eigenvalue. [Ans.: 2]
10.13 Properties of Eigenvectors 10.105
1 -2 4. Find the eigenvalues of the matrix . Hence, form the matrix -5 4 1 whose eigenvalues are and -1. 6 1 4 2 Ans. : 6, -1, - 6 5 1 5 3 10 5. If 2 and 3 are the eigenvalues of A = -2 -3 - 4 , find the eigenvalues 3 5 7 -1 3 of A and A . 1 1 1 3 3 3 Ans. : (i) 2 , 2 , 3 (ii) 2 , 2 , 3 6 6 4 6. Two eigenvalues of A = 1 3 2 are equal and are double the third. -1 -5 -2 Find the eigenvalues of A2. [Ans.: 1, 4, 4] 1 2 -3 7. If A = 0 3 2 , find the eigenvalues of 3A3 + 5A2 - 6A + 2I 0 0 -2 [Ans.: 4, 110, 10] 2 2 1 8. If A = 1 3 1 , find the eigenvalues of the following matrices: 1 2 2 (i) A3 + I
(ii) A-1
(iii) A2 - 2A + I
Ans. : (i) 2, 2, 126
(iv) A3 - 3A2 +A (ii) 1, 1,
1 5
(iii) 0, 0, 16
(iv) - 1, - 1, 55
9. Find the eigenvalues and eigenvectors for the following matrices:
(i)
(iv)
9 -1 9 3 -1 3 -7 1 -7 4 0 1 -2 1 0 -2 0 1
(ii)
1 1 -2 -1 2 1 0 1 -1
(iii)
1 2 -1 0 -2 1 0 0 -3
(v)
2 1 1 2 3 2 3 3 4
(vi)
2 2 1 1 3 1 1 2 2
10.106 Chapter 10
Matrices
(vii)
2 -3 -2 2 1 -6 -1 -2 0
(x)
0 -2 7 0 5 -2 -2 -2 6
(xiii)
6 -2 2 -2 3 -1 2 -1 3
2 0 1 (xvi) 0 3 0 1 0 2 -3 -7 -5 (xix) 2 4 3 1 2 2
(viii)
1 0 0 2 0 1 3 1 0
(xi)
7 -2 -2 -2 1 4 -2 4 1
(xiv)
1 7 -2 -2 10 -2 1 -2 7
(xvii)
1 2 3 0 2 3 0 0 2
(xx)
2 1 0 0 2 1 0 0 2
Ans. : 1 1 4 (i) - 1, 0, 2; 1 , 0 , 1 1 -1 -3 1 1 1 (iii) - 1, - 2, - 3; 0 , -1 , 2 0 0 -2 1 0 1 (v) 1, 1, 7; 2 , 1 , 0 3 -1 -1 1 2 3 (vii) 5, - 3, - 3; 2 , -1 , 0 -1 0 1 0 2 1 - , 2 , 1 , , ; (ix) 2 9 18 1 0 1 4
(ix)
2 4 -6 4 2 -6 -6 -6 -15
(xii)
3 -1 1 -1 5 -1 1 -1 3
(xv)
2 -1 1 -1 2 -1 1 -1 2
(xviii)
5 3 10 -2 -3 -4 3 5 7
1 1 3 (ii) - 1, 2, 1; 0 , 3 , 2 1 1 1 0 -1 -1 (iv) 1, 2, 3; 1 , 2 , 1 0 2 1 1 2 1 (vi) 5, 1, 1; 1 , -1 , 0 1 0 -1 0 0 1 (viii) - 1, 1, 1; -1 , 1 , 0 1 1 0 1 2 2 (x) 3, 6, 9; 2 , -2 , 1 2 1 -2
10.13 Properties of Eigenvectors 10.107
(xi) (xiii) (xv) (xvii) (xix)
0 1 -2 - 3, 3, 9; 1 , 1 , 1 -1 1 1 2 0 1 8, 2, 2; -1 , 1 , 1 1 1 -1 1 1 -1 4, 1, 1; -1 , 1 , 1 1 0 2 1 2 1, 2, 2; 0 , 1 0 0 -3 1, 1, 1; 1 1
(xii)
(xiv)
(xvi)
(xviii)
(xx)
-1 1 1 2, 3, 6; 0 , 1 , -2 1 1 1 1 1 1 12, 6, 6; -2 , 0 , 1 1 -1 1 1 1 1 1, 3, 3; 0 , 1 , -2 -1 1 1 1 5 3, 2, 2; 1 , 2 -2 -5 1 2, 2, 2; 0 0
10. Verify that x = [2, 3, -2, -3]T is an eigenvector corresponding to the eigenvalue l = 2 of the matrix 1 - 4 -1 - 4 2 0 5 -4 A= -1 1 -2 3 1 4 1 6 3 -1 1 11. If A = -1 3 -1 then check whether eigenvectors of A are 1 -1 3 orthogonal. 5 3 10 12. If A = -2 -3 -4 then verify whether eigenvectors are linearly 3 5 7 independent or not.
10.108 Chapter 10
Matrices
10.14 cAyLey-hAMILton theoreM Statement: Every square matrix satisfies its own characteristic equation. Proof: Let A be an n-rowed square matrix. Its characteristic equation is A - λ I = ( -1) n ( λ n + a1λ n -1 + a2 λ n - 2 + p + an ) ( A - λ I ) adj ( A - λ I ) = A - λ I I
…(10.16)
{A adj ( A) = A I Since adj (A –l I ) has element as cofactors of elements of |A – lI |, the elements of adj (A – l I ) are polynomials in l of degree n – 1 or less. Hence, adj (A – l I ) can be written as a matrix polynomial in l. adj (A –l I ) = B0ln – 1 + B1ln – 2 + ⋅⋅⋅ + Bn –2l + Bn – 1 where
B0, B1,…, Bn –1 are matrices of order n.
( A - λ I ) adj ( A - λI ) = ( A - λ I )[ B0 λ n -1 + B1λ n - 2 + + Bn - 2 λ + Bn -1 ] A - λ I I = ( A - λ I )[ B0 λ n -1 + B1λ n - 2 + + Bn - 2 λ + Bn -1 ] ( -1) n [ Iλ n + a1 Iλ n -1 + a2 Iλ n - 2 + + an -1 Iλ + an I ] = ( - IB0 )λ n + ( AB0 - IB1 )λ n -1 + ( AB1 - IB2 )λ n - 2 + + ( ABn - 2 - IBn -1 )λ + ABn -1 Equating corresponding coefficients, - IB0 = ( -1) n I AB0 - IB1 = ( -1) n a1 I AB1 - IB2 = ( -1) n a2 I
ABn - 2 - IBn -1 = ( -1) an -1 I n
ABn -1 = ( -1) n an I Premultiplying the above equations successively by An, A n–1, A n–2,… I and adding, ( -1) n [ An + a1 An -1 + a2 An - 2 + + an I ] = 0 Hence,
An + a1 An -1 + a2 An - 2 + + an I = 0
…(10.17)
Corollary: If A is a non-singular matrix, i.e. det (A) ≠ 0 then premultiplying equation (1) by A–1, we get An -1 + a1 An - 2 + a2 An - 3 + … + an A-1 = 0 1 A-1 = - n [ An -1 + a1 An - 2 + … + an -1 I ] a
10.14 Cayley-Hamilton Theorem 10.109
example 1
1 2 Apply Cayley–Hamilton theorem to A = and deduce that 2 -1 8 A = 625I. Solution
1 2 A= 2 -1
The characteristic equation is
det( A - λ I ) = 0 1- λ 2
2 =0 -1 - λ
λ 2 - S1λ + S 2 = 0 where S1 = Sum of the principal diagonal elements of A = 1 – 1 = 0 S 2 = det( A) =
1 2 2 -1
= -1 - 4 = -5 Hence, the characteristic equation is
λ2 - 5 = 0 By Cayley–Hamilton theorem, the matrix A satisfies its own characteristic equation. A2 - 5 I = 0 A2 = 5 I A4 = 25 I A8 = 625 I
example 2 Verify Cayley–Hamilton theorem for the following matrix and hence, find A–1 and A4. 2 -1 1 A = -1 2 -1 1 -1 2 Solution
2 -1 1 A = -1 2 -1 1 -1 2
10.110 Chapter 10
Matrices
The characteristic equation is det( A - λ I ) = 0 2-λ -1 1
-1 2-λ -1
1 -1 = 0 2-λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 2 + 2 + 2 = 6 S2 = Sum of the minors of principal diagonal elements of A =
2 -1 2 1 2 -1 + + -1 2 1 2 -1 2
= (4 - 1) + (4 - 1) + (4 - 1) =9 2 -1 1 S3 = det( A) = -1 2 -1 1 -1
2
= 2(4 - 1) + 1( -2 + 1) + 1(1 - 2) = 6 -1-1 =4 Hence, the characteristic equation is
λ 3 - 6λ 2 + 9λ - 4 = 0 2 A2 = -1 1 6 A3 = -5 5
-1 1 2 2 -1 -1 -1 2 1 -5 5 2 6 -5 -1 -5 6 1
-1 1 6 2 -1 = -5 -1 2 5 -1 1 22 2 -1 = -21 -1 2 21
-5 5 6 -5 -5 6 -21 21 22 -21 -21 22
A3 - 6 A2 + 9 A - 4 I 22 = -21 21 0 0 = 0 0 0 0
9 4 0 0 -21 21 36 -30 30 18 -9 22 -21 - -30 36 -30 + -9 18 -9 - 0 4 0 36 9 -9 18 0 0 4 -21 22 30 -30 0 0 = 0 ...(1) 0
The matrix A satisfies its own characteristic equation. Hence, Cayley–Hamilton theorem is verified.
10.14 Cayley-Hamilton Theorem 10.111
Premultiplying Eq. (1) by A–1, A-1 ( A3 - 6 A2 + 9 A - 4 I ) = 0 A2 - 6 A + 9 I - 4 A-1 = 0 4 A-1 = ( A2 - 6 A + 9 I ) 6 9 0 0 6 -5 5 12 -6 = -5 6 -5 - -6 12 -6 + 0 9 0 5 -5 6 6 -6 12 0 0 9 3 1 -1 = 1 3 1 -1 1 3 A-1 = Multiplying Eq. (1) by A,
3 1 -1 1 1 3 1 4 -1 1 3
A( A3 - 6 A2 + 9 A - 4 I ) = 0 A4 - 6 A3 + 9 A2 - 4 A = 0
A4 = 6 A3 - 9 A2 + 4 A 45 8 - 4 4 132 -126 126 54 - 45 54 - 45 + - 4 8 - 4 = -126 132 -126 - - 45 126 -126 132 45 - 45 54 4 - 4 8 86 -85 85 = -85 86 -85 85 -85 86
example 3
c -b 0 0 Show that matrix A = -c a satisfies Cayley–Hamilton theorem b - a 0 and hence find A–1, if it exists. Solution
c -b 0 0 A = -c a b - a 0
10.112 Chapter 10
Matrices
The characteristic equation is
det( A - λ I ) = 0 -λ c -b -c - λ a =0 b -a -λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 0 S2 = Sum of the minors of principal diagonal elements of A 0 a 0 -b 0 c = + + -a 0 b 0 -c 0 = (0 + a 2 ) + (0 + b 2 ) + (0 + c 2 ) = a 2 + b2 + c2 0 c -b S3 = det( A) = -c 0 a b -a 0 = 0 - c(0 - ab) - b(ac - 0) = abc - abc =0 Hence, the characteristic equation is λ 3 + (a 2 + b 2 + c 2 )λ = 0 c -b 0 c -b - c 2 - b 2 0 A2 = -c 0 a -c 0 a = ab b - a 0 b - a 0 ac -c - b A = ab ac 2
3
2
ab -c - a 2 bc 2
ab -c - a 2 2
bc
0 c -b c 0 a -b 2 - a 2 b - a 0 ac bc
0 -c 3 - cb 2 - ca 2 3 2 2 = c + ca + cb 0 -bc 2 - b3 - a 2 b ac 2 + ab 2 + a 3 c -b 0 2 2 2 = -(a + b + c ) -c 0 a b - a 0
b3 + bc 2 + ba 2 - ab 2 - ac 2 - a 3 0
= -(a 2 + b 2 + c 2 ) A
A3 + (a 2 + b 2 + c 2 ) A = 0
-b 2 - a 2 ac bc
10.14 Cayley-Hamilton Theorem 10.113
The matrix A satisfies its own characteristic equation. Hence, Cayley–Hamilton theorem is verified. 0 c -b det( A) = - c 0 a = -c(0 - ab) - b(ac - 0) b -a 0 = abc - abc = 0 –1
Hence, A does not exist.
example 4
1 4 Find the characteristic roots of the matrix A = and verify 2 3 Cayley–Hamilton theorem for this matrix. Find A–1 and also express A5 – 4A4 – 7A3 + 11A2 – A – 10I as a linear polynomial in A. Solution
1 4 A= 2 3
The characteristic equation is
det( A - λ I ) = 0 1- λ 4 =0 2 3- λ
λ 2 - S1λ + S 2 = 0 where S1 = Sum of the principal diagonal elements of A = 1 + 3 = 4 1 4 S 2 = det( A) = 2 3 = 3-8 = -5 Hence, the characteristic equation is λ 2 - 4λ - 5 = 0 λ = -1, 5 1 A2 = 2 9 A2 - 4 A - 5 = 8
4 1 4 9 16 = 3 2 3 8 17 16 4 16 5 0 0 0 =0 = 17 8 12 0 0 0 0
... (1)
The matrix A satisfies its own characteristic equation. Hence, Cayley–Hamilton theorem is verified. Premultiplying Eq. (1) by A–1,
10.114 Chapter 10
Matrices
A-1 ( A2 - 4 A - 5) = 0 A - 4 I - 5 A-1 = 0 1 4 A-1 = ( A - 4 I ) 5 1 -3 4 = 5 2 -1 Now, A5 – 4A4 – 7A3 + 11A2 – A – 10I = A3(A2 – 4A – 5I) – 2A(A2 – 4A – 5I) + 3(A2 – 4A – 5I ) + A + 5I = ( A2 - 4A - 5I )( A3 - 2A + 3I ) + ( A + 5 I ) = 0 + ( A + 5I ) [Using Eq. (1)] = A + 5I which is a linear polynomial in A.
example 5
2 1 1 Find the characteristic equation of the matrix A = 0 1 0 and hence 1 1 2 find the matrix represented by A8 – 5A7 + 7A6 – 3A5 + A4 – 5A3 + 8A2 – 2A + I. Solution 2 1 1 A = 0 1 0 1 1 2
The characteristic equation is det( A - λ I ) = 0 2-λ 1 0 1- λ 1
1
1 0
=0
2-λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 2 + 1 + 2 = 5 S2 = Sum of the minors of principal diagonal elements of A =
1 0 2 1 2 1 + + 1 2 1 2 0 1
= (2 - 0) + (4 - 1) + (2 - 0) =7
10.14 Cayley-Hamilton Theorem 10.115
2 1 1 S3 = det( A) = 0 1 0 1 1 2 = 2(2 - 0) - 1(0 - 0) + 1(0 - 1) = 4 - 0 -1 =3 Hence, the characteristic equation is
λ 3 - 5λ 2 + 7 λ - 3 = 0 By Cayley–Hamilton theorem, A3 – 5A2 + 7A – 3I = 0 Now, A8 – 5A7 + 7A6 – 3A5 + A4 – 5A3 + 8A2 – 2A + I
…(1)
= A5 (A3 - 5A2 + 7A - 3I ) + A(A3 - 5A2 + 7A - 3I ) + ( A2 + A + I ) = (A3 - 5A2 + 7 A - 3I )( A5 + A) + ( A2 + A + I ) = 0 + ( A2 + A + I )
[Using Eq.(1)]
= A + A+ I 2
1 1 2 1 1 5 4 4 1 0 0 1 0 = 0 1 0 1 2 1 1 2 4 4 5 4 4 2 1 1 1 0 0 8 5 5 1 0 + 0 1 0 + 0 1 0 = 0 3 0 4 5 1 1 2 0 0 1 5 5 8
2 A2 = 0 1 5 2 A + A + I = 0 4
8 5 5 A - 5A + 7A - 3A + A - 5A + 8A - 2A + I = 0 3 0 5 5 8 8
7
6
5
4
3
2
example 6 1 0 0 If A = 1 0 1 , prove by induction that for every integer n ≥ 3, 0 1 0 An = An – 2 + A2 – I. Hence, find A50. Solution 1 0 0 A = 1 0 1 0 1 0
10.116 Chapter 10
Matrices
The characteristic equation is det( A - λ I ) = 0 1- λ 1 0
0 -λ 1
0 1 =0 -λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 1 + 0 + 0 = 1 S2 = Sum of the minors of principal diagonal elements of A =
0 1 1 0 1 0 + + 1 0 0 0 1 0
= (0 - 1) + (0 - 0) + (0 - 0) = -1 1 0 0 S3 = det( A) = 1 0 1 0 1 0 = 1(0 - 1) + 0 + 0 = -1 Hence, the characteristic equation is
λ3 - λ2 - λ +1 = 0 By Cayley–Hamilton theorem, A3 - A2 - A + I = 0 A3 = A2 + A - I = A3- 2 + A2 - I Hence, An = An – 2 + A2 – I is true for n = 3. Assuming that Eq. (1) is true for n = k, Ak = Ak – 2 + A2 – I Multiplying both the sides by A Ak + 1 = Ak + 1 + A3 – A Substituting the value of A3 Ak + 1 = Ak + 1 + (A2 + A – I) – A A(k + 1) – 2 + A2 – I Hence, An = An – 2 + A2 – I is true for n = k + 1 Thus, by mathematical induction, it is true for n ≥ 3. We have, An = An – 2 + A2 – I
…(1)
10.14 Cayley-Hamilton Theorem 10.117
= ( An - 4 + A2 - I ) + A2 - I = A n - 4 + 2( A 2 - I ) = ( A n - 6 + A 2 - I ) + 2( A 2 - I ) = An–6 + 3(A2 – I ) .......................... An = An–2r + r(A2 – I ) Putting n = 50 and r = 24, A50 = A50 - 2(24) + 24( A2 - I ) = A2 + 24 A2 - 24 I = 25 A2 1 2 A = 1 0
- 24 I 0 0 1 0 0 0 1 1 0 1 1 0 0 1 0 1 0 0 = 1 1 0 1 0 1 25 0 0 24 0 0 50 A = 25 25 0 - 0 24 0 25 0 25 0 0 24 1 0 0 = 25 1 0 25 0 1
eXercISe 10.6 1. Verify Cayley-Hamilton theorem for the matrix A and hence, find A-1 and A4. (i)
(iii)
1 2 -2 -1 3 0 0 -2 1 2 0 -1 0 2 0 -1 0 2
(ii)
1 2 3 2 -1 4 3 1 -1
(iv)
1 2 -2 -1 3 0 0 -2 1
10.118 Chapter 10
Ans. : 3 2 (i) 1 1 2 2 4 1 (iii) 0 6 2
Matrices
6 -55 104 24 2 , -20 -15 32 5 32 - 42 13
-3 1 (ii) 14 40 5
0 2 41 0 - 40 3 2 3 0 , 0 16 0 (iv) 1 1 2 2 0 4 - 40 0 41
5 11 248 101 218 -10 2 , 272 109 50 5 -5 104 98 204 6 -55 104 24 2 , -20 -15 32 5 32 - 40 -23
1 2 0 2. Verify that the matrix A = 2 -1 0 satisfies its characteristic equation 0 0 -1 and hence, find A-2. 1 0 0 1 3 2 Ans. : A + A - 5 A - 5I = 0, 5 0 1 0 0 0 1 3. Use Cayley-Hamilton theorem to find 2A5 - 3A4 + A2 3 1 A= . -1 2 Ans. : 138 A - 403I = 1 4 7 2 4. If A = , find A - 9A + I. 1 1
- 4I, where 11 138 -138 127
[Ans. : 609A + 640I]
1 2 5. Verify Cayley-Hamilton theorem for (i) A = (ii) A = 3 4 -1 3 2 hence, find A and A - 5A .
2 4 1 2 and
1 -2 -1 2 Ans. : (i) , 2 A (ii) A does not exist, A 1.5 -0.5 1 2 3 6. Compute A9 - 6A8 + 10A7 - 3A6 + A + I, where A = -1 3 1 . 1 0 2 2 2 3 Ans. : -1 4 1 1 0 3
10.16 Diagonalization of a Matrix 10.119
1 2 7. Verify Cayley–Hamilton theorem for A = and evaluate 2 2 2A4-5A3 - 7A - 6I. 36 32 Ans. : 32 52
10.15 SIMILArIty trAnSforMAtIon If A and B are two square matrices of order n then B is said to be similar to A, if there exists a nonsingular matrix P such that B = P -1AP
Note: (1) Similarity of matrices is an equivalence relation. (2) Similar matrices have the same determinant. (3) Similar matrices have the same characteristic polynomial and hence the same eigenvalues. If x is an eigenvector of A corresponding to the eigenvalue l, then P -1x is an eigenvector of B corresponding to the eigenvalue l where B = P -1AP.
10.16 dIAGonALIZAtIon of A MAtrIX A matrix A is said to be diagonalizable if it is similar to a diagonal matrix. A matrix A is diagonalizable if there exists an invertible matrix P such that P -1AP = D where D is a diagonal matrix, also known as spectral matrix. The matrix P is then said to diagonalize A or transform A to a diagonal form. P is known as the modal matrix. Note (1) An n × n matrix is diagonalizable if and only if it possesses n linearly independent eigenvectors. (2) If the eigenvalues of an n × n matrix are all distinct then it is always similar to a diagonal matrix. (3) If A is similar to a diagonal matrix D, the diagonal elements of D are the eigenvalues of A.
10.16.1
orthogonally Similar Matrices
If A and B are two square matrices of order n then B is said to be orthogonally similar to A if there exists an orthogonal matrix P such that B = P -1AP -1 T Since P is orthogonal, P = P B = P -1AP = PTAP Note: (1) Every real symmetric matrix is orthogonally similar to a diagonal matrix with real elements. (2) A real symmetric matrix of order n has n mutually orthogonal real eigenvectors.
10.120 Chapter 10
Matrices
(3) Any two eigenvectors corresponding to two distinct eigenvalues of a real symmetric matrix are orthogonal. Note: To find the orthogonal matrix P, each element of the eigenvector is divided by its norm (length).
working rule for diagonalization of Square Matrix A (i) (ii) (iii) (iv)
Find the eigenvalues of the square matrix A. Find the eigenvectors corresponding to each eigenvalue. Find the modal matrix P having the normalized eigenvectors as its column vectors. Find the diagonal matrix D = PTAP. The diagonal matrix D has eigenvalues as its diagonal elements.
example 1
2 1 0 Show that the matrix 0 2 1 is not diagonalizable. 0 0 2 Solution 2 1 0 A = 0 2 1 0 0 2
Let The characteristic equation is
det( A - λ I ) = 0 2-λ 0 0
1 2-λ
0 1
0
2-λ
=0
λ - S1λ + S2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 2 + 2 + 2 = 6 S2 = Sum of the minors of principal diagonal elements of A =
2
1
0 2
+
2 0 0 2
+
2
1
0 2
= ( 4 - 0) + ( 4 - 0 ) + ( 4 - 0 ) = 12 2 1 0 S3 = det( A) = 0 2 1 0 0 2 = 2(4 - 0) - 1(0 - 0) + 0 = 8-0+0 =8
10.16 Diagonalization of a Matrix 10.121
Hence, the characteristic equation is
For l = 2,
λ 3 - 6λ 2 + 12λ - 8 = 0 λ = 2, 2, 2 [A – l I ]x = 0 0 1 0 x 0 0 0 1 y = 0 0 0 0 z 0 0x + y + 0z = 0 0x + 0 y + z = 0
By Cramer’s rule,
x 1 0 0 1
=-
y 0 0 0 1
=
z 0 1 0 0
=t
x y z = = =t 1 0 0 1 x t x = y = 0 = t 0 = tx1 where x1 is an eigenvector corresponding to l = 1. 0 z 0 Since the matrix A has only one linearly independent eigenvector which is less than its order 3, matrix A is not diagonalizable.
example 2
1 -2 0 Show that the matrix 1 2 2 is not diagonalizable. 1 2 3 Solution
1 -2 0 A = 1 2 2 1 2 3
Let The characteristic equation is
det( A - λ I ) = 0 1- λ 1 1
-2 0 2-λ 2 =0 2 3-λ
λ 3 - S1λ 2 + S2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 1 + 2 + 3 = 6
10.122 Chapter 10
Matrices
S2 = Sum of the minors of principal diagonal elements of A 2 2 1 0 1 -2 = + + 2 3 1 3 1 2 = (6 - 4) + (3 - 0) + (2 + 2) =9 1 -2 0 S3 = det( A) = 1 2 2 1 2 3 = 1(6 - 4) + 2(3 - 2) + 0 =4 Hence, the characteristic equation is
λ 3 - 6λ 2 + 9λ - 4 = 0 λ = 1, 1, 4 (a) For l = 1,
By Cramer’s rule,
[A – lI ]x = 0 0 -2 0 x 0 1 1 2 y = 0 0 2 2 z 0 0x - 2 y + 0z = 0 x + y + 2z = 0 x + 2 y + 2z = 0 x y z == =t -2 0 0 0 0 -2 1 2 1 2 1 1 x y z = = =t -4 0 2 x y z = = =t 2 0 -1
2 x 2t x = y = 0 = t 0 = tx1 where x1 is an eigenvector corresponding to l = 1. -1 z -t (b) For l = 4,
[A – l I ]x = 0 -3 -2 0 x 0 1 -2 2 y = 0 1 2 -1 z 0 -3 x - 2 y + 0 z = 0 x - 2 y + 2z = 0 x + 2y - z = 0
10.16 Diagonalization of a Matrix 10.123
By Cramer’s rule, x y z == =t -2 0 -3 0 -3 -2 -2 2 1 2 1 -2 x = -4 x = -2
y z = =t 6 8 y z = =t 3 4
-2 x -2t x = y = 3t = t 3 = tx 2 where x2 is an eigenvector corresponding to l = 1. 4 z 4t Since the matrix A has two linearly independent eigenvectors which is less than its order 3, the matrix A is not diagonalizable.
example 3 Determine a diagonal matrix orthogonally similar to the real symmetric 3 -1 1 matrix -1 5 -1 . Also find the modal matrix. 1 -1 3 Solution
3 -1 1 A = -1 5 -1 1 -1 3
Let The characteristic equation is
det( A - λ I ) = 0 3-λ -1 -1 5 - λ 1
-1
1 -1 = 0 3-λ
λ - S1λ + S2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 3 + 5 + 3 = 11 S2 = Sum of the minors of principal diagonal elements of A =
5 -1 3 1 3 -1 + + -1 3 1 3 -1 5
= (15 - 1) + (9 - 1) + (15 - 1) = 36
10.124 Chapter 10
Matrices
3 -1 1 S3 = det( A) = -1 5 -1 1 -1 3 = 3(15 - 1) + 1( -3 + 1) + 1(1 - 5) = 42 - 2 - 4 = 36 Hence, the characteristic equation is
λ 3 - 11λ 2 + 36λ - 36 = 0 λ = 2, 3, 6 (a) For l = 2,
[A – l I ]x = 0 1 -1 1 x 0 -1 3 -1 y = 0 1 -1 1 z 0 x - y+z=0 - x + 3y - z = 0
By Cramer’s rule, x y z == =t -1 1 1 1 1 -1 3 -1 -1 -1 -1 3 x = -2 x = -1
y z = =t 0 2 y z = =t 0 1
-1 x -t x = y = 0 = t 0 = t x1 where x1 is an eigenvector corresponding to l = 2. 1 z t (b) For l = 3,
[A – l I ]x = 0 0 -1 1 x 0 -1 2 -1 y = 0 1 -1 0 z 0 0x - y + z = 0 -x + 2y - z = 0 x - y + 0z = 0
10.16 Diagonalization of a Matrix 10.125
By Cramer’s rule, x y z == =t -1 1 0 1 0 -1 2 -1 -1 -1 -1 2 x y z = = =t -1 -1 -1 x y z = = =t 1 1 1 1 x t x = y = t = t 1 = tx 2 where x2 is an eigenvector corresponding to l = 3. 1 z t (c) For l = 6,
[A – l I ]x = 0 -3 -1 1 x 0 -1 -1 -1 y = 0 1 -1 -3 z 0 - 3x - y + z = 0 x+ y - z=0 x - y - 3z = 0
By Cramer’s rule, x
-1 1 -1 -1
=-
y -3 1 -1 -1
=
z -3 -1 -1 -1
=t
x y z = = =t 2 -4 2 x y z = = =t 1 -2 1
1 x t x = y = -2t = t -2 = tx3 where x3 is an eigenvector corresponding to l = 6. 1 z t Since matrix A has three linearly independent eigenvectors which is same as its order, the matrix A is diagonalizable. Length of eigenvector x1 = ( -1) 2 + 02 + 12 = 2 Length of eigenvector x 2 = 12 + 12 + 12 = 3 Length of eigenvector x3 = 12 + ( -2) 2 + 12 = 6
10.126 Chapter 10
Matrices
The normalized eigenvectors are 1 1 1 6 3 2 1 2 x1 = 0 , x 2 = x3 = 6 1 3 1 1 2 6 3 The modal matrix P has normalized eigenvectors as its column vectors. 1 2 P= 0 1 2 1 2 1 P -1 = PT = 3 1 6 1 2 1 D = PT AP = 3 1 6
1 3 6 1 2 3 6 1 1 3 6 1 0 2 1 1 3 3 2 1 6 6 1
0 1 -
3 2 6
1 1 2 2 3 -1 1 1 -1 5 -1 0 3 1 -1 3 1 1 6 2
1 3 1 3 1 3
1 6 2 6 1 6
2 0 0 = 0 3 0 0 0 6 Hence, the diagonal matrix D has eigenvalues as its diagonal elements.
example 4 Determine a diagonal matrix orthogonally similar to the real symmetric 2 8 -6 7 - 4 . Also find the modal matrix. matrix -6 2 - 4 3
10.16 Diagonalization of a Matrix 10.127
Solution
2 8 -6 7 - 4 A = -6 2 - 4 3
Let The characteristic equation is
det( A - λ I ) = 0 8-λ -6 2 -6 7 - λ - 4 = 0 2 -4 3- λ
λ 3 - S1λ 2 + S 2 λ - S3 = 0 where S1 = Sum of the principal diagonal elements of A = 8 + 7 + 3 = 18 S2 = Sum of the minors of principal diagonal elements of A =
7 -4 -4
3
+
8 2 2 3
+
8 -6 -6
7
= (21 - 16) + (24 - 4) + (56 - 36) = 5 + 20 + 20 = 45 8 -6 2 7 -4 S3 = det ( A) = -6 2 -4 3 = 8(21 - 16) + 6( -18 + 8) + 2(24 - 14) = 40 - 60 + 20 =0 Hence, the characteristic equation is
λ 3 - 18λ 2 + 45λ = 0 λ = 0, 3, 15 (a) For l = 0,
[A – l I ]x = 0 2 x 0 8 -6 -6 7 - 4 y = 0 2 - 4 3 z 0 8x - 6 y + 2 z = 0 -6 x + 7 y - 4 z = 0 2 x - 4 y + 3z = 0
10.128 Chapter 10
Matrices
By Cramer’s rule, x -6 2 7 -4
=-
y 8 2 -6 - 4
=
z 8 -6 -6 7
=t
x y z = = =t 10 20 20 x y z = = =t 1 2 2 1 x t x = y = 2t = t 2 = tx1 where x1 is an eigenvector corresponding to l = 0. 2 z 2t (b) For l = 3,
[A – l I ]x = 0 2 x 0 5 -6 -6 4 - 4 y = 0 2 -4 0 z 0 5x - 6 y + 2 z = 0 -6 x + 4 y - 4 z = 0 2x - 4 y + 0z = 0
By Cramer’s rule, x -6 2 4 -4
=-
y 5 2 -6 - 4
=
z 5 -6 -6 4
=t
x y z = = =t 16 8 -16 x y z = = =t 2 1 -2 2 x 2t x = y = t = t 1 = tx 2 where x2 is an eigenvector corresponding to l = 3. -2 z -2t (c) For l = 15,
[A – l I ]x = 0 2 x 0 -7 -6 -6 -8 -4 y = 0 2 -4 -12 z 0
10.16 Diagonalization of a Matrix 10.129
-7 x - 6 y + 2 z = 0 -6 x - 8 y - 4 z = 0 2 x - 4 y - 12 z = 0 By Cramer’s rule, x -6 2 -8 -4
=-
y
=
z
=t -7 2 -7 -6 -6 -4 -6 -8 x y z = = =t 40 - 40 20 x y z = = =t 2 -2 1
2 x 2t x = y = -2t = t -2 = tx3 where x3 is an eigenvector corresponding to l = 15. 1 z t Since the matrix A has three linearly independent eigenvectors which is same as its order, the matrix A is diagonalizable. Length of eigenvector x1 = 12 + 22 + 22 = 3 Length of eigenvector x 2 = 22 + 12 + ( -2) 2 = 3 Length of eigenvector x3 = 22 + ( -2) 2 + 12 = 3 The normalized eigenvectors are 1 2 2 3 3 3 1 2 2 x1 = x3 = - , x2 = 3 3 3 2 - 2 1 3 3 3 The modal matrix P has normalized eigenvectors as its column vectors. 1 3 2 P= 3 2 3
2 2 3 3 1 2 3 3 2 1 3 3
10.130 Chapter 10
Matrices
2 2 1 3 3 3 2 1 2 -1 T P =P = 3 3 3 1 2 - 2 3 3 3 2 2 2 1 1 3 3 3 3 3 8 -6 2 2 1 2 1 2 T - -6 7 -4 D = P AP = 3 3 3 3 3 2 -4 3 1 2 - 2 2 - 2 3 3 3 3 3 0 0 0 = 0 3 0 0 0 15
2 3 2 3 1 3
Hence, the diagonal matrix D has eigenvalues as its diagonal elements.
example 5 Determine a diagonal matrix orthogonally similar to the real symmetric 6 -2 2 matrix -2 3 -1 . Also find the modal matrix. 2 -1 3 Solution Let
6 -2 2 3 -1 A = -2 2 -1 3
The characteristic equation is det( A - λ I ) = 0
λ - S1λ + S 2 λ - S3 = 0 3
2
where S1 = Sum of the principal diagonal elements of A = 6 + 3 + 3 = 12 S2 = Sum of the minors of principal diagonal elements of A 3 -1 6 2 6 -2 = + + -1 3 2 3 -2 3 = (9 - 1) + (18 - 4) + (18 - 4) = 8 + 14 + 14 = 36
10.16 Diagonalization of a Matrix 10.131
6 -2 2 3 -1 S3 = det( A) = -2 2 -1 3 = 6(9 - 1) + 2( -6 + 2) + 2(2 - 6) = 48 - 8 - 8 = 32 Hence, the characteristic equation is λ 3 - 12λ 2 + 36λ - 32 = 0 λ = 2, 2, 8 (a) For l = 8,
[A – lI ]x = 0 -2 -2 2 x 0 -2 -5 -1 y = 0 2 -1 -5 z 0 -2 x - 2 y + 2 z = 0 -2 x - 5 y - z = 0 2 x - y - 5z = 0
By Cramer’s rule, x y z == =t -2 2 -2 2 -2 -2 -5 -1 -2 -1 -2 -5 x y = = 12 -6 x y = = 2 -1
z =t 6 z =t 1
2 x 2t x = y = -t = t -1 = tx1 where x1 is an eigenvector corresponding to l = 8. 1 z t Since matrix A has three linearly independent eigenvectors which is same as its order, matrix A is diagonalizable. (b) For l = 2,
[A – l I ]x = 0 4 -2 2 x 0 -2 1 -1 y = 0 2 -1 1 z 0 4x - 2 y + 2z = 0
10.132 Chapter 10
Matrices
Let
y = t1 and z = t2 1 1 t1 - t2 2 2 1 1 1 1 1 1 t - t t - t - 2 2 x 2 1 2 2 2 1 2 2 t1 x = y = = t1 + 0 = t1 1 + t2 0 = t1x 2 + t2 x3 0 t 1 0 z t2 2 x=
where x2 and x3 are linearly independent eigenvectors corresponding to l = 2. The orthogonal matrix P has mutually orthogonal eigenvectors. Since x2 and x3 are not orthogonal, we must choose x3 such that x1, x2, x3 are orthogonal. l Let x3 = m n For orthogonality of eigenvectors, x1T x3 = 0
and
l [ 2 -1 1] m = 0 n
and
2l - m + n = 0
and
xT2 x3 = 0 l 1 2 1 0 m = 0 n 1 l + m + 0n = 0 2
By Cramer’s rule, l m == -1 1 2 1 1 0 1 0 2 l = -1
n =t 2 -1 1 1 2 m n = =t 1 5 2 2 l m n = = =t -2 1 5
-2 l -2t x = m = t = t 1 = t x3 where x3 is an eigenvector corresponding to l = 2. 5 n 5t Length of eigenvector x1 =
( 2)2 + ( -1)2 + (1)2
= 6
10.16 Diagonalization of a Matrix 10.133
2
1 Length of eigenvector x 2 = + 12 + 02 = 2
( -2)2 + (1)2 + (5)2
Length of eigenvector x3 =
5 2 = 30
The normalized eigenvectors are 2 1 6 5 1 2 x1 = , x2 = , x3 = 5 6 1 0 6
2 30 1 30 5 30
The modal matrix P has normalized eigenvectors as its column vectors. 2 6 1 P = 6 1 6 -1 T P =P = T D = P AP = 8 = 0 0
2 6 1
1 5 2 5
-
0 1
-
6 2
5 2
5 1
30
30
2 6 1
-
1 6 2
5 2
5 1
30
30
2 30 1 30 5 30 1 6 0 5 30 2 1 6 6 6 -2 2 1 0 -2 3 -1 6 2 -1 3 5 1 6 30
0 0 2 0 0 2
Hence, the diagonal matrix D has eigenvalues as its diagonal elements.
1 5 2 5 0
-
2 30 1 30 5 30
10.134 Chapter 10
Matrices
example 6 For a symmetric matrix A, the eigenvectors are [1, 1, 1]T, [1, –2, 1]T corresponding to l1 = 6 and l 2 = 12. Find the eigenvector corresponding to l3 = 6 and find the matrix A. Solution Let x3 = [x3, y3, z3]T be the eigenvector corresponding to l 3 = 6. x1 = [1, 1, 1]T,
x2 = [1, –2, 1]T
Since A is real symmetric matrix, x1, x2 and x3 are orthogonal. i.e.,
x1T x3 = 0 and xT2 x3 = 0 x3 x3 [1, 1, 1] y3 = 0 and [1, - 2, 1] y3 = 0 z3 z3 x3 + y3 + z3 = 0 x3 + 2 y3 + z3 = 0
By Cramer’s rule, x3 y z3 =- 3 = =t 1 1 1 1 1 1 -2 1 1 1 1 -2 x3 =3 x3 = 1
y3 z3 = =t 0 -3 y3 z3 = =t 0 -1
x3 t 1 x = y3 = 0 = t 0 = t x3 where x3 is an eigenvector correspponding to l = 8. z3 -t -1 1 x3 = 0 -1 Length of eigenvector x1 = 12 + 12 + 12 = 3
10.16 Diagonalization of a Matrix 10.135
Length of eigenvector x 2 = 12 + ( -2) 2 + 12 = 6 Length of eigenvector x3 = 12 + 02 + ( -1) 2 = 2 The normalized eigenvectors are x1 =
1 3 1 , x2 = 3 1 3
P=
1 3 1 3 1 3
1 1 6 2 2 , x3 = 0 6 1 1 2 6
1 2 0 1 2
1 6 2 6 1 6
6 0 0 D = 0 12 0 = PT AP 0 0 6 A = IAI = PPT A PPT = P(PT AP) PT = PDPT =
1
1
3 1
6 2
3 1 3
-
6 1 6
1 1 2 3 6 0 0 1 0 0 12 0 6 0 0 6 1 1 2 2
1 7 -2 = -2 10 -2 1 -2 7
1 -
3 2 6 0
1 3 1 6 1 2
10.136 Chapter 10
Matrices
eXercISe 1.9 1. Show that the following matrices are not similar to diagonal matrices.
(i)
2 3 4 0 2 -1 0 0 1
(iii)
1 2 3 0 2 0 0 0 2
(ii)
2 -1 1 2 2 -1 1 2 -1
(iv)
5 3 10 -2 -3 -4 3 5 7
2. Determine diagonal matrices orthogonally similar to the following real symmetric matrices. Also, find the modal matrix in each case. (i)
4 -4 7 4 -8 -1 - 4 -1 8
(ii)
0 -2 7 5 -2 0 -2 -2 6 Ans. : 9 (i) D = 0 0 3 (ii) D = 0 0
-1 7 -1 3. Find A , where A = 0 1 0 0 15 -2
0 -9 0
0 6 0
0 , P = 0 -9 1 3 0 2 0 , P = 3 9 2 3
4 8 1 18 1 18 2 3 2 3 1 3
1 0 3 1 2 - 3 2 1 2 3 2 2 3 1 3 2 3
11
-1 10237 -2047 1 0 Ans. : 0 0 10245 -2048
Multiple Choice Questions 10.137
Multiple choice questions Select the most appropriate response out of the various alternatives given in each of the following questions: 1. If A is an orthogonal matrix, then A-1 equals (a) A (b) AT (c) A2
(d) none of these
2. If A is a symmetric matrix and n ∈ N, then An is (a) symmetric (b) skew symmetric (c) diagonal matrix (d) none of these 1 -5 7 7 9 , then trace of the matrix A is 3. If A = 0 11 8 9 (a) 17
(b) 25
(c) 3
(d) 12
P11 P12 4. All the four entries of the 2 × 2 matrix P = are non-zero, and one of P21 P22 its eigenvalues is zero. Which of the following statements are true? (a) P11 P22 - P12 P21 = 1 (b) P11 P22 - P12 P21 = -1 (c) P11 P22 - P12 .P21 = 0 (d) P11 P22 + P12 P21 = 0 5. The system of linear equations 4x + 2y = 7, 2x + y = 6 has (a) a unique solution (b) no solution (c) an infinite number of solution (d) exactly two distinct solutions 1 1 1 6. The rank of the matrix 1 -1 0 is 1 1 1 (a) 0 (b) 1 (c) 2 (d) 3 7. The eigenvalues and the corresponding eigenvectors of a 2 × 2 matrix are given by eigenvalue eigenvector 1 l1 = 8 x1 = 1 l2 = 4
1 x2 = -1
The matrix is 6 2 (a) 2 6
4 6 (b) 6 4
2 4 (c) 4 2
4 8 (d) 8 4
10.138 Chapter 10
Matrices
2 - 0.1 -1 8. Let A = and A = 0 3 (a)
7 20
(b)
1 2 0
3 20
(c)
a . Then (a + b) is b 19 60
(d)
11 20
1 1 1 1 1 1 -1 -1 , [AA-1] is 9. Given an orthogonal matrix A = 1 -1 0 0 1 1 0 0 1 4 0 0 0 0 1 0 0 4 (a) 0 0 1 0 2 0 0 0 1
1 0 (c) 0 0
0 0 0 1 0 0 0 1 0 0 0 1
1 2 0 0 0 1 0 2 (b) 0 0 1 2 0 0 0
0 0 0 1 2
1 4 0 (d) 0 0
0 0 0 1 4
0
0
1 4
0
0
1 4
0
0
10. The characteristic equation of a 3 × 3 matrix P is defined as a (l) = | l I - P | = l3 + l2 + 2l + 1 = 0. If I denotes identity matrix, then the inverse of matrix P will be (a) P2 + P + 2I (b) P2 + P + I 2 (c) - (P + P + I) (d) - (P2 + P + 2I) 3 - 2 2 11. For the matrix P = 0 - 2 1 , one of the eigenvalue is equal to -2. Which of 0 0 1 the following is an eigenvector? 3 (a) - 2 1
- 3 (b) 2 -1
1 (c) - 2 3
2 (d) 5 0
Multiple Choice Questions 10.139
12. Consider the following system of equations in three real variables x1, x2 and x3. 2x1 - x2 + 3x3 = 1 3x1 + 2x2 + 5x3 = 2 - x1 + 4x2 + x3 = 3 The system equation has (a) no solution (b) a unique solution (c) more than one but a finite number of solutions (d) an infinite number of solutions 3 2 13. Eigenvalues of a matrix S = are 5 and 1. What are the eigenvalues of the 2 3 matrix S2? (a) 1 and 25 (b) 6 and 4 (c) 5 and 1 (d) 2 and 10 14. Match the items in columns I and II I II (P) Singular matrix (1) Determinant is not defined (Q) Non-square matrix (2) Determinant is always ±1 (R) Real symmetric matrix (3) Determinant is zero (S) Orthogonal matrix (4) Eigenvalues are always real (5) Eigenvalues are not defined (a) P-3, Q-1, R-4, S-2 (b) P-2, Q-3, R-4, S-1 (c) P-3, Q-2, R-5, S-4 (d) P-3, Q-4, R-2, S-1 15. A is a 3 × 4 real matrix and Ax = b is an inconsistent system of equations. The highest possible rank of A is (a) 1 (b) 2 (c) 3 (d) 4 1 1 3 16. The sum of eigenvalues of the matrix 1 5 1 is 3 1 1 (a) 5 (b) 7 (c) 9 (d) 18 17. Let A and B be real symmetric matrices of size n × n. Then which one of the following is true? (a) AA′ = I (b) A = A-1 (c) AB = BA (d) (AB)′ = BA 18. Among the following, the pair of vectors orthogonal to each other is (a) [3, 4, 7], [3, 4, 7] (b) [1, 0, 0], [1, 1, 0] (c) [1, 0, 2], [0, 5, 0] (d) [1, 1, 1], [-1, -1, -1]
10.140 Chapter 10
Matrices
2 -1 0 0 3 0 19. The eigenvalues of the matrix 0 0 -2 0 0 1 (a) 2, -2, 1, -1 (c) 2, 3, 1, 4
0 0 are 0 4
(b) 2, 3, -2, 4 (d) none of these
1 1 1 20. The eigenvalues of 1 1 1 are 1 1 1 (a) 0, 0, 0
(b) 0, 0, 1
(c) 0, 0, 3
(d) 1, 1, 1
1 1 3 21. The minimum and maximum eigenvalues of the matrix 1 5 1 3 1 1 are -2 and 6 respectively. What is the other eigenvalue? (a) 5 (b) 3 (c) 1 (d) -1 1 0 0 22. The inverse of the matrix 0 2 0 is 0 0 3 1 0 (a) 0
0 1 2 0
0 0 1 3
1 2 0 0 (c) 0 1 0 1 0 0 3
1 0 (b) 0
0 1 3 0
0 0 1 2
(d) none of these
23. Are the following vectors linearly dependent? X1 = [3, 2, 7], X2 = [2, 4, 1] and X3 = [1, -2, 6] (a) Dependent (b) Independent (c) Cannot say (d) None of these
Multiple Choice Questions 10.141
1 0 2 24. If A = , then the value of k for which A = 8A + kI is 1 7 (a) 5 (b) -5 (c) 7 (d) -7 a + i c -b + id 25. A = is unitary matrix if and only if b + id a - i c (a) a2 + b2 + c2 = 0 (b) b2 + c2 + d2 = 0 (c) a2 + b2 + c2 + d2 = 1 (d) a2 + b2 + c2 + d2 = 0 26. Which of the following matrices is not diagonalizable? 1 1 (a) 1 2
1 0 (b) 3 2
0 -1 (c) 1 0
1 1 (d) 0 1
3 4 27. The sum of the eigenvalues of the matrix for real and negative value of x 1 x is (a) greater than zero (b) less than zero (c) zero (d) dependent values of x. 28. A is a singular matrix of order 3 with eigenvalues 2 and 3. The third eigenvalue is (a) 1 (b) 0 (c) 4 (d) -1 29. The eigenvalues and the corresponding eigenvectors of a 2 × 2 matrix are given by l1 = 8
1 x1 = 1
l2 = 4
1 x2 = -1
The matrix is 6 2 (a) 2 6
4 6 (b) 6 4
2 4 (c) 4 2
4 8 (d) 8 4
Answers 1. (b) 9. (c) 17. (d) 25. (c)
2. (a) 10. (d) 18. (c) 26. (d)
3. (a) 11. (d) 19. (b) 27. (a)
4. (c) 12. (b) 20. (b) 28. (b)
5. (b) 13. (a) 21. (b) 29. (a)
6. (c) 14. (a) 22. (a)
7. (a) 15. (b) 23. (a)
8. (a) 16. (b) 24. (d)
Index A Absolute Convergence of a Series 5.94 Alternating Series 5.87 Area as Double Integral 9.141 Area in Cartesian Coordinates 9.141 Area in Polar Coordinates 9.154 Area of Surface of Solid of Revolution 4.46 Area of Surface of Solid of Revolution in Cartesian Form 4.46 Area of Surface of Solid of Revolution in Parametric Form 4.53 Area of Surface of Solid of Revolution in Polar Form 4.60
B Beta Function 3.11 Beta Function as Improper Integral 3.28 Bounded Sequence 5.3
C Cauchy’s Integral Test 5.82 Cauchy’s Root Test 5.73 Cayley-Hamilton Theorem 10.108 Chain Rule 8.59 Change of Order of Integration 9.31 Change of Variables from Cartesian to Other Coordinates 9.95 Change of Variables from Cartesian to Polar Coordinates 9.77 Characteristic Equation 10.65 Characteristic Matrix 10.64 Characteristic Polynomial 10.65 Coefficient Matrix 10.23
Cofactor of an Element of a Determinant 10.6 Column Matrix 10.2 Combining Series 5.18 Comparison Test 5.20 Composite Function of One Variable 8.59 Composite Function of Two Variables 8.66 Conditional Convergence of a Series 5.94 Conjugate of a Matrix 10.4 Continuous Function Theorem for Sequences 5.2 Convergence and Divergence of Improper Integrals 2.17 Convergence, Divergence and Oscillation of Infinite Series 5.9
D Determinant of a Matrix 10.5 Diagonal Matrix 10.3 Diagonalization of a Matrix 10.119 Direct Comparison Test 2.17 Directional Derivative 8.103 Double Integrals Over Rectangles 9.1 and General Regions 9.2 Double Integrals as Volume 9.3 Double Integrals in Polar Coordinates 9.66 Duplication Formula 3.13 D’Alembert’s Ratio Test 5.40
E Echelon Form of a Matrix 10.7
I.2
Index
Eigenspace 10.65 Eigenvalues and Eigenvectors 10.64 Elementary Matrices 10.7 Elementary Transformations 10.6 Equivalence of Matrices 10.7 Euler’s formulae 7.8 Extreme Values with Constrained Variables 8.134
F Fourier Coefficients 7.8 Fourier Series 7.5 Fourier Series of Even and Odd Functions 7.66 Fourier Series of Functions of Any Period 7.7 Functions of Two or More Variables 8.2
G Gamma Function 3.1 Gauss Elimination Method 10.24 Gauss–Jordan Method 10.24 Geometric Series 5.10 Gradient 8.103
H Half-Range Cosine Series 7.93 Half-Range Fourier Series 7.93 Half-Range Sine Series 7.93 Harmonic Analysis 7.1 Harmonic Series 5.19 Hermitian Matrix 10.5 Higher-Order Partial Derivatives 8.11
I Implicit Differentiation 8.94 Improper Integral of the Third Kind 2.16 Improper Integrals 2.1 Improper Integrals of the First Kind 2.2 Improper Integrals of the Second Kind 2.9 Infinite Series 5.8 Infinite Solutions 10.49 Interval and Radius of
Convergence 5.101 Inverse of a Matrix 10.6 Inverse of a Matrix by Gauss–Jordan Method 10.18
L Leibnitz’s Test for Alternating Series 5.87 Length of Arc in Cartesian Form 4.6 Length of Arc in Parametric Form 4.20 Length of Arc in Polar Form 4.33 Length of Plane Curves 4.6 Limit Comparison Test 2.18, 2.23 Limit and Continuity of Functions of Several Variables 8.2 Limit of a Sequence 5.2 Linear Dependence and Independence of Eigenvectors 10.76 Local Extreme Values (Maximum and Minimum Values) 8.116 Lower Triangular Matrix 10.3 L’Hospital’s Rule 1.1
M Maclaurin’s Series 6.27 Matrix 10.2 Method of Lagrange Multipliers 8.145 Minor of an Element of a Determinant 10.6 Mixed Derivative Theorem 8.11 Monotonic Sequence 5.3 Multiple Integrals by Substitution 9.77
N Non-Trivial Solution 10.49 nth Term Test for Divergence 5.9
O Orthogonal Matrix 10.5 Orthogonality of Functions 7.2 Orthogonality of Trigonometric Functions 7.2 Orthogonally Similar Matrices 10.119
Index
P p-Series 5.20 Partial Derivatives 8.10 Periodic Functions 7.1 Power Series 5.101 Properties of Beta Functions 3.12 Properties of Double Integrals 9.3 Properties of Eigenvalues 10.65 Properties of Eigenvectors 10.76 Properties of Even and Odd Functions 7.66 Properties of Gamma Function 3.2 Properties of Infinite Series 5.9
R Rank of a Matrix 10.13 Raabe’s Test (Higher Ratio Test) 5.67 Rank of a Matrix by Echelon Form 10.14 Relation between Beta and Gamma Functions 3.12 Row Matrix 10.2
S Sandwich Theorem for Sequences 5.7 Scalar Matrix 10.3 Sequence 5.2 Similarity Transformation 10.119 Singular and Non-Singular Matrices 10.6 Skew-Hermitian Matrix 10.5 Skew-Symmetric Matrix 10.4 Solutions of System of Linear Equations 10.23 Solutions of a System of Linear Equations 10.49 Spectrum 10.65 Square Matrix 10.2 Symmetric Matrix 10.4
I.3
System of Homogeneous Linear Equations 10.48 System of Non-Homogeneous Linear Equations 10.22
T Tangent Plane and Normal to a Surface 8.107 Taylor’s Series 6.1 Telescoping Series 5.15 Test for Non-existence of a Limit 8.2 Total Derivatives 8.59 Trace of a Matrix 10.4 Transpose of a Matrix 10.4 Transposed Conjugate of a Matrix 10.5 Trigonometric Form of Beta Function 3.11 Trigonometric Fourier Series 7.6 Triple Integrals 9.109 Triple Integrals in Cartesian Coordinates 9.109 Triple Integrals in Cylindrical Coordinates 9.109 Triple Integrals in Spherical Coordinates 9.110 Trivial Solution 10.49
U Unit or Identity Matrix 10.3 Unitary Matrix 10.5 Upper Triangular Matrix 10.3
V Volume Using Cross-sections 4.1
Z Zero or Null Matrix 10.2