Calculus II Workbook For Dummies [1 ed.] 1394188021, 9781394188024
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Table of contents :
Title Page
Copyright Page
Table of Contents
Introduction
About This Book
Foolish Assumptions
Icons Used in This Book
Beyond the Book
Where to Go from Here
Part 1 Introduction to Integration
Chapter 1 An Aerial View of the Area Problem
Measuring Area on the xy-Graph
Defining Area Problems with the Definite Integral
Calculating Area Defined by Functions and Curves on the xy-Graph
Answers and Explanations
Chapter 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
Fractions
Factorials
Negative and Fractional Exponents
Simplifying Rational Functions
Trigonometry
Parent Functions
Parent Function Transformations
Sigma Notation for Series
Answers and Explanations
Chapter 3 Recent Memories: Calculus Review
Evaluating Limits
Derivatives of Common Functions and the Constant Multiple Rule
The Power Rule
The Sum Rule
The Product Rule and Quotient Rule
The Chain Rule
Answers and Explanations
Part 2 From Definite to Indefinite Integrals
Chapter 4 Approximating Area with Riemann Sums
Calculating Riemann Sums with Left and Right Rectangles
Getting a Better Estimate with Midpoint Rectangles
Improving Your Estimate with the Trapezoid Rule
Using Simpson’s Rule to Further Improve Your Approximation
Answers and Explanations
Chapter 5 The Fundamental Theorem of Calculus and Indefinite Integrals
Evaluating Definite Integrals Using FTC2
Anti-differentiation and Indefinite Integrals
Signed and Unsigned Area
Answers and Explanations
Part 3 Evaluating Indefinite Integrals
Chapter 6 Instant Integration
Antiderivatives of Common Functions and the Constant Multiple Rule
The Power Rule
The Sum Rule
Answers and Explanations
Chapter 7 Sharpening Your Integration Moves
Integrating Rational and Radical Functions
Using Algebra to Prepare Functions for Integration
Integrating Using Inverse Trig Functions
Using Trig Identities to Prepare Functions for Integration
Integrating Compositions of Functions with Linear Inputs
Answers and Explanations
Chapter 8 Here’s Looking at u-Substitution
Understanding the How of u-Substitution
When to Use u-Sub: The Simpler Case
When to Use u-Sub: The More Complex Case
Using Variable Substitution with Definite Integrals
Answers and Explanations
Part 4 Advanced Integration Techniques
Chapter 9 Integration by Parts
Using the Formula for Integration by Parts
Knowing How to Assign u and dv
Applying Integration by Parts More Than Once
Answers and Explanations
Chapter 10 Trig Substitution
Integrating the Six Basic Trig Functions
Integrating Powers of Sines and Cosines
Integrating Powers of Tangents and Secants
Using Trig Substitution
Answers and Explanations
Chapter 11 Integration with Partial Fractions
Understanding Partial Fractions
Integrating Case 1: Rational Expressions That Have Distinct Linear Factors
Integrating Case 2: Repeated Linear Factors
Integrating Case 3: Distinct Quadratic or Higher-Degree Factors
Integrating Case 4: Repeated Quadratic Factors
Answers and Explanations
Part 5 Applications of Integrals
Chapter 12 Solving Area Problems
Breaking Definite Integrals in Two
Looking at Improper Integrals
Finding the Unsigned Area between Curves
Using the Mean Value Formula for Integration
Calculating Arclength
Answers and Explanations
Chapter 13 Pump up the Volume — Solving 3-D Problems
Using Disk and Washer Methods to Find the Volume of Revolution around a Horizontal Axis
Using Inverse Functions to Find Volume of Revolution around a Vertical Axis
Surface Area of Revolution
Shell Method
Answers and Explanations
Chapter 14 Differential Equations
Understanding Differential Equations
Solving Separable Differential Equations
Answers and Explanations
Part 6 Infinite Series
Chapter 15 Sequences and Series
Understanding Sequences
Understanding Series
Finding the Sequence of Partial Sums for a Series
Understanding and Evaluating Geometric Series
Understanding p-series
Answers and Explanations
Chapter 16 Convergent and Divergent Series
Nth-Term Test for Divergence
The Direct Comparison Test
The Limit Comparison Test
Integral Test
Ratio Test
Root Test
Answers and Explanations
Chapter 17 Taylor and Maclaurin Series
Expressing Functions as Maclaurin Series
Expressing Functions as Taylor Series
Answers and Explanations
Part 7 The Part of Tens
Chapter 18 Ten Mathematicians Who Anticipated Calculus before Newton and Leibniz
Zeno of Elea (495–430 BCE)
Eudoxus of Cnidus (408–355 BCE)
Archimedes (287–212 BCE)
Lui Hui (3rd Century CE)
Zu Chongzhi (425–500)
Hasan Ibn al-Haytham (965–1040)
Madhava of Sangamagrama (1340–1425)
Johannes Kepler (1571–1630)
Bonaventura Cavalieri (1598–1647)
Isaac Barrow (1630–1677)
Chapter 19 10 Skills from Pre-Calculus and Calculus I That Will Help You to Do Well in Calculus II
Expressing Functions as Exponents Whenever Possible
Knowing the Basic Trig Identities Cold
Recalling the Most Common Functions and Their Basic Transformations
Using Sigma Notation for Series
Evaluating Limits
Differentiating the Most Common Functions
Knowing the Power Rule for Differentiation
Applying the Product Rule
Using the Chain Rule
Calculating Common Derivatives Quickly in Your Head
Index
EULA
Citation preview
Calculus II Workbook
Calculus II Workbook by Mark Zegarelli
Calculus II Workbook For Dummies® Published by: John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, www.wiley.com Copyright © 2023 by John Wiley & Sons, Inc., Hoboken, New Jersey Published simultaneously in Canada No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the Publisher. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permissions. Trademarks: Wiley, For Dummies, the Dummies Man logo, Dummies.com, Making Everything Easier, and related trade dress are trademarks or registered trademarks of John Wiley & Sons, Inc. and may not be used without written permission. All other trademarks are the property of their respective owners. John Wiley & Sons, Inc. is not associated with any product or vendor mentioned in this book.
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Contents at a Glance Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Part 1: Introduction to Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 CHAPTER 1: CHAPTER 2: CHAPTER 3:
An Aerial View of the Area Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Forgotten but Not Gone: Review of Algebra and Pre-Calculus. . . . . . . . . . . . . . . . . . . . 15 Recent Memories: Calculus Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Part 2: From Definite to Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 4: CHAPTER 5:
Approximating Area with Riemann Sums. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 The Fundamental Theorem of Calculus and Indefinite Integrals. . . . . . . . . . . . . . . . . . 69
Part 3: Evaluating Indefinite Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 6: CHAPTER 7: CHAPTER 8:
51
81
Instant Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Sharpening Your Integration Moves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Here’s Looking at u-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Part 4: Advanced Integration Techniques. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 CHAPTER 11: Integration with Partial Fractions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 CHAPTER 9:
CHAPTER 10: Trig
Part 5: Applications of Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
167
CHAPTER 12: Solving
Area Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 up the Volume — Solving 3-D Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 CHAPTER 14: Differential Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 CHAPTER 13: Pump
Part 6: Infinite Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 15: Sequences
219
and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
CHAPTER 16: Convergent
and Divergent Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .235 CHAPTER 17: Taylor and Maclaurin Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
Part 7: The Part of Tens. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
267
CHAPTER 18: Ten
Mathematicians Who Anticipated Calculus before Newton and Leibniz �������������269 CHAPTER 19: 10 Skills from Pre-Calculus and Calculus I That Will Help You to Do Well in Calculus II���������������������������������������������������������������������������������������������������������������������������273
Index. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
279
Table of Contents INTRODUCTION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 About This Book. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Foolish Assumptions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Icons Used in This Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Beyond the Book. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Where to Go from Here . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
PART 1: INTRODUCTION TO INTEGRATION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 CHAPTER 1:
An Aerial View of the Area Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Measuring Area on the xy-Graph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Defining Area Problems with the Definite Integral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Calculating Area Defined by Functions and Curves on the xy-Graph. . . . . . . . . . . . . 10 Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Forgotten but Not Gone: Review of Algebra and Pre-Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
Fractions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Factorials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Negative and Fractional Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simplifying Rational Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Trigonometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parent Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parent Function Transformations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sigma Notation for Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 17 18 20 20 25 28 31 33
Recent Memories: Calculus Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
Evaluating Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivatives of Common Functions and the Constant Multiple Rule . . . . . . . . . . . . . The Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Sum Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Product Rule and Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Chain Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 40 41 42 43 44 46
PART 2: FROM DEFINITE TO INDEFINITE INTEGRALS. . . . . . . . . . . . . . . . .
51
CHAPTER 2:
CHAPTER 3:
CHAPTER 4:
Approximating Area with Riemann Sums. . . . . . . . . . . . . . . . . . . . . . .
53
Calculating Riemann Sums with Left and Right Rectangles. . . . . . . . . . . . . . . . . . . . . Getting a Better Estimate with Midpoint Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . Improving Your Estimate with the Trapezoid Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . Using Simpson’s Rule to Further Improve Your Approximation. . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54 57 59 61 63
Table of Contents
vii
The Fundamental Theorem of Calculus and Indefinite Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
Evaluating Definite Integrals Using FTC2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Anti-differentiation and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Signed and Unsigned Area. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70 72 75 78
PART 3: EVALUATING INDEFINITE INTEGRALS . . . . . . . . . . . . . . . . . . . . . . . .
81
CHAPTER 5:
CHAPTER 6:
CHAPTER 7:
Instant Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
Antiderivatives of Common Functions and the Constant Multiple Rule . . . . . . . . . . The Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Sum Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83 85 86 88
Sharpening Your Integration Moves . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
Integrating Rational and Radical Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Using Algebra to Prepare Functions for Integration. . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Integrating Using Inverse Trig Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Using Trig Identities to Prepare Functions for Integration . . . . . . . . . . . . . . . . . . . . . 96 Integrating Compositions of Functions with Linear Inputs. . . . . . . . . . . . . . . . . . . . . 98 Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
Here’s Looking at u-Substitution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103
Understanding the How of u-Substitution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . When to Use u-Sub: The Simpler Case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . When to Use u-Sub: The More Complex Case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Using Variable Substitution with Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103 105 107 109 112
PART 4: ADVANCED INTEGRATION TECHNIQUES. . . . . . . . . . . . . . . . . . . .
115
Integration by Parts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117
Using the Formula for Integration by Parts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Knowing How to Assign u and dv. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Applying Integration by Parts More Than Once . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117 119 121 125
Trig Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131
Integrating the Six Basic Trig Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Integrating Powers of Sines and Cosines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Integrating Powers of Tangents and Secants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Using Trig Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131 132 134 137 142
Integration with Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
151
CHAPTER 8:
CHAPTER 9:
CHAPTER 10:
CHAPTER 11:
Understanding Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 Integrating Case 1: Rational Expressions That Have Distinct Linear Factors . . . . . 154 Integrating Case 2: Repeated Linear Factors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
viii
Calculus II Workbook For Dummies
Integrating Case 3: Distinct Quadratic or Higher-Degree Factors . . . . . . . . . . . . . . 157 Integrating Case 4: Repeated Quadratic Factors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
PART 5: APPLICATIONS OF INTEGRALS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
167
Solving Area Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169
Breaking Definite Integrals in Two . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Looking at Improper Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Finding the Unsigned Area between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Using the Mean Value Formula for Integration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Calculating Arclength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169 172 174 177 178 181
Pump up the Volume — Solving 3-D Problems . . . . . . . . . . . . . . . .
189
CHAPTER 12:
CHAPTER 13:
Using Disk and Washer Methods to Find the Volume of Revolution around a Horizontal Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 Using Inverse Functions to Find Volume of Revolution around a Vertical Axis . . . 195 Surface Area of Revolution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 Shell Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 CHAPTER 14:
Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
211
Understanding Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Solving Separable Differential Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
PART 6: INFINITE SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
219
Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
221
Understanding Sequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Understanding Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Finding the Sequence of Partial Sums for a Series. . . . . . . . . . . . . . . . . . . . . . . . . . . Understanding and Evaluating Geometric Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . Understanding p-series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
221 223 225 227 230 231
Convergent and Divergent Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
235
CHAPTER 15:
CHAPTER 16:
Nth-Term Test for Divergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 The Direct Comparison Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 The Limit Comparison Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 Integral Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 Root Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 CHAPTER 17:
Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
255
Expressing Functions as Maclaurin Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Expressing Functions as Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Answers and Explanations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
Table of Contents
ix
PART 7: THE PART OF TENS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
267
Ten Mathematicians Who Anticipated Calculus before Newton and Leibniz. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
269
CHAPTER 18:
Zeno of Elea (495–430 BCE) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Eudoxus of Cnidus (408–355 BCE). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Archimedes (287–212 BCE). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Lui Hui (3rd Century CE) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Zu Chongzhi (425–500) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Hasan Ibn al-Haytham (965–1040) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Madhava of Sangamagrama (1340–1425). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Johannes Kepler (1571–1630). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 Bonaventura Cavalieri (1598–1647) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 Isaac Barrow (1630–1677) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
10 Skills from Pre-Calculus and Calculus I That Will Help You to Do Well in Calculus II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
273
Expressing Functions as Exponents Whenever Possible. . . . . . . . . . . . . . . . . . . . . . Knowing the Basic Trig Identities Cold. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Recalling the Most Common Functions and Their Basic Transformations. . . . . . . Using Sigma Notation for Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Evaluating Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Differentiating the Most Common Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Knowing the Power Rule for Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Applying the Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Using the Chain Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Calculating Common Derivatives Quickly in Your Head. . . . . . . . . . . . . . . . . . . . . . .
274 274 274 275 275 275 276 276 276 277
INDEX. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
279
CHAPTER 19:
x
Calculus II Workbook For Dummies
Introduction
C
alculus II is tough stuff by just about any standard.
In this book, I’ve done my best to explain the key topics that tend to hang students up as clearly and concisely as possible. I’ve also provided practice problems that I think will help you to make sense of this information, so you’ll be ready to take on similar problems that your teacher or professor might throw at you.
About This Book Chapter by chapter, this workbook follows the typical order of topics in a standard Calculus II course. But please feel free to jump back and forth throughout the book in whatever order makes sense for you. Chapter 1 provides an overview of the course, and then Chapters 2 and 3 give you a review of the Pre-Calculus and Calculus I topics you’ll need to know to move forward. In Parts 2 through 4, you learn a variety of integration methods, all of which are commonly taught in Calculus II. Part 5 focuses on basic applications of integration. Part 6 covers sequences and series. And in Part 7, I give you a couple of Top Ten lists related to calculus. Additionally, for convenience, this workbook follows the same chapter-by-chapter format as Calculus II For Dummies, 3rd edition. Of course, you don’t have to buy that book to make good use of this one. But if you do want to use both books, you’ll find that corresponding numbered chapters cover the same topics, so you’ll be able to flip back and forth between them easily.
Foolish Assumptions I assume that you either want or need to learn Calculus II. So, either you’re interested in the topic and want to study it on your own, or like many people, you’re taking a high-school or college course in the subject.
Icons Used in This Book In this book, I use a variety of icons to give you a heads-up about what information is important and what you can safely skip over when you’re in a rush.
Introduction
1
The example icon accompanies a sample question followed by a step-by-step solution. In most cases, examples should help you get a handle on difficult material in a way that makes sense.
This icon alerts you to key information that you may need to pay special attention to, especially if you’re currently studying for a test.
Tips provide you with a quick and easy way to work on a problem. Try them out as you work your way through the book, so you can use them in assignments and on tests.
The Technical Stuff icon marks information of a highly technical nature that you can normally skip if you are in a hurry.
This icon warns you of typical errors that students tend to fall into. Keep an eye on these little traps so that they don’t get you, too!
Beyond the Book If you want to explore Calculus II in greater depth, or get an additional perspective on it, look no further than my book Calculus II For Dummies, 3rd edition. This workbook has been written in conjunction with that book to provide an even more complete picture of the topics taught here. This book also includes accompanying online material in the form of a Cheat Sheet. To access it, go to www.dummies.com and type Calculus II Workbook For Dummies in the Search box.
Where to Go from Here Success in a Calculus II class is built on a foundation of a whole lot of other math you’ve been studying since you learned how to count. To help you shore up this foundation, here are a few additional resources:
»» If you need more help with some of the foundational math than this book provides, an easy place to start is my book, Basic Math and Pre-Algebra For Dummies.
»» If you need a refresher on any algebra concepts not covered here, check out Algebra I For Dummies and Algebra II For Dummies, both by Mary Jane Sterling.
»» If you find that your trig skills are in need of a makeover, Trigonometry For Dummies, also by Mary Jane Sterling, gives you wider and deeper coverage of the topic than you’ll find here.
»» Even if you’ve passed Calculus I, you may want a refresher. Check out Calculus For Dummies and its accompanying workbook by Mark Ryan for a closer look at basic calculus topics.
2
Calculus II Workbook For Dummies
1
Introduction to Integration
IN THIS PART . . .
See Calculus II as an ordered approach to finding the area of unusual shapes on the xy-graph Use the definite integral to clearly define an area problem Slice an irregularly shaped area into rectangles to approximate area Review the math you need from Pre-Algebra, Algebra, Pre-Calculus, and Calculus I
IN THIS CHAPTER
»» Measuring the area of shapes with classical geometry »» Finding the area of shapes on the xy-plane »» Using integration to frame the area problem
1 An Aerial View of the Area Problem Chapter
I
n Calculus I, you discovered how to use math to solve a single problem: the tangent problem or slope problem, which involved finding the slope of a tangent line to a function on the xy-graph. Calculus II is also devoted to solving a single problem: the area problem — finding the area of a region of the xy-graph under a function. In this chapter, you use a variety of simple methods to frame and solve area problems. First, you use formulas from classical geometry to measure the area of familiar shapes on the xy-graph. Next, you discover how to define the area of a region under a function as a definite integral. To finish up, you put those two skills together, both setting up and solving definite integrals to solve area problems for relatively simple functions. Ready to get started?
Measuring Area on the xy-Graph People have been calculating the area of shapes for thousands of years. More than 2,000 years ago, Euclid developed a thorough system of geometry that included methods for finding the area of squares, rectangles, triangles, circles, and so on. Descartes’s invention of the xy-graph in the early 17th century made the study of geometry possible from an algebraic perspective, laying the foundation for calculus.
CHAPTER 1 An Aerial View of the Area Problem
5
But you don’t need calculus to measure the area of many basic shapes on the xy-graph. Here are a few formulas you know from geometry that are just as useful when measuring area on the xy-graph. Square:
Area
Rectangle:
s2
Area
Trapezoid:
Area
Q. A.
A.
6
Circle:
b1 b 2 h 2
Area
Area
1 bh 2
Semi-Circle:
r2
Area
1 2 r 2
What is the area of the shape in this figure? 54. The figure shows a trapezoid with bases of lengths 12 and 6, and a height of 6. Plug these values into the formula for a trapezoid:
b1 b2 h 2
Q.
bh
Triangle:
12 6 6 2
54
Find the area inside the semi-circle shown in the figure.
12.5 . The semi-circle has a radius of 5, so plug this value into the formula for a semi1 2 1 r 5 2 12.5 circle: 2 2
PART 1 Introduction to Integration
In each of the following xy-graphs, find the area of the shape.
1
2
3
4
5
6
CHAPTER 1 An Aerial View of the Area Problem
7
Defining Area Problems with the Definite Integral The definite integral provides a systematic way to define an area on the xy-graph. For functions that reside entirely above the x-axis, a definite integral
b a
f ( x ) dx defines an area that lies:
»» Vertically between the function f(x) and the x-axis »» Horizontally between two x-values, a and b, called the limits of integration
Q. A.
What definite integral describes the shaded area in the figure? 4 1
1 1 dx . The shaded area resides vertically between the function and the x-axis, and x x
horizontally between x = 1 and x = 4.
8
PART 1 Introduction to Integration
Q. A.
Write a definite integral that defines the area of the shaded region in the figure. 2
2 cos x dx . The shaded region is vertically between f ( x ) 2 cos x and the x-axis, and
2
horizontally between x
2
and x
2
.
In each of the following xy-graphs, write the definite integral that represents the shaded area.
7
8
9
10
CHAPTER 1 An Aerial View of the Area Problem
9
11
12
Calculating Area Defined by Functions and Curves on the xy-Graph In the previous section, you discover how to use the definite integral to frame an area problem on the xy-graph. In Calculus II, you’ll learn a ton of different ways to evaluate definite integrals to solve area problems. But for many simple functions, you don’t need calculus to evaluate a definite integral. For example:
10
PART 1 Introduction to Integration
Q. A.
Write a definite integral to describe the shaded area in the figure, and then use geometry to evaluate it. 3 1
3 x 3 dx
6. The shaded area is the triangle that resides vertically between the
function 3 x
3 and the x-axis and horizontally between x
1 and x
3. The base of this
triangle is 3 – 1 2. To find the height, plug in f ( 3 ) 3( 3 ) 3 6 . Now, to find the area, plug in 2 for the base and 6 for the height into the formula for a triangle:
1 bh 2
Area
Q. A.
1 ( 2 )( 6 ) 6 2
Use a definite integral to define the shaded region shown, and then find the area of that region without calculus. 6 6
x 2 dx
36
18 . The shaded area resides vertically between the function 36 x 2
and the x-axis. To find the limits of integration, set this function to 0 and solve:
36 x 2
0 0 x2 x
36 x 2 36 6 and
6
Thus, the limits of integration are x –6 and x 6. To find the area, plug in 6 for the radius into the formula for the area of a semi-circle:
1 2 r 2
1 2
6
2
18
CHAPTER 1 An Aerial View of the Area Problem
11
In each of the following xy-graphs, write and evaluate the definite integral that represents the shaded area.
13
14
15
16
17
18
12
PART 1 Introduction to Integration
Answers and Explanations 1
16. s 2
2
60. bh
42
16
10 6
60
1 1 bh 8 4 16 2 2 4 16.5. b1 b2 h 7 4 3 2 2 3
16.
5 25 . r 2
5
6 32 . 1 r 2
1 2
2
7 8 9 10 11 12 13 14 15 16 17 18
4 1
3 x 4
4
2
25 3
9 2
2
x 1 dx
0 2
2
16.5
3 dx
3 x 2 dx sin x dx
0 1
e x dx
0 4 1 7 1 6 0
ln x dx 4 dx
6
x dx
1 6 2
4
24 6
18
1 2 6 8 32 x 2 dx 2 2 9 1 1 x 3 dx 9 3 13.5 0 3 2 3 1 2 3 4 .5 9 x 2 dx 3 2 8
0
1 0
1
1 x 2 dx
1
4
CHAPTER 1 An Aerial View of the Area Problem
13
IN THIS CHAPTER
»» Calculating with fractions and factorials »» Working with exponents and simplifying rational expressions »» Remembering radian measure »» Proving trig identities »» Understanding important parent functions and their transformations »» Converting an infinite series from sigma notation to expanded notation
2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus Chapter
M
ost students have been studying math for at least 10 years before they enter their first calculus classroom. This fact leaves many students overwhelmed by all the math they should know, and perhaps did know at one time, but can’t quite recall.
Fortunately, you don’t need another 10 years of review to be ready for Calculus II. In this chapter, I get you back up to speed on the key topics from your Pre-Algebra, Algebra, and PreCalculus classes that will help you the most this semester. To begin, you go all the way back to middle school for a quick review of fractions. I also give you some practice calculating factorials.
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
15
After that, I remind you how to work with exponents, and especially how to use negative and fractional exponents to express rational and radical functions. Then I cover a few important ideas from trigonometry that you’re sure to need, such as radian measure and trig identities. Next, I give you an overview of how to sketch the most important parent functions on the xy-graph: polynomials, exponentials, radicals, logarithmic functions, and the sine and cosine functions. You use these to work with a variety of function transformations, such as vertical and horizontal transformations, as well as stretch, compress, and reflect transformations.
Fractions When finding derivatives in Calculus I and integrals in Calculus II, you’ll often need to add 1 to (or subtract 1 from) a fraction. Here’s a trick for doing both of those operations quickly in your head without getting a common denominator:
Q.
What is
A.
12 . To do this calculation in your head, add the numerator and denominator, and then 5
7 1? 5
keep the denominator of 5:
7 1 5
Q. A.
7 5 5
What is
12 5
5 1? 6
1 . To calculate this value in your head, subtract the numerator minus the denomina6
tor, and then keep the denominator of 6.
5 1 6 1
c.
1 6
3 1 5 2 1 3
b. d.
11 1 6 13 1 5
Subtract 1 from the following fractions and express each answer as a proper or improper fraction (no mixed numbers). a. c.
16
1 6
Add 1 to the following fractions and express each answer as a proper or improper fraction (no mixed numbers). a.
2
5 6 6
1 1 4 1 1 5
PART 1 Introduction to Integration
b. d.
7 1 3 21 1 2
Factorials In Calculus II, when working with infinite series, you also may need to make use of factorials. Recall that the symbol for factorial is an exclamation point (!). The factorial of any positive integer is that number multiplied by every positive integer less than it. Thus:
1! 1 4!
2! 2 1 2
4 3 2 1 24 5 !
Also, by definition, 0 !
3!
3 2 1 6
5 4 3 2 1 120 6 ! 6 5 4 3 2 1 720
1.
When you know how to expand factorials in this way, simplifying rational expressions that include them is relatively straightforward. Always look for opportunities to cancel factors in both the numerator and denominator.
Q.
Simplify
A.
10. Begin by expanding the factorials:
6! 2! 4! 3!
6! 2! . 4! 3!
6 5 4 3 2 1 2 1 4 3 2 1 3 2 1
Now, cancel factors in both the numerator and denominator, and simplify the result:
6 5 4 3 2 1 2 1 4 3 2 1 3 2 1
Q. A.
Simplify
n 2 n 2! n n 1!
30 3
10
n 2! . n n 1!
n 1 . Expand the factorials as follows: n 2 n 1 n ... 3 2 1 n n 1 n 2 n 3 ... 3 2 1
Cancel factors in both the numerator and denominator, and simplify the result:
n 2 n 1 n ... 3 2 1 n n 1 n 2 n 3 ... 3 2 1
n 2
n 1
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
17
3
4
Simplify each of the following factorial expressions. a.
5! 3!
b.
c.
100 ! 101!
d.
7! 9! 19 ! 8 ! 11! 17 !
Simplify each expression in terms of n. a.
n! n 2 !
b. n 1 !
c.
n 1 ! n 1 ! n!
d.
n 3 !
n n 2 n 2 ! n 3 4n n 1 !
Negative and Fractional Exponents When an expression has a negative exponent, you can rewrite it with a positive exponent and place it in the denominator of a fraction. For example:
x
1
1 x
x
1 x2
2
x
3
1 x3
x
1 x4
4
When an expression has a fractional exponent, you can rewrite it as a radical. For example: 1
1
x2
x
3
x3
1
x
4
x4
1
x
5
x5
x
More complicated fractional exponents can be written in two separate and equally valid ways as a combination of a radical and an exponent. For example: 5
3
x3 x
5 3
3
x5 3
x
4
x4 5
x
3 4
5
x3 4
4
x4 3
x
x
5 4
11 5
x5
x
4
11 x5
5
x
5
x 11 5
x
11
When a fractional exponent is negative, you can rewrite it as a radical in the denominator of a fraction:
x
18
1 2
1 x
x
3 2
PART 1 Introduction to Integration
1 x
3
x
5 3
1 3
x
5
x
7 8
1 8
x7
Keep track of these types of conversions! In Calculus II, you’ll need to change expressions back and forth between these two forms just about every day.
Q.
Rewrite the expression x
A.
5
1 x8
8 5
without using a negative fractional exponent.
. First, rewrite the expression with a positive exponent in the denominator of a
fraction, then change this to radical form:
x
8 5
1 x
Q. A.
Express
1
1 4
9 4.
using a negative fractional exponent.
x9
1
1
x9
x4
9
x
9 4
5
b. x
3
c. x 2
7
x8
Express each of the following without using negative or fractional exponents. a. x 6
6
5
Begin by changing the radical to a fractional exponent, then change this value from negative to positive by bringing the expression out of the denominator:
x
4
5
1
8 5
4 3
d. x
Express each of the following as a negative or fractional exponent of x. a.
3
x
b.
c.
5
x4
d.
1 x8 1 x9
Express each of the following using negative or fractional exponents of x. a. 3x x
c.
b.
10 x x
23
x
2
d.
2 3
5 x2 3x 4 x 4 x3
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
19
Simplifying Rational Functions Another key algebra skill you’ll need to recall is simplifying rational expressions by factoring and then canceling factors.
Q. A.
Simplify the rational expression
5 x 3 45 x . 10 x 2 50 x 60
Always begin by factoring out the greatest common factor (GCF) in both the numerator and denominator, if possible, then simplify:
5 x 3 45 x 10 x 2 50 x 60
5x x 2 10 x
2
x x2
9
5x 6
2 x
2
9
5x 6
Now, factor the numerator as a difference of squares and the denominator as a quadratic expression:
x x 3 x 3 2 x 2 x 3
x x 3 2 x 2
Depending on the problem, you may also need to remove all parentheses:
x 2 3x 2x 4 8
Fully simplify each of the following rational functions by factoring, cancelling factors, and then removing all parentheses: a.
x2 2x 3
c.
x 2 2 x 15 x 2 4x 3
3x 6x 2
b.
3x 3 3x 4x 4 4x 3
d.
2 x 4 26 x 2 72 9 x 2 45 x 54
Trigonometry In Calculus II, you’ll be working with trigonometric functions constantly. Most students are at least a little shy about their trig skills. Fortunately, you don’t have to remember every last trig fact you were ever tested on. Here’s a quick review of the skills you’ll need most to do well in your current class.
20
PART 1 Introduction to Integration
When you draw a right triangle and identify one of the two small angles as x, you can distinguish the three sides of the right triangle as the opposite (O), the adjacent (A), and the hypotenuse (H). The first three trig functions (sine, cosine, and tangent) are all defined in terms of this angle x (recall the mnemonic SOH-CAH-TOA):
O H
sin x
cos x
A H
tan x
O A
The three reciprocal trig functions (cosecant, secant, and cotangent) all follow from these definitions:
H O
csc x
sec x
H A
cot x
A O
Given any trig function of x, you can deduce the other five functions by drawing a triangle, labeling it properly, and finding the remaining side with the Pythagorean Theorem.
Q.
If sin x
A.
3 . Recall that sin x 4
3 , what is the value of tan x? 5
O , so draw a right triangle with angle x, an opposite side of 3, H
and a hypotenuse of 5. Now, use the Pythagorean Theorem to find the length of the adjacent side:
A2 A
32 2
52
9 25 A2
16
A 4 Thus, the adjacent side has a length of 4, so tan x
O A
3 . 4
In Calculus II, you’ll almost always use radian measure for angles instead of degrees. Here are the most common conversions from degrees to radians:
0 90
0
30
2
180
6
45 270
4 3 2
60 360
3 2
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
21
To convert any number of degrees to radians, multiply by the conversion factor To change any number of radians to degrees, multiply by the conversion factor
Q.
How do you express 40 in radian measure?
A.
2 . Multiply by the conversion factor , then simplify: 9 180 40
180
180 180
. .
2 9
Q.
What is the equivalent of
A.
7 . Multiply by the conversion factor 10 7 180 18
7 in degrees? 18 180
, then simplify:
7 10
Another important trig skill you’ll need in Calculus II is a basic understanding of trig identities. In Pre-Calculus, teachers sometimes go way overboard on this topic. Here, I give you a review of the identites you need to survive Calculus II. These come in three groups. The first group of identities are the five basic trig identities:
sin x
1 csc x
cos x
1 sec x
tan x
1 cot x
tan x
sin x cos x
cot x
cos x sin x
The next group are the three Pythagorean identities:
sin 2 x cos 2 x
1
1 tan 2 x
sec 2 x
1 cot 2 x
csc 2 x
The final group are the following useful versions of the double-angle identity (for sines) and the two half-angle identities (for sines and cosines):
sin 2 x
2 sin x cos x
sin 2 x
1 cos 2 x 2
cos 2 x
1 cos 2 x 2
If needed, you can always check online to remember the double-angle and half-angle identities for the remaining trig functions. But this group of 11 identities is enough to get you started. In Calculus II, one important use for the trig identities is to rewrite a complicated-looking trig function as the product of two trig functions raised to powers of either positive or negative integers.
22
PART 1 Introduction to Integration
Q. A.
Show that
sin x tan x sec x
1 cos 2 x . 2
To begin, use the reciprocal identities to get rid of the denominator on the left side of the equation:
sin x tan x sec x
sin x cot x cos x
Next, rewrite cot x in terms of sines and cosines, then simplify:
sin x cos x cos x sin x
cos 2 x
To finish up, use the half-angle formula for cosines:
cos 2 x
1 cos 2 x 2
This chain of inference proves that
sin x tan x sec x
Q.
Show that
A.
To begin, factor out tan x in the denominator:
cos 2 x tan x cos 2 x tan x
cos 2 x tan x cos 2 x tan x
1 cos 2 x . 2
cot 3 x .
cos 2 x tan x 1 cos 2 x
Next, use the first Pythagorean identity to rewrite 1 cos 2 x :
cos 2 x tan x sin 2 x cos 2 x
Use the two basic identities for cot x to rewrite tan x and in terms of cotangents, sin 2 x then simplify:
cot x cos 2 x sin 2 x
cot x cot 2 x
cot 3 x
This chain of inference proves that
cos 2 x tan x cos 2 x tan x
cot 3 x .
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
23
9
If sin x
11
If cot x
13
5 , what is cos x ? 13
a. 25
If tan x
12
If sec x
14
Convert each of the following angles from radians to degrees:
21 , what is csc x ?
Convert each of the following angles from degrees to radians: b. 80
c. 144
7 , what is cos x ? 6
d. 200 a.
24
2 , what is sec x ? 3
10
PART 1 Introduction to Integration
5
b.
11 8
c.
7 12
d.
21 16
tan x sec x csc x
15
Demonstrate that
17
Prove the following: 2
cos x csc x cos x sin 2 x
1 cos 2 x .
16
Show that
tan 2 x 1 cos x cot x
18
Show that
sin 2 x 2 sin x
tan x sec 3 x .
cos x.
3
cot x csc x
Parent Functions A parent function is the simplest form of a family of functions — that is, a set of functions whose equations and graphs all resemble one another. For example, f ( x ) x 2 is the parent function 3 x 2 4 x 11, all of which are for the family of more complicated quadratic functions, f ( x ) shaped like parabolas. In Algebra I and II, and then in Pre-Calculus, students study the behavior of about a dozen families of functions. I’m sure you’ll recognize the ones that I introduce here. Knowing these common parent functions well enough to sketch their graphs can give you visual information that’s very useful when solving Calculus II problems.
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
25
Arguably, the most important family of functions is the polynomials, which include infinitely many sub-families of functions. Figure 2-1 shows you the parent functions for the four most common polynomials: linear (degree 1), quadratic (degree 2), cubic (degree 3), and quartic (degree 4) functions. Note that the degree of a polynomial is the exponent of its leading term — that is, its greatest exponent. In each graph, I’ve identified an anchor point on that function — an important point that can help to improve the accuracy of your graph. Every parent function for the polynomials includes the anchor point ( 0, 0 ).
FIGURE 2-1:
Parent functions and graphs of four polynomial functions.
Figure 2-2 shows the parent functions and graphs for the two most common radical functions, the square root and cube root. As with the polynomials, these parent functions also include the anchor point ( 0, 0 ). Notice that the square root parent function, f ( x ) x , has a domain of ( 0, ). Thus, inputting a negative x-value to this function is not allowed. In contrast, the domain of the cube root parent function, f ( x ) 3 x , is ( , ), so you can input any real value into this function. In Figure 2-3, you see the parent functions for the exponential and logarithmic functions. Note that in Calculus II and beyond, the default base for these functions is the transcendental number e 2.718 . Accept no substitutes! Because this pair of functions are inverses of each other, their graphs are reflections along the diagonal line f ( x ) x . Correspondingly, their anchor points are ( 0, 1) and (1, 0 ).
26
PART 1 Introduction to Integration
FIGURE 2-2:
Parent functions and graphs for the square root and cube root functions.
FIGURE 2-3:
Parent functions and graphs for the exponential and logarithmic functions.
The two most important trig functions are shown in Figure 2-4. The sine and cosine functions are so important that you should be able to sketch these two graphs from memory. Note that the anchor point for the sine function is ( 0, 0 ) and the one for the cosine function is ( 0, 1) . As for the other four trig functions, they’re less important by several orders of magnitude. I’m not saying that you’ll never need to know what the graph of a tangent or secant function looks like. Rather, the sine and cosine functions model everything in the physical world that can be modeled using waves: light, sound, heat convection, weather, climate, the oceans, earthquakes, radiation, quantum events, and economics to name just a few. They’re also the heart and soul of Fourier transformations, which is one of the greatest and most useful achievements of 19thcentury math. Two additional parent functions that I think will be handy are shown in Figure 2-5: the absolute value and rational parent functions. As you can see, the anchor point for the absolute value function, ( 0, 0 ), has a sharp turn, so this is a non-differentiable point — in other words, it doesn’t have a measurable slope at this point. And the rational function has a vertical asymptote at x 0, which is also, clearly, non-differentiable.
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
27
FIGURE 2-4:
Parent functions and graphs for the sine and cosine functions.
FIGURE 2-5:
Parent functions and graphs for the absolute value and rational functions.
Parent Function Transformations Transformations are changes made to a function that predictably affect the way that function appears on the xy-graph. When a transformation changes the position of a function without altering its shape, it’s called a translation: The vertical translation f ( x ) n moves the function up or down on the graph: For example, starting with the function f ( x ) x 2:
f(x) 3
x2
f(x) 5
2
x
3 is three units up from the original parent function. 5 is five units down from the original parent function.
The horizontal translation f ( x the function f ( x ) x 3 :
n ) moves left or right on the graph: For example, starting with
f ( x 2 ) ( x 2 ) 3 is two units to the left from the original parent function. f ( x 4 ) ( x 4 ) 3 is four units to the right from the original parent function.
28
PART 1 Introduction to Integration
Another important transformation is the vertical stretch/compress/reflect transformation nf ( x ), which changes the shape of the function as it appears on the xy-graph:
When n 1, this transformation vertically stretches the function away from the x-axis. When 0
n 1, this transformation vertically compresses the function toward the x-axis.
When 1 n 0, this transformation vertically compresses the function toward the x-axis and reflects it across the x-axis.
1, this transformation vertically stretches the function away from the x-axis When n and reflects it across the x-axis. When working with trigonometric functions, especially the sine and cosine functions, a fourth transformation is the horizontal stretch/compress/flip function f(nx):
When n 1, this transformation compresses the function toward the y-axis. When 0
n 1, this transformation stretches the function away from the y-axis.
When 1 n 0, this transformation stretches the function away from the y-axis and reflects it across the y-axis.
1, this transformation compresses the function toward the y-axis and reflects When n it across the y-axis.
Q. A.
Identify how the transformations in f ( x ) 3 x use this information to sketch this graph.
1
4 affect its resulting graph, then
Right 1, Vertical stretch 3, Down 4.
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
29
Sketch each of the following functions on the xy-graph.
19
f(x) 2 x 1
21
f(x) 2 x
23
f(x) ex
30
2
3
3 1
1
PART 1 Introduction to Integration
x3
20
f(x)
22
f(x)
24
f ( x ) ln( x
2
1 1 x 2
3)
1 sin x 1 2
25
f(x)
27
f ( x ) 2 cos x
2
26 f ( x )
3 cos x
28
4 sin 2 x
f(x)
Sigma Notation for Series A Calculus II class usually includes an extensive look at infinite series. A common and compact way to write an infinite series is with sigma notation, which utilizes the Greek letter Σ (sigma) to specify an infinite sum. Sigma notation looks impressive, and when you use it, people will know that you’re really good at math (even if you don’t think so sometimes). On the downside, sigma notation can be confusing. However, when you change a series into its expanded form, you can usually begin to make sense of it. That’s why I recommend that students change sigma notation to expanded notation at the beginning of a tough-looking problem. This is a great skill to have going into a Calculus II class.
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
31
Q. A. Q. A.
Expand the series
i
to its first five terms, and then simplify the result.
4 5 1 1 3 1 1 ... 10, 000 100, 000 10 50 1, 000 2, 500 20, 000 2 Expand the series ( 1) i 1 to its first five terms, and then simplify the result. i ! i 1 1 1 1 2 1 .... Begin by substituting in the values 0 through 4 for i: 3 12 60 i
i 1 10
i
( 1) i i 0
1 10
1
2 100
i i 1 10
2 ( i 1)!
3 1, 000
( 1) 0
2 1!
( 1)1
2 2!
( 1) 2
2 3!
( 1) 3
2 4!
( 1) 4
...
2 ... 5!
Now, noticing that the powers of –1 create an alternating series of positive and negative terms, go ahead and simplify the result:
2 1
2 2
2 6
2 24
2 1 ... 2 1 120 3
1 12
1 ... 60
Expand each of the following series to its first five terms, and then simplify the result.
29 i
2i i 12
31
( 1) i i 1
32
30 i
2n i 3i
PART 1 Introduction to Integration
n i 0 i!
32
( 1) i i 1
1
1 i! i 1
Answers and Explanations 1
a.
3 1 5
2 a. 1
8 5 3 4
1
4
3 a. 5 !
n 1 ! n 1 ! n! 3
1
5 a. x 6 6 a.
3
2 1 3
b.
7 1 3
c.
1 1 5
6
1 9 8
n 2 ! 4n n 1 !
n n 2 n 2 n 3 4n
x
b. x
x3
7 a. 3 x x
7! 9!
1 72
c.
b. 3
3x 2
b.
1 x8
2
n3 n3
4n 4n
8
x
c.
2 3
d.
21 1 2
23 2
19 ! 8 ! 11! 17 !
n 3 !
x x2
3x x 2 1
3x x 1 x 1 4x 3 x 1
c.
x 2 2 x 15 x 2 4x 3
19 18 11 10 9
19 55
1 n 2 n 1 n
n 3
9 x
10 x
3 x 1 4x 2
d. x
4
x4
3
x2 x2
5 3
d. 4 d. 3 x x
4 x3
5
3x 4 4x
3 2
4 3
1 3
1 x
9
x4
x
5 3 2
3 4 x 4
9 2
3 x 4
1 4
3x 3 4x 2
x 5 x 1
2 x 4 13 x 2 2
x5
3
3x 3 3x 4x 4 4x 3
x 3 x 5 x 1 x 3
5
10
b.
4x 3 x 1
x3
10 x
1 2x
9x
8 5
d.
c. x 2
c.
x x 3 2x 2 x 3
26 x 2 72 45 x 54
13 1 5
b. n 1 !
n 1
3
3x 6x 2
2
d.
1
1 x5
5
3
4 d. 2 x
1 101
1
5 x2 x2 2x 3
6 5
100 ! 101!
n 2
2 x 5
8 a.
1 3
n 1 n 1 !
1
x
4 3
20 b.
d. n n 2
n
c.
n n 1 n 2 ... 3 2 1 n 2 n 1 n ... 3 2 1
n! n 2 !
c.
11 17 1 6 6
5 4 3 2 1 3 2 1
3!
4 a.
b.
36
5x 6
2 x 2 x 2 9 x 2
2 x 4 13 x 2 36 9 x 2 x 3
x 3 x 3
x 3
2 x2 4 9 x 2
x2 9 x 3
2 x 2 10 x 12 9
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
33
9
cos x
A H
sec x
H A
13 3
csc x
H O
22
1 sec x
6 7
12 13
10
11
12
cos x
13 a. 25
14 a.
34
180 180
5
5 36 36
b. 80
b.
11 8
180 180
15
tan x sec x csc x
tan x cos x sin x
16
tan 2 x 1 cos x cot x
sec 2 x cos x cot x
4 9
c. 144
247.5
c. 12
sin x cos x sin x cos x
sec 2 x sec x tan x
PART 1 Introduction to Integration
7
4 5
180 180
105
sin 2 x
d. 200
d.
1 cos 2 x
tan x sec 3 x
21 16
180 180
10 9 236.25
17
cos x csc 2 x cos x sin 2 x
18
sin 2 x 2 sin x
2 sin x cos x 2 sin x
cos x csc 2 x 1 2
sin x
cos x cot 2 x sin x sin x
cot x cot 2 x sin x
cot 3 x csc x
cos x
19
20
21
22
23
24
CHAPTER 2 Forgotten but Not Gone: Review of Algebra and Pre-Calculus
35
25
26
27
28
i 1
2i 2i
i 0
n i i!
n 0!
n 1 1!
( 1) i
2n i 3i
2n 3
( 1) i
1
29 30 31 i 1
32 i 1
36
2 2
4 4
6 8
1 i! i 1
8 16
10 32
n 2 2! 2n 2 6 1 1! 2
PART 1 Introduction to Integration
... 1 1 n 3 3! 2n 3 9 1 2! 3
3 4
n 4 4! 2n 4 12
1 2
5 16
... n n 1
25 ... 15
1 3! 4
...
1 4! 5
2n 3
n2 3
n 2 2 2n 3 9
1 ... 5! 6
1 2
n 3 6
n 4 24
n4 6
2n 5 ... 15
1 6
1 24
1 120
...
1 ... 720
IN THIS CHAPTER
»» Understanding and evaluating limits »» Knowing derivatives of the most common functions »» Using the Power rule, Sum rule, and Constant Multiple rule »» Making sense of the Chain rule
3 Recent Memories: Calculus Review Chapter
I
f you’re currently taking a Calculus II course, you probably just finished Calculus I. But even if you got a good grade in that class, you may be overwhelmed by all the information you had to cram into your brain in only 15 weeks.
Fortunately, you don’t need to remember everything you learned there to do well in Calculus II. In this chapter, I review the information that’s likely to help you the most in your current class. To begin, I show you a variety of ways to calculate limits, or to show that a limit doesn’t exist. Next, I remind you of 17 derivatives of important functions that you absolutely, positively need to remember. I show you how to apply the Power rule, the Sum rule, and the Constant Multiple rule to vastly increase the types of functions that you can differentiate. Beyond that, I give you practice working with the Product rule, the Quotient rule, and the infamous Chain rule to find the derivatives of just about any function you can name.
Evaluating Limits Limits are the heart of calculus. The limit enables both differentiation and integration, and the formulas for the derivative and the definite integral are both limits.
CHAPTER 3 Recent Memories: Calculus Review
37
You can evaluate some limits by direct substitution: Simply plug in the value in the limit for the variable in the function and evaluate using the order of operations (PEMDAS). Many rational functions turn into zero, positive infinity, or negative infinity when you apply direct substitution. In each of the following cases, k stands for any real constant value:
0 k
k 0
0
k
0
Limit Does Not Exist Limit Does Not Exist
k
Technically speaking, positive and negative infinity aren’t values that you can plug into a function and calculate with. However, this type of notation is common when working with limits.
Q.
Evaluate the limit lim 4 x 2
A.
–1. Solve by direct substitution: Plug in 1 for x into the function 4 x 2
x
4 1
Q. A.
2
5
4 5
1
5 , or show why it doesn’t exist. 5 and simplify:
1 x3 2 , or show why it doesn’t exist. 1 x 1
Evaluate the limit lim x
x3
Limit does not exist. Solve by direct substitution: Plug in 1 for x into the function x and simplify:
13 2 1 1
2 1
3 0
Thus, the limit does not exist. As you can see if you plug in values that are either from the left and slightly less than or slightly greater than 1, the limit approaches from the right. Often, when you try direct substitution with rational functions (functions with a numerator and denominator), your first result is either
0 or 0
. Results like these are called indeterminate
results, and they require more work to evaluate the limit. To evaluate, try factoring both the numerator and denominator, and then cancel common factors.
Q.
Evaluate the limit lim
A.
4 0 . When you try direct substitution, the result is the indeterminate form : 3 0
x
22 4 3( 2 ) 6
4 4 6 6
2
x2 4 , or show why it doesn’t exist. 3x 6
0 0
To evaluate, factor both the numerator and denominator, and then cancel the common factor.
x2 4 2 3x 6
lim x
38
lim x
2
x 2 x 2 3 x 2
PART 1 Introduction to Integration
lim x
2
x 2 3
Now you can evaluate using direct substitution:
2 2 3
4 3
Another tactic, especially when the function includes an expression that contains a radical, is to multiply both the numerator and denominator by the conjugate of that expression — in other words, the same expression with the second term negated.
Q. A.
x 9 , or show why it doesn’t exist. x 3 0 6. Direct substitution leads to the indeterminate form : 0 Evaluate the limit lim x
9 9 9 3
9
0 0
Instead, multiply both the numerator and the denominator by the conjugate of which is x 3 :
lim x
9
x 9 x 3
lim x
9
x 9 x 3
x x
3 3
lim x
9
x 9 x x 9
3
lim x
9
x
x
3,
3
Now, direct substitution solves the problem:
9 1
6
b. lim e x
a. lim x 3
1
x c. lim 1 10 x
2
10
x
0
4x 2
d. lim sin 2 x x
cos 2 x
2
Simplify each of the following limits either to evaluate it or show that it doesn’t exist. a. lim
x x 1 b. lim x 1x 1 2
c. lim
1 x 4 d. xlim x 1, 000, 000 9
x
0
x
3
3 3
Evaluate each of the following limits by direct substitution.
x
2
3
Find each limit by simplifying the function and then applying direct substitution. a. lim
x 2 100 x 2 7 x 10 b. lim x 5 x 10 x 5
c. lim
x 1 x
x
x
10
0
3
1
d. lim x
0
x 2 16 x2
4
CHAPTER 3 Recent Memories: Calculus Review
39
Derivatives of Common Functions and the Constant Multiple Rule Knowing the derivatives of the most basic functions is essential to solving more difficult problems. Here are five important derivative formulas:
d n 0 dx
d x dx
d x e dx
1
d x n dx
ex
d ln x dx
n x ln x
1 x
Here are the derivatives of the six trig functions:
d sin x dx
d cos x dx
cos x
d cot x dx
d sec x dx
csc 2 x
d tan x dx
sin x
sec 2 x
d csc x dx
sec x tan x
csc x cot x
And here are the derivatives of the six inverse trig functions:
d arcsin x dx
1 x
2
d arccos x dx
d arccot x dx
1 1 x2
d arcsec x dx
1
1 1 x
2
1 x x
2
1
d arctan x dx d arccsc x dx
1 1 x2 1 x x2 1
The Constant Multiple rule enables you to move a constant outside the derivative:
d d f(x) cf ( x ) c dx dx Although the notation may seem a little opaque, using the Constant Multiple rule is fairly straightforward:
Q.
What is
A.
5 cos x. You already know that
d 5sin x ? dx d sin x dx
cos x , so use the Constant Multiple rule to keep
the same coefficient as you differentiate.
d 5 sin x dx 4
5 cos x
Evaluate the following derivatives using the Constant Multiple rule:
d d 3e x b. 8 ln x dx dx d d c. 100cos x d. 6tan x dx dx
a.
40
PART 1 Introduction to Integration
The Power Rule The Power rule enables you to calculate the derivative of x to the power of any real value of n as follows:
d n x dx
nx n
Q.
What is
A.
8x 7
Q.
What is
A.
1 2 x3
1
d 8 x ? dx
d 1 dx x
?
. To evaluate this derivative using the Power rule, first rewrite it as a power and
then apply the rule:
d 1 dx x
d x dx
1 2
1 x 2
3 2
Optionally, you can rewrite this result using radical form:
1 2 x3 The Power rule is very often used in conjunction with the Constant Multiple rule, especially when differentiating polynomials with a coefficient not equal to 1:
Q.
What is
A.
15x2.
d 5x 3 dx 5
6
d 5 x 3? dx
5 3 x2
15 x 2
Evaluate the following derivatives using the Power rule: a.
d 5 d x b. 3x 7 dx dx
c.
d dx
2x 9
d.
d 1 13 x dx 3
Evaluate the following derivatives of negative and fractional powers using the Power rule: 1
a.
d x dx
c.
5 d d 1 4 x 4 d. x dx dx 4
2
b.
d 5 x dx 6 5
CHAPTER 3 Recent Memories: Calculus Review
41
7
Rewrite each of the following derivatives as a power of x and then differentiate using the Power rule, simplifying in each case: a.
d d 3 x b. dx dx x 5
c.
d 4 3 6 x dx
d.
d 2 dx 5 x 3
The Sum Rule The Sum rule (sometimes called the Sum and Difference rule) enables you to break up functions into terms and differentiate each term separately:
d f ( x ) g( x ) dx
d f(x) dx
d g( x ) dx
This may look a little complicated until you see how simple it is in practice. Combined with the Constant Multiple rule and Power rule, the Sum rule allows you to differentiate any polynomial, regardless of complexity:
Q.
What is
A.
8 x 3 18 x 2
d 2x 4 dx
6x 3
5x ?
5. Begin by breaking the derivative up into the sum of three separate derivatives, and then evaluate each separately. d 2x 4 dx
6x 3
5x
d 2x 4 dx
d 6x 3 dx
d 5x dx
8 x 3 18 x 2
5
The Sum rule works equally well for differentiating the sums of other types of functions whose derivatives you know:
Q.
What is
A.
8 x 3 18 x 2
d cot x 3e x dx
5. Break the derivative up into the sum of three separate derivatives, and then evaluate each separately. d cot x 3e x dx
8
42
2x ?
2x
d cot x dx
d 3e x dx
d x 2 dx
csc 2 x 3e x
2 x ln 2
Evaluate the following polynomials using the Sum rule and Constant Multiple rule in conjunction with the Power rule. a.
d dx
c.
d 4 x 11 13 x 4 dx
x2
b.
4x 9 x
d.
PART 1 Introduction to Integration
d 2x 3 dx d dx
6 x 2 7 x 11
1 20 x 2
1 12 x 3
1 8 x 100 4
9
10
11
Evaluate the following functions using the Sum rule and Constant Multiple rule in conjunction with the Power rule. (You can leave your answers in exponential form.)
a.
d 3x dx
c.
d dx
1
1
6x
4x 2 1 2
b.
x 3
10 x 2
d.
5x
d 7x dx d dx
x
3
3
2x 5
1
9x 3
5
8
2x
5 7
Evaluate each of the following functions. (You can leave your answers in exponential form.) a.
d 1 dx x
c.
d dx
2 x
5 x3
b.
7 8x 6
14 x
d.
7
d 12 3 x dx
9 x2
d dx
2 x3
1 2x
d dx
8 x
7 cos x
4
4 5x 8
2 5
x9
Evaluate each of the following functions. a.
d 2 sin x dx
c.
d 2 tan x 2 sec x dx
3e x
arctan x 1 x
3
15 x
b.
d.
d 4 cot x dx 5
1 ln x 2
2 csc x arcsin x 2 arcsec x 3
The Product Rule and Quotient Rule The Product rule is a formula that enables you to calculate the derivative of the product of two functions:
d f ( x ) g( x ) dx
d f ( x ) g( x ) dx
d g( x ) f ( x ) dx
A mantra you can use to keep this rule clear is: “The derivative of the first function times the second, plus the derivative of the second function times the first.” The Quotient rule is a formula that enables you to calculate the derivative of the quotient of two functions:
d f(x) dx g ( x )
d d f ( x ) g( x ) g( x ) f ( x ) dx dx 2 g( x )
The mantra in this case is: “The derivative of the top function times the bottom, plus the derivative of the bottom function times the top, divided by the bottom function squared.”
CHAPTER 3 Recent Memories: Calculus Review
43
Q.
Calculate
A.
2x ln x
d 2 x ln x . dx x. Begin by differentiating
d 2 x ln x dx
Q. A.
12
d 2 x dx
d ln x dx
ln x
What is the derivative of
d 2 x dx x2
2 x and 2 x ln x
d ln x dx
x2 x
1 . So: x
2 x ln x
x
sin x ? x3
x cos x 3 sin x d d 3 x . Begin by differentiating sin x cos x and 2 x 2 . Thus: dx dx x4 d d 3 sin x x 3 sin x x cos x x 3 3 x 2 sin x d sin x x cos x 3 sin x dx dx 6 3 2 dx x 3 x x4 x
Evaluate each of the following functions using the Product rule.
d 3 x d x e b. sin x ln x dx dx d d c. 2 x 5 cos x d. tan x arctan x dx dx a.
13
Evaluate each of the following functions using the Quotient rule. a.
d arcsec x d cot x b. dx dx 4 x 5 ex
c.
d 3sec x d 5 csc x d. dx ln x dx 6 arcsin x
The Chain Rule The Chain rule enables you to differentiate compositions of functions (also known as nested functions, or functions within functions):
d [f ( g ( x ))] dx
d d g( x ) f ( g ( x )) dx dx
To help you apply the Chain rule, repeat these words: “The derivative of the outer function with respect to the inner function, times the derivative of the inner function.” In practice, to use the Chain rule, you’ll need to be able to identify functions starting from the outside and working your way in.
Q. A.
What is the derivative of cos 10 x 4 ?
40 x 3 sin 10 x 4 . Begin by noticing that the outer function is f ( x ) cos x and that the inner function is g ( x ) 10 x 4 . So: d cos 10 x 4 dx
44
sin 10 x 4
PART 1 Introduction to Integration
40 x 3
40 x 3 sin 10 x 4
Q. A.
Evaluate
d ln sin x 2 dx
2 x cot x 2 . In this case, the nested function is of the form f ( g ( h( x ))) with f ( x ) ln x , g ( x ) sin x , and h( x ) x 2. Thus: d ln sin x 2 dx
1 sin x 2
d sin x 2 dx
Apply the Chain rule two more times to complete the evaluation:
1 sin x 2
cos x 2
d 2 x dx
1 sin x 2
cos x 2
2x
Simplify:
2 x cos x 2 sin x 2 14
2 x cot x 2
Evaluate each of the following functions using the Chain rule.
d 3x d e b. sin5 x dx dx d 1 d c. cos x d. ln 11x dx 2 dx
a.
15
Evaluate each of the following functions by applying the Chain rule twice.
d cos 4 x 3 b. dx d c. 3e sin x d. dx
a.
16
d 5x 2 e dx d arctan x 3 dx
Evaluate each of the following functions by applying the Chain rule multiple times.
d d sec 2 x 3 tan e 2 x b. e dx dx d d c. ln arcsin cot x d. sec e 5 sin x dx dx
a.
17
Evaluate each of the following functions by applying the Product rule, the Quotient rule, and the Chain rule as needed: a.
d 10 x 5e sin x dx
b.
d sin x cos e 2 x dx
CHAPTER 3 Recent Memories: Calculus Review
45
Answers and Explanations 1
a. lim x 3 x
2 3 1 7 b. lim e x
1
2
x
1 10 10
x c. lim 1 10 x 10
d. lim sin 2 x x
2
x 2
c. lim
x 4 9
d. lim
1 x 1, 000, 000
x
x
3
0 2
x
x
5
0
x 1 x x
x
0
lim x
46
3
0
x
x
a.
d 3e x dx
c.
d 100 cos x dx
a.
d 5 x dx
5
lim x
x3
x 1 x 1
4
1
2
1 2
3
2 2
1 1 1 1
1 2
2 0
1
Does Not Exist
0
3x
lim x
0
2
16
4
lim x 10 10 10 0 20
x
x 2 x 5 x 5
x3
0
3x 2 x
2
1
2
2 2
2
x 10 x 10 x 10
10
lim
1
x x
2
1 1, 000, 000
lim
0
2
2 1 2
Does Not Exist
x
x 2 16 x2
d. lim
5
4 9
x 2 7 x 10 x 5
lim
cos 2
x
x 2 100 10 x 10
c. lim x
sin 2
0 b. lim
a. lim b. lim
4
cos 2 x
2
0
4 0
1 100
a. lim x
e0
4x 2
0
lim x 2 x
5 3
5
3
3x 2
3x 1 1 x x x 2 3x 3 lim lim x 2 x 0 x 0 x x 2 16 x2 lim x
0
4
x
PART 1 Introduction to Integration
4
x 2 16
4
16
100 sin x d.
d 3x 7 dx
x 2 16
1 2
3e x b.
5 x 4 b.
10
d 8 ln x dx d dx
lim x
1 0 16
4
8
1 x
6 tan x
21x 6 c.
d dx
3x
2x 9
0
0 0 3
3
3
x 2 16 16 x2
x 2 16 1 4 4
4
4
1 8
8 x 6 sec 2 x
18 x 8 d.
d 1 13 x dx 3
13 12 x 3
6
7
8
a.
d x dx
c.
5 d 4x 4 dx
a.
d x dx
c.
d 4 3 6 x dx
a.
d dx
2
2 x 3 b. 1
5 x 4 d.
1
d 2 x dx
1 x 2 3
d 6x 4 dx
x2
1 2
4 5
1 x 5 6 5
d 1 x dx 4
1 4
1 4
3 x 4
2x
11 5
6 x 5
1 d 3 b. dx x 5 2 x
6
4x 9
1
d 5 x dx
d 3x dx
d 2 9 d. dx 5 x 3 2 x
4 b.
d 4 x 11 13 x 4 x 44 x 10 52 x 3 1 dx d 1 20 1 12 1 8 d. x x x 100 10 x 19 dx 2 3 4
d 2x 3 dx
5
15 x
d 2 x dx 5
4
11 5
3 x 10
3 2
15 x6
6
2 5
6 x 2 7 x 11
5 2
3 x 2
3 5 x5
6 x 2 12 x 7
c.
9
10
1
a.
d 3x dx
c.
d dx
6x
d.
d dx
x
a.
d 1 dx x
b.
d 12 3 x dx
c.
d dx
5 x3
d.
d dx
2 x3
1
4x 2 1 2
1
3
10 x 2
5x 2x
7 8x 6 1 2x
x
1
3 2
1
15 x 2
1
x
1
9x d dx
x7 2 5
x9
2
2
3
2 3
2x 4
3 x 2
3
2x 5
5
21x
4
6 x 5
2 5
1 4
12 7
18 x
7 x 8 3
d dx
3
1 2
x
4x
5x 2
d 7x dx
5 10 x 7
24 x 3
2x 2
14
1 b.
5
2
d 12 x 3 dx
4 5x 8
1 2
2x
3x
5 7
d x dx
9 x2
2
3x
8
9x 3
2 x
4
x
4 x 11 2 x 7
6
3
14 x 1 x 2
1 x 2
2
1
7 2
1
15 2 x 2
4 x 5
32 x 5
9
8
2x 18 x 5
21 x 4
7
49 x
9 2
9 5 14 5
CHAPTER 3 Recent Memories: Calculus Review
47
11
12
a.
d 2 sin x dx
3e x
b.
d dx
7 cos x
c.
d 2 tan x 2 sec x dx
d.
d 4 cot x dx 5
8 x
arctan x
4 x
7 sin x
x
1 2x 3
2 sec 2 x 2 sec x tan x
15 x
3
2 csc x arcsin x 2 arcsec x 3
4 csc 2 x 5
2 x5
2 csc x cot x 3
d arcsec x a. dx ex d cot x dx 4 x 5
x x2 1 5
e x arcsec x
1
e2x
2
xe 4
x
x
2
2 x x2 1
1
arcsec x ex
2
4 x csc x 20 x cot x x csc x 5 cot x 16 x 10 4x 6 3 sec x d 3 sec x 3 sec x tan x ln x 3 sec x tan x 3 sec x x c. dx ln x ln x x ln nx 2 ln x 2 d. d 5 csc x dx 6 arcsin x
36 arcsin x
dx
ln x
2
d sec e 5 sin x dx
c. d sin x cos x
dx
x ln x
sec e 5 sin x tan e 5 sin x
e x sin x
e x sin x
cos 2 x sin 2 x x ln x 1 e7x
PART 1 Introduction to Integration
2
e x cos x ln x ln x
e 5 sin x
5 cos x
2
1 x2
1 x
e x sin x
2
5e 5 sin x cos x sec e 5 sin x tan e 5 sin x
d d x ln x sin x cos x sin x cos x x ln x dx dx 2 x ln x
1
6 arcsin x
e x sin x x ln x 2
x 2 ln x
d arcsin e 7 x dx
5 csc x
1 x2
cos 2 x sin 2 x x ln x
d.
5 csc x cot x 6 arcsin x
1 x2
d d ln x e x sin x ln x dx dx 2 ln x e x sin x e x cos x ln x
b.
2
1 x 2 cot x arcsin x 1
6 arcsin x
x a. d e sin x
3 x sec x tan x ln x 3 sec x x ln x 2
30 csc x
30 csc x cot x arcsin x
5 csc x
48
1 1 x2
d 3 x d sin x x e 3 x 2e x x 3e x b. sin x ln x cos x ln x dx dx x d d tan x c. 2 x 5 cos x 10 x 4 cos x 2 x 5 sin x d. tan x arctan x sec 2 x arctan x dx dx 1 x2
b.
14
15
a.
ex 13
3e x
1 ln x 2 1
1 1 x2
2 cos x
e7x
ln x 1 sin x cos x 2
ln x 1 sin x cos x x 2 ln x 2 7e 7 x 7 1 e 14 x
15
d 3x d e 3e 3 x b. sin 5 x 5 cos 5 x dx dx d 1 1 1 d 11 1 c. cos x sin x d. ln 11x dx 2 2 2 dx 11x x
16
a.
17
a.
d tan e 2 x dx
b.
d sec 2 x 3 e dx
c.
d ln arcsin cot x dx
d.
d sec e 5 sin x dx
a.
d cos 4 x 3 sin 4 x 3 12 x 2 12 x 2 sin 4 x 3 dx 2 2 d 5x 2 b. e e 5 x 10 x 10 xe 5 x dx d c. 3e sin x 3e sin x cos x dx d 1 3x 2 d. arctan x 3 3x 2 6 dx 1 x 1 x6 sec 2 e 2 x
2e 2 x
2e 2 x sec 2 e 2 x
3
e sec 2 x sec 2 x 3 tan 2 x 3 6 x 2 1 arcsin cot x
3
6 x 2e sec 2 x sec 2 x 3 tan 2 x 3 1 1 cot 2 x
sec e 5 sin x tan e 5 sin x
csc 2 x
csc 2 x arcsin cot x
1 cot 2 x
5 cos x
5 cos( x )sec e 5 sin x tan e 5 sin x
18
a.
d 10 x 5e sin x dx
b. d
dx
sin x cos e 2 x
d 10 x 5 e sin x dx
d e sin x 10 x 5 dx
50 x 4 e sin x
d d sin x cos e 2 x cos e 2 x sin x dx dx cos x cos e 2 x 2e 2 x sin x sin e 2 x
10 x 5e sin x cos x
cos x cos e 2 x
sin e 2 x 2e 2 x sin x
CHAPTER 3 Recent Memories: Calculus Review
49
2
From Definite to Indefinite Integrals
IN THIS PART . . .
Approximate area using Riemann sums See how the Fundamental Theorem of Calculus (FTC) frames integration as the inverse of differentiation Calculate definite integrals using anti-differentiation
IN THIS CHAPTER
»» Approximating definite integrals with Riemann sums that use left and right rectangles »» Using midpoint integrals to improve Riemann sums »» Estimating area more precisely by replacing rectangles with trapezoids »» Applying Simpson’s rule for approximating definite integrals
4 Approximating Area with Riemann Sums Chapter
I
n Chapter 1, you used the definite integral to provide a precise way to state area problems on the xy-graph. You found that, in a few simple cases, you can use geometry to evaluate definite integrals. In most cases, however, solving an area problem requires more complex math than geometry offers. In this chapter, you use Riemann sums to approximate the area under a function. To begin, you use left and right rectangles to give you a basic estimate of area. Next, you use midpoint rectangles to improve these estimates. After that, you see how changing rectangles to trapezoids can further improve your area approximations. To finish, you write Riemann sums using Simpson’s rule to provide an even better approximation of the definite integral.
CHAPTER 4 Approximating Area with Riemann Sums
53
Calculating Riemann Sums with Left and Right Rectangles The most basic shape you can use to partition an area that you wish to measure is the rectangle. When you use left rectangles to construct a Riemann sum, you slice that area into n rectangles of equivalent width and then measure the heights of these rectangles where the left side of each rectangle intersects the function.
Q.
Estimate x 2
A.
136. To begin, calculate the width of each rectangle:
8
3 dx using a Riemann sum with n
4 left rectangles.
0
b a n
8 0 4
2
So each rectangle has a width of 2. When calculating with left rectangles, use the value of the lower bound (in this case, 0) as input to the function to find the height of the first rectangle. Next, keep adding 2 to this value to find the inputs for the other three rectangles: 8
x2
3 dx
2 f (0) f (2) f ( 4 ) f (6 )
0
Then, use the function that you’re integrating to calculate these heights:
2 (02
3) (22
2 3 7 19 39
3) (4 2
3) (6 2
3)
136
Thus, the approximate area you’re measuring is 136. You can also use right rectangles to build a Riemann sum, slicing that area into n rectangles of equivalent width, and then measuring the heights of these rectangles where the right side of each rectangle intersects the function.
54
PART 2 From Definite to Indefinite Integrals
Q.
Estimate
A.
38. To begin, calculate the width of each rectangle:
6
x2
6 x 1 dx using a Riemann sum with n
3 right rectangles.
0
b a n
6 0 3
2
So each rectangle has a width of 2. When calculating with right rectangles, use the value of the upper bound (in this case, 6) as input to the function to find the height of the first rectangle. Next, keep subtracting 2 from this value to find the inputs for the other five rectangles: 6
x2
6 x 1 dx
2 f (6 ) f ( 4 ) f (2)
0
Then, use the function that you’re integrating to calculate these heights:
2
62
6( 6 ) 1
21 9 9
42
6( 4 ) 1
22
6( 2 ) 1
38
Thus, the approximate area you’re measuring is 38.
1
Estimate
4 0
x2
5 dx using a Riemann sum
with four left rectangles.
2
Estimate
8 0
x2
8 x dx using a Riemann sum
with four left rectangles.
CHAPTER 4 Approximating Area with Riemann Sums
55
3
Estimate
10 2
x 2 12 x dx using a Riemann sum
4
with four right rectangles.
5
Estimate
0
sin x dx using a Riemann sum
with four left rectangles.
56
PART 2 From Definite to Indefinite Integrals
Estimate
10 0
x 3 100 x dx using a Riemann
sum with five right rectangles.
6
Estimate
2 2
3 cos x dx using a Riemann sum
with four right rectangles.
Getting a Better Estimate with Midpoint Rectangles Using midpoint rectangles to measure the area under a function usually provides a better estimate of area than either left or right rectangles. To build a Riemann sum using n midpoint rectangles, slice the area into n partitions of equal width, but measure the height of each rectangle at the midpoint of each partition.
Q.
Estimate x 2
A.
26. To begin, calculate the width of each rectangle:
8
1 dx using a Riemann sum with n
4 midpoint rectangles.
0
b a n
8 0 4
2
So each rectangle has a width of 2. These rectangles, as shown in the graph, have midpoints x 1, 3, 5, and 7, so use the function at these four x-values to find the heights of the four rectangles: 8
x 2 1 dx
2 f (1) f ( 3 ) f ( 5 ) f (7 )
0
Then, use the function that you’re integrating to calculate these heights.
2 (12 1) ( 3 2 1) ( 5 2 1) (7 2 1) 2 2 10 26 50
176
Thus, the approximate area you’re measuring is 26.
CHAPTER 4 Approximating Area with Riemann Sums
57
7
Estimate
9 1
x 2 dx using a Riemann sum with
8
four midpoint rectangles.
9
Estimate
4 0
cos x 2 dx using a Riemann
sum with four midpoint rectangles.
58
PART 2 From Definite to Indefinite Integrals
Estimate
4 4
x 2 16 dx using a Riemann sum
with four midpoint rectangles.
10
Estimate
3 2
sin x 1 dx using a Riemann
2
sum with four midpoint rectangles.
Improving Your Estimate with the Trapezoid Rule So far in this chapter, all of the methods you’ve used to approximate definite integrals have used rectangular partitions to slice up the area you’re trying to measure. Another relatively simple shape you can use is trapezoids. Cutting up an area into trapezoids is a bit more complex than cutting it up into rectangles, but this method usually provides a higher level of precision. The good news about the Trapezoid rule is that there’s a formula for using this method. So, after you divide the area you’re measuring into separate partitions, you don’t have to spend a lot of time calculating the areas of individual trapezoids. The Trapezoid rule: To estimate a definite integral using n trapezoids, use the following formula. b
b a f ( x 0 ) 2f ( x1 ) 2f ( x 2 ) ... 2f ( x n 1 ) f ( x n ) 2n
f ( x ) dx a
Q.
Estimate x 2
A.
216. As usual, begin by calculating the width of each partition:
8
5 dx using a Riemann sum with n 4 trapezoids.
0
b a n
8 0 4
2
Thus, to begin drawing trapezoids, you draw vertical lines at x 0 0, x 1 2, x 2 4 , x 3 6, and x 4 8 , as shown in the graph. Next, you plug these values into the formula for the Trapezoid rule: 8 0
x2
5 dx
2 f ( 0 ) 2f ( 2 ) 2f ( 4 ) 2f ( 6 ) f ( 8 ) 2
CHAPTER 4 Approximating Area with Riemann Sums
59
Before moving on, notice that the first and last terms inside the brackets — f ( 0 ) and f ( 8 ) — correspond to the two bounds of integration at 0 and 8. These are the only two terms in the formula that don’t have coefficients of 2. Now, simplify and plug in the values of the function:
1 (02
5 ) 2( 2 2
5 ) 2( 4 2
5 ) 2( 6 2
5) ( 8 2
5)
5 18 42 82 69 216 8
Therefore, the value of the definite integral x 2
5 dx is approximately 226.
0
11
13
60
7
Estimate x 2 dx using a Riemann sum with 3 n 4 trapezoids.
12
cos x 1 dx using a Riemann Estimate sum with n 4 trapezoids.
14
PART 2 From Definite to Indefinite Integrals
Estimate with n
3
x2
9 dx using a Riemann sum
6 trapezoids.
Estimate with n
3
5 0
1 3 x dx using a Riemann sum 2
5 trapezoids.
Using Simpson’s Rule to Further Improve Your Approximation Simpson’s rule allows you to measure a definite integral using strips that are rectangular on three sides but are topped with a parabolic curve. This curved feature makes measuring an area using Simpson’s rule even more precise than the Trapezoid rule. In fact, when the function you’re working with is quadratic, Simpson’s rule can provide you with the exact value of an integral. As with the Trapezoid rule, Simpson’s rule provides a formula. One proviso is that to use this formula, you must use an even number of partitions. Simpson’s rule: To estimate a definite integral using n partitions, where n is an even number, use the following formula. b
f ( x ) dx a
b a f ( x 0 ) 4 f ( x1 ) 2f ( x 2 ) 4 f ( x 3 ) ... 4 f ( x n 3 ) 2f ( x n 2 ) 4 f ( x n 1 ) f ( x n ) 3n
Q.
Estimate x 2 dx using a Riemann sum with n
A.
170.6 . As usual, begin by calculating the width of each partition:
8
8 partitions.
0
b a n
8 0 8
1
Thus, to begin drawing partitions, you draw vertical lines at x 0 0, x 1 1, x 2 2, x 3 3 , x 4 4 , x 5 5, x 6 6, x 7 7 , and x 8 8 , as shown in the graph. Next, you plug these values into the formula for Simpson’s rule: 8 0
x 2 dx
1 f ( 0 ) 4 f (1) 2f ( 2 ) 4 f ( 3 ) 2f ( 4 ) 4 f ( 5 ) 2f ( 6 ) 4 f (7 ) f ( 8 ) 3
CHAPTER 4 Approximating Area with Riemann Sums
61
Before moving on, check your work to make sure that all the coefficients are correct. Next, simplify and plug in the values of the function:
1 ( 0 2 ) 4(12 ) 2( 2 2 ) 4( 3 2 ) 2( 4 2 ) 4( 5 2 ) 2( 6 2 ) 4(7 2 ) ( 8 2 ) 3 1 512 0 4 8 36 32 100 72 196 64 170.6 3 3 8
Therefore, the value of the definite integral x 2 dx is approximately 170.6 . This is, 0
in fact, the exact value of the integral. As you can see, Simpson’s rule provides 100-percent precision when the function you’re working with is quadratic.
15
Use Simpson’s rule to estimate using a Riemann sum with n
17
3 3
x2
2 dx
6 partitions.
Use Simpson’s rule to estimate 2
2 cos 2 x dx using a Riemann sum
2
with n
62
16
6 partitions.
PART 2 From Definite to Indefinite Integrals
Use Simpson’s rule to estimate a Riemann sum with n
18
8 0
x 3 dx using
8 partitions.
Use Simpson’s rule to estimate using a Riemann sum with n
12 0
x 2 dx
4 partitions.
Answers and Explanations 1
34. To begin, calculate the width of each rectangle:
b a n
4 0 4
1
So each rectangle has a width of 1. When calculating with left rectangles, use the value of the lower bound (in this case, 0) as input to the function to find the height of the first rectangle. Next, keep adding 1 to this value to find the inputs for the remaining rectangles: 4
x2
5 dx
1 f ( 0 ) f (1) f ( 2 ) f ( 3 )
0
Now, use the function f ( x )
1 (0 2
2
2
5 ) (1
5) (2
2
x2
5 to calculate these heights:
5) ( 3
2
5)
34
80. Calculate the width of each rectangle as
b a n
8 0 4
2. So each rectangle has a width of 2.
When calculating with left rectangles, use the value of the lower bound (in this case, 0) as input to the function to find the height of the first rectangle. Next, keep adding 2 to this value to find the inputs for the remaining rectangles:
x2
Now, use the function f ( x )
2 0 12 16 12 3
8 0
x2
8 x dx
2 f ( 0) f ( 2) f ( 4 ) f (6 )
8 x to calculate these heights:
80
1,104. Calculate the width of each rectangle as
b a n
8 0 4
2 . So each rectangle has a width
of 2. When calculating with right rectangles, use the value of the upper bound (in this case, 10) as input to the function to find the height of the first rectangle. Next, keep subtracting 2 from this value to find the inputs for the remaining rectangles: 10 2
x 2 12 x dx
2 f (10 ) f ( 8 ) f ( 6 ) f ( 4 )
Now, use the function f ( x )
2 220 160 108 64 4
x 2 12 x to calculate these heights:
1,104
2,400. Calculate the width of each rectangle as
b a n
10 0 5
2. So each rectangle has a
width of 2. When calculating with right rectangles, use the value of the upper bound (in this case, 10) as input to the function to find the height of the first rectangle. Next, keep subtracting 2 from this value to find the inputs for the remaining rectangles: 10 0
x 3 100 x dx
2 f (10 ) f ( 8 ) f ( 6 ) f ( 4 ) f ( 2 )
Now, use the function f ( x )
2 0 288 384 336 192
x 3 100 x to calculate these heights: 2, 400
CHAPTER 4 Approximating Area with Riemann Sums
63
2
5
4
. Calculate the width of each rectangle as
width of
4
b a n
0 4
4
. So each rectangle has a
. When calculating with left rectangles, use the value of the lower bound (in this
case, 0) as input to the function to find the height of the first rectangle. Next, keep adding to this value to find the inputs for the remaining rectangles:
sin x dx
0
f 0
4
f
f
4
Now, use the function f ( x )
4
6
3
0
2 2
2 2
1
sin x to calculate these heights: 4
2 3 b a . Calculate the width of each rectangle as 4 n
has a width of (in this case, subtracting
4
2
0
2 4
4
. So each rectangle
. When calculating with right rectangles, use the value of the upper bound
) as input to the function to find the height of the first rectangle. Next, keep
4
f
2
f
4
Now, use the function f ( x )
4
2
from this value to find the inputs for the remaining rectangles:
4
3 cos x dx
0
7
2
2 1
4
3 4
f
2
3 2 2
3
3 2 2
4
f 0
f
4
3 cos x to calculate these heights: 3 2
3
3
2 3 4
240. To begin, calculate the width of each rectangle as a width of 2. These rectangles have midpoints x
b a n
9 1 4
2. So each rectangle has
2, 4, 6, and 8, so use the function at these
four x-values to find the heights of the four rectangles: 9 1
x 2 dx
2 f (2) f ( 4 ) f (6 ) f ( 8 )
Then, use the function f ( x )
2 22
42
62
82
x 2 to calculate these heights:
2 4 16 36 64
240
Thus, the approximate area you’re measuring is 240. 8
88. To begin, calculate the width of each rectangle as
b a n
4
4 4
2. So each rectangle
has a width of 2. These rectangles have midpoints x 2, 4, 6, and 8, so use the function at these four x-values to find the heights of the four rectangles: 4 4
64
x 2 16 dx
4
2 f ( 3 ) f ( 1) f (1) f ( 3 )
PART 2 From Definite to Indefinite Integrals
x 2 16 to calculate these heights:
Then, use the function f ( x )
2
2
3
16
1
2
16
1
2
16
3
2
9 8 . To begin, calculate the width of each rectangle as b a width of
. These rectangles have midpoints x
2
,
16 a
2 7 15 15 7 4
n
0
88
. So each rectangle has
4
3 5 7 , and , , so use the function at 2 2 2
these four x-values to find the heights of the four rectangles: 4
cos x 2 dx
0
f
3 2
f
2
Then, use the function f ( x )
cos
2
2
cos
2 2 2 2
3 2
5 2
f
f
7 2
cos x 2 to calculate these heights.
2
2 cos
5 2
2
cos
7 2
2
8
This estimated area is, in fact, the exact value of the integral. 10 2 . To begin, calculate the width of each rectangle as b gle has a width of
2
3 2
a
2
n
. These rectangles have midpoints x
4 4
,
4
,
2
. So each rectan-
3 5 , and , so use the 4 4
function at these four x-values to find the heights of the four rectangles: 3 2
sin x 1 dx
2
2
f
f
4
Then, use the function f ( x )
2
sin 2 2
2
1
4
sin 2 2
1
1
4
f
4
3 4
f
5 4
sin x 1 to calculate these heights. 1 2 2
sin 1
3 4 2 2
1
sin
1
2
5 4
1
This estimated area is, in fact, the exact value of the integral. 11 106. Begin by calculating the width of each partition:
b a n
7 3 4
1
Thus, to begin drawing trapezoids, you draw vertical lines at x 0 3, x 1 4 , x 2 x 4 7. Next, you plug these values into the formula for the Trapezoid rule: 7 3
x 2 dx
5, x 3
6, and
1 f ( 3 ) 2f ( 4 ) 2f ( 5 ) 2f ( 6 ) f ( 7 ) 2
CHAPTER 4 Approximating Area with Riemann Sums
65
Before moving on, notice that the first and last terms inside the brackets — f ( 3 ) and f (7 ) — correspond to the two bounds of integration at 0 and 8. These are the only two terms in the formula that don’t have coefficients of 2. Now, simplify and plug in the values of the function:
1 ( 3 2 ) 2( 4 2 ) 2( 5 2 ) 2( 6 2 ) (7 2 ) 2 1 9 32 50 72 49 106 2 Therefore, the value of the definite integral
7 3
x 2 dx is approximately 106.
12 35. Begin by calculating the width of each partition:
3
b a n
3
1
6
Thus, to begin drawing trapezoids, you draw vertical lines at x 0 3, x1 2, x 2 1, x 3 0, x 4 1, x 5 2, and x 6 3. Next, you plug these values into the formula for the Trapezoid rule: 3
x2
3
9 dx
1 f ( 3 ) 2f ( 2 ) 2f ( 1) 2f ( 0 ) 2f (1) 2f ( 2 ) f ( 3 ) 2
Now, plug in the values of the function and simplify:
( 3)2
1 2
(3)
9
2
2
( 2)2
9
( 1) 2
2
9
2
3
x2
(0)2
9
2
(1) 2
9
2
(2)2
9
9
1 0 10 16 18 16 10 0 2
35
Therefore, the value of the definite integral
3
9 dx is approximately 35.
13 2 . Begin by calculating the width of each partition:
b a n
4
2
Thus, to begin drawing trapezoids, you draw vertical lines at x 0 and x 4
, x1
2
, x2
. Next, you plug these values into the formula for the Trapezoid rule:
cos x 1 dx
4
f
2f
2
2f 0
2f
2
f
Now, plug in the values of the function and simplify:
4 4
1 1
2 0 1
0 2 4 2 0
2 1 1
2 0 1
1 1
2
Therefore, the value of the definite integral
cos x 1 dx is approximately 2 .
14 81.25. Begin by calculating the width of each partition:
b a n
66
5 0 5
1
PART 2 From Definite to Indefinite Integrals
0, x 3
2
,
Thus, to begin drawing trapezoids, you draw vertical lines at x 0 0, x 1 1, x 2 2, x 3 3 , x 4 4 , x 5 5, and x 6 3. Next, you plug these values into the formula for the Trapezoid rule: 5 0
1 3 x dx 2
1 f ( 0 ) 2f (1) 2f ( 2 ) 2f ( 3 ) 2f ( 4 ) f ( 5 ) 2
Now, plug in the values of the function and simplify:
1 1 1 1 1 2 (1) 3 2 ( 2) 3 2 (3)3 ( 0) 3 2 2 2 2 2 1 325 125 0 1 8 27 64 81.25 2 2 4 Therefore, the value of the definite integral
5 0
2
1 (4)3 2
1 ( 5) 3 2
1 3 x dx is approximately 81.25. 2
15 30. Begin by calculating the width of each partition:
b a n
3
3
1
6
Thus, to begin drawing partitions, you draw vertical lines at x 0 3, x1 2, x 2 1, x 3 0, x 4 1, x 5 2, and x 6 3. Next, you plug these values into the formula for Simpson’s rule: 3 3
x2
2 dx
1 f ( 3 ) 4 f ( 2 ) 2f ( 1) 4 f ( 0 ) 2f (1) 4 f ( 2 ) f ( 3 ) 3
Before moving on, check your work to make sure that all the coefficients are correct. Then simplify and plug in the values of the function:
1 ( 3 )2 2 4 ( 2)2 2 3 1 11 24 6 8 6 24 11 3
2 ( 1) 2
2
4 ( 0)2
2
2 (1) 2
2
4 ( 2)2
2
( 3)2
2
30 3
Therefore, the value of the definite integral x 2 2 dx is approximately 30. And because the 3 function you’re integrating is quadratic, this is an exact value for the definite integral. 16 1,024. Begin by calculating the width of each partition:
b a n
8 0 8
1
Thus, to begin drawing partitions, you draw vertical lines at x 0 0, x 1 1, x 2 2, x 3 3 , x 4 4 , x 5 5, x 6 6, x 7 7 , and x 8 8 . Next, you plug these values into the formula for Simpson’s rule: 8 0
x 3 dx
1 f ( 0 ) 4 f (1) 2f ( 2 ) 4 f ( 3 ) 2f ( 4 ) 4 f ( 5 ) 2f ( 6 ) 4 f (7 ) f ( 8 ) 3
Then simplify and plug in the values of the function:
1 ( 0 3 ) 4(13 ) 2( 2 3 ) 4( 3 3 ) 2( 4 3 ) 4( 5 3 ) 2( 6 3 ) 4(7 3 ) ( 8 3 ) 3 1 3, 072 0 4 16 108 128 500 432 1, 372 512 1, 024 3 3
CHAPTER 4 Approximating Area with Riemann Sums
67
17 0. Begin by calculating the width of each partition:
b a n
2
2 6
6
Thus, to begin drawing partitions, you draw vertical lines at x 0
x3
0, x 4
6
, x5
Simpson’s rule: 2
2 cos 2 x dx
2
3
, and x 6
1 f 3
2
2
4f
2
, x1
3
, x2
. Next, you plug these values into the formula for
2f
3
4f 0
6
2f
4f
6
f
3
6
,
2
Before moving on, check your work to make sure that all the coefficients are correct. Then simplify and plug in the values of the function:
1 1 1 2 1 4 2 2 2 3 2 2 1 2 4 2 8 2 4 2 0 3
4 2 1
Therefore, the value of the definite integral
2 2
2
1 2
4 2
1 2
2
1
2 cos 2 x dx is approximately 0. This is, in
2
fact, the exact value of the integral.
18 576. Begin by calculating the width of each partition:
b a n
12 0 4
3
Thus, to begin drawing partitions, you draw vertical lines at x 0 0, x 1 3 , x 2 x 4 12. Next, you plug these values into the formula for Simpson’s rule: 12 0
x 2 dx
6, x 3
9, and
3 f ( 0 ) 4 f ( 3 ) 2f ( 6 ) 4 f ( 9 ) f (12 ) 3
Before moving on, check your work to make sure that all the coefficients are correct. Then simplify and plug in the values of the function:
1 ( 0 2 ) 4( 3 2 ) 2( 6 2 ) 4( 9 2 ) (12 2 ) 1 0 36 72 324 144
576
Therefore, the value of the definite integral
12 0
x 2 dx is approximately 576. This is, in fact, the
exact area under the function. As you can see, even with only 4 partitions, Simpson’s rule provides 100-percent precision when the function you’re working with is quadratic.
68
PART 2 From Definite to Indefinite Integrals
IN THIS CHAPTER
»» Understanding how the Fundamental Theorem of Calculus, Part 2 (FTC2) connects differentiation and integration »» Applying FTC2 to solve definite integrals using indefinite integrals »» Making sense of signed area »» Seeing how the Fundamental Theorem of Calculus, Part 1 (FTC1) allows you to evaluate area functions
5 The Fundamental Theorem of Calculus and Indefinite Integrals Chapter
I
n this chapter, you discover and work with the Fundamental Theorem of Calculus, which provides a practical way to calculate integrals that isn’t dependent on Reimann sums.
To begin, you see how one part of this theorem, the Fundamental Theorem of Calculus, Part 2 (FTC2), connects the two main branches of calculus (differentiation and integration). Then, you begin to work with indefinite integrals, which are expressions that help you to calculate definite integrals. Next, you discover how signed area on the xy-plane occurs when a definite integral produces a negative value. To finish up, you see how the Fundamental Theorem of Calculus, Part 1 (FTC1) makes sense of area functions.
CHAPTER 5 The Fundamental Theorem of Calculus and Indefinite Integrals
69
Evaluating Definite Integrals Using FTC2 The Fundamental Theorem of Calculus is a two-part theorem demonstrating that differentiation and integration are inverse operations. The more useful part of the theorem is the second part, abbreviated FTC2:
If f '( x ) is continuous on the closed interval [a, b], b
f '( x ) dx
f (b ) f (a)
a
In practice, FTC2 means that you can integrate any function, provided that you can find a way to anti-differentiate that function — that is, think of it as a derivative and then undo the differentiation. A few examples should be helpful.
Q.
Evaluate the definite integral 3 x 2 dx using the Fundamental Theorem of Calculus.
A.
56. Begin by noticing that when you differentiate the function f ( x )
4
2
x 3 , the result is:
f '( x ) 3 x 2 b
Thus, by FTC2, f '( x ) dx
f ( b ) f ( a ), which allows you to make the following mathe-
a
matical inference: 4
3 x 2 dx
f ( 4 ) f ( 2) 4 3
23
64 8
56
2 4
Therefore, the area on the xy-graph represented by the definite integral 3 x 2 dx is 56. 2
As you can see, applying the FTC in this way requires that you find a clever way to represent the function you’re trying to integrate as the derivative of another function. At times, this may take some ingenuity. For example:
Q.
Evaluate the definite integral x 5 dx using the Fundamental Theorem of Calculus.
A.
21 . Begin by noticing that when you differentiate the function f ( x ) 2
2 1
f '( x ) 6 x 5 So, when you differentiate the function f ( x )
f '( x )
1 6x 5 6
1 6 x , the result is: 6
x5 b
Now, use FTC2, f '( x ) dx
f ( b ) f ( a ), as follows:
a 2 1
70
x 5 dx
f ( 2 ) f (1)
1 2 6
6
1 1 6
PART 2 From Definite to Indefinite Integrals
6
64 6
1 6
63 6
21 2
x 6 , the result is:
4
21. 2 Therefore, the area on the xy-graph represented by the definite integral 3 x dx is 2
2
1
3
Evaluate the definite integral 4 x 3 dx using
2
1
1
Fundamental Theorem of Calculus, Part 2.
the Fundamental Theorem of Calculus, Part 2.
3
2
Evaluate the definite integral cos x dx using
2
Evaluate the definite integral x 7 dx using the
4
Evaluate the definite integral
4
sec 2 x dx
0
the Fundamental Theorem of Calculus, Part 2.
5
1
Evaluate the definite integral e 2 x dx using 0
the Fundamental Theorem of Calculus, Part 2.
4
using the Fundamental Theorem of Calculus, Part 2.
6
Evaluate the definite integral 3 sin x dx 0
using the Fundamental Theorem of Calculus, Part 2.
CHAPTER 5 The Fundamental Theorem of Calculus and Indefinite Integrals
71
Anti-differentiation and Indefinite Integrals As you saw in the previous section, you can evaluate any definite integral provided that you can find a way to cast it as the derivative of another function. Another way to think of this process is to see it as integrating a function by anti-differentiation — that is, reversing the operation of differentiation. One small complication to this process is that more than one function can result in the same derivative, because all constants differentiate to 0.
Q.
Differentiate the following three functions:
f ( x ) sin x
g ( x ) sin x 1
h( x ) sin x 100
In light of this exercise, what is the antiderivative of the function cos x?
A.
sinx
C . Here are the three derivatives:
f '( x ) cos x
g '( x ) cos x
h '( x ) cos x
As you can see, the antiderivative of cos x could be any of these three original functions. Thus, the most complete way to represent the antiderivative of cos x is sin x + C. This understanding of integration as anti-differentiation is captured in the formal notation for the indefinite integral, which superficially is simply a definite integral with the bounds of integration omitted:
f '( x ) dx
f(x) C
This formulation allows you to begin to collect and organize a limited but vital set of functions based on what you already know about differentiation: Derivatives
72
Indefinite integrals (antiderivatives)
d n dx
0
0 dx
C
d x dx
1
1 dx
x C
d nx dx
n
n dx
nx C
d 2 x dx
2x
x dx
1 2 x C 2
d 3 x dx
3x 2
x 2 dx
1 3 x C 3
PART 2 From Definite to Indefinite Integrals
d x e dx
ex
e x dx
ex
d ln x dx
1 x
1 dx x
ln x C
d sin x dx
cos x
cos x dx
d cos x dx
sin x
d tan x dx
C
sin x C
sin x dx
cos x C
sec 2 x
sec 2 x dx
tan x C
d cot x dx
csc 2 x
csc 2 x dx
d sec x dx
sec x tan x
sec x tan x dx
d csc x dx
Q. A.
csc x cot x
cot x C
csc x cot x dx
sec x C csc x C
1 dx by anti-differentiating. Then use this information x 5 1 dx using FTC2. to evaluate the definite integral to x 2 1 0.916. The function f ( x ) ln x differentiates to f '( x ) , so: x
Evaluate the indefinite integral
1 dx x
ln x C
Therefore, you can evaluate the related definite integral as follows: 5 2
1 dx x
ln x
x 5 x 2
The notation ln x
x 5 x 2
is commonly used as a half-step after anti-differentiating but
before evaluating the definite integral as a real value. You can read it as “ln x evaluated from 2 to 5.” Next, simplify this notation using FTC2:
ln 5 C
ln 2 C
ln 5 ln 2 0.916
As you can see, the notation provides a compact way to encapsulate the information you’ll need to apply FTC2 without getting bogged down in details. Notice that the value C drops out of the result so, in practice, you can omit it whenever you evaluate definite integrals.
CHAPTER 5 The Fundamental Theorem of Calculus and Indefinite Integrals
73
7
9
11
74
Evaluate the definite integral anti-differentiation.
10
x dx using
1
4
x Evaluate the definite integral e dx using 0 anti-differentiation.
Evaluate the definite integral anti-differentiation.
8
2
sin x dx using
0
PART 2 From Definite to Indefinite Integrals
10
12
2
Evaluate the definite integral anti-differentiation.
x 2 dx using
2
e3
Evaluate the definite integral anti-differentiation.
Evaluate the definite integral using anti-differentiation.
e
4
3
1 dx using x
sec 2 x dx
Signed and Unsigned Area In the real world, area is always positive. In the context of integration on the xy-plane, however, area below the x-axis is considered negative. Thus, a definite integral can be evaluated as any real value: positive, negative, or zero. This concept of area that includes non-positive values is called signed area.
Q. A.
What is the signed area of the shaded region in this figure?
3 . The shaded region is bounded below by the function f ( x ) x , between x 1 and 2 x 2. You can express the area of this function using the following definite integral: 1
x dx 2
Solve using anti-differentiation:
1 2 x 2
x
1
x
2
1 2
1
2
1 2
2
2
1 2
2
3 2
When a shaded region includes components that are equally positive and negative, its signed area balances out to 0.
CHAPTER 5 The Fundamental Theorem of Calculus and Indefinite Integrals
75
Q. A.
What is the signed area of the shaded region in this figure? 0. The shaded region is defined by the function f ( x ) cos x , between x 0 and x . Express the area of this function as a definite integral and then evaluate it as follows:
cos x dx
sin x
0
x x 0
sin
sin 0 0 0 0
When a shaded region includes parts that are unequally positive and negative, its signed area results in a value that reflects this imbalance.
Q.
What is the signed area of the shaded region in this figure?
A.
0.260. The shaded region is defined by the function f ( x ) sec x tan x , between x and x follows: 6
6
sec x tan x dx
sec x
4
x x
sec
6 4
6
sec
4
Now, evaluate this trig expression using what you know about the unit circle:
2 3
76
2
4
. Express the area of this function as a definite integral and then evaluate it as
2 3
3 2 3
0.260
PART 2 From Definite to Indefinite Integrals
13
What is the signed area of the shaded region shown here?
14
What is the signed area of the shaded region shown here?
15
What is the signed area of the shaded region shown here?
16
What is the signed area of the shaded region shown here?
CHAPTER 5 The Fundamental Theorem of Calculus and Indefinite Integrals
77
Answers and Explanations 1
x 4 , the result is f '( x ) 4 x 3. Thus, by the FTC2:
80. When you differentiate the function f ( x ) 3
4 x 3 dx
f ( 3 ) f (1) 3 4 14
81 1 80
1
2
x 8 , the result is f '( x ) 8 x 7 . So, when you
31.875. When you differentiate the function f ( x ) differentiate f ( x )
1 8x7 8
f '( x )
1 8 x , the result is: 8
x7
Thus, by the FTC2: 2
x 7dx
1 2 8
f ( 2 ) f (1)
1
3
8
32
1 8
31.875 sin x , the result is f '( x ) cos x . Thus,
1. When you differentiate the function f ( x ) by the FTC2: 2
cos x dx
f ( ) f ( 0 ) sin 2 2
0
4
1 1 8
8
sin 0 1 0 1 tan x , the result is f '( x ) sec 2 x . Thus,
2. When you differentiate the function f ( x ) by the FTC2: 4
sec 2 x dx
f( ) f( ) tan 4 4 4
tan
4
1
1
2
4
5
1 2 e 2
1 2
3.195. When you differentiate the function f ( x ) e 2 x, the result is f '( x ) 2e x . So, 1 2x when you differentiate f ( x ) e , the result is: 2 f '( x )
1 2e 2 x 2
e2x
Thus, by the FTC2: 1
e 2 x dx
f (1) f ( 0 )
0
6
1 2( 1 ) e 2
1 2( 0 ) e 2
1 2 e 2
1 2
6. When you differentiate the function f ( x ) 3 cos x , the result is: differentiate f ( x )
f '( x )
3sin x
3.195 cos x , the result is f '( x )
3sin x
Thus, by the FTC2:
3 sin x dx
f ( ) f (0)
3 cos
0
78
PART 2 From Definite to Indefinite Integrals
3 cos 0
3
3
6
sin x . So, when you
7
10
x dx 1
8
1 2 x 2
49.5. The indefinite integral x dx as follows:
1 2 x 2
x 10
1 10 2
x 1
1 1 2
2
2
16 . The indefinite integral x 2 dx 3
1 3 x 3
as follows: 2
x 2
1 3 x 3
x 2 dx
2
x
1 2 3
2
3
1 3
2
1 2
50
3
8 3
C , so you can evaluate the related definite integral
49.5 C , so you can evaluate the related definite integral
8 3
16 3
9 e 4 1 53.598 . The indefinite integral e x dx nite integral as follows: 4
e x dx
ex
0
x 4 x 0
e4
e0
e
1 dx x
ln x
x e3 x e
ln e 3
ln e
ex
sin x dx
cos x
0
x 2 x 0
C , so you can evaluate the related definite integral as
3 ln e ln e
11 0. The indefinite integral sin x dx gral as follows: 2
C , so you can evaluate the related defi-
e 4 1 53.598
10 2. The indefinite integral e x dx follows: e3
ex
cos 2
2 ln e
2
cos x C , so you can evaluate the related definite inte-
cos 0
1
1
12 1 3 2.732. The indefinite integral sec 2 x dx definite integral as follows: 4
sec 2 x dx
tan x
x
tan
4
x
3
3
4
tan
3
0
tan x C , so you can evaluate the related
1
3
1
3
2.732
13 –20. The shaded region is defined by the function f ( x ) 2 , between x 3 and x 7. Express the area of this function as a definite integral and then evaluate it as follows: 7
2 dx 3
2x
x 7 x 3
2 7
2( 3 )
14 6
20
14 –2. The shaded region is defined by the function f ( x ) sin x , between x and x 4 . Express the area of this function as a definite integral and then evaluate it as follows: 4
sin x dx
cos x
x 4 x
cos 4
cos
1
1
2
CHAPTER 5 The Fundamental Theorem of Calculus and Indefinite Integrals
79
15
1.609 . The shaded region is defined by the function f ( x )
1 , between x x
5 and x
1.
Express the area of this function as a definite integral and then evaluate it as follows: 1 5
1 dx x
ln x
x x
1 5
ln 1
ln 5
As you can see, you need to use absolute value bars to input negative values into the natural log function. Now you can complete the evaluation:
ln1 ln 5 0 ln 5
ln 5
1.609
16 –2. The shaded region is defined by the function f ( x ) x , between x 2.5 and x 1.5 . Express the area of this function as a definite integral and then evaluate it as follows: 1.5
x dx 2.5
80
1 2 x 2
x 1.5 x
2.5
1 1.5 2
2
1 2
2.5
PART 2 From Definite to Indefinite Integrals
2
1.125 3.125
2
3
Evaluating Indefinite Integrals
IN THIS PART . . .
Evaluate 17 indefinite integrals as anti-derivatives Apply the Sum rule, Constant Multiple rule, and Power rule for integrals Use algebra and trigonometry to prepare functions to be integrated Integrate functions using variable substitution
IN THIS CHAPTER
»» Knowing 17 antiderivatives that are the inverses of the 17 derivative formulas from Calculus I »» Understanding the Constant Multiple rule for integration »» Evaluating powers, roots, and rational functions using the Power rule for integration »» Breaking up complicated-looking functions with the Sum rule for integration
6 Instant Integration Chapter
I
ntegration doesn’t always have to be difficult. In many cases, you can integrate complexlooking functions using antiderivatives plus a few basic rules.
In this chapter, you discover how to reverse the basic derivative formulas you know from Calculus I to find indefinite integrals. Then, you see how the Constant Multiple rule, the Power rule, and the Sum rule that you know from Calculus I can be applied in reverse to evaluate integrals. By the end of this chapter, you’ll have a powerful set of tools for integrating that you can use confidently throughout your entire Calculus II course!
Antiderivatives of Common Functions and the Constant Multiple Rule In Chapter 3, I provide a list of 17 derivatives of the most basic functions. Reversing these formulas provides you with 17 antiderivative formulas that are essential for evaluating indefinite integrals.
CHAPTER 6 Instant Integration
83
Here are five important antiderivatives:
0 dx
C 1 dx
x C e x dx
ex
C a x dx
ax ln a
C
1 dx x
ln x C
Here are the antiderivatives of the six trig functions:
sin x dx
cos x C cos x dx
csc 2 x dx
sin x C sec 2 x dx
cot x C sec x tan x dx
tan x C
sec x C csc x cot x dx
csc x C
And here are the antiderivatives of the six inverse trig functions:
1 1 x2
dx
1 dx 1 x2 1 x x2 1
1
arcsin x C
1 x2
1 dx 1 x2
arctan x C dx
dx
1
arcsec x C
x x2 1
arccos x C
arccot x C dx
arccsc x C
As you can see, each of these three antiderivatives can be evaluated in two ways. In practice, you’ll most likely use only the arcsin, arctan, and arcsec forms. The Constant Multiple rule for integration enables you to move a coefficient outside the integral:
cf ( x ) dx
Q. A.
c f ( x ) dx
7cos x dx 7 sin x
C . Use the Constant Multiple rule to move the coefficient 7 outside the integral, then evaluate the resulting integral: 7 cos x dx
Q.
3 4 1 x2
A.
3 arctan x 4
7 cos x dx
7 sin x C
dx
C . Use the Constant Multiple rule to move the coefficient
integral, then evaluate the resulting integral:
3 4 1 x2
84
dx
3 1 dx 4 1 x2
PART 3 Evaluating Indefinite Integrals
3 arctan x C 4
3 outside the 4
1
Evaluate each of the following integrals. a. 5 dx c.
2
5 dx 6x
b.
3e x dx
d.
2x dx 7
Evaluate each of these integrals as a different trig function: a. 2sin x dx c.
e.
cos x dx
b.
1 d. sec 2 x dx 2 2 3 1 x
2
dx f.
sec x tan x dx 99 100 x x 2 1
dx
The Power Rule The Power rule for integration enables you to integrate a variable raised to any power except –1:
xn 1 C n 1
x n dx
The value of n may be 0, negative numbers, rational numbers, and irrational numbers.
Q. A.
x 6 dx 1 7 x 7
C . Apply the formula as follows:
x 6 dx
Q. A.
2 4
5 x3 84 x 5
x6 1 C 6 1
1 7 x 7
C
dx C . Use the Constant Multiple rule to move the coefficient
gral, and express the remaining function as an exponent of x:
2 4
5 x3
dx
2 x 5
3 4
2 outside the inte5
dx
Now, apply the Power rule and simplify the result:
2 x 5
3 4
dx
1 2 4 x4 5
C
84 x 5
C
CHAPTER 6 Instant Integration
85
3
Evaluate the following integrals using the Power rule: a. x dx d. x
4
4
2
c. x dx
dx
9
f. x
e. x dx
2
dx
b. x
d. x 3 dx
e. x
3
1
dx
1 2
1
dx
c. x 2 dx
dx
f. x
2 3
dx
Rewrite each of the following integrals as a power of x and then integrate using the Power rule, simplifying in each case: a.
d.
6
62
Evaluate the following integrals of negative and fractional powers using the Power rule: a. x
5
3
b. x dx
3
x dx
b.
1 dx x2
c.
x 2 dx
e.
1 dx x
f.
1 dx x5 1 3
x4
dx
Rewrite each of the following integrals as a power of x and then integrate using the Power rule and Constant Multiple rule, simplifying in each case: a. d.
2 dx x3
b.
7 dx 10 x 9
e.
1 x dx 4
c.
3
f.
5
4 x
2
dx
54 x dx 7 27 dx 5 x
The Sum Rule The Sum rule for integration (sometimes called the Sum and Difference rule) lets you integrate the sum of two or more functions as a sum of integrals:
f ( x ) g ( x ) dx
f ( x ) dx
g ( x ) dx
In practice, this rule allows you to break up a function with multiple terms and integrate it, term by term, using the other methods discussed in this chapter.
86
PART 3 Evaluating Indefinite Integrals
Q.
6x 2
A.
2x 3
8x
5 dx
4x 2
5 x C . Use the Sum rule to break up the integral into the sum of three integrals, and use the Constant Multiple rule to move the coefficient outside the integral in each case: 6x 2
8x
5 dx
6 x 2 dx 8 x dx
5 1 dx
Now, evaluate each integral separately and then simplify the result:
6 x 2 dx 8 x dx
Q.
4 sin x
A.
4 cos x 2 ln x
2 x
5 1 dx
6 3 x 3
8 2 x 2
5x C
2x 3
4x 2
5x C
1 dx 7 x 2 x 7
C . Use the Sum rule to break up the integral into the sum of
three integrals and then use the Constant Multiple rule to move the coefficient outside the integral in each case:
4 sin x
2 x
1 dx 7 x
4 sin x dx 2
1 dx x
1 7
1 dx x
Express the third term as an exponent:
4 sin x dx 2
1 dx x
1 x 7
1 2
dx
Evaluate all three integrals and simplify:
4 cos x 2 ln x
7
C
2 x 7
4 cos x 2 ln x
C
Evaluate the following polynomials using the Sum rule and Constant Multiple rule in conjunction with the Power rule. a.
x2
c. 6 x
8
1
1 2 x2 7
8
5 x 3 dx
b. 6 x
12 x 5
d.
2 x dx
3
5x 2
1 15 x 4
9 x 1 dx
1 14 x 5
1 13 x 7
1 dx 8
Evaluate the following functions using the Sum rule and Constant Multiple rule in conjunction with the Power rule. (You can leave your answers in exponential form.) a.
2
5x
c. 4 x
1 2
1
4x 3 3
5x 2
2 x dx 6 x 7 dx
3
b. 2 x
6
10 x 7
x
1
8x 3
d.
19 dx
5
6x
5 7
x dx
CHAPTER 6 Instant Integration
87
Answers and Explanations 1
a. 5 dx c.
2
5 dx 6x
c.
1 sec 2 x dx 2
1 sec 2 x dx 2
sec x tan x dx 2 3 1 x2 99 100 x x
a. x dx 4
2
a. x
1 2 x 2
dx
x
dx
1
1
2x 2
x dx
c.
1 dx x5
x
e.
1 dx x
x
9
e. x dx
1 x
1 10 x 10
C
b. x
C
3
f. x
1 x 2
dx
f. x
5
dx 1 2
dx
2 x3 3
C
1 x 4 1
2x 2
4
C
C
PART 3 Evaluating Indefinite Integrals
C
1 4x 4 2 x
C
C
2 3
2
62
1 63 x 63
dx
1 2x 2
C
C C
C
4
C 2 32 x 3
1 4 x 4
c. x dx
d. x 3 dx
x 2 dx
C
3
C
C
1
a.
1 3 x 3
1
2 32 x 3
2x 7 ln 2
99 arcsec x C 100
2
C
1 x 2 dx 7
C
sec x C
b. x dx
C
1
2x dx 7
3e x
2 arcsin x C 3
dx
1 x2
C
1 5 x 5
1
1 2
1
d.
3 e x dx
1 tan x C 2
99 1 dx 100 x x 2 1
dx
dx
c. x 2 dx e. x
2
3e x dx
sin x C
sec x tan x dx
2 3
dx
b.
2 cos x C
cos x dx
d. x
88
2 sin x dx
cos x dx
f.
5
5 ln x C 6
b.
e.
4
5x C
5 1 dx 6 x
a. 2 sin x dx
d.
3
5 1 dx
3 3 x 4
C
1
dx
3x 3
b.
1 dx x2
d.
f.
3
x 2 dx
1 3
x
4
dx
C
x
2
dx
2
1
1 x
C
5
x 3 dx
x
x
4 3
dx
3 3 x 5
3x
33 5 x 5
C 1 3
C
C
3
3 x
C
C
6
a.
b.
c.
d.
e.
7
8
2 dx x3
2 x
3
1
54 x dx 7
4 x
7 x 10 3 x 4
dx
2
f.
27 dx 5 x
a.
x2
2 5
9
1 2
5x 2
9 x 1 dx
c. 6 x
8
12 x 5
2 x dx
a.
5x
b. 2 x
c. 4 x d.
x
2
6
1 2
1
1 14 x 5
5 2 x 2
3 4 x 2 2 9 x 3
1 13 x 7
2 x dx
3
5x 2
6 x 7 dx
5
8x 3
6x
5 7
2x 6
2 x 5
19 dx
5 3 x 3
1
8x 2
x dx
55 3 x 4
C
C
54 x 5
C
x2
C
7x
x2 10 7
5
2x 2 ln x
x C
C
1 16 x 64
3x 3 5
7 80 x 8
9 2 x 2
4
1
C
3x C
1 dx 8
5x
3
10 x 7
C
C
C
8
x
8
27 2 12 x 5
dx
1
4x 3
10
3 5 5 x 4 3
1 3 x 3
3
1 15 x 4
44 5 x 7
C 7
dx
1 x3 6
C
3
b. 6 x
d.
C
1 2 32 x 4 3
dx
27 x 5
5 x 3 dx
1 x2
C
5 4 45 x 7 5
5 x 4 dx 7
7 dx 10 x 9 3
2
1 1 x 2 dx 4
1 x dx 4
5
2 x 2
dx
1 15 x 75
C
5 x
19 x C
1 14 x 98 4
3x 3
x2
2 5x 5
7x
1 x C 8
C 10 7
19 x C
3x 2 7x C 8
2
3x 3
21x 7
1 2 x 2
C
CHAPTER 6 Instant Integration
89
IN THIS CHAPTER
»» Integrating rational and radical functions with the Power rule »» Using algebra to simplify rational functions so you can integrate them »» Identifying rational functions that integrate easily to inverse trig functions »» Applying trig identities to make trig functions easier to integrate »» Integrating relatively simple compositions of functions
7 Sharpening Your Integration Moves Chapter
I
n this chapter, you extend your ability to evaluate indefinite integrals, building on the skills you gained from Chapter 7.
To begin, you use the Power rule to integrate rational and radical functions by framing them as negative and fractional powers. You then discover how to evaluate integrals of some rational functions by first simplifying them. You also get some practice identifying the rational functions that integrate easily as inverse trig functions. Next, you turn your attention to integrating trig functions by using trig identities to simplify or change their form. And to finish up, you discover how to integrate relatively simple compositions of functions using a simple formula.
CHAPTER 7 Sharpening Your Integration Moves
91
Integrating Rational and Radical Functions In Chapter 2, I show you how to express rational and radical functions of x as negative and fractional powers of x. And in Chapter 3, I reacquaint you with the Power rule for derivatives, which enables you to differentiate any power of x. Then in Chapter 7, you discover the Power rule for integration:
1 xn n 1
x n dx
1
C
You can combine these skills to integrate a lot of difficult-looking functions.
Q. A.
1 dx . x3
Evaluate
1 2x 2
C . Rewrite the function as a negative power of x, and then integrate it using the
Power rule:
1 dx x3
x
3
dx
1 x 2
2
C
This problem was originally stated in rational form, so put it back into this form:
1 2x 2
C
Q.
Integrate:
A.
33 5 x 5
3
x 2 dx .
C . Rewrite the function as a negative power of x, then integrate it using the
Power rule and put it back into radical form: 3
1
92
x 2 dx
2
x 3 dx
5
3 3 x 5
1 dx x7
PART 3 Evaluating Indefinite Integrals
C
33 5 x 5
C
2
3
9 x 4 dx
3
10 x3
2 x3
4
dx
4 7
x3
dx
Using Algebra to Prepare Functions for Integration You can often integrate a rational function by simplifying it first and then integrating the result. This opportunity arises surprisingly often in Calculus II problem sets, as well as on tests, so be sure to take advantage of it.
Q.
Integrate:
A.
1 2 x 10
x2 9 dx . 5 x 15
3 x C . Factor the numerator and the denominator of the rational expression, 5
then cancel out common factors and express the result as two rational expressions:
x2 9 dx 5 x 15
x
3 x 3 dx 5 x 3
x
3 5
dx
x 3 + dx 5 5
Now, split this result into two separate integrals, move the coefficients outside each integral, and evaluate each integral separately:
x dx 5
3 dx 5
1 x dx 5
3 1 dx 5
1 2 x 10
3 x C 5
CHAPTER 7 Sharpening Your Integration Moves
93
5
3x 3 6x 2 dx 4x 8
6
2 x 2 50 dx 3 x 15
7
3x 2 3x 6 dx 5 x 10
8
3x 3 4x 2 3x 4 dx 3x 2 x 4
Integrating Using Inverse Trig Functions In Chapter 7, I show you six rational functions that are actually the derivatives of the six inverse trig functions. Here they are again:
1 1 x2
dx
1 dx 1 x2 1 x x2 1
94
arcsin x C
arctan x C
dx
arcsec x C
PART 3 Evaluating Indefinite Integrals
1 1 x2
dx
1 dx 1 x2 1 x x2 1
arccos x C
arccot x C
dx
arccsc x C
Notice that each pair of these functions differs by only a minus sign. Thus, you only need to memorize the three positive functions in the left-hand column. Although variations on these three functions may not occur all that frequently, they tend to arise a lot on Calculus II quizzes and tests!
Q.
Evaluate the indefinite integral
A.
5 arctan x 3
5 dx . 3 3x 2
C . Begin by factoring in the numerator, and then bring the resulting coeffi-
cient outside the integral:
5 dx 3 3x 2
5 3 1 x
2
dx
5 1 dx 3 1 x2
Now, integrate this result:
5 arctan x C 3
9
11
2 3x x
2
1
5 dx 8 8x 2
dx
10
12
8 17 1 x 2
dx
7 10 x 9 x 2
9
dx
CHAPTER 7 Sharpening Your Integration Moves
95
Using Trig Identities to Prepare Functions for Integration In Chapter 2, I give you a relatively short list of trig identities that are indispensable for Calculus II. Here they are again for easy reference. The five basic trig identites:
sin x
1 csc x
cos x
1 sec x
tan x
1 cot x
tan x
sin x cos x
cot x
1 cot 2 x
csc 2 x
cos x sin x
The three Pythagorean identities:
sin 2 x cos 2 x
1 tan 2 x
1
sec 2 x
Useful double-angle and half-angle identities:
sin 2 x
sin 2 x
2 sin x cos x
1 cos 2 x 2
cos 2 x
1 cos 2 x 2
These identities are immediately useful for transforming some complex-looking trig functions into functions that are easy to integrate. And you can use them again in Chapter 10, where you work more in depth with trig functions and explore the more advanced concept of trig substitution. Then in Chapter 7, I show you how to integrate the derivatives of the six basic trig functions:
cos x dx sec 2 x dx
sin x C
sin x dx csc 2 x dx
tan x C
sec x tan x dx
sec x C
cos x C cot x C
csc x cot x dx
csc x C
When you combine these six integrals with the trig identities, you have a very useful set of tools for integrating a variety of less friendly looking trig functions. Notice that sines and cosines combine well together, as do tangents and secants, and cotangents and cosecants.
Q.
Integrate the following: tan 2 x dx
A.
tan x
x
C . To begin, add 1 and subtract 1 to this integral, and then split the integral
in two:
tan 2 x dx
tan 2 x 1 1 dx
tan 2 x 1 dx
1 dx
Now, apply the second Pythagorean identity to the first integral, and then evaluate both integrals:
sec 2 x dx
96
1 dx
tan x
PART 3 Evaluating Indefinite Integrals
x C
Q.
Evaluate
A.
1 cot x 2
1 dx . 1 cos 2 x C . Notice that the function you’re trying to integrate looks a bit like the
identity sin 2 x
1 cos 2 x 1 . Begin by multiplying the entire function by both and 2, 2 2
then move the 2 inside the function:
1 dx 1 cos 2 x
1 1 2 dx 2 1 cos 2 x
1 2 dx 2 1 cos 2 x
Now, the function is the reciprocal of
1 cos 2 x , so you can rewrite it as the reciprocal 2
of sin 2 x , which is csc 2 x , which integrates easily:
1 csc 2 x dx 2
1 cot x C 2
13
cos x sec x dx csc x
14
15
cot x sec x dx csc x sin x
16
cos x dx 1 cos 2 x
sin x
sin x cos 2 x dx sin 2 x
CHAPTER 7 Sharpening Your Integration Moves
97
Integrating Compositions of Functions with Linear Inputs Unlike differentiation, integration doesn’t have a Chain rule for integrating functions within functions. However, in certain cases, you can integrate relatively simple functions within functions when the inner function is linear — that is, when it’s of the form ax + b. To begin, here’s a formula you can use for integrating functions of the form e ax b:
e ax
Q. A.
b
dx
1 ax e a
b
C
Integrate e 5 x dx .
1 5x e 5
e 5 x dx e 5 x dx
1 5x e 5
C . Use the formula e ax
b
dx
1 ax e a
b
C:
C
Similar formulas exist for integrating linear inputs to the derivatives of the six basic trig functions:
cos ax b dx
1 sin ax b a
sec 2 ax b dx sec ax
1 tan ax b a
b tan ax
b dx
Q.
Integrate sec 2
A.
2 tan sec 2
Q. A.
1 x 2
sin ax b dx
C
1 sec ax a
b
C
csc 2 ax b dx
1 cot ax b a
csc ax
b dx
C . Use the formula sec 2 ax b dx
1 x dx 2
2 tan
1 x 2
b cot ax
C C
1 csc ax a
b
1 tan ax b a
1 cot 5 x 1 dx .
C . Use the formula csc ax b cot ax b dx
PART 3 Evaluating Indefinite Integrals
C:
C
Evaluate the integral csc 5 x
1 csc 5 x 1 5
1 cos ax b a
1 x dx . 2
csc 5 x 1 cot 5 x 1 dx
98
C
1 csc 5 x 1 5
C
1 csc ax b a
C:
C
You can also use this strategy to integrate linear inputs to any power function using a modification to the Power rule:
For all n ax b
Q. A.
For n
1: n
1 a n 1
dx
n 1
ax b
1: 1
ax b
C
2
What is the result when you evaluate the integral
3
1 ln ax b a
dx
9x 1
2
C
dx ?
23 9x 1 3
C . To begin, move the coefficient of –2 outside the function, then express the 1 2 n and use the formula ax b dx ax b n 1 C integral as an exponent of a n 1 3
to integrate:
2 3
Q. A.
9x 1
e
2x
dx
2
1 3
9x 1
2
dx
2 9x 1
2 3
dx
6 9x 1 9
1 3
C
23 9x 1 C 3
4 dx 5x 9 4 ln 5 x 9 5
C . Begin by moving the coefficient of 4 outside the function and express1 1 dx ln ax b C: ing the integral as a power of –1, then apply the formula ax b a 4 dx 5x 9
17
2
dx
4
1 dx 5x 9
4
5x 9
1
dx
4 ln 5 x 9 5
18
1 sec( 6 x 2
C
4 ) tan( 6 x
4 ) dx
CHAPTER 7 Sharpening Your Integration Moves
99
19
100
5 csc 2 ( 2 x 1)
11 2 x 3
2
dx
PART 3 Evaluating Indefinite Integrals
20
3 7x 5
4 sin( 2 x 1) dx 5
Answers and Explanations 1
1 dx x7
2
9 x 4 dx
3
10
x
3
x
7
4
2 x3
5
3x 3 6x 2 dx 4x 8
6
7
8
9 10
4
10 x
4
7
x
3
dx
6
1 6x 6
C 7
9 x 3 dx
dx
3
1 x 6
dx
2 x
9 3 2
3
3 3 x 7
dx
27 3 7 x 7
C
10 x
2
dx 4 x
3 7
dx
3x 2 x 2 dx 4 x 2
2 x2 2 x 2 50 dx 3 x 15 3 x 2 2 2 x dx 5 dx 3 3 3
C
1 2
2
C 20 x
C
1 x 2
3x 2 dx 4
C 4
2
4
7 7 x 4
3 2 x dx 4
3 1 3 x 4 3
17 1 x
dx
11
5 dx 8 8x 2
12
10 x 9 x 2
1 1 x
5 8 1 x2
7
Equivalently :
8 17
9
dx
cos x sec x dx csc x
14
cos x dx 1 cos 2 x
2
dx
C
5 1 dx 8 1 x2
dx
3 x 1 dx 5
8 arcsin x C 17
7 10 x 9 x 2 1
7 arccsc x 30
13
1 3 x 4
C
x 2 1 3x 4 dx x 1 3x 4
x 2 3x 4 1 3x 4 3x 3 4x 2 3x 4 dx dx 2 x 1 3x 4 3x x 4 x 1 x 1 1 2 dx x 1 dx x x C x 1 2 2 2 1 2 dx dx arcsec x C 3 x x2 1 3 3x x 2 1 2
C
7
7 x4
25 2 x 5 x 5 2 dx dx x 5 dx x 5 5 3 3 1 2 2 1 2 10 x x x C 5x C 2 3 3 3
3 x2 x 2 3x 2 3x 6 3 x 1 x 2 dx dx dx 5 x 10 5 x 2 5 x 2 3 3 3 2 3 3 1 2 3 x dx x x C x x C 1 dx 5 5 5 2 5 10 5
8
1 x2
C
dx
5 arctan x C 8 7 30 x x 2 1
dx
7 1 dx 30 x x 2 1
7 arcsec x 30
C
C
cos x sin x sec x dx cos x cos x dx sin 2 x
sin x dx
cos x 1 dx sin x sin x
cos x C
cot x csc x dx
csc x C
CHAPTER 7 Sharpening Your Integration Moves
101
15
cot x sec x dx csc x sin x csc x dx cos 2 x sin x
16
sin x 1 cos 2 x sin x cos 2 x dx dx sin 2 x 2 sin x cos x cos 2 x dx cos x dx sin x C cos x
sin x
2x
17
e
18
1 sec( 6 x 2
19
dx
102
1 e 2
2x
3 7x 5
1
csc x dx 1 sin 2 x sin x tan x C
1 cos 2 x dx 2 cos x
1 cos 2 x 1 dx 2 cos x
C
4 ) tan( 6 x
5 csc 2 ( 2 x 1) + 11 x 3 2
20
cos x 1 sin x cos x dx sin x dx 1 1 sin 2 x sin x sin x sin x sin x sin x csc x 1 dx dx sec 2 x dx cos 2 x cos 2 x
4 ) dx
11 2 x 3
2
1 1 sec( 6 x 2 6
4) C
1 sec( 6 x 12
5 csc 2 ( 2 x 1) dx +
dx
11 2
x 3
4) C 2
dx
5 cot( 2 x 1) 2
C
4 sin( 2 x 1) dx 5
3
1 dx 7x 5
PART 3 Evaluating Indefinite Integrals
4 sin( 2 x 1) dx 5
3 2 ln|7 x 5| cos 2 x 1 7 5
C
IN THIS CHAPTER
»» Understanding how to do variable substitution step by step »» Integrating the product of a function times its derivative »» Recognizing more complex integrals that you can evaluate with variable substitution »» Evaluating definite integrals by changing the limits of integration
8 Here’s Looking at u-Substitution Chapter
V
ariable substitution (also called u-substitution, or u-sub for short) is a relatively simple way to integrate complex functions, especially products of functions and compositions of functions.
In this chapter, you first discover the how of variable substitution. Then, I show you the simplest type of function that you can integrate using variable substitution: functions that are the product of a function and its derivative. Next, I show you the more complicated case: functions that are the product of a composition of functions and the derivative of the input function. After that, you see how when you evaluate definite integrals using variable substitution, you can often make your work easier by changing the limits of integration.
Understanding the How of u-Substitution In this section, I show you how to do variable substitution step by step. Later in the chapter, you see how to identify situations where variable substitution works best. For now, though, just get used to how the process works.
CHAPTER 8 Here’s Looking at u-Substitution
103
Q.
cos 4 x sin x dx
A.
1 cos 5 x 5
C . To begin, I declare the variable u to be equal to part of the function I’m
trying to integrate. I think of this as “sidework” to the integral:
u
cos x
This declaration allows me to make the following change to the integral I’m working with:
cos 4 x sin x dx
u 4 sin x dx
Next, more sidework: I differentiate u and then multiply both sides of the equation by dx :
du dx du
sin x sin x dx
Now, I substitute du for sin x dx into the integral:
u 4 du As you can see, here I applied the Constant Multiple rule, moving the minus sign outside the integral. And I’m ready to integrate as follows:
1 5 u 5
C
This result is in terms of u, so to complete the problem, I substitute cos x for u:
1 cos 5 x C 5
1
104
Use the variable substitution u = sin x to evaluate sin 7 x cos x dx .
PART 3 Evaluating Indefinite Integrals
2
Evaluate the integral 3 x 2
u
x
3
x3
5
5
5 as a variable substitution.
dx , using
3
2x 5
3 as a variable substitution, x4 dx. integrate the following: 2x 5 3 Using u
4
2
Evaluate 7 xe 3 x dx , using the variable substitution u
3 x 2.
When to Use u-Sub: The Simpler Case There are two common cases when variable substitution works well. The simpler case (when you’re lucky enough to get it!) is when you want to integrate the product of a function times its derivative — that is, integrals of this form:
f ( x )f '( x ) dx Here’s an example:
Q. A.
sin x cos x dx 1 sin 2 x 2 u du dx du
C . Here, notice that
d sin x dx
cos x . Thus, make your declaration as follows:
sin x cos x cos x dx
As a result, you can rewrite the integral as follows:
sin x cos x dx
u du
You can evaluate this integral easily:
1 2 u 2
C
1 sin 2 x C 2
When you get good at identifying cases where you want to integrate the product of a function and its derivative, you can skip the variable substitution step altogether and use the following formula:
f ( x )f '( x ) dx
1 f(x) 2
2
C
CHAPTER 8 Here’s Looking at u-Substitution
105
Q. A.
ln x dx x 1 ln x 2
2
ln x dx x
Q. A.
C . Here, f ( x ) ln x and f '( x ) 1 ln x 2
2
1 so you can use the formula: x
C
tan x sec 2 x dx 1 tan 2 x 2
C . The function f ( x ) tan x , and f '( x ) sec 2 x , so you can use the formula:
tan x sec 2 x dx
1 tan 2 x C 2
In some cases, if the integral doesn’t have the precise form that you need to use the formula, you can tweak it a bit before you integrate. For example:
Q. A.
sin 5 x cos 5 x dx 1 sin 2 5 x 10
C . In this case, f ( x ) sin5 x and f '( x ) 5cos5 x . You can make a small
change to the integral that will allow you to use the formula:
sin 5 x cos 5 x dx
1 5
sin 5 x
5 cos 5 x dx
Here, I multiplied the integral by
1 on the outside and by 5 on the inside, which is 5
equivalent to multiplying by 1. This technique is common in integration, so watch for opportunities to use it. Now, the integral takes the form that the formula requires:
1 1 sin 2 5 x C 5 2
5
106
1 sin 2 5 x C 10
arctan x dx 1 x2
PART 3 Evaluating Indefinite Integrals
6
x 2 sin x 3 cos x 3 dx
7
ln x 2 dx x
8
e x tan e x sec 2 e x dx
When to Use u-Sub: The More Complex Case The more complex (and far more common) case for using u-substitution occurs when you’re trying to integrate the product of a composition of functions times the derivative of the input function. Here’s the general form to look for:
g ( f ( x ))f '( x ) dx
When you know how to integrate g(x)!
This sounds more complicated than it really is, and in practice you’ll probably get good at spotting these opportunities when they arise. For example:
Q.
sec 2 x dx tan 7 x
A.
1 cot 6 x 6
C . The function you’re trying to integrate is actually a product of functions.
Rewriting it in an equivalent form highlights this point:
sec 2 x dx tan 7 x
sec 2 x tan
7
x dx
Before I move on, note that here, tan 7 x really does mean tangent and not arctangent. Situations like this are why I always, always use the notation arctan (rather than tan 1) to mean the inverse of the tangent function. This stuff is confusing enough without introducing ambiguous symbols into the mix. In this case, f ( x ) tan x , and f '( x ) sec 2 x . So far, this problem resembles the simpler case that I discuss in the previous section. However, a slight complication here is that the tan x function is taken to a power of 7. Watch how this plays out as I choose tan x for my u-substitution:
u du dx du
tan x sec 2 x sec 2 x dx CHAPTER 8 Here’s Looking at u-Substitution
107
Now, substitute u for tan x and du for sec 2 x dx into the integral:
u
7
du
This integral is easy to evaluate using the Power rule:
1 u 6
6
1 6u 6
C
C
As usual, the final step is simply replacing u with the original function, and then simplifying:
1 6 tan 6 x
Q. A.
C
1 cot 6 x C 6
tan x sec 5 x dx 1 sec 5 x 5
C . Here, the problematic composition of functions is sec 5 x . The inner func-
tion here is sec x , with a power of 5 applied to it. You want to set u results in the following:
u du dx du
sec x , but this
sec x sec x tan x sec x tan x dx
Thus, in order to rewrite the entire function in terms of u, you need to “peel off” a secant to enable the substitution (stay tuned for more of this strategy in Chapter 10):
sec x tan x sec 4 x dx Now, you’re ready to substitute u for sec x and du for sec x tan x dx:
u 4 du From here, the integral practically evaluates itself:
1 5 u 5
C
1 sec 5 x C 5
Before moving on to the practice problems, take a look back at Questions 1 through 4 earlier in this chapter. All of these are examples of this type of function. Notice how in each case, the choice of u-declaration is the inner function of the composition.
108
PART 3 Evaluating Indefinite Integrals
9
11
5
5 x 4 e x dx
ln x x
2
dx
10
cos x dx sin 3 x
12
cot x csc 5 x dx
Using Variable Substitution with Definite Integrals As you’ve seen throughout this chapter, when you’re using variable substitution to evaluate an indefinite integral, your final step always includes expressing the result in terms of the original variable (usually x) rather than the substitution variable u.
CHAPTER 8 Here’s Looking at u-Substitution
109
When using variable substitution to evaluate definite integrals, however, you have an opportunity to save a step that can sometimes make the calculation easier. This option is to leave the result in terms of u. An example should help make this process clear:
Q. A.
1
Evaluate
0
4 x 3 13
x 4 13 x 7
2
dx .
2400.6. In this problem, the opportunity to set u x 4 13 x 7 should be obvious. Then du 4 x 2 13 dx allows you to rewrite the integral in terms of u as follows: 1 0
4 x 3 13
x 4 13 x 7
2
dx
x 1 x 0
u 2 du
Before moving on, notice that I’ve changed the notation in the limits of integration to explicitly show that they’re in terms of x. In the next step, as I evaluate the integral using anti-differentiation, I retain this notation:
1 3 u 3
x 1 x 0
If you were solving an indefinite integral, you would now have no choice but to rewrite this result in terms of x, and you’d be done. Because this is a definite integral, however, another option open to you is to recast the two limits of integration x = 0 and x = 1 in terms of u. Here’s what that looks like in terms of sidework:
u ( 0 ) 4 13( 0 ) 7 7
u (1) 4 13(1) 7 1 13 7 19
As you can see, setting x to 0 and 1 respectively leads to values of u of –7 and –19. These values become the new limits of integration:
1 3 u 3
u
19
u
7
This evaluation isn’t too bad, and is perhaps a little easier than if you’d changed the problem back in terms of x:
1 3
19
3
1 3
7
3
1 6, 859 343 3
1 7, 202 3
2400.66
Although I describe this skill of rewriting the variables of the limits of integration in terms of u as a choice, your teacher may require you to demonstrate your knowledge of it on a test. For this reason, in Questions 15 to 18, evaluate all definite integrals using u-substitution, rewriting all the limits of integration in terms of u.
110
PART 3 Evaluating Indefinite Integrals
13
15
2 0
4 0
5 sin 3 x cos x dx
14
2 1
8x 3 dx 3x 4 1
2
2 xe x dx
CHAPTER 8 Here’s Looking at u-Substitution
111
Answers and Explanations 1
1 sin 8 x C . Using u 8 u sin x du cos x dx du cos x dx
sin x results in the following:
Now, substitute u 7 for sin 7 x and du for cos x dx into the integral:
sin 7 x cos x dx
u 7du
Evaluate the integral and then rewrite this result in terms of x:
2
1 8 u 8
C
1 x3 6
5
1 sin 8 x C 8 6
x3
C . Using u
5 results in
du dx
3 x and du
3 x dx . Rewrite and evaluate the
integral as follows:
3x 2 3
5
x3
5
1 ln 2 x 5 10
3
u
2x 5
dx
u 5du
C . Using u
1 6 u 6 2x 5
1 3 x 6
C
5
6
C
3 results in the following:
3 4
du 10 x dx 1 du x 4 dx 10 Substitute u for 2 x 5 it in terms of x:
x4 dx 2x 5 3 4
1 1 du 10 u
1 du for x 4 dx as follows, then evaluate the integral and rewrite 10
1 ln u 10
2
1 arctan 2 x 2
7 u e du 6
7 u e 6
C
1 ln 2 x 5 10
3
C
arctan x dx 1 x2
7 3x2 e 6
C
C . In this problem, f ( x ) arctan x and f '( x )
follows:
112
C
7 3x2 7 e C . Using u 3 x 2 results in du 6 x dx and du 7 x dx . Rewrite the integral, replacing 6 6 7 3x2 u e with e and 7x dx with du, then evaluate it as follows: 6 7 x e 3 x dx
5
3 and
1 arctan 2 x C 2
PART 3 Evaluating Indefinite Integrals
1 , so use the formula as 1 x2
6
1 sin 2 x 3 6
C . Here, f ( x ) sin x 3 and f '( x ) 3 x 2 cos x 3 . Thus, multiply the integral by
1 on 3
the outside and by 3 on the inside:
1 3 x 2 sin x 3 cos x 3 dx 3
x 2 sin x 3 cos x 3 dx Now, apply the formula:
1 1 sin 2 x 3 3 2 7
1 sin 2 x 3 6
C
C
2 1 ln x 2 C . In this case, f ( x ) ln x 2 and f '( x ) 4 1 by on the outside and by 2 on the inside: 2
ln x 2 dx x
1 2x x2
2 . Thus, multiply the integral x
1 2 ln x 2 dx 2 x
Now, apply the formula:
1 1 ln x 2 2 2 8
1 tan 2 e x 2
2
1 ln x 2 4
C
2
C
C . In this problem, f ( x ) tan e x and f ’( x ) e x sec 2 e x . Apply the formula as
follows:
1 tan 2 e x 2
e x tan e x sec 2 e x dx 9 e x
5
5
5 x 4 e x dx 10
x 5, so du
C . Let u
csc 2 x
e u du
C . Let u
cos x dx sin 3 x
5 x 4 dx . Thus:
eu
ex
C
sin x , so du
1 du u3
u
C
3
du
5
C cos x dx . Thus: 1 u 2
2
C
Substitute sin x for u and simplify:
1 sin 2 11 1 ln x
2
x C
3
3
ln x x
2
dx
1 2 sin 2 x
C . Let u x 2 du
C
ln x , so du 1 3 x 3
C
1 csc 2 x C 2 1 dx . Thus: x 1 ln x 3
3
C
CHAPTER 8 Here’s Looking at u-Substitution
113
1 csc 5 x 5
12
C . Let u
csc x , so du
csc x cot x dx . Thus, to put the integral into a form
that’s ready for this substitution, peel off a cosecant, and also multiply the outside and the inside by –1:
cot x csc 5 x dx
csc x cot x csc 4 x dx
Now, substitute and evaluate the integral:
1 5 u 5
u 4 du 13
2 0
5 sin 3 x cos x dx
2 0
5 sin 3 x cos x dx
Now, use u problem:
5 4 u 4 14
2 1
u 1
8x 3 dx 3x 4 1 8x 3 dx 3x 4 1
2 1
5 . Let u 4 5
x 2 x 0
5 0 4
4
2 3
x 2 x 1
2 ln 2 3
1 du u
15
4 0 4 0
u 2 2
2 xe x dx
eu
114
u 16 u 0
e 16
0 and u
8, 886,109 . Let u
02
eu
0 and u
e0
2 du 3
8 x 3 dx :
x 1
e 16 1 e u du
3 x 4 1, so du 12 x 3 dx , and
x 2
2 ln 2 2.105 3
x 0
Now, use u problem:
1 to change the limits of integration and complete the
2 ln 47 3
x 4
2
2 xe x dx
2
x 0
2.105. Let u
2 ln u 3
4
u 47
2
x
sin x dx:
5 4
Now, use u 3 1 1 1 1 and complete the problem:
2 ln u 3
5 4 u 4
sin
4
2 ln 47 3
sin x , so du
u 3 dx
sin0 0 and u
5 1 4
u 0
1 csc 5 x C 5
C
3 2
4
1 48 1 47 to change the limits of integration
x 2, so du
2 x dx :
x 4 x 0
42
16 to change the limits of integration and complete the
e 16 1 8, 886,109
PART 3 Evaluating Indefinite Integrals
4
Advanced Integration Techniques
IN THIS PART . . .
Apply integration by parts as a consequence of the Product rule Integrate a wide variety of trigonometric functions Use trig substitution to evaluate integrals Evaluate rational functions by integrating with partial fractions
IN THIS CHAPTER
»» Understanding the formula for integration by parts »» Knowing how to assign u and dv values »» Evaluating integrals by applying integration by parts multiple times
9 Integration by Parts Chapter
I
ntegration by parts provides a formula for rewriting integrals in a way that makes evaluating them more accessible. It allows you to integrate a function by splitting it into factors, differentiating one factor and integrating the other, and then piecing together the results.
In this chapter, I first show you the formula for integration by parts to give you practice using it. Next, I give you guidance on when and how to use this strategy. Finally, you see how to apply integration by parts more than once to evaluate a single integral.
Using the Formula for Integration by Parts Here’s the formula for integration by parts:
u dv
uv
v du
In this formula, both u and v are functions of x.
CHAPTER 9 Integration by Parts
117
Integration by parts is commonly used to integrate a product of two functions. For example:
Q.
Evaluate x 2 ln x dx .
A.
1 3 x ln x 3
1 3 x 9
C . This function is the product of two factors, x 2 and ln x . Set u
(because you don’t know how to integrate it) and set dv integrate dv as follows:
u du
ln x 1 dx x
dv v
ln x
2
x dx . Now, differentiate u and
x 2 dx 1 3 x 3
Next, use the formula for integration by parts to rewrite the integral:
u dv
uv
v du
1 3 x ln x 3
x 2 ln x dx
1 3 1 dx x 3 x
Simplify and solve the resulting integral:
1 3 x ln x 3
1 2 x dx 3
1 3 x ln x 3
1 3 x 9
C
You can also use integration by parts to evaluate integrals of functions that you know how to differentiate. For example:
Q. A.
ln x dx x ln x x C . As in the last example, set u ferentiate u and integrate dv: u du
ln x 1 dx x
ln x , but in this case, set dv
dv v
dx x
Plug these values into the formula for integration by parts:
u dv ln x dx
uv x ln x
v du x
1 dx x
Complete the evaluation:
x ln x
118
1 dx
x ln x
x C
PART 4 Advanced Integration Techniques
dx . Now, dif-
1
Evaluate x cos x dx , letting u dv cos x dx .
3
Evaluate
dv
x ln10 x dx , letting u
x and
ln10 x and
2
4
Evaluate xe x dx , letting u
x and dv
Evaluate arctan x dx , letting u dx .
e x dx .
arctan x and
dv
x dx .
Knowing How to Assign u and dv Integration by parts allows you to integrate problematic functions such as logarithms and inverse trig functions by differentiating them. When faced with an integral that includes either of these functions as a factor, let u equal this problematic factor and assign dv to the other factor.
Q. A.
ln x 4 dx x ln x 4
4x
C . This is a composition of functions, so set u
ln x 4 and assign dv
dx .
Use the Chain rule to differentiate u and integrate dv:
u du
ln x 4 1 4 x 3dx x4
4 dx x
dv v
dx x
CHAPTER 9 Integration by Parts
119
Plug these values into the formula for integration by parts:
u dv
uv
ln x 4 dx
v du
x ln x 4
x
4 dx x
Complete the evaluation:
x ln x 4
4 dx
x ln x 4
4x C
Polynomials tend to differentiate down to zero. So, when integrating the product of a polynomial and another function (especially a sine, cosine, or e function), let u equal the polynomial and let dv equal the other factor.
Q. A.
2 xe 3 x dx x ln x 4
4x
C . Let u
2 x and assign dv
e 3 x dx . As usual, your next step is to differ-
entiate u and integrate dv:
u du
2x 2 dx
e 3 x dx 1 3x e 3
dv v
Plug these values into the formula for integration by parts:
u dv 2 xe 3 x dx
uv
v du
2 3x xe 3
2 3x e dx 3
Complete the evaluation:
2 3x xe 3
5
120
2 3x e x C 9
arcsin5x dx
PART 4 Advanced Integration Techniques
6
3
ln x 2 dx
7
x 3 ln 2 x dx
9
x cos
1
1 x dx 4
x
8
8 xe 2 dx
10
2 x 3
sin 6 x dx
Applying Integration by Parts More Than Once In some cases when integrating by parts, you may need to repeat the procedure more than once to get an answer. This is especially true when you’re integrating a product of an algebraic function and a sine, a cosine, or an e function. In each of these cases, you’ll need to set u to the algebraic factor and dv to the remaining portion. Then, apply the formula for integration repeatedly until the algebraic factor differentiates down to 0. For example:
Q. A.
7 x 2 cos 4 x dx 7 2 x sin 4 x 4
7 x cos 4 x 8
7 sin 4 x 32
C . Let u 7 x 2 and assign dv
cos4 x dx . As usual,
your next step is to differentiate u and integrate dv:
u 7x 2 du 14 x dx
dv v
cos 4 x dx 1 sin 4 x 4
CHAPTER 9 Integration by Parts
121
Plug these values into the formula for integration by parts:
u dv
uv
v du
7 2 x sin 4 x 4
7 x 2 cos 4 x dx
7 x sin 4 x dx 2
Now, to evaluate the resulting integral, you need to apply integration by parts again, using u
u du
7 x and dv 2
sin4 x dx : dv
7 x 2 7 dx 2
v
sin 4 x dx 1 cos 4 x 4
Plug these values into the formula for integration by parts:
u dv 7 2 x sin 4 x 4
uv
7 x sin 4 x dx 2
v du
7 2 x sin 4 x 4
7 x cos 4 x 8
7 cos 4 x dx 8
This time, you can evaluate the resulting integral directly as follows:
7 2 x sin 4 x 4
7 x cos 4 x 8
7 sin 4 x C 32
Another common type of problem that you may encounter in your Calculus II class arises when integrating the product of an e function and either a sine or cosine function. In these cases, you may need to apply the integration-by-parts formula twice and then solve the resulting equation. For example:
Q.
e 2 x sin 5 x dx
A.
2 2x e sin 5 x 29
5 cos 5 xe 2 x 29
C . This time, let u
sin5 x and assign dv
differentiate u and integrate dv as follows:
u du
sin 5 x 5 cos 5 x dx
dv
e 2 x dx
v
1 2x e 2
Plug these values into the formula for integration by parts:
u dv e 2 x sin 5 x dx
122
uv
v du
1 2x e sin 5 x 2
5 2x e cos 5 x dx 2
PART 4 Advanced Integration Techniques
e 2 x dx . Next,
To evaluate the resulting integral, you need to apply integration by parts again, using
u
cos5 x and dv
u du
cos 5 x 5 sin x dx
5 2x e dx : 2 dv v
5 2x e dx 2 5 2x e 4
Plug these values into the formula for integration by parts:
u dv 1 2x e sin 5 x 2
5 2x e cos 5 x dx 2
uv
v du
1 2x e sin 5 x 2
5 cos 5 xe 2 x 4
25 2 x e sin 5 x dx 4
This time, the resulting integral is the same as the one that you started with. This looks like bad news, but you can now build a new equation as follows using the integral you started with and the most recent result:
e 2 x sin 5 x dx
1 2x e sin 5 x 2
5 cos 5 xe 2 x 4
To simplify the next few steps a bit, let I you’re trying to evaluate:
I
1 2x e sin 5 x 2
5 cos 5 xe 2 x 4
25 2 x e sin 5 x dx 4 e 2 x sin 5 x dx — in other words, the integral
25 I 4
Now use algebra to isolate I:
I
25 I 4 29 I 4 I
1 2x e sin 5 x 2 1 2x e sin 5 x 2 2 2x e sin 5 x 29
5 cos 5 xe 2 x 4 5 cos 5 xe 2 x 4 5 cos 5 xe 2 x 29
To complete the problem, substitute the original integral for I, and be sure to include + C in your answer:
e 2 x sin 5 x dx
2 2x e sin 5 x 29
5 cos 5 xe 2 x 29
C
CHAPTER 9 Integration by Parts
123
11
6 x 2 sin 7 x dx
13
e 2 sin 3 x dx
124
1
x
PART 4 Advanced Integration Techniques
12
e x cos x dx
Answers and Explanations 1
x sin x u du
cos x C . Let u
x and dv
cos x dx , then differentiate u and integrate dv:
dv v
x dx
cos x dx sin x
Substitute these values into the formula for integration by parts and complete the problem:
u dv
uv
x cos x dx 2
xe x u du
ex
v du
x sin x
sin x dx
C . Let u
x and dv
x sin x cos x C e x dx , then differentiate u and integrate dv:
x dx
dv v
e x dx ex
Substitute these values into the formula for integration by parts and complete the problem:
u dv
uv
xe x dx 3
xe x
2 x 3 ln 10 x 3 u du
v du e x dx 4 x3 9
xe x
ex
C . Let u
C ln10 x and dv dv
ln10 x 10 dx 10 x
1 dx x
v
x dx , then differentiate u and integrate dv:
x dx 2 32 x 3
Substitute these values into the formula for integration by parts and complete the problem:
u dv x ln10 x dx
uv
v du 3 2
2 x ln10 x 3
3 2 2 1 dx x 3 x
Simplify and solve:
2 32 x ln10 x 3 4
x arctan x u du
1
2 x 2 dx 3 1 ln 1 x 2 2
arctan x 1 dx 1 x2
2 32 x ln10 x 3 C . Let u dv v
3
4 2 x 9
C
2 x 3 ln10 x 3
arctan x and dv
4 x3 9
C
dx , then differentiate u and integrate dv:
dx x
CHAPTER 9 Integration by Parts
125
Substitute these values into the formula for integration by parts and complete the problem:
u dv
uv
arctan x dx
v du x dx 1 x2
x arctan x
Evaluate the resulting integral using variable substitution (see Chapter 8 for more on this method):
x arctan x 5
x arcsin 5 x u
1 ln 1 x 2 2
C
1 1 25 x 2 5
C . Let u
arcsin5 x 5
du
dv v
dx
1 25 x 2
arcsin5 x and dv
dx , then differentiate u and integrate dv:
dx x
Substitute these values into the formula for integration by parts:
u dv
uv
arcsin 5 x dx
v du 5x
x arcsin 5 x
1 25 x 2
dx
Evaluate the resulting integral using variable substitution (see Chapter 8 for more on this 50 x dx : method) with values u 1 25 x 2 and du
x arcsin 5 x
5 50
x arcsin 5 x
1 2u 2 10
1
x arcsin5 x 3 x 5
3
x ln x 2
C
1 u C 5 1 1 25 x 2 5
x arcsin 5 x
6
1 dx u
C . Let u
C 3
ln x 2 and dv
3
ln x 2
u
1
du
x
2 3
3 x 5
1 3
3 dx 5x
dx
dx , then differentiate u and integrate dv: dv v
dx x
Substitute these values into the formula for integration by parts and complete the problem:
u dv 3
ln x 2 dx
uv
v du 3
x ln x 2 3
x ln x 2 3
x ln x 2
126
3 x dx 5x 3 dx 5 2 x C 3
PART 4 Advanced Integration Techniques
7
1 4 x ln 2 x 4 u du
1 4 x 16
ln2 x 2 dx 2x
C . Let u
x 3 dx , then differentiate u and integrate dv:
ln2 x and dv dv
1 dx x
v
x 3 dx 1 4 x 4
Substitute these values into the formula for integration by parts:
u dv
uv
v du 1 4 1 dx x 4 x
1 4 x ln 2 x 4
x 3 ln 2 x dx
Simplify and evaluate the resulting integral:
1 4 x ln 2 x 4 8
1
16 xe 2 u du
1 3 x dx 4 1
x
32e 2
x
1 4 x ln 2 x 4
C . Let u
1 4 x 16
C 1
8 x and dv dv
8x 8 dx
v
1
x
e 2 dx , then differentiate u and integrate dv:
x
e 2 dx 1
2e 2
x
Substitute these values into the formula for integration by parts and complete the problem:
u dv 1 x 8 xe 2
4
v du 1
dx
9 4 x sin 1 x
u du
uv
16 xe 2 16 cos
1
x
1 x 4
1
x
16e 2 dx C . Let u
x and dv dv
x dx
16 xe 2
v
1
x
32e 2
x
C
1 cos x dx , then differentiate u and integrate dv: 4
1 cos x dx 4 1 4 sin x 4
Substitute these values into the formula for integration by parts and complete the problem:
u dv 1 x cos x dx 4 10
1 x cos 6 x 9
integrate dv:
u du
2 x 3 2 dx 3
uv
v du
4 x sin
1 x 4
1 sin 6 x 54
4 sin
1 x dx 4
4 x sin
1 1 x 16c cos x C 4 4
2 x and dv 3
C . Let u
dv v
sin6 x dx , then differentiate u and
sin 6 x dx 1 cos 6 x 6
CHAPTER 9 Integration by Parts
127
Substitute these values into the formula for integration by parts:
u dv 2 x 3
sin 6 x dx
uv
v du
2 1 x cos 6 x 3 6
2 1 cos 6 x dx 3 6
Simplify and evaluate:
11
1 x cos 6 x 9
1 cos 6 x dx 9
6 2 x cos 7 x 7
12 x sin 7 x 49
1 x cos 6 x 9 12 cos 7 x 343
1 sin 6 x C 54 C . Let u
6 x 2 and assign dv
sin7 x dx , then differ-
entiate u and integrate dv:
u 6x 2 du 12 x dx
dv v
sin 7 x dx 1 cos 7 x 7
Plug these values into the formula for integration by parts:
u dv 6 x 2 sin 7 x dx
uv
v du
6 2 x cos 7 x 7
12 x cos 7 x dx 7
To evaluate the resulting integral, apply integration by parts again, using u dv cos7 x dx :
u du
12 x 7 12 dx 7
dv v
12 x and 7
cos 7 x dx 1 sin 7 x 7
Plug these values into the formula for integration by parts:
u dv 6 2 x cos 7 x 7
uv
12 x cos 7 x dx 7
v du
6 2 x cos 7 x 7
12 7x x sin7 49
12 sin 7 x dx 49
Complete the problem by integrating the resulting integral as follows:
6 2 x cos 7 x 7 12 1 e x cos x
2
u du
128
cos x sin x
12 x sin 7 x 49
12 cos 7 x C 343
1 x e sin x C . Let u 2
cos x and let dv dv v
PART 4 Advanced Integration Techniques
e x dx ex
e x dx , then differentiate u and integrate dv:
Plug these values into the formula for integration by parts:
u dv
uv
e cos x dx
x
x
v du e x sin x dx
e cos x
To evaluate the resulting integral, apply integration by parts again, using u e x dx :
sin x and
dv
u du
dv
sin x cos x
v
e x dx ex
Plug these values into the formula for integration by parts:
x
u dv
uv
e sin x dx
x
x
e cos x
v du e x sin x
e cos x
e x cos x dx
This time, the resulting integral is the same as the one that you started with, so build a new equation as follows using the integral you started with and the most recent result:
e x cos x dx Let I
e x sin x
e x cos x dx
e x cos x dx , which is the integral you’re trying to evaluate:
e x cos x
I
e x cos x
e x sin x
I
Now use algebra to isolate I:
2I I
e x cos x e x sin x 1 x 1 x e cos x e sin x 2 2
To complete the problem, substitute the original integral for I, and be sure to include + C in your answer:
e x cos x dx 13
1 x e cos x 2
2 12 x e sin 3 x 35
1 x e sin x C 2
12 12 x e cos 3 x C . Let u 35
sin3 x and let dv
1
x
e 2 , then differentiate u and
integrate dv:
u du
sin 3 x 3 cos 3 x
dv v
1
x
e 2 dx 1
2e 2
x
Plug these values into the formula for integration by parts:
u dv 1 x e2
sin 3 x dx
uv 1 x 2e 2
v du sin 3 x
1
x
6e 2 cos 3 x dx
CHAPTER 9 Integration by Parts
129
To evaluate the resulting integral, apply integration by parts again, using u and dv
u
6 cos 3 x
du
6 cos 3 x
1 x e2 :
1
x
dv
e2
v
2e 2
18 sin 3 x
1
x
Plug these values into the formula for integration by parts:
u dv 1 x 2e 2
sin 3 x
6 cos 3 x
1 x e2
uv
v du
1 x 2e 2
dx
1
x
sin 3 x 12e 2 cos 3 x
1
x
36 e 2 sin 3 x dx
This time, the resulting integral is the same as the one that you started with, so build a new equation as follows using the integral you started with and the most recent result: 1
x
e 2 sin 3 x dx 1
Let I
I
1
1
x
x
2e 2 sin 3 x 12e 2 cos 3 x
1
x
36 e 2 sin 3 x dx
x
e 2 sin 3 x dx , which is the integral you’re trying to evaluate: 1
1
x
x
2e 2 sin 3 x 12e 2 cos 3 x
36 I
Now use algebra to isolate I:
35 I I
1
1
x
x
2e 2 sin 3 x 12e 2 cos 3 x 2 12 x e sin 3 x 35
1
12 2 x e cos 3 x C 35
To complete the problem, substitute the original integral for I, and be sure to include + C in your answer: 1
x
e 2 sin 3 x dx
130
1
2 2x e sin 3 x 35
1
12 2 x e cos 3 x C 35
PART 4 Advanced Integration Techniques
IN THIS CHAPTER
»» Integrating tan x, cot x, sec x, and csc x »» Evaluating integrals of common combinations of trig functions »» Knowing how to do the three cases of trig substitution
10 Trig Substitution Chapter
T
rig substitution is a somewhat sophisticated method of integration that allows you to integrate a variety of complicated rational and radical functions. Before you discover how to do trig substitution, you need to know a few more basics of integrating trig functions.
First, I show you how to integrate the four less commonly used trig functions (tan x, cot x, sec x, and csc x). After that, you get to use a variety of techniques integrating combinations of sines and cosines, then secants and tangents. With these techniques in place, you find out how to identify three cases when trig substitution makes sense and discover how to use this method.
Integrating the Six Basic Trig Functions In Chapter 6, you discover how to integrate sin x and cos x using anti-differentiation. For completeness, I repeat them below. When working more extensively with trig functions, you’ll also need to know how to integrate the remaining four trig functions:
sin x dx
cos x C
cos x dx
sin x C
tan x dx
ln sec x
C
sec x dx
ln sec x
tan x
C
cot x dx
ln sin x
C
csc x dx
ln csc x cot x
C
Okay, no peeking. When you think you can remember the integrals of these four trig functions, close the book and write them down.
CHAPTER 10 Trig Substitution
131
1
tan x dx
2
cot x dx
3
sec x dx
4
csc x dx
Integrating Powers of Sines and Cosines One case of trig substitution — the sine case — turns complicated-looking functions into powers of sines and cosines that you’ll need to integrate. When you’re working with at least one odd-numbered power of either sines or cosines, alone or in any combination, begin by “peeling off” a single sine or cosine from an odd power and then use trig identities to prepare the function to be integrated. For example:
Q.
sin 5 x cos 2 x dx
A.
1 cos 2 x 3
2 cos 5 x 5
1 cos 7 x 7
C . Here, the exponent of sin x is an odd number, so
begin by peeling off a sin x, leaving behind an even power of sine:
sin 5 x cos 2 x dx
132
sin 4 x cos 2 x sin x dx
PART 4 Advanced Integration Techniques
Next, use the Pythagorean identity sin 2 x
1 cos 2 x
2
2
2
1 to rewrite sin 4 x as 1 cos 2 x :
cos 2 x sin x dx
Now, use the variable substitution u result:
1 u2
cos 2 x
u 2 du
1 2u 2
cos x — so du
u 4 u 2 du
u2
2u 4
sin x dx — and simplify the
u 6 du
Integrate and rewrite in terms of x:
1 2 u 3
2 5 u 5
1 7 u 7
1 cos 2 x 3
C
2 cos 5 x 5
1 cos 7 x C 7
When integrating even powers of sines and cosines, alone or in any combination, use the two half-angle trig identities:
sin 2 x
1 cos 2 x 2
Q.
sin 2 x cos 2 x dx
A.
1 x 8
1 sin 4 x 32
cos 2 x
1 cos 2 x 2
C . Use the two half-angle identities to rewrite the integral and then
simplify:
sin 2 x cos 2 x dx
1 cos 2 x 1 cos 2 x dx 2 2
1 1 cos 2 2 x dx 4
Next, apply the half-angle identity for cosines to the second expression inside the integral:
1 1 cos 4 x 1 dx 4 2
1 1 1 4 2
cos 4 x dx 2
Simplify:
1 4
cos 4 x dx 2
1 2
1 8
1 cos 4 x dx
Integrate:
1 8
x
1 sin 4 x 4
C
1 x 8
1 sin 4 x C 32
CHAPTER 10 Trig Substitution
133
5
sin 3 x cos 7 x dx
6
cos 5 x
7
cos 7 x dx
8
cos 4 x dx
3
sin 8 x dx
Integrating Powers of Tangents and Secants The other two cases of trig substitution — the tangent case and the secant case — both produce powers of tangent and secant functions that you’ll need to integrate. When you’re working with at least one even-numbered power of secants in combination with any powers of tangent, begin by peeling off sec 2 x and placing it next to the dx. Then use trig identities to prepare the function to be integrated. For example:
134
PART 4 Advanced Integration Techniques
Q. A.
sec 6 x tan 3 x dx 5 2 tan 2 x 5
1 tan 4 x 2
1 tan 7 x C . Here, the exponent of sec x is an even number, so 7 begin by peeling off sec 2 x and expressing the tangent as an exponent: 3
sec 4 x tan 2 x sec 2 x dx
sec 4 x tan 3 x dx
Now, use the trig identity 1 tan 2 x tangents: 2
1 tan 2 x
3
tan 2 x sec 2 x dx
Use variable substitution with u
1 u2
2
sec 2 x to express the remaining secants in terms of
3
1 2u 2
u 2 du
sec 2 x dx and simplify:
tan x and du 3
u 4 u 2 du
3
7
u2
11
2u 2
u 2 du
Integrate and rewrite the result in terms of x:
2 25 u 5
1 92 u 9
2 132 u 13
C
5 2 tan 2 x 5
9 4 tan 2 x 9
13 2 tan 2 x C 13
When working with at least one odd-numbered power of tangents in combination with any powers of secant, begin by peeling off sec x tan x and placing it next to the dx. Then use trig identities to prepare the function to be integrated. For example:
Q. A.
1
tan 5 x sec 5 x dx 21 5 sec 5 x 21
11 1 10 sec 5 x 5 sec 5 x C . This time, the exponent of tan x is an odd number, 11 so peel off sec 2 x tan 2 x and place it next to the dx: 1
tan 5 x sec 5 x dx
tan 4 x sec
4 5
x sec x tan x dx
Now, use the trig identity tan 2 x secants: 4 5
2
sec 2 x 1 sec
x sec x tan x dx
Use variable substitution with u 2
u2 1 u
4 5
u4
du
sec 2 x 1 to express the remaining secants in terms of
sec x and du
2u 2 1 u
4 5
du
u
16 5
sec x tan x dx and simplify: 6
2u 5
u
4 5
du
Integrate and rewrite the result in terms of x:
5 u 21
21 5
11
10 5 u 11
1
5u 5
21
C
5 sec 5 x 21
11
10 sec 5 x 11
1
5 sec 5 x C
CHAPTER 10 Trig Substitution
135
To integrate all other cases of secants and tangents, alone or in combination, use integration by parts, as shown in Chapter 9. In some cases, this involves peeling off a factor, as I show you earlier in this chapter.
Q.
sec 3 x dx
A.
1 sec x tan x 2 sec 3 x dx
1 ln sec x 2
tan x
C . This time,
sec 2 x sec x dx
Now, integrate by parts as follows:
u du
sec x sec x tan x dx
dv v
sec 2 x dx
Next, use the trig identity tan 2 x
sec 3 x dx
sec x tan x
sec 3 x dx
sec x tan x
3
sec x dx
sec x tan x
u dv
tan x
3
sec x dx
uv
v du
sec x tan x
tan 2 x sec x dx
sec 2 x 1 to rewrite the integral:
sec 2 x 1 sec x dx sec 3 x dx x
sec x dx
3
sec x dx ln sec x
tan x
Notice now that the integral sec 3 x dx appears on both sides of the equation. Thus, you can add this expression to both sides of the equation and then divide both sides by 2 to complete the problem:
2 sec 3 x dx sec 3 x dx
sec x tan x ln sec x 1 sec x tan x 2
tan x
1 ln sec x 2
C
tan x
C
Note that C is an unknown constant, so there’s no need to divide it by 2 in the final step.
9
136
sec 4 x tan 8 x dx
PART 4 Advanced Integration Techniques
10
sec 6 x tan 5 x dx
11
sec 11 x tan 3 x dx
12
13
tan 3 x dx
14
tan 5 x sec 9 x
dx
tan 2 x sec x dx
Using Trig Substitution Trig substitution allows you to integrate a variety of complicated functions based on three basic forms, as shown in Table 10-1.
Table 10-1 Case
The Three Cases for Trig Substitution
Form
Example
Sine
a
2
n
Tangent
a 2 bx 2
n
Secant
2
bx
bx
2
a
2
x
5 sin 2
dx
x
7 tan 6
dx
x
10 sec 3
25 4 x dx 1 49 36 x
n
x-value
2
2
1 9 x 2 100
CHAPTER 10 Trig Substitution
137
In the examples that follow, I show you how to use trig substitution in all three of these cases.
Q.
25 4 x 2 dx
A.
25 2x arcsin 4 5
x 25 4 x 2 2
C . To begin, notice that the function you want to 4x 2
integrate has the following form: 25
1 2
a2
n
bx 2
Thus, this function is the sine case. Set up a right triangle as follows:
From these values, calculate sin
25 4 x 2 5
cos
25 4 x 2
5 cos
2x 5 x dx d dx
and cos , then find dx in terms of
2x 5 2x arcsin 5
sin 5 sin 2 5 cos 2 5 cos d 2
and
in terms of x:
sin
Now, rewrite the integral in terms of :
25 4 x 2 dx
5 cos
5 cos d 2
25 cos 2 2
d
Use the method described earlier in this chapter, in “Integrating Powers of Sines and Cosines,” to evaluate the integral of the even power of a cosine:
25 1 2 2
1 sin 2 4
25 4
C
Use the trig identity sin 2
25 4
25 2 sin cos 8
25 sin 2 8
C
2 sin cos to rewrite this result: C
25 4
25 sin cos 4
C
To complete the problem, substitute values in terms of x:
2x 25 arcsin 4 5
138
25 4
25 4 x 2 2 x 5 5
PART 4 Advanced Integration Techniques
C
25 2x arcsin 4 5
x 25 4 x 2 2
C
Q. A.
1
Use trig substitution to evaluate
1 6x ln 6
49 36 x 7
2
49 36 x 2
dx .
C . To begin, notice that the function that you want to integrate 1 2
36 x 2
has the following form: 49
a2
bx 2
n
Thus, this function is the tangent case. Set up a right triangle as follows:
From these values, calculate tan
49 36 x 2 7
6x 7
sec
49 36 x 2 1 49 36 x 2
and sec , then find dx in terms of :
x
7 sec 1 7 sec
dx d dx
tan 7 tan 6 7 sec 2 6 7 sec 2 6
d
Now, rewrite the integral in terms of :
1 49 36 x 2
dx
1 7 sec
7 sec 2 6
1 sec d 6
d
Use the method described earlier in this chapter, in “Integrating Powers of Tangents and Secants,” to evaluate the integral, and then rewrite this result using sec
1 ln sec 6
Q.
Evaluate
A.
1 ln 30
49 36 x 2 and tan 7 tan
49 36 x 2 7
1 ln 6
C
6x : 7 6x 7
C
1 6x ln 6
49 36 x 2 7
C
1 dx . 9 x 2 100
9 x 2 100 3 x 10
C . To begin, notice that the function you want to integrate has the
following form: 9 x 2
100
1
a2
bx 2
n
CHAPTER 10 Trig Substitution
139
Thus, this function is the tangent case. Set up a right triangle as follows:
From these values, calculate tan
9 x 2 100 10 9 x 2 100 1 9 x 2 100 1 9 x 2 100
and sec , then find dx in terms of :
3x 10
tan 10 tan
x
1 10 tan
dx d
1 100 ta an 2
dx
sec 10 sec 3 10 tan sec 3 10 tan sec d 3
Now, rewrite the integral in terms of :
1 9x
2
100
dx
1 100 tan 2
10 tan sec d 3
1 1 sec d 30 tan
1 csc d 30
Evaluate this integral and rewrite in terms of x using values of csc x and cot x from the previous right triangle:
1 ln csc x cot x 30
C
1 ln 30
3x
10
9 x 2 100
9 x 2 100
C
1 ln 30
3 x 10 9 x 2 100
C
Your professor might like you to rationalize and simplify this result:
3 x 10 9 x 2 100 1 ln 30 9 x 2 100
15
140
C
1 9 x 2 dx
PART 4 Advanced Integration Techniques
3 x 10 9 x 2 100 1 ln 30 3 x 10 3 x 10
16
1 4 25 x 2
dx
C
1 ln 30
9 x 2 100 3 x 10
C
17
19
1 49 x 2
81
dx
18
9 x 2 16
3 2
dx
1 dx 25 36 x 2
CHAPTER 10 Trig Substitution
141
Answers and Explanations 1
tan x dx
ln sec x
C
2
cot x dx
ln sin x
C
3
sec x dx
ln sec x
tan x
C
4
csc x dx
ln csc x cot x
C
5
1 cos 8 x 8
1 cos 10 x C . Peel off a sin x (the trig function with the smaller of the two odd 10 exponents) and place it next to the dx. Then use the trig identity sin 2 x cos 2 x 1 to express the remaining powers of sin x in terms of cos x:
sin 3 x cos 7 x dx
sin 2 x cos 7 x sin x dx
Use variable substitution with u
1 u 2 u 7 du
u7
1 cos 2 x cos7 x sin x dx
cos x and du
–sin x dx, and simplify:
u 9 du
Integrate and rewrite the result in terms of x:
1 8 u 8 6
1 10 u 10
11 3 sin 3 x 11
1 cos 8 x 8
C
17 6 sin 3 x 17
1 cos10 x C 10
23 3 sin 3 x 23
C . The cosine has an odd exponent, so peel off a cos x and
place it next to the dx. Then use the trig identity sin 2 x powers of cos x in terms of sin x, and simplify:
cos 5 x
3
sin 8 x dx
cos 4 x
3
sin 8 x cos x dx
Use variable substitution with u
1 u2
2
8
u 3 du
1 2u 2
1 sin 2 x
sin x and du 8
u3
2
1 to express the remaining 8
sin 3 x cos x dx
cos x dx:
8
u 4 u 3 du
cos 2 x
2u
14 3
u
20 3
du
Integrate and rewrite the result in terms of x: 11
3 3 u 11 7
6 u 17
3 u 11
23 3
11
C
3 sin 3 x 11
17
6 sin 3 x 17
3 sin 23
23 3
x C
3 1 sin 5 x sin 7 x C . Peel off a cos x and place it next to the dx. Then use 5 7 the trig identity sin 2 x cos 2 x 1 to express the remaining powers of cos x in terms of sin x: sin x
sin 3 x
cos 7 x dx
142
17 3
cos 7 x cos x dx
1 sin 2 x
PART 4 Advanced Integration Techniques
3
cos x dx
Use variable substitution with u 3
1 u2
1 3u 2
du
3u 4
sin x and du
cos x dx and simplify:
u 6 du
Integrate and rewrite the result in terms of x:
u u3 8
3 x 8
3 5 u 5
1 7 u 7
1 sin 2 x 4
C
sin x sin 3 x
1 sin 4 x 32
3 sin 5 x 5
1 sin 7 x C 7
C . Use the half-angle identity cos 2 x
1 cos 2 x to 2
rewrite the integral and simplify; then express as three separate integrals:
1 cos2 x 1 cos2 x 1 1 dx 1 2 cos2 x cos 2 2 x dx 1 dx 2 cos2 x dx cos 2 2 x dx 2 2 4 4 Evaluate the first and second integrals using methods you already know from earlier chapters: cos 4 x dx
1 x 4
cos 2 2 x dx
sin 2 x
Evaluate the remaining integral cos 2 2x dx using the trig identity cos 2 x
cos 2 2 x dx
1 cos 4 x dx 2
1 1 dx 2
1 cos 4 x dx 2
1 x 2
1 cos 2 x : 2
1 sin 4 x C 8
Substitute this result into the expression and simplify:
1 1 1 x sin 2 x x sin 4 x C 4 2 8 1 3 1 x sin 2 x sin 4 x C 4 2 8 3 1 1 x siin 2 x sin 4 x C 8 4 32 9
1 1 tan 9 x tan 11 x C . The exponent of sec x is an even number, so begin by peeling 9 11 off sec 2 x : sec 4 x tan 8 x dx
sec 2 x tan 8 x sec 2 x dx
Now, use the trig identity 1 tan 2 x tangents:
sec 2 x to express the remaining secants in terms of
1 tan 2 x tan 8 x sec 2 x dx Use variable substitution with u
1 u 2 u 8 du
tan x and du
sec 2 x dx and simplify:
u 8 + u 10 du
Integrate and rewrite the result in terms of x:
1 9 u 9
1 11 u C 11
1 tan 9 x 9
1 tan11 x C 11
CHAPTER 10 Trig Substitution
143
7
11 15 4 2 tan 2 x tan 2 x C . The exponent of sec x is an even number, so begin by 7 11 15 peeling off sec 2 x and expressing the tangent as an exponent:
10 2 tan 2 x
5
sec 4 x tan 2 x sec 2 x dx
sec 6 x tan 5 x dx
Now, use the trig identity 1 tan 2 x gent and simplify: 5
sec 4 x tan 2 x sec 2 x dx
1 tan 2 x
Use variable substitution with u
1 u2
2
5
2
5
tan 2 x sec 2 x dx sec 2 x dx and simplify:
tan x and du 5
1 2u 2
u 2 du
sec 2 x to express the remaining secants in terms of tan-
u 4 u 2 du
5
u2
9
2u 2
u
13 2
du
Integrate and rewrite the result in terms of x:
2 72 u 7 11
4 11 u2 11
2 152 u 15
7 2 tan 2 x 7
C
11 4 tan 2 x 11
15 2 tan 2 x C 15
1 1 sec 13 x sec 11 x C . This time, the exponent of tan x is an odd number, so peel off 13 11 sec 2 x tan 2 x and place it next to the dx: sec 11 x tan 3 x dx
sec 10 x tan 2 x sec x tan x dx
Now, use the trig identity tan 2 x secants:
sec 2 x 1 to express the remaining tangents in terms of
sec 10 x sec 2 x 1 sec x tan x dx Use variable substitution with u
u 10 u 2 1 du
u 12
sec x and du
sec x tan x dx and simplify:
u 10 du
Integrate and rewrite the result in terms of x:
1 13 u 13 12
2 sec
1 11 u C 11 1 2
x
4 sec 5
1 sec 13 x 13
1 sec 11 x C 11
5 2
9 2
x
2 sec 9
x
C . The exponent of tan x is an odd number, so express
the secant as an exponent and then peel off sec x tan x and place it next to the dx:
tan 5 x sec 9 x
144
dx
tan 5 x sec
9 2
x dx
tan 4 x sec
PART 4 Advanced Integration Techniques
11 2
x sec x tan x dx
Now, use the trig identity tan 2 x secants: 2
sec 2 x 1 sec
11 2
sec 2 x 1 to express the remaining tangents in terms of
x sec x tan x dx
Use variable substitution with u 2
u2 1 u
11 2 du
u4
sec x and du
2u 2 1 u
11 2 du
sec x tan x dx and simplify:
3 2
u
2u
7 2
u
11 2 du
Integrate and rewrite the result in terms of x:
2u
1 2
4 u 5
5 2
2 u 9
9 2
C
2 sec
1 2
4 sec 5
x
5 2
2 sec 9
x
9 2
x C
13 1 sec x tan x
1 ln sec x tan x C . The exponent of tan x is an odd number, so peel off 2 2 sec x tan x and place it next to the dx: tan 3 x dx
tan 2 x
1 sec x tan x dx sec x
Now, use the trig identity tan 2 x secants:
sec 2 x 1 to express the remaining tangents in terms of
sec 2 x 1 sec x tan x dx sec x Use variable substitution with u
u2 1 du u
u
sec x and du
sec x tan x dx and simplify:
1 du u
Integrate and rewrite the result in terms of x:
1 2 u 2
ln u
1 sec 2 x ln sec x 2
C
14 1 sec x tan x
2
1 ln sec x 2
tan x
C C . Use the trig identity tan 2 x
sec 2 x 1 to convert the
tangents to secants:
tan 2 x sec x dx
sec 2 x 1 sec x dx
sec 3 x dx
sec x dx
These are both integrals that you’ve seen before, so integrate them as shown earlier in the chapter and then combine like terms:
1 sec x tan x 2 1 sec x tan x 2
1 ln sec x tan x 2 1 ln sec x tan x 2
ln sec x
tan x
C
C
CHAPTER 10 Trig Substitution
145
x 1 9x 2 2
15 1 arcsin 3 x
6
C . To begin, notice that the function you want to integrate has
the following form:
1 9x 2
1 2
a2
bx 2
n
Thus, this function is the sine case. Set up a right triangle as follows:
From these values, calculate sin
3x 1 9x 2
sin 1 sin 3 1 cos 3 1 cos d 3
x
cos
and cos , then find dx in terms of
dx d dx
3x arcsin 3 x
and
in terms of x:
sin
Now, rewrite the integral in terms of :
1 9 x 2 dx
1 cos d 3
cos
1 cos 2 3
d
Use the method described earlier in this chapter, in “Integrating Powers of Sines and Cosines,” to evaluate the integral:
1 1 3 2
1 sin 2 4
C
1 6
Use the trig identity sin 2
1 6
1 sin 2 sin cos 12
1 sin 2 12
C
2 sin cos to rewrite this result: C
1 6
1 sin cos 6
To complete the problem, substitute values in terms of x:
1 arcsin 3 x 6
146
1 6
1 9x 2 3x
C
PART 4 Advanced Integration Techniques
1 arcsin 3 x 6
x 1 9x 2 2
C
16 1 ln 5 x
5
4 25 x 2 2
C . To begin, notice that the function you want to integrate has the
following form:
4 25 x 2
1 2
a2
bx 2
n
Thus, this function is the tangent case. Set up a right triangle as follows:
From these values, calculate tan
4 25 x 2 2 4 25 x 2 1 4 25 x 2
and sec , then find dx in terms of :
5x 2
sec
tan 2 tan 5 2 sec 2 5 2 sec 2 5
x
2 sec 1 2 sec
dx d dx
d
Now, rewrite the integral in terms of :
1 4 25 x 2
dx
1 2 sec
2 sec 2 5
d
1 sec d 5
Use the method described earlier in this chapter, in “Integrating Powers of Tangents and Secants,” to evaluate the integral:
1 ln sec 5
tan
C
1 ln 5
4 25 x 2 2
5x 2
C
1 5x ln 5
4 25 x 2 2
C
CHAPTER 10 Trig Substitution
147
17
1 ln 63
49 x 2 81 7x 9
C . To begin, notice that the function you want to integrate has the fol-
lowing form:
49 x 2
81
1
a2
bx 2
n
Thus, this function is the tangent case. Set up a right triangle as follows:
From these values, calculate tan
49 x 2 9
81
49 x 2 1
81
49 x 2 81 1 49 x 2 81
and sec , then find dx in terms of :
7x 9
tan
sec 9 sec 7 9 tan sec 7 9 tan sec d 7
x
9 tan 1 9 tan
dx d
1 81tan 2
dx
Now, rewrite the integral in terms of :
1 dx 49 x 2 81
1 81tan 2
9 tan sec d 7
1 1 63 tan
sec d
1 csc d 63
Evaluate this integral and rewrite in terms of x using the values of csc x and cot x from the right triangle:
1 ln csc x cot x 63
C
1 ln 63
7x
9
2
2
49 x
81
49 x
81
1 ln 63
C
7x 9 49 x 2
81
C
Here, the function inside the natural log can be rationalized and simplified as follows:
7 x 9 49 x 2 81 1 ln 63 49 x 2 81
148
C
7 x 9 49 x 2 81 1 ln 63 7x 9 7x 9
PART 4 Advanced Integration Techniques
C
1 ln 63
49 x 2 81 7x 9
C
18
x 9 x 2 16 144 x 2 256
C . To begin, notice that the function you want to integrate has the following
form: 3 2
9 x 2 16
a2
bx 2
n
Thus, this function is the tangent case. Set up a right triangle as follows:
From these values, calculate tan
9 x 2 16 4
tan
9 x 2 16
4 tan
9x
2
3x 4 x
16 16 tan
9 x 2 16
3 2
2
dx d
64 tan 3
3 2
9 x 2 16
and sec , then find dx in terms of :
dx
1 64 tan 3
sec 4 sec 3 4 tan sec 3 4 tan sec d 3
Now, rewrite the integral in terms of : 3 2
9 x 2 16
1 64 tan 3
dx
4 tan sec d 3
1 1 48 tan 2
sec d
1 cos 48 sin 2
d
Evaluate this integral as shown earlier in this chapter, in “Integrating Powers of Sines and Cosines,” using u sin and du cos d :
1 1 du 48 u 2
1 48u
C
1 48 sin
1 csc 48
C
C
Use the right triangle to rewrite csc in terms of x:
1 48
3x 9 x 2 16
C
x 16 9 x 2 16
C
x 9 x 2 16 16 9 x 2 16
C
x 9 x 2 16 144 x 2 256
C
CHAPTER 10 Trig Substitution
149
19
1 ln 750
25 36 x 2 25 6 x
C . To begin, notice that the function you want to integrate has the
following form:
25 36 x 2
1
a2
bx 2
n
Thus, this function is the sine case. Set up a right triangle as follows:
From these values, calculate sin
and cos , then find dx in terms of :
6x 5
25 36 x 2 25
cos
25 36 x 2
25 cos
25 36 x 2
625 cos 2
sin 5 sin 6 5 cos 6 5 cos d 6
x dx d dx
Now, rewrite the integral in terms of :
1 dx 25 36 x 2
1 625 cos 2
5 cos d 6
1 1 d 750 cos 25
Evaluate the integral and rewrite sec
1 ln sec 750
tan
C
1 ln 750
25
1 sec d 750
and tan
25 36 x 2 6x
25 36 x 2
25 36 x 2
C
6x
:
25 36 x 2 1 25 6 x ln 750 25 36 x 2
C
In this case, the value inside the natural log can be rationalized and simplified:
25 6 x 25 36 x 2 1 ln 750 25 36 x 2
150
C
25 6 x 25 36 x 2 1 ln 750 25 6 x 25 6 x
PART 4 Advanced Integration Techniques
C
1 ln 750
25 36 x 2 25 6 x
C
IN THIS CHAPTER
»» Breaking fractions into the sum of two or more partial fractions »» Extending partial fractions to rational expressions »» Using partial fractions to integrate rational expressions that have distinct linear factors »» Integrating with partial fractions for rational expressions that have repeated linear factors »» Persevering while integrating rational expressions that have distinct quadratic factors »» Setting up difficult partial sums that have repeated quadratic factors
11 Integration with Partial Fractions Chapter
N
o Calculus II class would be complete without a unit on partial fractions. Well, actually, you might get lucky: Some calculus teachers opt out of this particular form of torture, preferring to focus on a greater variety of applications of integrals (see Chapters 12 and 13). Either way, you’re still going to be doing a lot of integration for the foreseeable future. So, if partial fractions are in the cards for you (check your course syllabus, about halfway down the page), this chapter should help to get you through.
CHAPTER 11 Integration with Partial Fractions
151
Understanding Partial Fractions To start, just think about fractions: You can change any fraction to a sum of partial fractions based on the factors in the denominator of that fraction. For example:
Q.
Find any pair of integers A and B that make the equation
A.
One possible pair is A = 1 and B = 2. As you can see, the problem starts by splitting up the fraction
9 10
A 2
B true. 5
9 into a pair of fractions with denominators of 2 and 5, which are factors 10
of the denominator 10. If you spend a moment noodling around with this equation to find possible integer values for A and B, you’ll probably come up with A = 1 and B = 2 as the most obvious solution. Thus,
9 10
1 2
2 . Notice, however, that the solution A = –3 and B = 12 also 5
works just as well. In fact, there are infinitely many pairs of integer solutions to this equation. To show you why this is so, I’ll use a common denominator of 10 to add the two fractions:
9 10
A 2
B 5
5A 10
2B 10
5 A 2B 10
Next, I set the first and last versions of the fractions equal and multiply both sides by 10:
9 10 9
5 A 2B 10 5 A 2B
The resulting equation, 9 5 A 2 B, produces infinitely many pairs of A and B that satisfy the original equation, such as the two pairs that I’ve already given. This general idea of splitting a fraction into a sum of fractions carries over into rational expressions — in other words, expressions that include polynomials in both the numerator and denominator. For example:
Q. A.
2
Find the only pair of numbers A and B that make the equation x x for all values of x.
1
A x
B true x 1
A = 2 and B = –2. In this case, I’ve highlighted new elements of the problem in boldface.
2 is split into a sum of rational expressions, again x x 1 with denominators that are the two factors of the original denominator x x 1 . Here, the rational expression
To solve this problem, I use the same basic strategy of adding by using a common denominator:
2 x x 1
152
A x
B x 1
A x 1 x x 1
PART 4 Advanced Integration Techniques
Bx x x 1
Ax A Bx x x 1
And again, I set this result equal to the original expression and then multiply out the denominator:
2 x x 1 2
Ax A Bx x x 1 Ax A Bx
In this case, however, the equation is true for all values of x. Thus, when x = 0, the resulting equation is:
A 0
2
A B 0
A
You can now plug in 2 for A back into the equation and solve for B:
2 Ax A Bx 2 2 x 2 Bx 0 2 x Bx 0 2 B 2
B
To complete the problem, plug in 2 for A and –2 for B as the solution:
2 x x 1
2 x
2 x 1
Notice that I’ve simplified this answer a bit by placing the minus sign between the terms instead of in the numerator of the second term.
1
Find any pair of integers A and B that make the equation
32 35
A 5
B true. 7
2
Find a pair of fractions that add up to
17 . 21
Hint: Think about how the two factors of 21 might help you.
CHAPTER 11 Integration with Partial Fractions
153
3
4
Find the only pair of numbers A and B that
A x
4
make the equation x x for all values of x.
2
Express
B true x 2
2 as the sum of two rational x x 3
expressions. Hint: Think about how the two factors of x x 3 might help you.
Integrating Case 1: Rational Expressions That Have Distinct Linear Factors Your insights about partial sums from the previous section give you a reliable method for integrating rational expressions. For the moment, you need to limit this method to rational expressions whose denominators have distinct linear factors.
Q. A.
1 dx x x 1 ln x ln x 1 C . The rational expression you want to integrate has a denominator with two distinct linear factors, x and x – 1, so you can apply the method outlined in the previous section: 1 dx x x 1
A x
B dx x 1
1 x
1 dx x 1
Now, you can separate this integral into two separate integrals, and evaluate both using the method of integrating linear inputs that you already know from Chapter 7:
1 dx x
154
1 dx x 1
ln x
PART 4 Advanced Integration Techniques
ln x 1
C
5
7
1 dx x x 4
6
x 6 dx x 1 x 1 x 2
8
1
dx
x
3
x 2
x
3
x 3 dx x 1 x 4
Integrating Case 2: Repeated Linear Factors When the denominator of a rational expression has one or more pairs of repeated linear factors, you need to set up the partial fractions in a slightly different way. For each linear factor raised to the power of 2, set up the following sum:
A linear factor
B ( linear factor ) 2
For each linear factor raised to the power of 3, set up the following sum:
A linear factor
B ( linear factor ) 2
C ( linear factor ) 3
Generally speaking, for each linear factor raised to the power of n, set up a sum of n separate expressions using the pattern shown here.
CHAPTER 11 Integration with Partial Fractions
155
Q. A.
x 2 dx x 3 2 ln x
5
3
C . This rational expression has two repeated factors of x – 3, so set it
3
x
equal to the sum of two separate rational expressions, then add them and simplify, as in the example in the previous section:
x 2 x 3 2
A x 3
B x 3
A x 3 x 3
2
B
Ax 3 A B x 3 2
2
Pull out the usual equation from this result:
x 2
Ax 3 A B
You can solve this equation by noticing that x on the left side equals Ax on the right side, so A = 1. Thus, if you plug in 1 for A, the equation solves as B = 5. Now, rewrite the original integral using these values.
x 2 dx x 3 2
A x 3
B x 3
2
5 x 3
1 x 3
dx
2
dx
Evaluate the first term of the integral using the methods from the previous section, and the second term as a linear input to a power function with an exponent of –2, as I show you in Chapter 7:
1 dx x 3
2
x 3
5
dx
ln x 3
5 x 3
1
C
ln x 3
5 x 3
C
I’m sincerely hoping that your teacher won’t ask you to integrate a rational expression with a linear factor raised to a power of 3 or higher. More likely, you might get a question that asks you to do the setup for integrating repeated linear factors raised to a high-number power. For example:
Q.
How would you set up an equivalent sum of partial fractions to evaluate the integral
5 2
x 2
x
3
3
dx ? (Use as many capital letters as you need – A, B, C, and so on —
and then leave your answer in this form.)
A.
A x
2
x
B 2
C 2
x
3
x
D 3
2
x
E 3
3
. This question asks you only to set up the
partial fractions without worrying about the rest of the problem, so just do what your nice teacher is asking:
5 x 2
156
2
x
3
3
A x 2
B x 2
PART 4 Advanced Integration Techniques
C 2
x
3
x
D 3
E 2
x
3
3
9
11
x 5 x 1 2
10
Set up but don’t solve or simplify an equiva-
x 3 7x 5 lent form of the rational function x 4 5 as a sum of partial fractions.
12
2x 3 x 4 2
How would you prepare to write a sum of partial fractions to assist you in evaluating the integral
x 6
x5 x 7 x
8
2
x 9
3
?
(Don’t do anything I wouldn’t do, and I definitely wouldn’t try to evaluate this integral without an app, but please do show me the setup phase, using capital letters A, B, C, and so forth.)
Integrating Case 3: Distinct Quadratic or Higher-Degree Factors When the denominator of a rational expression has one or more distinct quadratic, cubic, or higher-degree factors, you need to set up the partial fractions with a bit more care. For example, for each distinct quadratic factor, add a term to your sum as follows:
A Bx quadratic factor
CHAPTER 11 Integration with Partial Fractions
157
For each distinct cubic factor, add a term that looks like this:
A Bx Cx 2 cubic factor Generally speaking, for each distinct higher-degree factor in the denominator, add a term corresponding to the pattern shown here.
Q. A.
3x 2 dx ( x 1) x 2 4 1 ln x 1 5
1 ln x 2 10
7 x arctan 5 n
4
C . The denominator of this rational expression
has two distinct repeated factors, one linear and the other quadratic, so set it equal to the sum of two separate rational expressions and then perform the usual calculations:
3x 2 ( x 1) x 2 4
A
Bx x2
x 1
C 4
A x2
4
Bx
( x 1) x
C ( x 1)
2
Ax 2
4 A Bx 2 Bx Cx C ( x 1) x 2 4
4
The result is the following equation:
3x 2
Ax 2
4 A Bx 2
Bx Cx C
Because this expression is true for all values of x, you can equate coefficients — that is, break it into three separate equations, one including only constants, another only x-terms, and a third with x 2-terms:
2
4A C
3x 3
0x 2 0
Bx Cx B C
The result is a system of equations with the solution A
Ax 2 Bx 2 A B 1 ,B 5
1 , and C 5
rewrite the original integral using these values and then simplify:
3x 2 dx ( x 1) x 2 4
A x 1
Bx C x2 4
1 5
x 1
1 14 x 5 5 dx x2 4
1 5 x 1
14 , so 5
x 14 dx 5 x2 4
Now, split the second rational expression into two separate terms and then split the resulting integral into three separate integrals:
1 5 x 1
x 5 x2 4
14 dx 5 x2 4
1 dx 5 x 1
x dx 5 x2 4
14 dx 5 x2 4
Evaluate the first term of the integral using the method from the previous section, and the second using variable substitution with u x 2 4 (see Chapter 8):
1 ln x 1 5
158
1 ln x 2 10
4
14 dx 5 x2 4
PART 4 Advanced Integration Techniques
To solve the third integral, move the coefficient outside the integral and then use the formula
x 1 dx arctan C: n n n2 1 14 1 ln x 2 4 dx 10 5 x2 4 x 1 7 ln x 2 4 arctan C 10 5 2
1 x2
1 ln x 1 5 1 ln x 1 5
I know, that’s a lot! A bit of good news at this juncture is that almost any teacher these days who asks you to set up partial fractions for a higher-degree denominator probably won’t force you to integrate the result. For example:
Q.
How would you set up an equivalent sum of partial fractions to evaluate the integral
1 x
2
x
2
3
x5
3
5
dx ? (Use as many capital letters as you need — A, B, C, and
so on — and then leave your answer in this form.)
A.
Ax x2
B 2
Cx 2 Dx E x3 3
Fx 4
Gx 3 Hx 2 x5 5
Ix
J
. This question specifically asks you
to set up the partial fractions without worrying about the rest of the problem, so here’s your answer:
1 x
13
x 12 dx x x2 4
2
2
x
3
3
x
5
5
Ax x2
B 2
Cx 2 Dx E x3 3
14
Fx 4
Gx 3 Hx 2 x5 5
Ix
J
Just do the partial sums setup step for the integral
17 x x 1 x 9
2
x3
5
x4
CHAPTER 11 Integration with Partial Fractions
3
dx .
159
Integrating Case 4: Repeated Quadratic Factors There’s no whitewashing it: Integration with partial fractions when the denominator includes one or more repeated quadratic factors is just plain long, hard, and tedious. You can probably tell this from the setup: For each squared quadratic factor, add the following pair of terms to your sum:
Ax B quadratic factor
Cx D ( quadratic factor ) 2
For each cubed quadratic factor, add these three terms:
Ax B quadratic factor
Cx D ( quadratic factor ) 2
Ex F ( quadratic factor) 3
And so on. In Calculus II For Dummies, I show you an example of how to evaluate an integral like this, and it requires a lot of algebra. Hopefully, you’ll only have to set up but not solve an integral with a repeated quadratic factor.
Q.
How would you set up a sum of partial fractions to evaluate
4x 3 3x 2 5 dx ? (Use as ( x 2 x 1) 3
many capital letters as you need — A, B, C, and so on — and then leave your answer in this form.)
A.
Ax B x2 x 1
Cx x
2
D x 1
Ex 2
x
2
F x 1
3
. Use the form for a quadratic raised to the third
power.
15
Do the partial sums setup step for
x3 5 dx . ( x 2 7) 2
160
PART 4 Advanced Integration Techniques
16
Set up partial sums for
4 x 3 2x 2 6x 5 x 1 x 2 2 x2 3 (x2
4 )2
dx .
Answers and Explanations 1
32 35
1 5
5 A . Begin by adding 7 5
32 35
A 5
B 7
B using a common denominator of 35: 7
7 A 5B 35
Now, set the first and last parts of this equation equal, and then simplify by multiplying both sides by 35:
32 7 A 5 B 35 35 32 7 A 5 B To find an integer solution, plug in 1 for A and see if the result produces an integer value for B:
32 7 1 25 5 B 5 B
5B
Therefore, the solution A = 1 and B = 5 is one of the infinitely many integer solutions. 2 17 21
2 3
1 . Use the values 3 and 7 as your denominators, and repeat the process used in 7
Question 1:
17 A B 7 A 3 B 21 3 7 21 17 7 A 3 B To find an integer solution, plugging in A = 1 doesn’t produce an integer value for B, but A = 2 does:
17 7 2 3 3B 1 B
3B
Therefore, the solution A = 2 and B = 1 is one of the infinitely many integer solutions. 3
4 x x 2 x x 2 : 4 x x 2
2 x
A x
2 . Begin by adding the two values using a common denominator of x 2
B x 2
A x 2 x x 2
Bx x x 2
Ax 2 A Bx x x 2
Set the first and last parts of this chain equivalent, and then multiply both by the denominator to simplify:
4 Ax 2 A Bx x x 2 x x 2 4 Ax 2 A Bx
CHAPTER 11 Integration with Partial Fractions
161
Let x = 0 and solve for A:
4 4 A
A 0 2A 2
2A B 0
Now, plug in –2 for A into the equation and solve for B:
4 2 x 2( 2 ) Bx 4 2 x 4 Bx 0 2 x Bx Bx 2x B 2 Thus, A = –2 and B = 2 is the only solution to this equation that works for all values of x, so plug these numbers into the original problem and simplify:
4
4 x x 2
2 x
2 x 2
2 x x 3
2 3x
2 x x 3
A x
2 x
2 x 2
2 . Set up the problem as follows: 3 x 3 A x 3 x x 3
B x
3
Bx x x 3
Ax 3 A Bx x x 3
Make an equation from the first and last expressions here, and multiply by the denominator to simplify:
2 x x 3 2
Ax 3 A Bx x x 3 Ax 3 A Bx
Solve this equation for A by plugging in 0 for x, then solve it for B by plugging in
2 A
2 2 x 3 3 3 2 B 3
2
3A 2 3
Thus, A
2 and B 3
2 for A: 3
Bx
2 is the only solution that works for all values of x, so plug these num3
bers into the original problem and simplify:
2 x x 3 5
4 dx x
2 3 x 3
2 3 x 4 x
4
2 3x
2 3 x 3
1 ln x 4
dx
1 ln x 4
4
C . Rewrite the rational expression you’re try-
ing to integrate as a sum of partial fractions:
1 dx x x 4
162
A x
B x
4
dx
1 4x
PART 4 Advanced Integration Techniques
1 dx 4 x 4
Separate this integral into two parts and integrate each using the method explained in the example for this section:
4 dx x 6
4 x
4
1 x
3
x 2
1 ln x 4
dx
1 ln x 4
1 ln x 5
dx
C
4
1 ln x 2 5
3
C . Rewrite the rational expression you’re
trying to integrate as a sum of partial fractions:
A 5 x 3
1 x
x 2
3
B dx 5 x 2
1 5 x 3
1 dx 5 x 2
Separate this integral and integrate both parts as in the previous problem:
1 5 7
x 1
1 x
3
dx
x 6 x 1
1 5
1 dx x 2 dx
x 2
1 ln x 5
5 ln x 1 6
3
1 ln x 2 5
7 ln x 1 2
C
3 ln x 2 8
C . Rewrite the rational
expression you’re trying to integrate as a sum of partial fractions using the common denominator x 1 x 1 x 2 :
x 6 x 1 x 1 x 2
A x 1
B x 1
C x 2
A x 1 x 2
B x 1 x 2 C x 1 x 1 x 1 x 1 x 2
Set the first and last parts equal and cancel the denominators, but leave the right side of the equation undistributed:
x 6 x 1 x 1 x 2 x 6
A x 1 x 2
B x 1 x 2 C x 1 x 1 x 1 x 1 x 2 A x 1 x 2 B x 1 x 2 C x 1 x 1
To solve for A, B, and C, set x successively to –1, 1, and 2 — that is, any of the three values that will zero out a bunch of terms:
1 6 A 1 1 5 6A 5 A 6
1 2
1 6 7 B
B 1 1 1 2 2B 7 2
2 6 C 2 1 2 1 8 3C 3 C 8
Plug these numbers for A, B, and C into the integral and simplify:
x 6 dx x 1 x 1 x 2
5 6
x 1
2 7 x 1
3 8
x 2
dx
5 6 x 1
2 7 x 1
3 dx 8 x 2
Separate this integral into three parts and integrate each using the method you’ve been using throughout this chapter:
5 1 dx 6 x 1
2 1 dx 7 x 1
3 1 dx 8 x 2
5 ln x 1 6
2 ln x 1 7
3 ln x 2 8
C
CHAPTER 11 Integration with Partial Fractions
163
8
x 3 x 1
3
x
3 ln x 14
dx
4
x
1 ln x 1 6
3
1 ln x 21
4
C . Rewrite the rational
expression you’re trying to integrate as a sum of partial fractions using the common denominator x 3 x 1 x 4 :
x
x 3 x 1 x 4
3
A x
B x 1
3
A x 1 x 4
C x 4
x
B x 3 x 4 C x 3 x 1 x 4
3
x 1
This equation simplifies as follows:
x 3
A x 1 x 4
B x
x 4
3
C x
3
x 1
To solve for A, B, and C, follow the procedure shown in the answer to Question 6, setting x successively to –3, 1, and 4. This results in the solution A these numbers for A, B, and C into the integral and simplify:
x
3 14 x 3
x 3 dx 3 x 1 x 4
1 6
1 21 dx x 4
x 1
3 ,B 14
3 14 x 3
1 , and C 6
1 6 x 1
1 , so plug 21
1 dx 21 x 4
Separate this integral into three parts and integrate each using the method you’ve been using throughout this chapter:
3 1 dx 14 x 3 9
x 5 2 x 1
1 1 dx 6 x 1
1 1 dx 21 x 4
6 x 1
ln x 1
3 ln x 14
3
1 ln x 1 6
1 ln x 4 21
C
C . Rewrite the rational expression you’re trying to integrate as
a sum of partial fractions:
x 5 x 1 2
A x 1
B x 1
A x 1 x 1
2
B
Ax A B x 1 2
2
Set the original expression equal to the result:
x 5 x 1 2 x 5
Ax A B x 1 2 Ax A B
This equation solves relatively easily by methods you already know as A = 1 and B = –6, so plug these values into the original integral and solve it as shown in the example for this section:
10
x 5 x 1 2
1 x 1
6 x 1
2x 3 2 x 4
2 ln x
4
2
dx 5
x
4
ln x 1
6 C x 1
C . Rewrite the rational expression you’re trying to integrate
as a sum of partial fractions:
2x 3 x 4 2
164
A x
4
x
B 4
2
A x 4 x 4
PART 4 Advanced Integration Techniques
B 2
Ax
4A B x 1 2
Set the original expression equal to the result:
2x 3 x 4 2 2x 3
Ax
4A B x 1 2 Ax 4 A B
Pull out two separate equations, one with the x-terms and the second with the constant terms:
2 x Ax A 2
4A B
3
Plug in 2 for A into the second equation:
B
3 4 2 B 5
To finish, plug in 2 for A and –5 for B into the original integral and solve it as shown in the example for this section:
2x 3 x 4 2 11
12
2 x
5 4
x
2
4
B C 2 3 x 4 x 4 x 3 7x 5 A B x 4 x 4 5 x 4
dx
A
A x
B 6
x
C 7
x
8
D 8
x
x
x
E 2
F 9
x
C
4
E . Set up the sum in this way: 5 4 D E x 4 4 x 4 5
x 2
5
4
D 4 4 C x 4 3
4
x
2 ln x
x
9
2
x
G 9
3
. Set up the partial fractions
using your knowledge of distinct and repeated linear factors:
x 6 13
x5 x 7 x
x 12 x x2 4
dx
8
2
x 9
3
A x 6
3 ln x 2 2
3 ln x
B x 7
C x
8
1 x arctan 2 2
4
x
D 8
2
E x 9
F x 9
2
G x 9
3
C . The denominator of this rational
expression has two distinct repeated factors, one linear and the other quadratic, so set it equal to the sum of two separate rational expressions and then perform the usual calculations:
x 12 x x2 4
A x
Bx C x2 4
A x2
Bx C x
4 x x
2
4
Ax 2
4 A Bx 2 x x2 4
Cx
The result is the following equation:
x 6
Ax 2
2 A Bx 2
Cx
CHAPTER 11 Integration with Partial Fractions
165
Equate coefficients as follows:
x C
4A 3
12 A
x 12 dx x x2 4
A x
0x 2 0 0 B
Cx 1
Bx C dx x2 4
Ax 2 Bx 2 A B 3 B 3
3x 1 dx x2 4
3 x
Split the second rational expression into two separate terms and then split the resulting integral into three separate integrals:
1 dx x
3
x dx x2 4
3
1 dx x2 4
Evaluate the first term of the integral as usual, the second using variable substitution with
x2
u
2 (see Chapter 8), and the third using the formula
3 ln x 14 A
x
3 ln x 2 2
B x 1
C x
1 x arctan 2 2
4
D x 9
9
2
1 x2
n2
dx
x 1 arctan n n
C:
C
Ex 2 Fx G x3 5
Hx 3
Ix 2 Jx x4 3
K
. This problem draws upon
all the skills you’ve learned in this chapter so far for setting up partial fractions. 15
16
Ax x2 A x 1
B 7
Cx
D
x2
7
B x 2
2
. This is the setup for the repeated quadratic case.
C x 2
2
Dx x2
E 3
Fx G x2 4
cases explained throughout the chapter.
166
PART 4 Advanced Integration Techniques
Hx x2
I 4
2
. This problem comprises all four setup
5
Applications of Integrals
IN THIS PART . . .
Calculate the area between curves on the xy-graph Find the arclength of functions Solve tricky volume of revolution problems Integrate to solve separable differential equations
IN THIS CHAPTER
»» Measuring area under two different functions »» Understanding both types of improper integrals »» Finding the unsigned area between curves »» Calculating the mean value of a region »» Using the Arclength formula
12 Solving Area Problems Chapter
I
n this chapter, you use your knowledge of integration in a variety of applied problems. To begin, you measure the area of a region on the xy-graph bounded above by two different functions. Next, you discover how to distinguish the two types of improper integrals and calculate each using a limit. After that, you use a simple formula for determining the unsigned area between a pair of functions. Next, you use the Mean Value formula to calculate the mean value of an integral on a given interval. To finish up, you measure the exact length along a curve using the Arclength formula.
Breaking Definite Integrals in Two When a region on an xy-graph is bounded above by two different functions, you need to take special care to measure the area of that region. For example:
CHAPTER 12 Solving Area Problems
169
Q. A.
What is the area of the shaded region shown in the graph?
2 e 1. The shaded region includes the area under the two functions f ( x ) e x and 3 g ( x ) e( x 2 ) 2 from x 0 to x 2. As shown in the graph, these two functions intersect when x 1. To find the shaded area, break the region into two separate regions and integrate their sum as follows: 1 0
2
e x dx
1
e x 2
2
dx
Evaluate to find the area (remember that in the second integral, e is just a constant):
ex
x 1 x 0
1 e x 2 3
3
x 2
e 1
x 1
1 e 2 2 3
3
1 e 1 2 3
3
e 1
1 e 3
1 e 3
5 e 1 3.530 3
In some cases, key elements of the shaded area (such as the x-intercepts or the intersection point of the two functions) are left out to make the problem more challenging. Break out your algebra skills to find this information before setting up your integrals.
Q.
What is the area of the shaded region shown in the graph?
A.
43 6 ? 0
7.167 . To begin, set up the problem as follows:
x2
3 x dx
? ?
x2
2x
3 dx
In this problem, you can see pretty easily that f(x) has an x-intercept at the origin, so this is the lower bound of the first integral. To find the upper bound of the second integral, set g(x) to 0 and solve. And to find the remaining bounds of integration, set the two functions equal and solve:
x2 x
170
2x
3
2
x2
0
2x 3 x 1 x 3
0 0
x
3,
1
PART 5 Applications of Integrals
x2
3x
2
2x x 3 x 1 2x 3
0 0
x
1,
2x
3 2
3
Thus, the upper bound of the second integral is 3 and the remaining bounds of integration are both 1: 1 0
x2
3
3 x dx
1
x2
2x
3 dx
Evaluate the integrals:
1 3 x 3
3 2 x 2
x 1
1 3 x 3
x 0
x2
3x
x 3 x 1
Simplify:
1 3
3 2
0
1 3 3
3
3
2
1 1 3 3
3 3
11 6
16 3
43 6
7.167
In the graphs below, find each of the shaded areas.
1
2
3
4
CHAPTER 12 Solving Area Problems
171
Looking at Improper Integrals An improper integral includes integration along either a horizontal or a vertical asymptote. These come in two varieties, often called Type 1 and Type 2. A Type 1 (or horizontally infinite) improper integral is easy to spot, because it has either ∞ or –∞ (or both) as a bound of integration. To evaluate this type of integral, replace ∞ with a constant such as k and find the limit of the integral.
Q.
Evaluate
A.
1 . To begin, rewrite this integral by substituting the constant k for ∞, placing the inte72
1 dx 4 3 2 x
gral inside a limit: k
1 dx 4 2 3x
lim
k
2
1 dx 3x 4
Now, evaluate the integral in terms of k:
1 9x 3
x k
1 9k 3
x 2
1 9 23
1 9k 3
1 72
Substitute this result into the limit and evaluate:
lim
k
1 9k 3
1 72
1 72
A Type 2 (or vertically infinite) improper integral includes a vertical asymptote either as one or both of its bounds of integration or lurking between them. These types of integrals can be harder to pick out because they masquerade as innocent proper integrals.
Q. A.
Evaluate
1
3 dx . x
0
6. This integral has a lower bound of 0, which is an asymptote, so this is a Type 2 improper integral. To fix this problem, let k 0 and place the integral inside a one-sided limit: 1 0
3 dx x
1
lim
k
0 k
3 dx x
Notice that when working with a lower bound, I use a one-sided limit from the left. Now, evaluate the integral in terms of k: 1 k
172
3 dx x
6 x
x 1 x k
6 6 k
PART 5 Applications of Integrals
Substitute this result into the limit and evaluate:
lim 6 6 k
k
Q. A.
6
0
8
Evaluate
8
1 3
dx
x2
4. In this case, neither bound of integration is an asymptote, but an asymptote resides at x 0. To fix this problem, break the integral into two separate integrals along this asymptote: 8
0
1 3
8
x
dx
2
8
8
1 3
x
2
dx 0
1 3
x2
dx
These two separate integrals are symmetrical halves of the original integral, so you can rewrite this sum as the product of 2 and the second integral: 8
2 0
1 3
x2
dx
The lower bound at x 0 is a vertical asymptote, so this is a Type 2 improper integral. Rewrite it by substituting the constant k for this lower bound, placing the limit inside a one-sided limit approaching 0 from the left: 8
2 lim k
0 k 3
1
dx
x2
Here, I moved the constant 2 outside the limit. Now, evaluate the integral in terms of k: 1 x 8
3x 3
3 8
1 3
1
1
3k 3
2 3k 3
x k
Substitute this result into the limit and evaluate: 1
2 lim 2 3k 3 k
5 5
1 dx x2
0
2 2
4
6
2
e 3 x dx
CHAPTER 12 Solving Area Problems
173
7
32 0
1 5
x
2
8
dx
4 0
x 3
x
2
4
dx
Finding the Unsigned Area between Curves Another common application of integration is finding the unsigned area between a pair of functions on the xy-graph. Here’s the formula:
Let f(x) and g(x) be two functions such that f ( x ) g ( x ) on an interval from a to b. Then, the unsigned area between these two functions on this interval is given by the following formula.
Unsigned area
b a
f ( x ) dx
b a
g ( x ) dx
Q.
What is the area of the shaded region shown in the graph?
A.
2
1 . To begin, set up the problem as follows, with the top function f(x) minus the 3
bottom function g(x): b a
174
f ( x ) dx
b a
g ( x ) dx
PART 5 Applications of Integrals
1 0
sin
2
x dx
1 0
x 2 dx
In this case, the integration is the easy part:
2
cos
2
x 1
x
x 0
1 3 x 3
x 1 x 0
Now evaluate this result carefully. The formula produces a lot of minus signs, so I set it up with extra parentheses to make sure I don’t flip a sign along the way:
2
cos
2 2
1 1 3
cos 0
1 0 3
3
3
2
1 3
You can use the same formula for finding the area between a pair of curves that are on opposite sides of the x-axis. The formula automatically produces the unsigned area, which is the sum of the positive areas both above and below the x-axis. In most cases, problems of this type ask for the unsigned area, but always make sure that’s what your professor wants you to find.
Q.
What is the unsigned area of the shaded region shown in the graph?
A.
9 2
1.432. Use the formula to set up the problem as follows, with the top function f(x)
minus the bottom function g(x): b a
f ( x ) dx
b a
3 2 0
g ( x ) dx
3 22 0
3 cos x dx 2 3
x 1
1 dx 2
2
Here again, the integration isn’t too bad:
9 sin x 2 3
x
3 2
2 x 1 3
x 0
3
1 x 2
x
3 2
x 0
Again, be careful when evaluating. Here, I take a few extra steps and write everything down to avoid making a careless mistake while adding or subtracting negative numbers:
9 sin 2 2 9 2
0
9 sin 0 2 2 1 3 2
3
3 4
2 3 1 3 2 2 3
1
3
3
3 4 9 2
2 0 1 3 2 3
2 3
3
0 9 2
1.432
CHAPTER 12 Solving Area Problems
175
9
10
11
12
176
PART 5 Applications of Integrals
Using the Mean Value Formula for Integration In Calculus I, you discovered the Mean Value Theorem for differentiation. This theorem ensures that for any differentiable function on a closed interval, there exists at least one point on that interval whose derivative equals the mean value of the function on that interval. Integration has its own Mean Value Theorem, which leads to the Mean Value formula for integration:
f (c)
1 b a
b a
f ( x ) dx
Here, note that f ( c ) is simply a numerical value that expresses the average value of the definite integral. In a nutshell, the formula tells you that if you divide the value of a definite integral b a
f ( x ) dx by its width, b – a, the result is the average height of the region expressed by that
definite integral. It also tells you that for at least one point c on the closed interval [a, b], the value of f(c) equals this average height. An example should help make this idea clear:
Q.
What is the mean value of the definite integral
A.
4
2 0
2 cos x dx ?
. Plug these values into the Mean Value formula:
f (c)
1 b a
b a
1
f ( x ) dx 2
0
2 0
2 cos x dx
Evaluate and simplify:
2
2 sin x
x
2 x 0
2
2 sin
2
2 sin 0
4
Note that this result, which equals about 1.273, is within the range of the function 2cos x on the given interval.
13
What is the mean value of the function f ( x )
x2
5 x on the interval [0, 5]?
CHAPTER 12 Solving Area Problems
177
14
What is the mean value of the function f ( x )
15
What is the mean value of the function
f(x)
3 ln x on the interval [1, 4]? x
2e x
1 on the interval [–5, 0]?
16
What is the mean value of the function
f ( x ) 3 cos
x on the interval [ 4
,
]?
Calculating Arclength The arclength of a function on a given interval is the length from the starting point to the ending point as measured along the graph of that function. You can use calculus to find the arclength of a function using the following formula:
Arclength =
178
b a
1
dy dx
PART 5 Applications of Integrals
2
dx
Q.
What is the arclength of the function y
A.
3 2 19 2 27
1
3
2 x 2 on the interval [0, 2]?
6.061. To begin, calculate the derivative of the function and then square
the result: 3
y
2x 2 1
dy dx dy 2 dx
3x 2 9x
Now, plug this value into the formula, along with the bounds of integration 0 and 2:
Arclength =
b a
dy dx
1
2
2
dx
1 9 x dx
0
You can solve this integral pretty easily with the u-substitution u u 19 1 u 1
9
1 9 x and du
9 dx , so:
1
u 2 du
Notice before moving on that I rewrite the bounds of integration explicitly in terms of u before integrating. Now, I’m ready to evaluate the integral: 2 0
3 u 19
2 2 u 27
1 9 x dx
3
2 19 2 27
u 1
1
6.061
In some cases, applying the Arclength formula may require you to break out those trig substitution skills (see Chapter 10), especially the tangent case:
Q.
What is the arclength of the function y
3 x 2 on the interval [–1, 1]?
A.
1 12
6 37
ln
37
6
6 37
ln
37
6
6.498 . Calculate the derivative of the
function and square it:
y dy dx dy 2 dx
3x 2 6x 36 x 2
Now, plug this value into the formula, along with the bounds of integration 0 and 2:
Arclength =
b a
1
dy dx
2
dx
1 1
1 36 x 2 dx
CHAPTER 12 Solving Area Problems
179
This result is a textbook opportunity to use the tangent case of trig substitution (see Chapter 10). Begin by calculating the tangent and secant functions as follows:
6x
1 36 x 2
sec
tan 1 tan 6 1 sec 2 6
x dx
d
Now, I rewrite the definite integral that I want to evaluate as an indefinite integral, and then change its terms from x to :
1 36 x 2 dx
1 sec 2 6
sec
1 sec 3 6
d
d
This is a hairy integration that I cover in Chapter 10. I evaluate it and then rewrite the result in terms of x:
1 ln sec 12
tan
C
tan sec
1 ln 1 36 x 2 12
6 x 1 36 x 2
6x
C
Now, I’m ready to evaluate this definite integral: 1 1
1 ln 1 36 x 2 12
1 36 x 2 dx
6 x 1 36 x 2
6x
x 1 x
1
Plug in the bounds of integration:
1 12
ln 1 36 1
2
6 1
6 1
1 36 1
2
ln 1 36
1
2
6
1
6
1
1 36
1
2
Simplify and approximate as a decimal:
1 12 17
180
ln 37
6
6 37
What is the arclength of the function y on the interval [0, 3]?
PART 5 Applications of Integrals
ln 37 3
4x 2
6
6 37
18
6.498
What is the arclength of the function y on the interval [0, 4]?
2x 2
Answers and Explanations 1
2
12
5.820. The shaded region includes area under the two functions f ( x ) 3 sin 3 x 4 4
g( x )
2
from x
0 to x
4
x and
4 . As shown in the graph, these two functions intersect
when x 2. To find the shaded area, break the region into two separate regions and integrate their sum as follows: 2
4
3 sin 0
4
x dx
3 x 4 2dx 4 2
Evaluate the integral:
12
cos
x
4
x 2 x 0
1 x 4 4
3
x 4 x 2
Simplify to find the area:
12
cos
12 2
1 4 4 4
cos 0
3
1 2 4 4
3
0
12
0 2 2
12
5.820
1 2 ln 3 2 ln 2 1.443. The shaded region includes the area under the two functions e 2 f ( x ) e x 1 and g ( x ) from x 0 to x 2. As shown in the graph, these two functions x 1 intersect when x 1. To find the shaded area, break the region into two separate regions and
2 1
integrate their sum as follows: 1
ex
1
2
dx
0
1
2 dx x 1
Evaluate to find the area:
ex
1 x 1 x 0
2 ln x 1
x 2 x 1
e0
e
1
2 ln 3 2 ln 2
1
1 e
2 ln 3 2 ln 2 1 0.368 2.197 1.386
1.443
CHAPTER 12 Solving Area Problems
181
3
40 . To begin, set up the problem as follows: 3 ?
5
1 2 x dx 2 0
2
8 x 5
dx
?
Clearly, f(x) has an x-intercept at the origin, so this is the lower bound of the first integral. And g(x) has an x-intercept at x 5, so this is the upper bound of the second integral. To find the remaining bounds of integration, set the two functions equal and solve:
1 2 x 2 1 2 x 2 x2 15 x
8 x 5 8x 2
2
80 x 200
16 x 2 160 x
400
2
160 x 400 0 3 x 32 x 80 0 x 4 3 x 20 0 x
20 3
4,
Thus, the remaining bounds of integration are both 4: 4
5
1 2 x dx 2 0
8 x 5
2
dx
4
Evaluate to find the area:
1 3 x 6 4
x 4 x 0
8 x 5 3
3
x 5 x 4
1 4 6
3
1 0 6
8 5 5 3
3
8 4 5 3
3
3
32 3
8 3
40 3
11 . To begin, set up the problem as follows: 12 ?
x 3 dx
0
?
x2
2 x dx
?
x2
2x
Clearly, f(x) has an x-intercept at the origin, so this is the lower bound of the first integral. To find the remaining bounds of integration, set g(x) to 0 and solve, then set the two functions equal and solve:
2
x 2x x x 2 x
x3
0 0 0, 2
x x
2
x2
2x
x3
x2
2x
0
x 2
x x 1 x 2 x
0 0 0, 1,
2
Thus, the upper bound of integration for g(x) is 2 and the remaining bounds of integration are both 1: 1 0
182
x 3 dx
2
x2
2 x dx
1
PART 5 Applications of Integrals
Evaluate to find the area:
1 4 x 4 5
x 1
1 3 x 3
x 0
x2
x 2
1 1 4
x 1
4
1 0 4
1 2 3
4
3
22
1 1 3
3
12
1 4
4 3
2 3
11 12
1 . Change this Type 1 improper integral to the limit of a proper integral: 5 1 dx x2
5
k
lim
k
5
1 dx x2
Evaluate this integral: k
1 dx x2
5
1 x
x k
1 k
x 5
1 5
1 k
1 5
Find the limit of this result:
lim
1 k
1 3e 6
0.001. Change this Type 1 improper integral to the limit of a proper integral:
k
6
2
e 3 x dx
1 5
1 5
2
lim
k
e 3 x dx
k
Evaluate this integral: 2
e 3 x dx
k
1 3x e 3
x
2
x k
1 3( e 3
2)
1 3k e 3
1 3e 6
3e 3 k
Find the limit of this result:
1 3e 6
lim
k
7
1 3e 6
3e 3 k
0.001
40 . This Type 2 improper integral has an asymptote at 0, so rewrite it as the limit of a proper 3
integral: 32 0
32
1 5
x
2
dx
lim
k
0
k
1 5
x2
dx
Evaluate the integral as follows: 32 k
1 5
x2
32
dx
x
2 5
3 x 32
dx
k
5 5 x 3
x k
5 32 3
3 5
3
5 5 k 3
40 3
3
5 5 k 3
Plug this result back into the limit:
lim
k
0
40 3
3
5 5 k 3
40 3
CHAPTER 12 Solving Area Problems
183
8
3 2
3
3
144
16 . This Type 2 improper integral has an asymptote at 2, so break it into two
improper integrals as follows: 4 0
2
x 3
x
2
dx
4
0
4
x 3
x
2
4
x
dx
3
2
x2
dx
4
Now, rewrite each of these improper integrals using limits: k
lim
k
2 0
4
x 3
x
2
4
dx
x
lim
2 k 3
k
x2
4
dx
To evaluate these two definite integrals, first evaluate the associated indefinite integral using variable substitution with u x2 – 4:
x 3
x
2
dx
4
u
1 3
2
3 3 u 2
dx
33 2 x 2
C
4
2
C
Use this result to evaluate the two definite integrals: k 0
x 3
x2
4 k
dx
4
x 3
x
2
dx
4
3 2 x 2
4
3 2 x 2
4
2 x k 3 x 0 2 x 4 3 x k
33 2 k 2
4
33 2 4 2
4
2
33 2 0 2
4
2
33 2 k 2
4
2
33 2 k 2
4
2
33 144 2
33 2 k 2
2
33 16 2
4
2
Plug these values into the two limits you found: k
x
lim
k
2 0 3
x
2
4
dx
4
x
2 k 3
2
lim
k
x
4
dx
lim
k
2
33 2 k 2
4
2
33 16 2
33 144 2
lim
k
2
33 2 k 2
4
In each case, the k term drops out of the limit:
33 16 2 9
33 144 2
3 2
3
144
3
16
8 . The two functions intersect at x 3
0 and x
1. Plug these results into the Unsigned Area
formula: b a
f ( x ) dx
b a
g ( x ) dx
2 0
2
sin x dx
0
2x 2
4 x dx
Evaluate the integral:
1
cos x
x 2 x 0
2 3 x 3
2x 2
x 2 x 0
Simplify:
1
184
cos 2
1
cos 0
PART 5 Applications of Integrals
2 2 3
3
2 2
2
1
1
16 3
8
8 3
2
10 9. To begin, set the two functions equal to find the upper and lower bounds of the integral you’re looking for:
x2
x2
6
2
2x 6x 2x x 3 x
6x 6
0 0 0, 3
Plug these results into the Unsigned Area formula: b a
b
f ( x ) dx
a
3
g ( x ) dx
x2
0
3
6 x 6 dx
0
x2
6 dx
Evaluate and simplify:
1 3 x 3
3x 2
x 3
6x
1 3 x 3
x 0
9 27 18 9 18
6x
x 3
1 3 3
x 0
3
3 3
2
6 3
1 3 3
3
6 3
9
11 5 . The bounds of integration are at x
4
0 and x
1. Plug these results into the Unsigned Area
formula: b a
b
f ( x ) dx
a
1
g ( x ) dx
0
cos 2
2
x dx
1 0
x 3 1 dx
Evaluate the first integral as I show you in Chapter 10, and the second using the Power rule:
1 x 2
1 sin x 2
x 1
1 4 x 4
x 0
x
x 1 x 0
Simplify:
1 2
1 sin 2
1 1 4
0
0
12 2. The bounds of integration are at x formula: b a
b
f ( x ) dx
a
1 2 2
3 4
5 4
and x
3 sin x cos 2 x dx
g ( x ) dx 2
. Plug these results into the Unsigned Area
cos 2 x 2
2
dx
Evaluate the first integral using u cos x and du –sin x dx, as I show you in Chapter 10, and evaluate the second integral using a linear input to the cosine function, as I show you in Chapter 6:
cos 3 x
x 1 x 0
1 sin 2 x 2
x 1
2
x 0
Simplify:
cos 3 1
1 2
cos 3 1 2
2
1 sin 2 2
2
1 sin 2
2
1 0
1 3 sin 2 2
1 sin 2 2
2
CHAPTER 12 Solving Area Problems
185
13 25 . Plug these values into the Mean Value formula:
6
f (c)
1 b a
b a
f ( x ) dx
1 5 0
0
1 5
1 3 5 3
5
x2
5 x dx
Evaluate and simplify:
1 5
1 3 x 3
5 2 x 2
x 5 x 0
5 2 5 2
1 5
0
125 3
125 2
1 125 5 6
25 6
Note that this result is within the range of the function –x2 + 5x on the given interval. 14 7
5
2e 5
f (c)
1.397 . Plug these values into the Mean Value formula:
1 b a
b a
1
f ( x ) dx
0
0
5
2e 0
0
5
2e x
1 dx
Evaluate and simplify:
1 2e x 5
x 0
x
x
5
1 5
2e
5
7 2e 5
5
5
1.397
Note that this result is approximately 60, which is within the range of the function 2ex + 1 on the given interval. 15 1 ln 4 2. Plug these values into the Mean Value formula:
2
f (c)
1 b a
b a
Evaluate using u 4 1 3 u du 1 3
3 ln x dx x 1 dx , and simplify: ln x and du x 1 4 1
f ( x ) dx
1 ln x 2
x 4
2
4
1
1 ln 4 2
x 1
2
1 ln1 2
2
1 ln 4 2
2
0
1 ln 4 2
2
Note that this result is approximately 0.961, which is within the range of the function on the given interval.
3ln x x
16 6 2 . Plug these values into the Mean Value formula:
f (c)
1 b a
b a
1
f ( x ) dx
3 cos
x dx 4
Evaluate and simplify:
6
sin
x 4
x x
6
sin
4
sin
6 4
2 2
2 2
6 2 x
Note that this result is approximately 2.7, which is within the range of the function 3 cos on 4 the given interval.
186
PART 5 Applications of Integrals
17
3 1 109 2 54
3
12
21.055. To begin, calculate the derivative of the function and then square
the result:
y dy dx dy 2 dx
3
4x 2 1
6x 2 36 x
Now, plug this value into the formula, along with the bounds of integration 0 and 3:
Arclength =
b a
1
dy dx
2
3
dx
0
1 36 x dx
To evaluate, use the u-substitution u u 109 u 1
1 + 36x and du
36 dx, so:
1 12 u du 36
Evaluate the original integral using these u-values: 2 0
3 u 109
1 2 u 54
1 9 x dx
18 1 ln
8
257
16
3
1 109 2 54
x 1
16 257
3
12
21.055
32.496. Calculate the derivative of the function and square it:
y 2x 2 dy 4x dx 2 dy 16 x 2 dx Now, plug this value into the formula, along with the bounds of integration 0 and 4:
Arclength
b a
1
dy dx
2
dx
4 0
1 16 x 2 dx
To evaluate this integral, use trig substitution, as I show you in Chapter 10. Begin by calculating the tangent and secant functions as follows:
4x sec
1 16 x 2
x dx
tan 1 tan 4 1 sec 2 4
d
CHAPTER 12 Solving Area Problems
187
Rewrite the original definite integral as an indefinite integral, and then change its terms from x to :
1 16 x 2 dx
sec
1 sec 2 4
1 sec 3 4
d
d
Integrate:
1 ln sec 8
tan
1 ln 1 16 x 2 8
C
tan sec
4x
4 x 1 16 x 2
C
Use this result to evaluate the definite integral: 4
1 16 x 2 dx
0
1 ln 1 16 x 2 8
4x
4 x 1 16 x 2
x 4 x 0
Plug in the bounds of integration:
1 8
ln 1 16 4
2
4 4
4 4
1 16 4
2
ln 1 16 0
2
4 0
4 0
1 16 0
Simplify and approximate as a decimal:
1 ln 257 16 8
188
16 257
PART 5 Applications of Integrals
1 ln 257 16 8
16 257
1 3.467 256.500 8
32.496
2
IN THIS CHAPTER
»» Using disk and washer methods to calculate volume of revolution around a horizontal axis »» Using inverse functions to calculate the volume of revolution around a vertical axis »» Finding the surface area of revolution »» Using the shell method to find vertical volume of revolution without inverse functions
13 Pump up the Volume — Solving 3-D Problems Chapter
A
common application of integration is finding the volume of a solid of revolution. These problems tend to come in a few main varieties, roughly sorted into three key categories: disk, washer, and shell. These methods build integrals from simple geometric formulas that enable you to accurately measure the volume of a complicated solid. In this chapter, I first show you how to use the disk and washer methods to find the volume of a solid of revolution around a horizontal axis. Next, I show you how to use the inverse of a function when working with solids of revolution around a vertical axis. After this, you use a formula for finding the surface area of a solid of revolution. To finish up, I show you how the shell method allows you to find the volume of a solid of revolution around a vertical axis without using inverse functions.
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
189
Using Disk and Washer Methods to Find the Volume of Revolution around a Horizontal Axis Volume of revolution problems are a staple of any Calculus II course. They tend to come in two basic varieties. Solving the first type of problem utilizes the disk method. To find the volume of a single function revolved around a horizontal axis, follow these steps:
1. Thin-slice the solid into disks of width dx. 2. Use the formula for the area of a circle (Area one disk.
r 2) to find the area of the face of
3. Set up the integral to find the total volume of all disks by using the formula b
Volume
r 2 dx , substituting the radius r in terms of x.
a
Q.
This figure shows the area bounded by the function f ( x )
1 x e , the vertical line x 3
4,
the x-axis, and the y-axis revolved around the x-axis. What is the volume of the resulting solid of revolution?
A.
18
e8 1
520.101. Begin by noticing that if you slice up the solid perpendicular to
the x-axis, each cross-section is a disk — that is, a thin cylinder with a radius that equals the height of the function. Begin by using the formula for a circle to help you find the area of the circle that this disk is based on:
r2
Area of disk
f(x)
2
Plug the function into this formula and simplify:
1 x e 3
190
2
9
e2x
PART 5 Applications of Integrals
Now, represent the volume by integrating this result from 0 to 4, using dx to represent the thickness of each circular disk: 4
Volume
9
e 2 x dx
0
Evaluate this definite integral using methods you already know:
1 2x e 9 2
x 4 x 0
18
e2x
x 4 x 0
18
e8
18
e0
18
e8 1
520.101
Thus, the volume of the solid of revolution is approximately 520.101. Solving the second type of volume of revolution problem utilizes the washer method. To find the volume between two functions revolved around a horizontal axis, follow these steps:
1. Thin-slice the solid into washers — that is, disks with a circular hole cut out of them — of width dx. 2. Use a modified formula for the area of a circle (Area face of one washer.
r1 2
r2 2 ) to find the area of the b
3. Integrate to find the total volume of all such washers: Volume
r1 2
r2 2 dx.
a
Q.
This figure shows the area bounded by functions f ( x ) tical line x
2
2 sin x , g ( x ) sin x , and the ver-
revolved around the x-axis. What is the volume of the resulting solid of
revolution?
A.
2
3 4
7.402. Notice that if you slice up the solid perpendicular to the x-axis, each
cross-section is a circular washer — that is, an outer circle with a smaller inner circle removed from it. You can find the area of this washer as follows:
Area of washer
Area of outer circle Area of inner circle
r1 2
r2 2
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
191
Here, I mark the radius of the outer circle as r1 and the radius of the inner circle as r2 . But just as in the previous example, the radius of each circle equals the value of the corresponding function:
f(x)
2
g( x )
2
Use this result as a general formula to measure the area of the washer, making sure to subtract the functions in the correct order — outer function minus inner:
2 sin x
2
sin x
2
4 sin 2 x
sin 2 x
3 sin 2 x
Now, represent the volume by integrating this result from 0 to , using dx to represent 2 the thickness of each circular disk:
Volume
3
2
sin 2 x dx
0
Evaluate this definite integral using the half-angle formula, as I show you in Chapter 10: 2
1 cos 2 x dx 2 0
3
2
3 4
1
192
1 sin 4
3 2
3 4
1 x 2
1 sin 2 x 4
x
2
x 0
3
1 2 2
1 2 sin 2 4
0
7.402
This figure shows the area bounded by the function f ( x ) 3 cos x , the x-axis, and the y-axis revolved around the x-axis. What is the volume of the resulting solid of revolution?
PART 5 Applications of Integrals
2
x , the two vertical lines x 4 and x 25, This figure shows the area bounded by the function f ( x ) and the x-axis revolved around the x-axis. What is the volume of the resulting solid of revolution? .
3
This figure shows the area bounded by the function f ( x ) 2 revolved around the axis y and the horizontal line y of revolution?
sin x , the vertical line x 2 , the y-axis, 2. What is the volume of the resulting solid
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
193
4
This figure shows the area bounded by the two functions f ( x )
x
5
194
3
x and g ( x )
x and the vertical line
1 revolved around the x-axis. What is the volume of the resulting solid of revolution? 2
This figure shows the area bounded by the two functions f ( x ) e x 1 and g ( x ) e x, the vertical line x 2, and the y-axis revolved around the x-axis. What is the volume of the resulting solid of revolution?
PART 5 Applications of Integrals
6
This figure shows the area between the two functions f ( x ) 3 x 2 and g ( x ) x 3 and the vertical line x revolved around the x-axis. What is the volume of the resulting solid of revolution?
1
Using Inverse Functions to Find Volume of Revolution around a Vertical Axis A common strategy to find the volume of revolution around a vertical axis is to turn the problem on its side. To do this, you change the function you’ve been given to its inverse and then frame the problem as the volume of revolution around a horizontal axis. Here’s an example to get you started:
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
195
Q. A.
This figure shows the area bounded by the function f ( x ) 3 x 2, y 4, and the y-axis revolved around the y-axis. What is the volume of the resulting solid of revolution?
8 3
8.378 . Begin by finding the inverse of the function f 1( x ) by changing x to y and
f(x) to x, and then solving for y:
x x 3 x 3
3y 2 y2 y
Now, use the resulting inverse function f the area formula:
Area
x 3
r2
1
(x)
x to build your integral, starting with 3
2
3
x
Now, represent the volume by integrating this result from 0 to 4, using dx to represent the thickness of each circular disk: 4
Volume
3
x dx 0
Simplify and evaluate this definite integral using methods you already know:
1 2 x 3 2
7
x 4 x 0
6
x2
x 4 x 0
6
4
2
6
0
2
This figure shows the area bounded by the function f ( x )
8 3
8.378
1 3 x ,y 2
the y-axis. What is the volume of the resulting solid of revolution?
196
PART 5 Applications of Integrals
4, and the y-axis revolved around
8
This figure shows the area bounded by the function f ( x ) 2 arccos x , the x-axis, and the y-axis revolved around the y-axis. Note that the function includes the two points ( 0, ) and (1, 0 ). What is the volume of the resulting solid of revolution?
9
This figure shows the area bounded by function f ( x ) ln x , the horizontal line y 1, and the vertical 1 revolved around x 1. What is the volume of the resulting solid of revolution? line x
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
197
10
1
This figure shows the area bounded by the two functions f ( x ) x 7 and g ( x ) x 2 , and the vertical line x 1 revolved around the y-axis. What is the volume of the resulting solid of revolution?
Surface Area of Revolution Another common type of application in Calculus II is finding the area of a surface of revolution. Fortunately, a formula exists for this type of problem: b
A 2
r 1
f’ x
2
dx
a
Here, the radius r is understood to be a function of x. Applying this formula may be more or less difficult depending on the function you’re working with. Fortunately, teachers tend to steer you toward problems that you’ll be able to evaluate using integration methods you already know.
198
PART 5 Applications of Integrals
Q.
This figure shows the area bounded by the function f ( x )
1 3 x , the vertical line x 2
1,
and the x-axis revolved around the x-axis. What is the surface area of the resulting solid of revolution?
A.
2 27
13 4
3 2
1
1.130 . Begin by finding the derivative of f ( x )
1 3 x : 2
3 2 x 2
f '( x )
Plug this into the formula: 1
A 2 0
1 3 x 2
1
3 2 x 2
2
1
dx
x3 1
0
9 4 x dx 4
Now evaluate using variable substitution, with u 1
9
0
1 u2
du
2 9 4 1 x 27 4
3 x 1 2
2 27
1
9 1 4
4
3 2
9 4 x and du 4
1 1
9 0 4
4
3 2
9 x 3 dx : 2 27
13 4
3 2
1
1.130
x 0
Therefore, the surface area of revolution is approximately 1.130.
11
This figure shows the area bounded by the function f ( x )
1 3 x , the vertical line x 8
2, and the x-axis
revolved around the x-axis. What is the surface area of the resulting solid of revolution?
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
199
12
9 x 2 , the vertical lines x 2 and x 2, This figure shows the area bounded by the function f ( x ) and the x-axis revolved around the x-axis. What is the surface area of the resulting solid of revolution? (Hint: After you set up the integral, see if you can simplify it to a form that you will be able to evaluate.)
Shell Method Earlier in this chapter, you used both the disk and the washer methods to measure the volume of a solid of revolution around a vertical axis. Recall that these methods required you to work with the inverse of the function, essentially turning it on its side. The shell method is an alternative way to perform these types of calculations without finding the inverse of the function. Instead of slicing a solid laterally from top to bottom into disks or washers, when you use the shell method, you slice it from the inside out into progressively larger cylindrical shells. A good way to begin picturing how the shell method works is to think about the rounded surface area of a can with the top and the bottom removed. This surface area forms the area of a single shell. It equals the circumference of the can multiplied by its height, which you can then represent in terms of the can’s radius:
Surface area of shell
Circumference height
2 rh
This formula forms the basis for the volume of a solid: b
Volume
rh dx
2 a
200
PART 5 Applications of Integrals
In this formula, both r and h are stated as functions of x. Here’s an example to help you make sense of this formula:
Q. A.
This figure shows the area bounded by the function f ( x ) x 2, the vertical line x 2, and the x-axis revolved around the y-axis. Use the shell method to find the resulting volume of the solid of revolution.
8
25.133 . Begin with the formula for the shell method: 2
Volume
rh dx
2 0
Here, I’ve replaced the bounds of integration a and b with the x-values 0 and 2. Now, the radius of each shell in terms of x is simply the distance from the y-axis to the shell — that is, x. And the height of each shell is simply the height of the function x 2. Substitute these two values into the formula: 2
2
x x 2 dx
2
2
0
x 3 dx
0
Simplify and evaluate this definite integral using methods you already know:
2
1 4 x 4
x 2 x 0
2
1 2 4
4
0
8
25.133
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
201
3
13
x 2 , the vertical line x 8 , and the x-axis This figure shows the area bounded by the function f ( x ) revolved around the y-axis. Use the shell method to find the resulting volume of the solid of revolution.
14
This figure shows the area bounded by the function f ( x ) e x , the vertical line x 1, the x-axis, and the y-axis revolved around the y-axis. Use the shell method to find the resulting volume of the solid of revolution.
202
2
PART 5 Applications of Integrals
15
x , the vertical line x 4 , and the x-axis This figure shows the area bounded by the function f ( x ) 1. Use the shell method to find the resulting volume of the solid revolved around the vertical axis x of revolution.
16
This figure shows the area bounded by the function f ( x ) e x , g ( x ) x , the y-axis, and the vertical lin x 1, revolved around the y-axis. Use the shell method to find the resulting volume of the solid of revolution.
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
203
Answers and Explanations 1
2
9
22.207. When you slice up the solid perpendicular to the x-axis, each cross-section is a
4
circular disk with a radius that equals the height of the function. Thus, you can use the formula for a circle to begin writing the integral to represent the volume of this solid:
r2
Area
3 cos x
2
9 cos 2 x
Now, represent the volume by integrating this result from 0 to , using dx to represent the 2 thickness of each circular disk: 2
Volume
9 cos 2 x dx
0
Integrate using methods you already know: 2
9
cos 2 x dx
1 x 2
9
0
x
1 sin 2 x 4
2
x 0
9 x 2
9 sin 2 x 4
x
2
x 0
Now, simplify this result:
9 2 2
9 sin 4
2
609 2
9 0 2
9 sin 0 4
2
9 4
9 sin 4
2
9 4
22.207
956.614. When you slice up the solid perpendicular to the x-axis, each cross-section
is a circular disk with a radius that equals the height of the function. Thus, you can use the formula for a circle to begin writing the integral to represent the volume of this solid:
r2
Area
x
2
x
Now, represent the volume by integrating this result from 0 to 9, using dx to represent the thickness of each circular disk: 25
x dx
Volume 4
Integrate using methods you already know: 25
x dx 4
3
9
x 25 x 4
2
25
2
2
4
2
625 2
16 2
2
609 2
956.614
88.826. When you slice up the solid perpendicular to the x-axis, each cross-section is a circular disk with a radius that equals the height of the function. Thus, you can use the formula for a circle to begin writing the integral to represent the volume of this solid. (Note 2, the radius of the solid is sin x 2 .) here that because the vertical axis of revolution is x
Area
204
2
x2
r2
sin x 2
2
PART 5 Applications of Integrals
sin 2 x
4 sin x
4
Now, represent the volume by integrating this result from 0 to 2 , using dx to represent the thickness of each circular disk: 2
sin 2 x
Volume
4 sin x
4 dx
0
Integrate using methods you already know: 2
sin 2 x
4 sin x
4 dx
0
1 x 2
1 sin 2 x 4 cos x 4
1 2 2
1 sin 4 4
4x
x 2 x 0
4 cos 2
1 0 2
4 2
1 sin 0 4 cos 0 4 0 4
Now, simplify this result:
4 8 4
3 5 3 32
4
9
2
88.826
0.201. Use the following formula:
8
f(x)
Area of washer
2
g( x )
2
3
x
2
x
2
2
x3
x 1
Now, represent the volume by integrating this result from 0 to , using dx to represent the 2 thickness of each circular disk: 1 2
Volume
2
x3
x dx
0
Evaluate: x
5
3 x3 5 5 2 e 2
2
x
2
1 2
x 0
2 2
3 5
1 2
5 3
1 2 2
2
3 5 3 32
0
0.201
8
42.144 . Use the following formula:
Area of washer
f(x)
2
g( x )
2
ex
1
2
ex
2
e2x
2 ex
1
e2x
2 ex
1
Now, represent the volume by integrating this result from 0 to 2, using dx to represent the thickness of each circular disk: 2
Volume
2 ex
1 dx
0
Evaluate:
2 ex
x
x 2 x 0
2 e2
2
2 e0
0
2 e2
2 2
46.427 2 6.283
42.144
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
205
6
58 35
5.206 . Use the following formula: 2
f(x)
Area of washer
g( x )
2
3x 2
2
x3
2
9 x4
x6
Now, represent the volume by integrating this result from 0 to 1, using dx to represent the thickness of each circular disk: 1
Volume
9 x4
x 6 dx
0
Evaluate:
9 x5 5 7
3
7
3 4 5
4
x 1
x7
9 1 5
x 0
5 3
5
7
1
7
9 5
0
7
58 35
5.206
30.159 . Begin by finding the inverse of the function f ( x )
1 3 x by changing x 2
to y and f(x) to x, and then solving for y:
2x
1 3 y 2 y3
2x
y
x 3
Now, use the resulting inverse function f area formula:
r2
Area
3
2x
2
3
1
3
(x)
2 x to build your integral, starting with the
2
4 x3
Now, represent the volume by integrating this result from 0 to 4, using dx to represent the thickness of each circular disk: 4
Volume
3
2
4 x 3 dx
0
Simplify and evaluate this definite integral using methods you already know: 3
4
2
x 3 dx
4
3
5
4
0
8
2
x 4 3 x 0
4
3 4 5
5 3
0
30.159
9.870. Begin by finding the inverse of the function f ( x ) and f(x) to x, and then solving for y: x x 2 x cos 2
206
3 3 x 5
2 arccos y arccos y y
PART 5 Applications of Integrals
2 arccos x by changing x to y
Now, using the resulting inverse function f the area formula:
r2
Area
cos 2
1
x ( x ) cos , build your integral, starting with 2
x 2
Now, represent the volume by integrating this result from 0 to 2 , using dx to represent the thickness of each circular disk:
cos 2
Volume 0
x dx 2
Simplify and evaluate this definite integral using methods you already know: 2
cos 2
0
9
e2
2
x dx 2
2 e
2
3 2
x
2
sin x
x 2 x 0
2
2
2
sin 2
2
0
9.870
23.974. Begin by finding the inverse of the function f ( x ) ln x by changing
x to y and f(x) to x, and then solving for y:
e
x
ln y
x
y
Now, using the resulting inverse function f 1( x ) e x , build your integral, starting with the 1, the radius of area formula. (Note here that because the vertical axis of revolution is x the solid is e x 1.)
ex
r2
Area
1
2
e2x
2 ex
Now, represent the volume by integrating this result from 0 to 4, using dx to represent the thickness of each circular disk: 1
Volume
e2x
2 ex
dx
0
Evaluate this definite integral using methods you already know:
2
e2x
2 ex
x
x 1 x 0
2
e2
2 e1
e2
2 e
2
e0
2 e0
0
Simplify:
2
e2
10 26
2 e
2
2
2
3 2
11.607 17.079 4.712 23.974
1.815 . Begin by finding the inverses of the two functions f ( x )
45
x 7 and g ( x )
1
x2
by changing x to y and f(x) to x, and then solving for y:
7
x
y7
x
y
1
x
y2
x2
y
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
207
Now, using the resulting inverse function f 1( x ) starting with the area formula for a washer:
f(x)
Area of washer
2
2
g( x )
7
7
2
x
x and g 1( x )
x2
x 2, build your integral,
2
2
x4
x7
Now, represent the volume by integrating this result from 0 to 1, using dx to represent the thickness of each washer: 1
2
x 4 dx
x7
Volume 0
Evaluate this definite integral using methods you already know: 1
2
9
7 7 x 9
x 4 dx
x7 0
11 8
13 4
27
3 2
x 1
1 5 x 5
7 1 9
x 0
9 7
1 1 5
7 9
0
4.523. Begin by finding the derivative of f ( x )
1
1 5
26 45
1.815
1 3 x : 8
3 2 x 8
f ’( x )
Plug this into the formula, and use the function 2
1 3 x 8
A 2 0
2
2
3 2 x 8
1
dx
4
1 3 x as the radius: 8
9 4 x dx 64
x3 1
0
Now evaluate using variable substitution, with u 2
4 9
5
1 u2
3 x 2 2
8 9 4 1 x 27 64
du
0
8 27
1
9 2 64
9 4 x and du 64
1 4
3 2
1
9 0 64
4
9 3 x dx : 16 3 2
8 27
13 4
x 0
Therefore, the surface area of revolution is approximately 4.523. 12 24
9 x2 :
75.398. Begin by finding the derivative of f ( x ) x
f '( x )
9 x2
Plug this into the formula, and use the function 9 2
A 2
x
9 x2 1
2
9 x2
2
2
dx
2
9 x2 1
2
x 2 as the radius: x2 dx 9 x2
You can further simplify this integral by multiplying the two radical expressions: 2
2 2
208
9 x2
x2 9 x2 9 x
2
2
dx
PART 5 Applications of Integrals
2 2
9 x2
x 2 dx
2
2
9 dx
2 2
3 dx
2 2
3 2
1
4.523
Now evaluate using variable substitution, with u x 2 x 2
6 x
12
12
24
1
9 4 x and du 64
9 3 x dx : 16
75.398
Therefore, the surface area of revolution is approximately 75.398. 13 192
603.186. Begin with the formula for the shell method: 8
Volume
rh dx
2 0
Here, I’ve replaced the bounds of integration a and b with the x-values 0 and 8. Now, the radius of each shell in terms of x is simply the distance from the y-axis to the shell — x, in 3 other words. And the height of each shell is simply the height of the function x 2 . Substitute these two values into the formula: 8
x
2
3
8
x2
dx
0
5
x 3 dx
2 0
Simplify and evaluate this definite integral using methods you already know: x 8
8
3 3 x 8
2 14
e 1
2 x 0
3 8 8
8 3
0
2
3 256 8
192
603.186
5.398 . Begin with the formula for the shell method: 1
Volume
rh dx
2 0
Here, I’ve replaced the bounds of integration a and b with the x-values 0 and 1. Now, the radius of each shell in terms of x is simply the distance from the y-axis to the shell — x, in 2
other words. And the height of each shell is simply the height of the function e x . Substitute these two values into the formula: 1
2
2
xe x dx
0
Evaluate this definite integral using variable substitution with u
2 2
1
2 x2 e 2
e u du
0
15 544
15
x 1
2
e1
e0
2
e 1
x 2 and du
2 x dx :
5.398
x 0
113.935. Begin with the formula for the shell method: 4
Volume
rh dx
2 0
CHAPTER 13 Pump up the Volume — Solving 3-D Problems
209
Here, I’ve replaced the bounds of integration a and b with the x-values 0 and 4. Now, the 1 to the radius of each shell in terms of x is simply the distance from the vertical axis y shell — x + 1, in other words. And the height of each shell is simply the height of the function x . Substitute these two values into the formula: 4
4
x 1
2
x dx
3
x2
2
0
1
x 2 dx
0
Evaluate the integral: 4
3
5
1
x2
2
4 x2 5
x 2 dx
0
3
4 x2 3
x 4
x 0
Complete the problem as follows:
4 4 5 16 4
4 3
5 2
4
3 2
4 32 5
0
4 3
128 5
8
32 3
384
160 15
544 15
113.935
4.189 . Begin with the formula for the shell method:
3
1
Volume
rh dx
2 0
Here, I’ve replaced the bounds of integration a and b with the x-values 0 and 1. Now, the radius of each shell in terms of x is simply the distance from the y-axis to the shell — x, in other words. And the height of each shell is the distance from the top function f ( x ) e x to the bottom function g ( x ) x — that is, of the function e x x . Substitute these two values into the formula: 1
2
x ex
1
x dx
0
2
xe x
x 2 dx
0
Evaluate this definite integral using a combination of integration by parts and the Power rule:
2
210
xe x e x
1 3 x 3
x 1
2
x 0
PART 5 Applications of Integrals
1e 1 e 1
1 1 3
3
0e 0 e 0
1 0 3
3
2
e e
1 1 3
4 3
4.189
IN THIS CHAPTER
»» Understanding how differential equations arise and how to begin thinking about them »» Solving separable differential equations
14 Differential Equations Chapter
D
ifferential equations (DEs for short) are complicated-looking equations that arise when you combine functions and their derivatives. DEs are useful for modeling a wide variety of phenomena in both physical sciences (such as physics, chemistry, and biology) and social sciences (such as economics and sociology). And although some DEs are extremely difficult to solve, others yield readily to a few methods that you’re already well prepared to learn in a Calculus II class. In this chapter, I show you how the most basic DE arises naturally whenever you find a derivative, and how it can be solved by integration. After that, I show you how even slightly more complicated DEs get a lot trickier to solve. I also provide you with a way to begin building DEs and then testing solutions to them. This work provides a basis for solving separable equations, which are the first type of DE most students work with because they’re the easiest to solve. Solving separable equations leans heavily on integration, which is why some Calculus II courses have begun to include a unit on this basic type of DE.
Understanding Differential Equations A differential equation is simply an equation that includes at least one derivative. The most basic DE arises whenever you find the derivative of a function. For example:
y 5x 3 dy 15 x 2 dx
y dy dx
sec 7 x 7 sec 7 x tan 7 x
y dy dx
ln x 1 x
CHAPTER 14 Differential Equations
211
The second equation in each case includes a derivative
dy , so each is a DE. When the derivative dx
in a DE is isolated on one side of the equals sign with a function of the independent variable (usually x) on the other side, you can solve it by integrating both sides of the equation.
Q.
Solve
A.
y
dy dx
sin x cos x by finding the value of y in terms of x.
1 sin 2 x 2 dy dx dx y
C . Integrate both sides of the equation:
sin x cos x dx 1 sin 2 x C 2
In most cases, however, a DE will include at least one complication that prevents you from solving the equation using integration. Here are some examples:
dy dx
2e 2 x
y
6e 3 x
y
dy dx
3x 5
8x 3
dy dx
4x
y cos2 x
A good way to start understanding how more complicated DEs arise is by building them yourself. To do this, start with a function and differentiate it, then combine these two elements to create a differential equation:
Q.
Given y
x2
A.
x2
7x
5 . Begin by differentiating y:
y dy dx
x2
5x
2x
5
5 x , what is the value of y
Now, add y and
dy dx
y
x2
dy ? dx
dy , and simplify: dx
5x 2x
5
x 2 7x
5
Thus, you can build a DE by setting the original expression equal to the simplified value of the expression you found:
y
Q. A.
212
dy dx
x 2 7x
Given y sign.
dy dx y dy dx
y
5
sin x cos x , write a DE with the expression
cos 2 x
sin 2 x
sin x cos x cos 2 x sin 2 x
PART 5 Applications of Integrals
dy dx
y on one side of the equals
sin x cos x. Begin by differentiating y using the Product rule:
Now, combine these results to build a DE:
dy dx
cos 2 x sin 2 x sin x cos x
y
When you understand how DEs can be built by combining the elements y and
dy , you can dx
move one step forward in your understanding by checking the proposed solution to a DE. For example:
Q.
Show that y
A.
Begin by differentiating y:
y dy dx
ex
2e 2 x
ex
2e 2 x
3e 3 x
ex
4e 2 x
9e 3 x
3e 3 x is a solution to the equation
dy dx
y
2e 2 x
6e 3 x .
Therefore:
dy dx
ex
y
4e 2 x
9e 3 x
ex
2e 2 x
3e 3 x
ex
2e 2 x
6e 3 x
Thus, the proposed solution is correct.
Q.
Show that y
A.
Begin by differentiating y:
y dy dx
x3 3x
x3
2 x is a solution to the equation y
dy dx
3x 5
8x 3
4x.
2x 2
2
Therefore:
y
1
Solve
dy dx
x3
dy dx
e2x
x3
2x
3x 2
2
cos x by finding the
value of y in terms of x.
3x 5
2x 3
6x 3
2
4x
3x 5
Given y
y
2
8x 3
3x 2
4x
4 x , what is the value of
dy ? dx
CHAPTER 14 Differential Equations
213
2x 3
dy dx
3
Show that y
4
Match each of the following four DEs with their solutions:
3 x 1 is a solution to the equation
dy 2x 3 6x 2 1 dx dy b) 3 y 3 x 2 10 x 4 dx dy 2x 3 6x c) y dx dy d) y 2 2 1 4 x 2 12 x 12 dx a) y
2y
i) y
x2
ii) y
2x 3 1
4x 3
6x 2
6 x 1.
3
iii) y
2x
3
iv) y
x2
4x
Solving Separable Differential Equations The simplest type of differential equation to solve is called a separable equation, because you can solve it by separating the x terms and y terms onto opposite sides of the equals sign and then integrating both sides. For example:
Q.
Solve
A.
1 sin 2 x Ce 2 .
dy dx
y cos2 x for y in terms of x.
Begin by multiplying both sides by dx, then use algebra to isolate terms of y on the left side of the equation and terms of x on the right side:
dy dx dy 1 dy y
y cos 2 x y cos 2 x dx cos 2 x dx
Now, integrate both sides of the equation:
214
1 dy y
cos 2 x dx
ln y
1 sin 2 x C 1 2
PART 5 Applications of Integrals
Note that I choose C 1 as a constant because I anticipate changing it before the end of the problem. Next, solve for y by using both sides of the equation as an exponent of e, and then simplify:
e ln y
e
y
1 sin 2 x C 1 2 1
e2
sin 2 x
eC 1
Now, because e C 1 is simply a constant, set C the final answer:
y
5
dy dx
y sin5 x
7
dy dx
e x cos 2 y
1
Ce 2
e C 1 and use this value as a coefficient in
sin 2 x
6
dy dx
x3 y2
CHAPTER 14 Differential Equations
215
Answers and Explanations 1
1 4 x 4
y
dy dx
1 2x e 2 x3
dy dx dx
2
9x 4 y dy dx
e2x e2x
x3 1 4 x 4
y
sin x
3x 2 6x
cos x cos x dx
1 2x e 2
24 x 3 16 x 2
C . Integrate both sides of the equation:
sin x C
6x
4 . Begin by finding the derivative of y:
4x 4
Now, build the expression:
y2 3
dy dx
3x 2
4x
2
6x
4
9x 4
24 x 3 16 x 2
Begin by finding the derivative of y:
y dy dx
2x 3 6x
3x 1
2
3
Now, build the DE:
dy dx
2y
6x 2
3 2 2x 3
3x 1
4x 3
Therefore:
dy dx
2y
4x 3
6x 2
6x 1
4 a. ii) y
2x 3 1
Find the derivative of y:
y dy dx
2x 3 1 6x 2
Now build the DE:
y
216
dy dx
2x 3 1 6x 2
PART 5 Applications of Integrals
2x 3
6x 2 1
6x 2
6x 1
6x 4
x2
b. iv) y
4x
Find the derivative of y:
y dy dx
x2
4x
2x
4
Now build the DE:
dy dx
3y
x2
c. i) y
3 x2
4x
2x
4
3 x 2 10 x 4
3
Find the derivative of y:
x2
y dy dx
3
2x
Now build the DE:
y
dy dx
d. iii) y
2x x 2
2x
3
2x 3
6x
3
Find the derivative of y:
y dy dx
2x 2
y2 1
3
2
dy dx
1
2x
3
2
2 2
1 4 x 2 12 x 12
cos 5 x
5 Ce 5 . Multiply both sides by dx, then isolate the x and y terms on opposite sides of the equation:
dy dx dy 1 dy y
y sin 5 x y sin 5 x dx sin 5 x dx
Now, integrate both sides of the equation:
1 dy y
sin 5 x dx
ln y
1 cos 2 x C 1 5
CHAPTER 14 Differential Equations
217
To solve for y, use both sides as an exponent of e and simplify, using C further:
6
1 cos 5 x C 1 5
e ln y
e
y
e
y
Ce
3
3 4 x 4
e C 1 to simplify
1 cos 5 x C 1 5 e 1 cos 5 x 5
C . Multiply both sides by dx, then isolate the x and y terms on opposite sides of the
equation:
dy dx
x3 y2
dy
x3 dx y2
y 2 dy
x 3 dx
Integrate both sides of the equation:
y 2 dy 1 3 y 3
x 3 dx 1 4 x 4
C1
Solve for y, rewriting the value of the constant as C:
y3 y 7
3 4 3C 1 x 4 3 3 x4 C 4
arctan e x
C . Multiply both sides by dx, then isolate the x and y terms on opposite sides of
the equation:
dy dx dy
e x cos 2 y e x cos 2 y dx
1 dy cos 2 y
e x dx
sec 2 y dy
e x dx
Integrate both sides of the equation, then solve for y:
sec 2 y dy tan y y
218
e x dx ex
C
arctan e x
C
PART 5 Applications of Integrals
6
Infinite Series
IN THIS PART . . .
Understand the relationships among an infinite series, its generating sequence, and its sequence of partial sums Distinguish convergent and divergent series Apply a variety of tests for convergence or divergence Express functions as Taylor and McLaurin series
IN THIS CHAPTER
»» Understanding both expanded and contracted notations for sequences »» Working with sigma notation for series »» Identifying the sequence of partial sums for a series »» Calculating the value of a convergent geometric series »» Knowing when a p-series is convergent or divergent
15 Sequences and Series Chapter
I
n this chapter, you get an introduction to the important Calculus II topic of sequences and series. To begin, you work with sequences in both expanded and contracted forms. Next, you use this understanding to get a grip on how applying sigma notation to a defining sequence turns it into a series. Once you gain some experience working with series, you apply this knowledge to see how every series produces a sequence of partial sums. Then, you work with geometric series, discovering how to distinguish between convergent and divergent series of this type as well as how to calculate the value of a convergent geometric series. To finish up, you work with p-series, learning to identify when this type of series is convergent and when it’s divergent.
Understanding Sequences A sequence is a set of numbers in a particular order. In Calculus II, you work mostly with infinite sequences, which are sequences that never end. For example:
an
2n
2, 4, 6, 8, ...
CHAPTER 15 Sequences and Series
221
bn
1 2n
1 1 1 1 , , , , ... 2 4 8 16
cn
1 n3
1,
1 1 1 , , , ... 8 27 64
As you can see, you can define a sequence either by using a function of n (compact form) or by stating the first few elements of the sequence (expanded form). By convention, the value n (called the index of the sequence) starts at 1, allowing you to both define and then reference any element of the sequence. For example:
a1 2
Q. A. Q. A.
1
2
222
1 16
b4
Write the first four terms of the sequence an
1 5n
an
1,
1 1 1 , , , ... 3 9 27
Q.
Given the sequence an
A.
a7
1 10 7
1 in expanded form. 5n
1 1 1 1 , , , , ... 5 10 15 20
Write the sequence an
an
1 1, 000
c10
1
1 1, 000, 000
1,
1 1 1 , , , ... in compact form. 3 9 27 1 3n 1
1 10 n
1
, what is the value of a7?
Write the first four terms of each of the following sequences in expanded form: a. an
1 3n
b. an
c. an
n2 2n 2
d. an
n 1 3n 1 n 1 n 1 !
3
1
Write each of the following sequences in compact form: a. an
2 4 8 16 , , , , ... 5 25 125 625
c. an
3,
5 7 9 , , , ... 4 16 64
PART 6 Infinite Series
b. an
1 2 3 4 , , , , ... 4 9 16 25
d. an
4 7 10 14 , , , , ... 5 11 29 83
3
2n 1 , what is n 1 !
Given the sequence an the value of a6?
4
What is the value of a15 in the sequence
n 1 ? 10 n 12
an
Understanding Series A series is the sum of a sequence of numbers. In Calculus II, an important topic is infinite series — that is, the sum of an infinite sequence. Every infinite series has a defining sequence, which is simply the sequence of terms for that series. For example:
an
2n
2n 2 4 6 8 ...
2, 4, 6, 8, ... n 1
bn
1 2n
1 1 1 1 , , , , ... 2 4 8 16
cn
1 n3
1,
1 n 12
1 1 1 , , , ... 8 27 64
n
1 n 1n
3
1 2 1
1 4 1 8
1 8
1 16
...
1 27
1 64
...
As you can see, you can use sigma notation to represent a series compactly. You can also expand a series out, calculating its first few terms to get an intuitive sense of how that series behaves.
Q. A. Q. A.
Expand out the first four terms of the series
n
n2 n 15
1 5
4 25
Write the series
9 125 7 8
11 16
n
16 625
...
15 32
19 64
n2 . n 15
... using sigma notation.
4n 3 . Here, you want to find functions in the numerator and denominator such 2n 2 7 11 that n 1 produces a value of , n 2 produces , and so forth. 8 16
n 1
CHAPTER 15 Sequences and Series
223
In contrast with sequences, the index n of a series isn’t required to start at 1. This provides some flexibility to define a single series in a variety of ways.
Q. A.
Rewrite the series
... using the sigma notation
. n 0
11 15 19 .... You 16 32 64 n can adjust this series to sigma notation starting with an index of n 0 as follows: 4n 7 2n 3
7 8
11 16
15 32
19 64
7 8
...
b.
n
d.
n
4n 1 3 2 2n
n
n2 1 1 n!
n
n4 7 2 n 1 1n 2
Write each of the following series in sigma notation using an index of n a. 0 c. 243
224
19 64
2n n 07
c.
7
15 32
Write the first four terms of each of the following series in expanded form. a.
6
11 16
4n 7 4n 3 . Recall from the previous example that n 3 n 2 2 n 1 2 0
n 0
5
7 8
3 5
6 25 81 2
9 125 27 3
... 9 2
...
b.
2 21
d.
2 9
3 28 4 25
4 35 8 49
5 42
0. ...
16 ... 81
Rewrite each of the series in Question 6 in sigma notation using an index of n 1.
PART 6 Infinite Series
Finding the Sequence of Partial Sums for a Series In the previous section, you saw how every series has a defining sequence, which is simply the list of numbers being added in that series. Another important tool for understanding a series involves finding its sequence of partial sums — that is, a running total of the value of that series when you add one, two, three, or more terms.
Q.
Calculate the first four partial sums for the series n 0
1 2
n
1
1 2
1 4
1 8
..., then use
them to write the sequence of partial sums for this series in both expanded and compact form.
A.
2n 1 2n 1
Sn
1,
3 7 15 , , , ... . Calculate the first four partial sums for this series 2 4 8
as follows:
S1 S2 S3 S4
0 n 0 1 n 0
1 2
n
1 2
n
1 2
n
1 2
n
2 n 0 3 n 0
1 1
1 2
3 2
1
1 2
1 4
7 4
1
1 2
1 4
1 8
15 8
Express these four values as the expanded form of a sequence Sn :
Sn
1,
3 7 15 , , , ... 2 4 8
Now, find a function of n that produces this sequence (remember that, by convention, the index of a series starts with n 1):
Sn
2n 1 2n 1
CHAPTER 15 Sequences and Series
225
8
9
Write the first four partial sums for the series n 2
1 2
n
1 4
1 8
1 16
1 32
n 1 2 3 4 ..., then write the
series
..., then
n 1
sequence of partial sums for this series in expanded form. Can you think of a way to express this sequence of partial sums in compact form? (Hint: This is called the sequence of triangular numbers, so if you get stuck, look for this formula online.)
write the sequence of partial sums for this series in both expanded and compact form.
10
Write the first four partial sums for the
n3
Write the first four partial sums for the series
1 8 27 64 ..., then write the sequence of par-
n 1
tial sums for this series in expanded form. Can you think of a way to express this sequence of partial sums in compact form? (Hint: Think about how the answer to Question 9 can help you answer this question.)
226
PART 6 Infinite Series
Understanding and Evaluating Geometric Series A geometric series is any series of the following form:
ar n
ar 2
a ar
ar 3
...
n 0
Here are a few examples of geometric series, with their respective a and r values:
3n
1 3 9 27 ...
7 2n
7
a 1, r
3
a 7, r
1 2
a
4 5
n 0
n 0
2 n 0
Q.
4 5
n
7 2
7 4
7 8
2
8 5
32 25
128 125
...
2, r
Write both the sigma notation and expanded notation for a geometric series such that
a 10 and r
A.
...
1 . 3
Use the formula for the geometric series:
10 n 0
n
1 3
10
10 3
10 9
10 27
...
You can test a series to see if it’s geometric in the following way:
1. Let a equal the first term of the series. 2. Let r equal the second term of the series divided by the first term. 3. Check to see whether the series fits the form a
Q. A.
1
2 5
2 5
8 25
Write the series 2 it’s geometric.
n
1 4 02 5
n
1 2
8 25
32 125
32 125
ar
ar 2
ar 3
... .
... in sigma notation, and then check to see whether
...; the series is geometric. Set a
1 , then find r as 2
follows:
r
2 5 1 2
2 2 5
4 5
Now, check the terms of the series:
a
1 2
ar
1 4 2 5
2 5
ar 2
1 2
4 5
2
8 25
ar 3
1 2
4 5
3
32 125
CHAPTER 15 Sequences and Series
227
All the given terms of the series match the form of a geometric series, so the series is geometric. To finish up, use the form for a geometric series to express the series using sigma notation:
ar n n 0
n
1 4 02 5
n
1 2
2 5
8 25
32 125
...
An important topic that you will focus on in depth in Chapter 16 is showing whether a series is convergent or divergent:
»» A convergent series equals a real value. »» A divergent series doesn’t equal a real value. A relatively simple test exists to discover whether a geometric series is convergent or divergent:
»» When 1 r 1, that geometric series is convergent. »» When r 1 or r 1, that geometric series is divergent. Q. A.
Is the geometric series
1 2
2 5
8 25
32 125
... convergent or divergent?
The series is convergent. In the previous example, you discovered that r the series is convergent.
4 ; therefore, 5
You can calculate the value of any convergent geometric series using the following formula:
ar n n 0
a 1 r
This value equals the infinite sum of all the terms in that series.
Q. A.
What is the value of the convergent geometric series
2 5
8 25
32 125
...?
5 . This is the series from the previous example, so you already know that a 2 4 r . Thus: 5 1 2
228
1 2
2 5
8 25
PART 6 Infinite Series
32 125
...
a 1 r
1 2
4 1 5
1 2 1 5
1 5 2
5 2
1 and 2
11
Write both the sigma notation and expanded notation for geometric series with the following values of a and r: a. a
2, r
b. a
1, r
c. a
3, r
d. a
2 ,r 7
12
Test each of the following series to see if it’s a geometric series. For each series that is geometric, find the values of a and r:
5
a. 1 4 16
2 3 2 5
1 b. 9
1 3
64 ... 1 25
c.
3 4
3 5
12 25
d.
1 2
1 3
1 4
e.
1 6
5 18
f.
13
1 16
2 3
1 2
1 36
...
48 125
...
1 ... 5
25 54
125 162
3 8
9 32
... ...
For each of the geometric series in Question 12, identify whether each is convergent or divergent. Then, for each convergent series, calculate the sum of the series.
CHAPTER 15 Sequences and Series
229
Understanding p-series A p-series is any series of the following form:
1 n 1n
p
1 2p
1
1 3p
1 4p
...
Here are a few examples of p-series, with their respective p values:
n
1 2 n 1 1
n 1n
1
1 n 1
Q. A.
1 4
1
1 n2
1 9
1 16
n
p
1 2 3 4 ... 1 2
1
1 3
1 4
Express the series 1
1 4 n 1
p
...
1
1 16
1 81
...
p
2
1 1 2
1 1 1 ... in sigma notation to show that it’s a p-series. 16 81 256 1 .... This is a p-series with p 4 . 256
Unlike with geometric series, no simple formula exists for finding the value of a convergent p-series. However, a simple convergence test does exist:
Any p-series with p
Q. A.
14
1 2
One important series is referred to as the harmonic series: 1 harmonic series is divergent.
1
Begin by expressing the harmonic series in sigma form: n 1n p-series with p 1; therefore, it’s a divergent series.
1 3 1
1 4 1 2
.... Show that the 1 3
1 4
... . This is a
Write the sigma notation and then expand out the first four terms for the p-series with the following values of p: a. p b. p c. p d. p
230
1 is convergent; any p-series with p 1 is divergent.
3 2 1 3 1 2
PART 6 Infinite Series
Answers and Explanations 1
2
3
a.
an
1 3n
b.
an
n 1 3n 1
c.
an
n2 2n 2
d.
an
a.
an
2 4 8 16 , , , , ... 5 25 125 625
b.
an
1 2 3 4 , , , , ... 4 9 16 25
c.
an
3,
d.
an
4 7 10 14 , , , , ... 5 11 29 83
3
25 5!
32 120
15 1 1015 12
16 10 3
5
n
2n n 7 0
2 7
4 49
n
n2 1 1 n!
2 1
5 2
n
4n 1 3 n 2 2
9 6
13 25
n
n4 7 2 n 1 1n 2
b. c. d. 6
a. n 0
b. n
8 3
1 1 9 1 , , , , ... 8 4 32 4
2n 5n
2 5
n n 1
n
2
2n 1 4n 1 3n 1 3n 2
4 15
4 1, 000
1 250
6 343 10 6
23 17
1 2 1 , , , ... 3 9 9
8 27 64 125 , , , , ... 1 5 23 119
1
5 7 9 , , , ... 4 16 64
0
0,
9 16 1 4 , , , , ... 8 16 32 64
4 a15
a.
1 2 3 , , , ... 3 9 27
0,
n 1 n 1 !
26 1 6 1 !
a6
1 1 1 1 , , , , ... 3 6 9 12
...
17 24
17 62 88 73
0.004
... 2
21 123 263 257
...
5 2
5 3
17 24
3 2
13 25
17 62
... 7 ... 41
...
3n 5n
n 2 7 0 n 21
n 0
35 n n 1
n 0
2n 1 2n 3
c. d.
2
CHAPTER 15 Sequences and Series
231
7
a. n 1
b.
3 n 1 5n 1 n 1 14
n 1 7n
c. n
36 n n 1 2n 2n 1
d. n 1
8
2
2n 1 2n 1
Sn
1 3 7 15 , , , , ... . Calculate the first four partial sums for this series as 4 8 16 32
follows:
S1 S2 S3 S4
2 n 2 3 n 2 4 n 2 5 n 2
1 2
n
1 4
1 2
n
1 4
1 8
3 8
1 2
n
1 4
1 8
1 16
7 16
1 2
n
1 4
1 8
1 16
1 32
15 32
Express these four values as the expanded form of a sequence Sn :
1 3 7 15 , , , , ... 4 8 16 32
Sn
Now, find a function of n that produces this sequence (remember that, by convention, the index of a series starts with n 1):
9
Sn
2n 1 2n 1
Sn
n n 1 2
1, 3, 6, 10, ... . Calculate the first four partial sums for this series as
follows:
S1 S2 S3 S4
232
1
n 1
n 1 2
n 1 2
3
n 1 3
n 1 2 3
6
n 1 4
n 1 2 3 4 10
n 1
PART 6 Infinite Series
Express these four values as the expanded form of a sequence Sn :
Sn
1, 3, 6, 10, ...
This result, called the sequence of triangular numbers, has the following compact form:
n n 1 2
Sn 10
2
n n 1 2
Sn
1, 9, 36, 100, ... . Calculate the first four partial sums for this series
as follows:
S1
1
n3
1
n3
1 8
n3
1 8 27
n3
1 8 27 64 100
n 1 2
S2
9
n 1 3
S3
36
n 1 4
S4
n 1
Express these four values as the expanded form of a sequence Sn :
Sn
1, 9, 36, 100, ...
These four numbers are square numbers, respectively 12 , 3 2, 6 2, and 10 2. In each case, the result is the square of the sequence of triangular numbers that you found in Question 9. Thus, it has the following compact form:
n n 1 2
Sn
11 a.
2
2( 5 ) n
2 10 50 250 ...
2 1( ) n 3
1
2 3
4 9
2 3( ) n 5 0
3
6 5
12 25
n 0
b. n 0
c. n
d.
2 n 07
(
1 n ) 3
2 7
2 21
8 27
...
24 125 2 63
... 2 189
...
CHAPTER 15 Sequences and Series
233
12 a. Geometric. a
4 1
1 and r
1 and r 9
b. Not geometric: a
64, which is correct.
9 , which is incorrect. 256
12 and ar 3 25
2 . Thus, ar 2 3
5 . Thus, ar 2 3 1 2 2 3
2 and r 3
f. Geometric: a
9 . Thus, ar 2 16
1 3 1 2
5 18 1 6
1 and r 6
16 and ar 3
4 . Thus, ar 2 5
1 and r 2
d. Not geometric: a
e. Geometric: a
1 16 1 9
3 5 3 4
3 and r 4
c. Geometric: a
4 . Thus, ar 2
48 , which is correct. 125
2 , which is incorrect. 9
25 and ar 3 54
3 . Thus, ar 2 4
125 , which is correct. 162
3 and ar 3 8
9 , which is correct. 32
13 In Question 12, only a, c, e, and f are geometric series; b and d are not geometric, so the question isn’t applicable.
1.
a. Divergent: r
c. Convergent: 1 r
1. In this series, r
is convergent. Additionally, a
1.
f. Convergent:
7 . In this series, r 6
14 a.
1 n 1n
b. n
1 n 1
n 1
n
1 2
1 27
1 64
...
1 4 9 16... 1
1 n3
1
d.
1 8
1
3
1 2 n 1
c.
234
a 3 , so 4 1 r
e. Divergent: r
convergent. Additionally, a
1 2
3
1
PART 6 Infinite Series
2
1 3
3
3
1 ... 4
3
2...
3 5 3 4
1 2 2 3
2 a , so 3 1 r
4 , so 1 r 5 3 4 1 5
3 4
4 1 5
3 , so 1 r 4 2 3 1
3 4
1; therefore, this geometric series
3 5 4
15 . 4
1; therefore, this geometric series is 2 3 7 4
2 4 3 7
8 . 21
IN THIS CHAPTER
»» Understanding the nth-term test for divergence »» Using the direct comparison test to show convergence and divergence »» Applying the limit comparison test to a series »» Using the integral test to decide whether a series is convergent or divergent »» Applying the ratio test to a series with exponents and factorials »» Using the root test when a series has an exponent of n
16 Convergent and Divergent Series Chapter
I
n this chapter, you expand your understanding of infinite series with a variety of tests for convergence or divergence.
To begin, you see how the nth-term test provides an easy and intuitive way to show that many series are divergent. Then, you see how two comparison tests — the direct comparison test and the limit comparison test — allow you to use information about a series that you know to find out whether a new series is convergent or divergent. Next, you break out your integration skills and apply it to series with the integral test. And to finish up, you use the ratio test and the root test to find out whether more complex series are convergent or divergent.
CHAPTER 16 Convergent and Divergent Series
235
Nth-Term Test for Divergence A simple but helpful test for divergence is the nth-term test. This test tells you that if the defining sequence a n for a series Here’s the formal statement:
If lim a n n
0, then n 1
n 1
a n doesn’t ultimately approach 0, then the series is divergent.
a n is a divergent series.
To get an intuition as to why this test makes sense, consider the following series:
n 1
n 1 n
0
1 2
2 3
3 4
...
As you can see, the defining sequence of the series is
n 1 , which approaches 1. Thus, the n
series sums up infinitely many values that get closer and closer to 1, so it’s no surprise that it increases without bound. When using the nth-term test, use a limit to show divergence formally. For example:
Q. A.
Use the nth-term test to show that n
To prove that n
n3 1 3 2n 7n 3
n3 1 is a divergent series. 7n 3 1 2n 3
n3 1 is divergent, you want to demonstrate that the sequence 3 7n 3 1 2n
doesn’t approach 0. That is, you want to show that the following state-
ment is true:
lim
n
n3 1 2n 7n 3 3
0.
To do this, notice that
n3 1 is a rational function with a third-degree polynomial 2n 7n 3 3
in both the numerator and denominator. Thus, only the two third-degree terms affect the limit, so you can simplify it as follows:
lim
n
n3 1 2n 7n 3 3
lim
n
n3 2n 3
lim
n
1 2
1 2 n3 1 doesn’t approach 0, so the 7n 3
Therefore, the defining sequence of the series 3 n 1 2n sequence diverges.
1
236
n4
Show that the series 4 n 1 3n divergent.
PART 6 Infinite Series
n2 3 is 2n 3 n 1
2
Show that the series n
7n 7 is divergent. n 99 100 1
3
4
Use the nth-term test to show that the series
1 2
2 3
3 4
4 5
Use the nth-term test to show that the series
1 2 3 4 ... is divergent.
... is divergent.
The Direct Comparison Test A good way to test a series for convergence or divergence is called the direct comparison test. This test depends upon a series whose behavior you already know, called a benchmark series. To prove that a series
a n is convergent using the direct comparison test:
1. Choose a benchmark
bn set that’s convergent.
2. For every value of n (remember, n is a positive integer!), show that a n
bn.
When testing for convergence, if the test fails — that is, if a n b n — then the test is inconclusive, so you can draw no conclusion about the series you’re testing.
Q.
Use your knowledge that the p-series n
that the series
A.
To prove that
1 n 1n
2
1 n 1n
1 4
3
2
3
1 7
1
1 12
1 19
1
1 4
1 9
1 16
... is convergent to show
... is also convergent.
is convergent, you want to show that for every value of n, the
corresponding value of 2 n 1n example:
1 4
1 7
1 2 1n
1 4
1 3
is less than that of the benchmark series n
1 12
1 9
1 19
1 . For 2 n 1
1 16
As you can see, each term of the given series is less than that of the convergent benchmark series. To begin developing the proof, start with the formal statement you want to prove:
1 n2
3
1 n2
Remember that n is always positive, so cross-multiply and then subtract n 2 :
n2 n2 0 3
3
CHAPTER 16 Convergent and Divergent Series
237
Now, to write the proof formally, reverse the steps you just wrote, starting with the obvious premise that 0 is less than or equal to 3:
0 3 n
2
n2
1 n2
3
3 1 n2
As you can see, the resulting proof begins with a premise that's obviously true, and then uses algebra to arrive at the statement you want to prove. This formally demonstrates that every term of the given series is less than the corresponding term of the benchmark series, so the given series is convergent.
a n is divergent using the direct comparison test:
To prove that a series
1. Choose a benchmark
bn set that’s divergent.
2. For every value of n (remember, n is a positive integer!), show that a n
bn.
When testing for divergence, if the test fails — that is, if a n b n — then the test is inconclusive, so you can draw no conclusion about the series you’re testing.
Q.
Use your knowledge that the harmonic series that the series n
A.
To prove that
2 1n 2
n 1n
2 2 1 3 2 1
2 3
1 2 1 2
of n, the corresponding value of For example:
2 1
1
1 2
2 3
1 3
1 n 1n
1
1 2
1 3
1 4
... is divergent to show
... is also divergent. ... is divergent, you want to show that for every value 2 n 1n
is greater than that of the benchmark series
1 2
1 n 1n
.
1 4
Here is the formal statement you want to prove:
2 n
1 n
Remember that n is always positive, so multiply both sides by n:
2 1 To write the proof, reverse these steps:
2 1 2 1 n n This time, starting with a premise that's clearly true, the proof formally demonstrates that every term of the given series is greater than the corresponding term of the divergent benchmark series. Therefore, the given series is divergent.
238
PART 6 Infinite Series
5
6
Use your knowledge that the geometric series n
1 n 02
1 1 2
1 4
1 8
... is
p-series
convergent to show that the series
1 n 0
2n
2
1 3
1 4
1 6
1 10
1 n 1n
1 8
1
3
1 27
1 64
... is
convergent to show that the series
1
... is also
n 1 2n
convergent.
7
Use your knowledge that the
3
1
1 3
1 17
1 55
1 129
... is
also convergent.
2
Show that the series n 1n 1 is divergent.
1
2 3
1 2
2 ... 5
8
Show that the series
1 n
2 1n
1 6
5
1 9
1 14
1 ... is 21
convergent.
9
Is the series n
1 n 2 1 0
1 2
convergent or divergent?
1 3
1 5
1 ... 9
10
Is the series
2 n 11
3
n
2
1
2 1
3
2 4
1
3
2 9
1
3
16
...
convergent or divergent?
CHAPTER 16 Convergent and Divergent Series
239
The Limit Comparison Test As with the direct comparison test, which I introduced in the previous section, the limit comparison test allows you to use a benchmark series whose behavior you already know to test whether a new series is convergent or divergent.
Given a test series
lim
n
an bn
a n and a benchmark series
or
lim
n
bn , find either of the following limits:
bn an
If the limit evaluates as a positive number, then either both series converge or both diverge. As with the direct comparison test, if this test fails, it’s inconclusive and you can draw no conclusion about the series you’re testing.
Q.
Use your knowledge that the harmonic series that the series n
A.
To prove that n
n 3 2 5 2 1 n
2 7
1 22
0
1 82
1 n 1n
1
1 2
1 3
1 4
... is divergent to show
... also diverges.
n 3 is divergent, you want to show that the following limit equals a 2 2 1 5n
positive number:
n 3 2 lim 5n 2 n 1 n Simplify the limit as follows:
lim
n
n 3 n 5n 2 2 1
lim
n
n 2 3n 5n 2 2
As you can see, the numerator and denominator inside the limit are both second-degree polynomials, so this limit equals the ratio of the coefficients of the two leading terms of these polynomials:
1 5 Thus, the limit equals a positive number, so the behavior of the test series matches that of the benchmark series; therefore, n
240
PART 6 Infinite Series
n 3 diverges. 2 2 1n
11
Use your knowledge that the geometric series n
1 n 12
1 2
1 4
1 8
1 16
12
... is
p-series
convergent to show that the series
1 n 12
n
1
1
1 3
1 7
1 15
Use your knowledge that the
1 2 n 1 3 n
1
1 4
1 9
3
3
3
1 ... is 16
divergent to show that the series
... is also
1 2 n 1 3 n
convergent.
2 3
3
1 4 3
3
1 9 3
3
1 ... 16 3
is also divergent.
13
Use the limit comparison test to show that the series n
n 8 is divergent. 2 5n 7 1n
14
Use the limit comparison test to decide
n
3
whether the series n 1n 1 4 or divergent.
n
is convergent
CHAPTER 16 Convergent and Divergent Series
241
Integral Test The integral test allows you to discover whether a series converges or diverges by integrating a modified form of that series:
For any series of the form
f ( n ), consider its associated integral f ( x ) dx . n a
a
If this integral converges, the series also converges; if this integral diverges, the series also diverges.
Q.
Use the integral test to find out whether the series
A.
Diverges. You want to evaluate the following improper integral:
n
6x 2 5 dx 3 5x 1 2x
6n 2 5 converges or diverges. 3 5n 1 2n
c
lim
c
6x 2 5 dx 3 5x 1 2x
Apply the u-substitution u
2x 3
5 x and du
6x 2
5:
u c
lim
c
u
1 du u 3
Evaluate the integral and the limit:
lim ln u
c
u c u 3
lim ln c
c
ln 3
Now, you can see that as c increases without bound, the limit diverges to infinity; therefore, the series n
15
Use the integral test to show whether the series n
242
6n 2 5 is also divergent. 3 5n 1 2n
1 converges or diverges. n 0e
PART 6 Infinite Series
16
Use the integral test to show whether the series n 2
1 converges or diverges. n ln n
17
18
Use the integral test to show whether the series n
n2 1 converges or diverges. 3 3n 7 0n
Use the integral test to show whether the series n
ln n converges or diverges. 2 1 n
Ratio Test Another indispensable test for convergence and divergence is the ratio test: Given a sequence
a n , consider lim n
an 1 : an
If lim
an 1 an
1, the series converges.
If lim
an 1 an
1 or divergent, the series diverges.
If lim
an 1 an
1, the test is inconclusive.
n
n
n
The ratio test works especially well with series based on exponents and/or factorials. For example:
Q. A.
n!
Use the ratio test to discover whether the series n n 15 gent or divergent.
1 5
2 25
6 125
24 625
... is conver-
Divergent. To use the ratio test, set up the following limit:
lim
n
n 1 ! 5n 1 n! 5n
lim
n
n 1 ! 5n 5n
1
n!
CHAPTER 16 Convergent and Divergent Series
243
As you can see, you can remove the absolute value bars because all values are positive. n 1 n ! into the limit and Now, simplify by substituting 5 n 1 5 5 n and n 1 ! then cancelling factors:
lim
n
n 1 n! 5 n 5 5
n
n!
lim
n 1
n
5 5
5n
n! n
lim
n
n!
n 1 5
Thus, the series is divergent.
19
Use the ratio test to show whether the series n
21
2 49
6 343
24 2, 401
120 16, 807
diverges.
244
PART 6 Infinite Series
Use the ratio test to show whether the series n
Use the ratio test to show whether the series
1 7
20
3n converges or diverges. 1 n!
... converges or
22
n! converges or diverges. n 1n
Why is the ratio test inconclusive if you try to use it to test the series
3 4
5 9
7 14
9 19
or divergence?
11 24
... for convergence
Root Test The root test is similar to the ratio test, allowing you to rewrite a sequence as a limit and then evaluate it to test for convergence or divergence.
a n , consider lim n a n :
Given a sequence
n
If lim n a n
1, the series converges.
If lim n a n
1 or divergent, the series diverges.
If lim n a n
1, the test is inconclusive.
n
n n
The root test works especially well with series that are raised to the power of n. For example: n
Q.
Use the root test to discover whether the series
A.
Convergent. To use the root test, set up the following limit:
n 1
lim n
n
n2 5 3n 2 1
n
lim
n
n2 5 3n 2 1
is convergent or divergent.
n2 5 3n 2 1
Here, you can remove the absolute value bars because all values are positive. As you can see, the root and the exponent cancel out, leaving you with a limit that you can evaluate easily:
lim
n
n2 3n 2
1 3
Thus, the series is convergent.
23
Use the root test to show whether the
2n
series n 1
3
2
4n 5n 4
10n 7 8
n
converges
24
Use the root test to show whether the series n 1
3n 1 n2 2
n
converges or diverges.
or diverges.
CHAPTER 16 Convergent and Divergent Series
245
Answers and Explanations 1
n4
n2 3 is divergent, you want to show that the following statement 2n 3 n 1
To prove that 4 n 1 3n is true:
lim
n
n4 3n
4
n2 3 2n 3 n 1
0
To do this, notice that
n4 3n
4
n2 3 is a rational function with fourth-degree polynomials 2n 3 n 1
in both the numerator and denominator. Thus, only the two fourth-degree terms affect the limit, so you can simplify it as follows:
lim
n
n4 3n
4
n2 3 2n 3 n 1
lim
n
n4 3n 4
lim
n
1 3
1 3
Therefore, the defining sequence of the series converges to 2
7n 7
1 , so the series diverges. 3
To prove that is divergent, you want to show that the following statement is n 1 100n 99 true:
lim
n
7n 7 100n 100
0
To do this, notice that
7n 7 is a rational function with first-degree polynomials in both 100n 99
the numerator and denominator. Thus, only the two first-degree terms affect the limit, so you can simplify it as follows:
lim
n
7n 7 100n 99
lim
n
7n 100n
lim
n
7 100
7 100
Therefore, the defining sequence of the series converges to 3
To prove that
n
1 2
2 3
3 4
4 5
7 , so the series diverges. 100
... is divergent, first write it in sigma notation:
n n 1 1
Now, you want to show the following:
lim
n
n n 1
0
To do this, notice that
n is a rational function with first-degree polynomials in both the n 1
numerator and denominator. Thus, only the two first-degree terms affect the limit, so you can simplify it as follows:
lim
n
n n 1
lim
n
n n
lim 1 1
n
Therefore, the defining sequence of the series converges to 1, so the series diverges.
246
PART 6 Infinite Series
4
3 4 ... is divergent, first write it in sigma notation:
To prove that 1 2
n n 1
Now, you want to show the following:
lim n 0
n
To do this, evaluate the limit by substituting
for n:
lim n
n
Therefore, the defining sequence of the series diverges, so the series diverges. 5
1
To prove that n 0
2n
2
is convergent, you want to show that for every value of n, the corre-
1
sponding value of
n 02
n
2
is less than that of the benchmark series
proof, beginning with the premise 2 n
n
2 2 n:
1 . Here’s a formal n 2 0
2n 2 2n 1 1 2n 2 2n 1
Thus, by the direct comparison test, n 0
6
To prove that
1 n 1 2n
sponding value of
3
1
2
is convergent.
is convergent, you want to show that for every value of n, the corre-
1 n 1 2n
2
n
3
1
is less than that of the benchmark series
statement and proof, beginning with the premise n 3
n3 n
3
1 n 1n
3
. Here’s a formal
1, because n is always positive:
1
1 0
2n 3 1 n 3 1 1 3 2n 1 n 3 Thus, by the direct comparison test, 7
To prove that
2 n 1n
1
1 n 1 2n
3
1
is convergent.
is divergent, you want to choose a similar benchmark series that you
know is divergent, and then show that every term in
2 is greater. The harmonic series n 1 1
n 1 is a divergent p-series, so this is a good choice. Here’s a formal proof, beginning with n 1n the premise that n 1:
n 1 n n n 1 2n n 1 1 1 2n n 1 1 2 n n 1 CHAPTER 16 Convergent and Divergent Series
247
Thus, by the direct comparison test, 8
To prove that
1 n 1n
2
5
2 n 1n
1
is divergent.
is convergent, you want to choose a similar benchmark series that you
1 is a convergent p-series, so this is a good choice. Here’s 2 1n
know is convergent. The series n
a formal statement and proof:
0 5 n2 1 n2
n2
5 1
n2
5
Thus, by the direct comparison test, 9
2
5
is convergent.
1 . Thus, see n 1 n 02 n 02 1 1 if you can show that every term of is less than the corresponding term of . n n 2 1 2 n 0 n 0 Here’s a formal proof, beginning with the premise 0 1: Convergent. The series
1
1 n 1n
is similar to the convergent geometric series
n
0 1 2n 1 2n
2n 1 1 n 2 1
Thus, by the direct comparison test, n 0
10 Divergent. The series
2 n 11
3
n2
1 is convergent. 2n 1 1
is similar to the p-series n 1
3
n2
, in which p
therefore, divergent. Thus, see if you can show that every term of
1
the corresponding term of n 1
3
n2
3
n2
1
3
n2
1
3
n2
2 n2 1
1
3
n2
3
3
2 n 1 3
2
n2
3
n
3
3
n2 n2
Thus, by the direct comparison test,
248
PART 6 Infinite Series
n2
is greater than
. Here’s a formal proof, beginning with the premise n 2
2 1
3
3
2
1 1
2 n 11
2 and is, 3
2 n 11
3
n2
is divergent.
1:
11 To prove that tive number:
1 n 12
n
1
is convergent, you want to show that the following limit equals a posi-
1 2n 1
lim
n
2n 1 Evaluate the limit as follows:
lim
n
1 2n 1 1 2n
lim
n
2n 1 2n
2n 2n
lim
n
1 2n
lim 1
n
1 2n
1
Thus, the limit equals a positive number, so the behavior of the test series matches that of the benchmark series; therefore,
1 n 12
n
converges.
1
1 12 To prove that is divergent, you want to show that the following limit equals a posi2 n 1 3 tive number: n 3 1 2
n3 1
lim
n
2
n3
3
Evaluate the limit as follows: 2
lim
n
1 2 n3
2
n3
3
lim
1
n
n3
2
3
2 n3
lim
n
n3
3
2 n3
2 n3
3
lim 1
n
2
1
n3
Thus, the limit equals a positive number, so the behavior of the test series matches that of the benchmark series; therefore,
1 2 n 1 3 n
diverges.
3
13 The series has a polynomial in the numerator that’s one degree less than the polynomial in
1
, which is divergent, so try this as your the denominator. This is similar to the series n 1n benchmark:
2 lim n
n
n 8 5n 7 1 n
lim
n
n n 8 n 2 5n 7
lim
n
n2 n2
8n 5n 7
lim
n
n2 n2
1
Thus, the limit equals a positive number, so the behavior of the test series matches that of the benchmark series; therefore,
n 1n
2
n 8 diverges. 5n 7
CHAPTER 16 Convergent and Divergent Series
249
n
14 Convergent. To test n
3 1n 1 4
n
, you can try using either the series
gent by the nth-term test, or the convergent geometric series n 1
series. First, try n
3 4
n n
n , which is divern 1 1 as your benchmark
n : n 1 1
n
n
3 n 1 4 lim n n n 1
lim
n
3 4
n
0
This limit converges to 0, so the test is inconclusive. Next, try a benchmark of n 1
3 4
n
:
n
n
3 n 1 4 lim n 3 n 4
lim
n
n n 1
lim
n
n n
1
Thus, the limit equals a positive number, so the behavior of the test series matches that of the benchmark series; therefore, n
n 3 1n 1 4
n
converges.
15 Convergent. You want to evaluate the following improper integral:
1 dx x 0 e
c
lim
c
0
1 dx ex
Rewrite the integral using a negative exponent and evaluate it as a linear input to the e x function, as shown in Chapter 7, then evaluate the limit: c
lim e x dx
c
0
lim
c
e
x c 0
lim
c
1 ec
1 e0
1
The limit converges, so the series also converges. 16 Divergent. You want to evaluate the following improper integral:
2
1 dx x ln x
c
lim
c
2
1 dx x ln x
Rewrite the integral using u-substitution with u c
lim
c
ln 2
1 du lim ln u c u
c ln 2
lim ln c ln(ln 2 )
c
The limit diverges, so the series diverges.
250
PART 6 Infinite Series
ln x and du
1 dx : x
17 Divergent. You want to evaluate the following improper integral:
0
x2 1 dx x 3x 7 3
c
x2 1 dx x 3x 7
lim
c
3
0
Rewrite the integral using u-substitution with u
du 3
x
2
3 x 7 and du
3x 2
3 dx , so
1 dx :
c
lim
c
x3
c
1 du 3 u 0
1 1 du lim 3c 7u
1 lim ln u 3c
1 lim ln c ln7 3c
c 7
The limit diverges, so the series diverges. 18 Convergent. You want to evaluate the following improper integral: c
ln x dx 2 1 x
lim
c
ln x dx 2 1 x
Rewrite the related indefinite integral using integration by parts as follows:
u
ln x 1 dx x
du
dv v
1 dx x2 1 dx x
Now, use the formula u dv
ln x dx x2
1 ln x x
v du:
uv
1 dx x2
1 ln x x
1 x
C
Use this result to evaluate the definite integral: c
lim
c
ln x dx 2 1 x
ln x x
lim
c
1 x
c
ln c c
lim
c
1
1 c
ln 1 1
1 1
lim
c
ln c c
1 1 c
Break this into three separate limits and evaluate the second and third:
lim
c
ln c c
lim
c
1 c
lim 1
lim
c
c
ln c c
1
Evaluate the remaining limit using L’Hôpital’s rule:
d ln c dc lim c d c dc
1
lim
c
1 c
1 1
The limit converges, so the series converges.
CHAPTER 16 Convergent and Divergent Series
251
19 Convergent. To use the ratio test, set up the following limit:
lim
n
3n 1 n 1 ! 3n n!
3n
lim
1
n!
n 1 ! 3n
n
You can remove the absolute value bars because all values are positive. Simplify by substitutn 1 n ! into the limit and then cancelling factors: ing 3 n 1 3 3 n and n 1 !
3 3n
lim
n! n
n 1 n! 3
n
lim
n
3 3n n 1
n!
lim
3n
n!
n
3 n 1
0
Thus, the series is convergent. 20 Convergent. To use the ratio test, set up the following limit:
lim
n
n 1 ! n 1 n1 n! nn
n 1 ! nn
lim
n 1
n
n 1
n!
You can remove the absolute value bars because all values are positive. Simplify by substituting n 1 n1 n 1 n 1 n and n 1 ! n 1 n ! into the limit and then cancelling factors:
n 1 n! nn
lim
n 1 n 1
n
n
n 1
lim
n!
n 1
n
nn
n! n 1
n
Now, use the equivalent reciprocal form of
1 n 1 n
lim
n
n! n n 1
lim
n n
nn n 1
n
lim
n
n
n
n 1
:
n
Here, you can rearrange the denominator and then split it into the division of two limits:
lim
n
1 n n
1 n
n
lim
n
1 1 1 n
lim 1
n
n
lim 1
n
1 n
n
Now, evaluate the limit in the numerator as 1, and the limit in the numerator as e by the identity e
1 e
lim 1
n
1 n
n
:
0.368
This value is less than 1, so by the ratio test, the series is convergent.
252
PART 6 Infinite Series
21 Divergent. To begin, put the series in sigma notation:
1 7
2 49
6 343
24 2, 401
120 16, 807
... n
n! n 17
To use the ratio test, set up the limit for the ratio test, then simplify:
n 1 ! 7n 1 n! 7n
lim
n
n 1 ! n 1 lim 7 n 2n 3 5n 4
lim
n 1 ! 7n 7n
n
1
n!
You can remove the absolute value bars because all values are positive. Simplify by substitutn 1 n ! into the limit and then cancelling factors: ing 7 n 1 7 7 n and n 1 !
n 1 n! 7n
lim
7 7n
n
lim
n!
n 1 7 7n
n
7n
n! n!
lim
n
n 1 7
The limit is divergent, so the original series is also divergent. 22 To begin, put the series in sigma notation:
3 4
5 9
7 14
9 19
11 24
... n 0
2n 3 5n 4
To use the ratio test, set up the limit for the ratio test, then simplify and evaluate:
2 n 1 3 5 n 1 4 lim n 2n 3 5n 4
2n lim 5n n 2n 5n
5 9 3 4
lim
n
2n 5 5n 4 5n 9 2n 3
lim
n
10n 2 10n 2
33n 20 33n 27
liim
n
10n 2 10n 2
1
The result is 1, therefore the ratio test is inconclusive. 23 Convergent. To use the root test, set up and evaluate the following limit:
lim n
2n 3
n
4 n 2 10n 7 5n 4 8
n
lim
n
2n 3
4 n 2 10n 7 5n 4 8
Here, you can remove the absolute value bars because all values are positive. As you can see, the root and the exponent cancel out, leaving you with a limit that you can evaluate easily. Now, evaluate the limit using only the two leading terms of the rational function:
lim
n
2n 3 5n 4
2 5n
0
Thus, the value of this limit is less than 1, so the series is convergent.
CHAPTER 16 Convergent and Divergent Series
253
24 Convergent. To use the root test, set up and evaluate the following limit:
lim n
n
3n 1 n2 2
n
lim
n
3n 1 n2 2
lim
n
3n n2
lim
n
3 n
0
Here, you can remove the absolute value bars because all values are positive. As you can see, the root and the exponent cancel out, leaving you with a limit that you can evaluate easily. Thus, the series is convergent.
254
PART 6 Infinite Series
IN THIS CHAPTER
»» Estimating functions using Maclaurin series »» Improving estimates using Taylor series
17 Taylor and Maclaurin Series Chapter
I
have to admit that, even as a math teacher, rewriting functions as infinite series seems magical to me. In this chapter, you get to see how some of this magic happens.
To begin, you discover how to express functions as Maclaurin series, which is a power series that’s a limited form of the more versatile but more complex Taylor series. Then, you proceed to the main event, which is expressing functions as Taylor series.
Expressing Functions as Maclaurin Series You can express any infinitely differentiable function as a Maclaurin series using the following formula:
f(x) n
f ( n) ( 0) n x n! 0
CHAPTER 17 Taylor and Maclaurin Series
255
The notation f ( n ) means “the nth derivative of f.” To help make sense of this notation, here’s the expanded version of the Maclaurin series:
f(x)
f ''(0) 2 x 2!
f (0) f '(0) x
f '''(0) 3 x 3!
...
Q.
Express the function sin x as a Maclaurin series expanded out to the first four terms.
A.
sin x
1 3 x 3
x
1 x5 120
1 x7 5, 040
... . Begin by finding the first three derivatives of
sin x, and substitute 0 into each of these derivatives. To keep this work clear, I use a table:
f ( x ) sin x
f ( 0 ) sin 0 0
f '( x ) cos x
f '(0) cos0 1
f ''( x )
sin x
f ''(0)
sin0 0
f '''( x )
cos x
f '''(0)
cos0
1
Plug the values in the second column into the expanded formula for the Maclaurin series:
sin x
0 1x x
0 2 x 2!
1 3 x 3
1 3 x 3!
...
...
Here, because two of the first four terms evaluate to 0, you need to find the next four derivatives. Fortunately, these values repeat the cycle shown in the previous table:
f(x)
f ( 4 ) ( x ) sin x
f (0)
f ( 4 ) ( 0) 0
f '( x )
f ( 5) ( x ) cos x
f '(0)
f ( 5) (0) 1
f ''( x )
f (6) ( x )
sin x
f ''(0)
f (6) (0) 0
f '''( x )
f (7) ( x )
cos x
f '''(0)
f (7) (0)
1
So here is the series taken out to enough terms to complete the problem:
sin x
x
256
0 2 1 3 0 4 1 5 x x x x 2! 3! 4! 5! 1 1 3 1 5 x x x 7 ... 3 120 5, 040
0 1x
PART 6 Infinite Series
0 6 x 6!
1 7 x 7!
...
Q. A.
Using only the four terms you found in the previous question, estimate the value of sin 2.
sin x 0.908. To begin, write an estimate of sin x using the first four terms you’ve already found: sin x
x
1 3 x 3
1 5 x 120
1 x7 5, 040
As you can see, in changing from an infinite series to a series with four terms, I change the equation to an approximation. Now, use this approximation to estimate sin 2:
sin x
2
1 2 3
3
1 2 120
5
1 2 5, 040
7
2
8 3
32 120
128 5, 040
0.908
The actual value of sin x is a little closer to 0.909, but this is still a good estimate.
1
3
Express the function cos x as a Maclaurin series expanded out to the first four terms.
2
Express the function e 2 x as a Maclaurin series expanded out to the first six terms.
4
Using only the four terms you found in
1 2
Question 1, estimate the value of cos .
Using only the six terms you found in Question 2, estimate the value of e 2.
CHAPTER 17 Taylor and Maclaurin Series
257
5
Express the function 5 sin
x as a Maclaurin 3
6
series expanded out to the first three terms.
7
258
Express the function 2 x as a Maclaurin series expanded out to the first four terms.
PART 6 Infinite Series
Using only the three terms you found in Question 3, estimate the value of 5 sin
8
Using only the four terms you found in Question 4, estimate the value of 2 5.
6
.
Expressing Functions as Taylor Series In the previous section, you discovered how to express functions as Maclaurin series. The Taylor series is a more general version of the Maclaurin series. Recall that when using the Maclaurin series, you evaluated a function and a set of its derivatives at the value x 0. In this way, the Maclaurin is centered at 0 — that is, 0 is “hardwired” into the Maclaurin series. This feature makes the Maclaurin series simpler to use than the Taylor series. Unfortunately, when working with values of x that are not very close to 0, the Maclaurin series tends to produce imprecise estimates. In contrast, the Taylor series is a bit more difficult to use, but it allows you to specify a value a where the series is centered. This feature generally makes it possible to use fewer terms to calculate to a greater level of precision using Taylor series. You can express any infinitely differentiable function as a Taylor series using the following formula:
f ( n) ( a ) x a n! 0
f(x) n
n
As with the Maclaurin series, the Taylor series utilizes the notation f ( n ) , which means “the nth derivative of f.” Here’s the expanded version of the Taylor series:
f(x)
Q. A.
f ''( a ) x a 2!
f ( a ) f '( a ) x a
Express the function f ( x ) first four terms.
e
3x
6
e
6
3e
x 2 3x
e
3x
f '''( a ) x a 3!
2
3
...
as a Taylor series centered at a
9e 6 x 2 2
6
27 e 6
2
x 2
3
2 expanded out to the
.... Begin by finding the first
three derivatives of e , and substitute 2 into each of these derivatives. To keep this work clear, I use a table:
f(x) e
3x
f '( x )
3e
f ''( x ) 9e f '''( x )
3( 2 )
e
f '(2) e 3
3(2)
f ( 2) e 3x
3x
27e
f ''(2) 9 3x
3(2)
f '''(2) 27e
6
9e
3(2)
3e
6
27e
6
6
Plug the values in the second column into the expanded formula for the Maclaurin series:
e
3x
e e
6
6
3e 3e
6
6
x 2 x 2
9e 6 x 2 2! 6 9e x 2 2
2
2
27e 3! 27e 6
6
6
x 2
3
...
x 2
3
...
CHAPTER 17 Taylor and Maclaurin Series
259
9
Recall that in Question 8, you used a Maclaurin series to express the function 2 x . Now, use a Taylor series centered at −5, expanded out to the first four terms, to express this same function.
10
Using only the four terms you found in Question 9, estimate the value of 2 5.
11
Use a Taylor series centered at 3, expanded out to the first four terms, to express e x .
12
Using only the four terms you found in Question 11, estimate the value of e .
260
PART 6 Infinite Series
13
Use a Taylor series centered at 6, expanded out to the first four terms, to express sin x .
14
Using only the four terms you found in Question 13, estimate the value of sin2 .
CHAPTER 17 Taylor and Maclaurin Series
261
Answers and Explanations 1
cos x
1
1 2 x 2
1 4 x 24
1 x6 720
.... Begin by finding the first three derivatives of cos x, and
then substitute 0 into each of these derivatives:
f ( x ) cos x
f ( 0 ) cos 0 1
f '( x )
sin x
f '(0)
sin0 0
f ''( x )
cos x
f ''(0)
cos0
f '''( x ) sin x
1
f '''(0) sin0 0
Plug the values in the second column into the expanded formula for the Maclaurin series:
cos x
1 0x 1
1 2 x 2!
1 2 x 2
0 3 x 3!
...
...
Because two of the first four terms evaluate to 0, you need to find the next four derivatives. Fortunately, these values repeat the cycle shown in the previous table:
f(x)
f ( 4 ) ( x ) cos x
f (0)
f ( 4 ) ( 0) 1
f '( x )
f ( 5) ( x )
sin x
f '(0)
f ( 5) (0) 0
f ''( x )
f (6) ( x )
cos x
f ''(0)
f (6) (0)
f '''( x )
f (7) ( x ) sin x
f '''(0)
f (7) (0) 0
1
So here is the series taken out to enough terms to complete the problem:
cos x
2
cos
1 2
1 2 0 3 1 4 x x x 2! 3! 4! 1 6 1 2 1 4 1 x x x ... 2 24 720 1 0x
0 5 x 5!
1 6 x 6!
0 7 x 7!
...
0.878 . To begin, write an estimate of cos x using the first four terms you’ve already
found:
cos x
1
1 2 x 2
1 4 x 24
1 6 x 720 1 2
Now, use this approximation to estimate cos :
cos
1 2
1
1 1 2 2
2
1 1 24 2
4
1 1 720 2
6
1
1 8
1 384
1 46, 080
0.878
(This estimate is the correct value of the expression to 3 decimal places.)
262
PART 6 Infinite Series
3
e2x
4 3 x 3
2x 2
1 2x
2 5 x 3
4 6 x 15
.... Begin by finding the first five derivatives of e 2 x
and then substitute 0 into each of these derivatives:
f ( x ) e2x
f ( 0 ) e 2( 0 )
f '( x ) 2e 2 x
f '(0) 2e 2(0)
2
f ''( x ) 4e 2 x
f ''(0) 4e 2(0)
4
f '''( x ) 8e 2 x
f '''(0) 8e 2(0)
8
f ( 4 ) ( x ) 16e 2 x
f ( 4 ) ( 0 ) 16e 2( 0 )
16
f ( 5 ) ( x ) 32e 2 x
f ( 5 ) ( 0 ) 32e 2( 0 )
32
1
Plug the values in the second column into the expanded formula for the Maclaurin series:
e2x
1 2x
4 2 x 2!
8 3 x 3! 4 3 x 3
1 2x 2x 2 4 e 2
16 2 x 4! 2 2 x 3
32 3 x 5!
4 3 x 15
...
...
7.26 . To begin, write an estimate of e 2 x using the first six terms you’ve already found:
e2x
4 3 x 3
1 2x 2x 2
2 4 x 3
4 5 x 15
Now, use this approximation to estimate e 2:
e2
1 2 1
2 1
2
4 1 3
3
2 1 3
4
4 1 15
5
7.26
(The correct value of the expression to 3 decimal places is 7.389.) 5 5 sin
x 3
of 5 sin
5 x 3
5 x5 29,160
.... Begin by finding the first five derivatives
x , and then substitute 0 into each of these derivatives: 3
f ( x ) 5 sin f '( x )
5 x3 162
x 3
5 x cos 3 3
f ( 0 ) 5 sin
0 3
0
f '(0)
0 5 cos 3 3 0 3
f ''( x )
5 x sin 9 3
f ''(0)
5 sin 9
f '''( x )
5 x cos 27 3
f '''(0)
0 5 cos 27 3
f (4)( x )
5 x sin 81 3
f ( 4 ) ( 0)
5 sin 81
f ( 5)( x )
5 x cos 243 3
f ( 5) ( 0)
0 5 cos 243 3
0 3
5 3 0 5 27 0 5 243
CHAPTER 17 Taylor and Maclaurin Series
263
Plug the values in the second column into the expanded formula for the Maclaurin series:
x 3
5 sin
6 5 sin
5 x 3 5 x 3
5 5 x3 x5 27 3 ! 243 5 ! 5 5 x3 x 5 ... 162 29,160
...
2.325. To begin, write an estimate of 5 sin
6
x using the first three terms 3
you’ve already found:
5 sin
x 3
5 x 3
5 3
6
5 x5 29,160
1 with this approximation to estimate 5 sin 2
Now, use x
5 sin
5 x3 162
1 2
5 162
1 2
3
5 29,160
5
1 2
5 6
:
6
5 1, 296
5 933,120
2.325
(The correct value of the expression is -2.5.) 7 2 x
1
ln 2 2
ln 2 x
2
x2
ln 2 6
3
x3
.... Begin by finding the first three derivatives of 2 x ,
and substitute 0 into each of these derivatives:
f (x) 2x
f ( 0) 2 0
f '( x ) 2 x ln2
f '(0) 2 0 ln2
f ''( x ) 2 x ln2 f '''( x ) 2 x ln2
2
1
f ''(0) 2 0 ln2
3
f '''(0) 2 0 ln2
ln2 2
3
ln2 ln2
2
3
Plug the values in the second column into the expanded formula for the Maclaurin series:
2x
8 2
1
ln 2 x
1
ln 2 x
5
2x
ln 2 2! ln 2 2
2
2
ln 2 3! n2 ln 6
x2 x2
3
3
x3
...
x3
...
3.403. To begin, write an estimate of 2 x using the first four terms you’ve already found: 1
ln 2 2
ln 2 x
2
ln 2 6
x2
3
x3
Now, use this approximation to estimate 2 5:
2
5
ln 2 2 ln 2 5 2 2 6 1 3.466 6.001 6.938 3.403 1
ln 2
5
(The correct value is 2
5
1 32
3
5
3
1 5 ln 2 25
ln 2 2
PART 6 Infinite Series
125
ln 2 6
3
0.03125 , so this estimate is particularly poor. See Questions 9
and 14 for a re-evaluation of this expression using Taylor series.)
264
2
2
ln 2 ln 2 x 5 x 32 32 2 ! x derivatives of 2 , and substitute x
9 2 x
1 32
5
3
ln 2 32 3 !
2
5
x
3
. Begin by finding the first three
5 into each of these derivatives. To keep this work clear,
I use a table:
f (x) 2x
f ( 5) 2
f '( x ) 2 x ln2
f '( 5) 2
f ''( x ) 2 x ln2 f '''( x ) 2 x ln2
2
f ''( 5) 2
3
1 32
5
5
ln2
5
f '''( 5) 2
2
ln2
5
ln2
3
1 ln2 32 1 ln2 2 32 1 ln2 3 32
Plug the values in the second column into the expanded formula for the Taylor series:
1 32
2x 10 2
5
2x
ln 2 x 32
ln 2 2 x 32 2 !
5
5
ln 2 3 x 32 3 !
2
5
3
....
1 . To begin, write an estimate of 2 x using the first four terms you’ve already found: 32 1 32
ln 2 x 32
ln 2 2 x 32 2 !
5
5
ln 2 3 x 32 3 !
2
5
3
Now, use this approximation to estimate 2 5:
2
5
1 32
ln 2 32
5 5
ln 2 2 32 2 !
5 5
2
ln 2 3 32 3 !
5 5
3
1 32
(As you can see, the Taylor series centered at –5 automatically produces the correct value of the function.)
e3 e3 2 3 x 3 x 3 ... . Begin by finding the first three derivatives 2 6 x of e , and substitute x 3 into each of these derivatives. To keep this work clear, I use a table:
11 e x
e3
e3 x 3
f(x) ex
f (3) e 3
f '( x ) e x
f '(3) e 3
f ''( x ) e x
f ''(3) e 3
f '''( x ) e x
f '''(3) e 3
Plug the values in the second column into the expanded formula for the Taylor series:
ex
e3
e3 x 3
e3
e3 x 3
e3 x 3 2! e3 x 3 2
2
2
e3 x 3 3! e3 x 3 6
3
...
3
...
CHAPTER 17 Taylor and Maclaurin Series
265
12 23.151. To begin, write an estimate of e x using the first four terms you’ve already found:
ex
e3
e3 x 3
e3 x 3 2
2
e3 x 3 6
3
Now, use this approximation to estimate e :
e
e3
e3
3
e3 2
3
2
e3 6
3
3
20.096 2.844 0.201 0.010 23.151
(This estimate is slightly higher than the actual value of e rounding errors.)
23.141, because of accumulated
sin 6 cos 6 2 3 .... Begin by finding the first three x 6 x 6 2 6 derivatives of sin x , and substitute x 6 into each of these derivatives. To keep this work
13 sin x
sin 6 cos 6 x
6
clear, I use a table:
f ( x ) sin x
f ( 6 ) sin 6
f '( x ) cos x
f '(6) cos6
f ''( x )
sin x
f ''(6)
sin6
f '''( x )
cos x
f '''(6)
cos6
Plug the values in the second column into the expanded formula for the Taylor series:
sin x
sin 6 x 6 2! sin 6 x 6 2
sin 6 cos 6 x 6 sin 6 cos 6 x 6
2
2
cos 6 x 6 3! cos 6 x 6 6
3
...
3
...
14 0.001. To begin, write an estimate of sin x using the first four terms you’ve already found:
sin x
sin 6 x 6 2
sin 6 cos 6 x 6
2
cos 6 x 6 6
3
Now, use this approximation to estimate sin2 :
sin 2
sin 6 2 6 2 2 0.276 0.279 0.960 0.283 0.283 2 0.279 0.272 0.011 0.003 0.001 sin 6 cos 6 2
6
cos 6 2 6 3 6 0.960 2 0.283 6
3
(This estimate is slightly higher than the actual value of sin2 rounding errors.)
266
PART 6 Infinite Series
0 , because of accumulated
7
The Part of Tens
IN THIS PART . . .
Discover how early mathematicians solved problems using principles of calculus See how skills from Calculus I transfer to Calculus II
IN THIS CHAPTER
»» Greek and Turkish mathematicians grapple with measuring the area under a curve »» Chinese mathematicians independently develop similar methods, improving on earlier measurements »» In India and the Middle East, mathematicians leverage HinduArabic numbers to anticipate methods of calculus »» European mathematicians pick up the gauntlet and calculus is born
18 Ten Mathematicians Who Anticipated Calculus before Newton and Leibniz Chapter
I
saac Newton and Gottfried Leibniz are widely acclaimed as the 16th-century inventors of a unified theory of calculus. But long before they lived, other mathematicians were making observations about the nature of infinity and employing these insights to perform calculations that anticipated calculus. We may never know all of their names, but here are ten mathematicians who lived before Newton and Leibniz, yet found surprisingly innovative ways to do calculus before calculus was actually invented.
CHAPTER 18 Ten Mathematicians Who Anticipated Calculus before Newton and Leibniz
269
Zeno of Elea (495–430 BCE) Zeno of Elea famously proposed several paradoxes showing the difficulty of adding up infinitely many finite values such as
1 2
1 4
1 8
1 16
... . In one version, known as the Dichotomy paradox,
he “proved” that motion is impossible, because traveling the distance between any two points necessarily involves traveling half the distance, plus one-fourth the distance, and so forth. These paradoxes are among the earliest known formulations of infinite series.
Eudoxus of Cnidus (408–355 BCE) Born in what is now southern Turkey, Eudoxus of Cnidus worked extensively as a geometer, providing much of the groundbreaking work later gathered by Euclid. Among his many accomplishments, he derived formulas for the cone and the pyramid using the “method of exhaustion,” a precursor to integration. This method measured progressively smaller areas in a way that heralded the later concept of a limit.
Archimedes (287–212 BCE) Archimedes is commonly remembered as the Greek mathematician who, upon settling into a bath and noticing how the water spilled over the edge, shouted “Eureka!” and ran to the king to show him how to apply this observation to discover whether his crown was made of pure gold. Between baths, Archimedes also applied the method of exhaustion discovered by Eudoxus to measure the area under a parabola with triangles of increasingly small size.
Lui Hui (3rd Century CE) As mathematics started to decline in Europe during the first few centuries of our current era (CE), Chinese mathematicians were making independent discoveries that, in many cases, surpassed those of the West. In China, Lui Hui, in his attempts to measure the area of a circle, developed an algorithm for calculating the value of pi, improving on the best approximations of pi that Archimedes had calculated. His algorithm used polygons of increasingly many sides inscribed inside a circle (topping out at 3,072!), then rearranged these shapes into triangles that could be assembled as a single rectangle of known area. This method anticipated integration methods used centuries later.
270
PART 7 The Part of Tens
Zu Chongzhi (425–500) Among his many other mathematical accomplishments, Chinese mathematician Zu Chongzhi improved upon Lui Hui’s method for calculating pi. These improvements enabled him to inscribe a polygon with 12,288 sides inside a circle to calculate pi accurately to six decimal places. This calculation would stand as the most accurate of its kind for another eight centuries.
Hasan Ibn al-Haytham (965–1040) While progress in math remained relatively stalled in Medieval Europe, Middle Eastern and Indian mathematicians were benefiting greatly from the most important mathematical innovation of the first millennium CE: the number 0. The Hindu-Arabic numeral system (also called decimal numbers) included this value, which the Egyptian, Greek, and Roman systems didn’t possess. Additionally, the Mayans also developed a numerical system that included 0, enabling them to independently produce mathematics still not fully understood in Europe. In the Middle East, mathematicians invented and developed algebra, which opened up new possibilities and presented more advanced ways of framing and solving problems. One such mathematician was Hasan Ibn al-Haytham. In his work on the behavior of light when reflected within a paraboloid (a parabolic solid whose concave surface can be used as a telescope), al-Haytham found a way to measure the volume inside such a structure. This process involved integrating quartic (4th-degree algebraic) equations, propelling al-Haytham to approximate such measurement using algebraic methods that anticipated the later, more general Power rule for integration.
Madhava of Sangamagrama (1340–1425) In India, many of Madhava’s sophisticated mathematical contributions predated similar European results by centuries. He was the first to use infinite series to precisely define values of trigonometric functions (for example, calculating sine and cosine functions as power series) and inverse trig functions (for example, calculating the arctan function as a series based on powers of tangents). These results anticipated the theory that would later be unified as the Taylor series. Additionally, Madhava applied similar methods to finally improve upon Zu Chongzhi’s earlier calculations of pi.
CHAPTER 18 Ten Mathematicians Who Anticipated Calculus before Newton and Leibniz
271
Johannes Kepler (1571–1630) Astronomer Johannes Kepler is well known for having discovered formulas in physics that Isaac Newton later built upon when formulating his theory of universal gravitation. Less famously, Kepler’s work on measuring volume created a similar foundation for the development of calculus. Although his treatise Nova stereometria doliorum vinariorum sounds far more impressive in Latin than when translated into its English equivalent (New Solid Geometry of Wine Barrels), this work set a new standard for thoroughness and coherence in the study of measuring area and volume. Kepler brought together methods that anticipated the use of infinite series and numerical analysis. Additionally, he independently discovered Simpson’s rule, which in German is still referred to as Keplersche Fassregel (again, losing some of its gravitas in its English translation as Kepler’s Barrel Rule).
Bonaventura Cavalieri (1598–1647) A monk born in Milan, Bonaventura Cavalieri developed a geometric technique he called the method of indivisibles. In this method, he used an infinite set of parallel lines inside a shape to measure the area of that shape to any level of precision. Applying it to a square dissected by a parabolic line enabled him to precisely evaluate what would later be termed the definite 1
1 . Later, he used a similar method in three dimensions to 3 0 1 demonstrate that a cone inscribed inside a cylinder must have the volume of that cylinder. 3
integral
x 2 dx as equivalent to
Isaac Barrow (1630–1677) His role as Isaac Newton’s most knowledgeable and influential mathematics professor places Isaac Barrow as the “grandfather” of calculus (though he lived to be only 46). Barrow, drawing from earlier work by Scottish mathematician James Gregory, proved the Fundamental Theorem of Calculus, which connects the slope problem of differential calculus and the area problem of integral calculus as a unified theory. Barrow’s foundation made possible Newton’s formal theory, which Gottfried Leibniz later added to and streamlined.
272
PART 7 The Part of Tens
IN THIS CHAPTER
»» Expressing rational and radical functions as exponents »» Remembering trig identities »» Sketching the most common parent functions and their transformations »» Working with sigma notation for series and calculating limits »» Knowing the Power rule, Product rule, and Chain rule from Calculus I »» Getting comfortable doing basic derivatives in your head
19 10 Skills from Pre-Calculus and Calculus I That Will Help You to Do Well in Calculus II Chapter
W
ith all the math you need to understand in order to pass Pre-Calculus and Calculus I, you may not be sure exactly what you’ll need to know when you finally walk into your Calculus II class for the first time.
CHAPTER 19 10 Skills from Pre-Calculus and Calculus I That Will Help You to Do Well in Calculus II
273
Here, I provide you with a list of ten key skills from your most recent year or so of math. If you’re sketchy on any of this stuff, flip to the recommended chapter for a quick explanation and instant practice.
Expressing Functions as Exponents Whenever Possible Can you just look at like
5
x 3 or
1 4x 7
1 and x and rewrite them as exponents of x? How about hairier functions x 2
?
If you’ve taken Calculus 1, you already know how handy expressing functions as exponents can be when differentiating. This goes double in Calculus II, where one quick algebra manoeuvre can turn a seemingly impossible problem on your midterm into a no-brainer. Check out Chapter 2 for a quick refresher on negative and fractional exponents. Then, see Chapter 7 for a bunch of ways to leverage these tricks when evaluating integrals.
Knowing the Basic Trig Identities Cold Trig identities are almost no student’s idea of a good time. In the first place, they’re often taught in a way that seems like a meaningless exercise. And then, once you’ve learned them, it’s often hard to see the point. But Calculus II is where trig identities begin to pay off. And as difficult as these identities seemed when you were learning them, integration without them is way harder. At a minimum, you should know the five most basic trig identities and all three versions of the Pythagorean identity. Beyond these, there’s a relatively short list of identities that get used very commonly when integrating. In Chapter 2, I review the relatively short list of trig identities that you absolutely need to know to do well in Calculus II. Then in Chapter 10, I show you how to use them to integrate a bunch of trig functions that you’re sure to see, especially when doing trig substitution.
Recalling the Most Common Functions and Their Basic Transformations Although you may feel like there are dozens of functions that you’ve worked with in one math class or another, in reality there are only about ten that you need to focus on: polynomial functions (such as linear, quadratic, cubic, and quartic), exponentials, logarithms, radicals, rationals, plus the two important trig wave functions (sine and cosine).
274
PART 7 The Part of Tens
Knowing how to sketch the parent (most basic) function for each of these cases will clarify a whole lot of what happens in your Calculus II class. And if you can also sketch the most common types of transformations of these functions (such as vertical and horizontal displacement, and stretch/compress/flip transformations), you’ll be set up for success. See Chapter 2 for practice with all these functions and their transformations.
Using Sigma Notation for Series Sigma notation, which uses to represent infinite series, can be very difficult to use until you get used to it. Unfortunately, this notation gets used a lot in the final third of Calculus II, which focuses on sequences and series. When your teacher first introduces this topic, don’t wait: Flip to Chapter 2 for a quick refresher on sigma notation. Then check out Chapter 15 to get a handle on sequences and series before the pace of your course goes into hyperdrive.
Evaluating Limits The limit is the very cornerstone of calculus. Both derivatives and integrals are formally defined as limits. Even so, many students (myself included, when I was a student) get confused or discouraged when a limit appears in their path. The thing is, your path through Calculus II is bound to include limits. For example, improper integrals (see Chapter 12) require limits to make sense of them. L’Hôpital’s rule (see Chapter 3), which shows up in a lot of Calculus II courses, is all about limits. And evaluating series often leans heavily on your understanding of limits. In Chapter 3, I give you a relatively quick review of limits. And if you need further review, check out Calculus All-in-One For Dummies (John Wiley & Sons) by Mark Ryan.
Differentiating the Most Common Functions In Chapter 3, I give you a list of derivatives of the most common functions, all of which you learned in Calculus I. Knowing these by heart as you start Calculus II is essential because, at its core, integration is simply the reverse of differentiation. If you’re at all shaky on these derivatives, don’t skip over the practice problems in Chapter 3.
CHAPTER 19 10 Skills from Pre-Calculus and Calculus I That Will Help You to Do Well in Calculus II
275
Knowing the Power Rule for Differentiation The Power rule allows you to differentiate any power of x:
d n x dx
nx n
1
Fortunately, most students find this rule pretty easy to work with, at least when a function includes a power that’s a positive integer. When applying this rule to rational functions (negative integer powers) and radical functions (fractional powers), however, some students get a little confused. This confusion can get magnified in Calculus II when you start working with the Power rule for integration, which essentially reverses this process:
x ndx
xn 1 C n 1
Chapter 3 provides you with practice working with the Power rule for differentiation. Check out that chapter’s problems to make sure you’re comfortable with this rule before you have to reverse it. Then, in Chapters 6 and 7, I give you practice using the Power rule for integration.
Applying the Product Rule The Product rule allows you to differentiate the product of two functions as follows:
d f ( x ) g( x ) dx
f '( x ) g ( x ) g ( x )' f ( x )
Unfortunately, no simple rule like this exists for integrating the product of two functions. However, integration by parts (IBP for short), an integration technique directly based on the Product rule, partially fills the gap. It allows you to integrate some products of functions when you know how to differentiate one function and integrate the other. It also allows you to integrate functions (such as the natural log function and certain inverse trig functions) by differentiating and then restructuring the integration. See Chapter 3 for a refresher on how to use the Product rule.
Using the Chain Rule The Chain rule allows you to differentiate compositions of functions — that is, functions that are nested inside other functions:
d f ( g ( x )) dx
276
PART 7 The Part of Tens
f '( g ( x )) g '( x )
As with the Product rule, no formula such as the Chain rule exists for integrating compositions of functions. However, variable substitution (also called u-sub) allows you to integrate some compositions of functions in a way that mimics the Chain rule. See Chapter 8 for more on variable substitution.
Calculating Common Derivatives Quickly in Your Head So you know that the
d sin x dx
cos x . But what about
d d sin2 x ? How about e x? dx dx
Knowing how to calculate common variations on the basic derivatives confidently and quickly can really make your life a lot easier when you start integrating. A lot of integration tricks lean heavily on differentiation. At times, you may feel like your Calculus II teacher is asking you to juggle while balancing on a basketball. So, to stretch the analogy to its breaking point, my advice is to get really, really comfortable balancing on a basketball (Calculus I) before you have to start juggling (Calculus II).
CHAPTER 19 10 Skills from Pre-Calculus and Calculus I That Will Help You to Do Well in Calculus II
277
Index A
algebra about, 15–16 example questions, 16–18, 19–20, 21–25, 29–31, 32 factorials, 17–18 fractional exponents, 18–19 fractions, 16 negative exponents, 18–19 parent functions about, 25–28 transformations, 28–31 sigma notation for series, 31–32 simplifying rational functions, 20 trigonometry, 20–25 using to prepare functions for integration, 93–94 Algebra I For Dummies (Sterling), 2 Algebra II For Dummies (Sterling), 2 al-Haytham, Hasan Ibn (mathematician), 271 answers and explanations area, 13, 181–188 calculus, 46–49 convergent series, 246–254 differential equations, 216–218 divergent series, 246–254 Fundamental Theorem of Calculus (FTC2), 78–80 indefinite integrals, 78–80 integration, 88–89, 101–102, 125–130 Maclaurin series, 262–266 partial fractions, 161–166 pre-calculus, 33–36 Riemann sums, 63–68 sequences, 231–234
series, 231–234 Taylor series, 262–266 3-D problems, 204–210 trig substitution, 142–150 u-substitution, 112–114 antiderivatives, of common functions, 83–85 anti-differentiation, 70, 72 applying integration by parts more than once, 121–124 Product rule, 276 approximating area, with Riemann sums, 53–68 Archimedes (mathematician), 270 arclength, calculating, 178–180 area about, 5, 169 answers and explanations, 13, 181–188 approximating with Riemann sums, 53–68 breaking definite integrals in two, 169–171 calculating arclength, 178–180 defined by functions and curves on xy-graph, 10–12 defining problems with definite integral, 8–10 example questions, 6–10, 11–12, 170–176 finding unsigned area between curves, 174–176 improper integrals, 172–174 measuring on xy-graph, 5–7 signed, 75–77 solving problems, 169–188 unsigned, 75–77 using Mean Value Theorem for integration, 177–178 assigning dv/u, 119–121
Index
279
B
Barrow, Issac (mathematician), 272 Basic Math and Pre-Algebra For Dummies (Zegarelli), 2 benchmark series, 237
C
calculating arclength, 178–180 area defined by functions and curves on xygraph, 10–12 on xy-graph, 5–7 common derivatives, 277 Riemann sums with left and right rectangles, 54–56 calculus about, 37 answers and explanations, 46–49 Chain rule, 44–45 Constant Multiple rule, 40 derivatives of common functions, 40 evaluating limits, 37–39 example questions, 38–43, 44–45 invention of, 269 Power rule, 41–42 Product rule, 43–44 Quotient rule, 43–44 Sum rule, 42–43 Calculus For Dummies (Ryan), 2 calculus I skills, 273–277 Calculus II For Dummies, 3rd Edition (Zegarelli), 2 Cavalieri, Bonaventura (mathematician), 272 centered at 0, 259 Chain rule about, 44–45 using, 276–277 common functions, derivatives of, 40
280
Calculus II Workbook For Dummies
compositions, integrating of functions with linear inputs, 98–100 Constant Multiple rule about, 40 for integration, 83–85 convergent series about, 235 answers and explanations, 246–254 direct comparison test, 237–239 example questions, 236–237, 238–245 integral test, 242–243 limit comparison test, 240–241 ratio test, 243–244 root test, 245 cosines, integrating powers of, 132–134 curves calculating area defined by, on xy-graph, 10–12 finding unsigned area between, 174–176
D
definite integrals breaking in two, 169–171 defining area problems with, 8–10 evaluating using Fundamental Theorem of Calculus (FTC2), 70–71 using variable substitution with, 109–111 derivatives calculating common, 277 of common functions, 40 differential equations about, 211–214 answers and explanations, 216–218 example questions, 212–215 solving separable, 214–215 differentiating functions, 275 Power rule for, 276 direct comparison test, 237–239 direct substitution, 38
disk method, finding volume of revolution around horizontal axis using, 190–195 divergent series about, 235 answers and explanations, 246–254 direct comparison test, 237–239 example questions, 236–237, 238–245 integral test, 242–243 limit comparison test, 240–241 nth-term test for divergence, 236–237 ratio test, 243–244 root test, 245 double-angle identities, 96 dv, assigning, 119–121
E
equations, differential, 214–215 estimating with midpoint rectangles, 57–58 with Simpsons rule, 61–62 with Trapezoid rule, 59–60 Eudoxus of Cnidus (mathematician), 270 evaluating definite integrals using Fundamental Theorem of Calculus (FTC2), 70–71 geometric series, 227–229 limits, 37–39, 275 Example icon, 2 example questions algebra, 16–18, 19–20, 21–25, 29–31, 32 area, 6–10, 11–12, 170–180 calculus, 38–43, 44–45 convergent series, 236–237, 238–245 differential equations, 212–215 divergent series, 236–237, 238–245 Fundamental Theorem of Calculus (FTC2), 70–77 indefinite integrals, 70–77 integration, 84–86, 87, 92–94, 95–100, 118–124
Maclaurin series, 256–258 partial fractions, 152–155, 156–157, 158–160 pre-calculus, 16–18, 19–20, 21–25, 29–31, 32 Riemann sums, 54–62 sequences, 222–223, 225–226 series, 223–230 Taylor series, 259–261 3-D problems, 190–195, 196–198, 199–200, 201–203 trig substitution, 132–134, 135–137, 138–141 u-substitution, 104–111 explanations and answers area, 13, 181–188 calculus, 46–49 convergent series, 246–253 differential equations, 216–218 divergent series, 246–254 Fundamental Theorem of Calculus (FTC2), 78–80 indefinite integrals, 78–80 integration, 88–89, 101–102, 125–130 Maclaurin series, 262–266 partial fractions, 161–166 pre-calculus, 33–36 Riemann sums, 63–68 sequences, 231–236 series, 231–236 Taylor series, 262–266 3-D problems, 204–210 trig substitution, 142–150 u-substitution, 112–114 exponents expressing functions as, 274 fractional, 18–19 negative, 18–19 expressing functions as exponents, 274 functions as Maclaurin series, 255–258 functions as Taylor series, 259–261
Index
281
F
factorials, 17–18 finding sequences of partial sums for series, 225–226 unsigned area between curves, 174–176 volume of revolutions around horizontal axis, 190–195 around vertical axis, 195–198 formula, for integration by parts, 117–119 fractional exponents, 18–19 fractions about, 16 partial (See partial fractions) functions antiderivatives of common, 83–85 calculating arclength of, 178–180 area defined by, on xy-graph, 10–12 derivatives of common, 40 differentiating, 275 expressing as exponents, 274 as Maclaurin series, 255–258 as Taylor series, 259–261 integrating basic trig, 131–132 compositions of, with linear inputs, 98–100 inverse, 195–198 inverse trig, 94–95 parent, 25–31 radical, 92–93 rational, 20, 92–93 transformations of, 274–275 using algebra to prepare for integration, 93–94 using trig identities to prepare for integration, 96–97 Fundamental Theorem of Calculus (FTC2) about, 69 answers and explanations, 78–80
282
Calculus II Workbook For Dummies
anti-differentiation and indefinite integrals, 72–74 evaluating definite integrals using, 70–71 example questions, 70–77 signed area, 75–77 unsigned area, 75–77
G
geometric series, evaluating, 227–229
H
half-angle identities, 96 higher-degree factors, 157–159 horizontal axis, finding volume of revolutions around, 190–195 horizontally infinite (Type 1) integral, 172–174
I
icons, explained, 1–2 improper integrals, 172–174 indefinite integrals about, 69 answers and explanations, 78–80 anti-differentiation and, 72–74 example questions, 70–77 signed area, 75–77 unsigned area, 75–77 indeterminate results, 38 inner function, 44 integral test, 242–243 integrals breaking in two, 169–171 definite, 109–111 improper, 172–174 integration about, 83, 91, 117 answers and explanations, 88–89, 101–102, 125–130 antiderivatives of common functions, 83–85 applying integration by parts more than once, 121–124
assigning u and dv, 119–121 of basic trig functions, 131–132 of compositions of functions with linear inputs, 98–100 Constant Multiple rule for, 83–85 example questions, 84–86, 87, 92–94, 95–97, 118–124 of partial fractions, 151–166 by parts, 117–130 Power rule for, 85–86 of powers of sines and cosines, 132–133 of tangents/secants, 133–137 of rational and radical functions, 92–93 Sum rule for, 86–87 using algebra to prepare functions for, 93–94 using formula for integration by parts, 117–119 using inverse trig functions, 94–95 using Mean Value Theorem for, 177–178 using trig identities to prepare functions for, 96–97 inverse functions, finding volume of revolutions around vertical axis using, 195–198 inverse trig functions, integrating using, 94–95
K
Kepler, Johannes (mathematician), 272
L
left rectangles, calculating Riemann sums with, 54–56 Leibniz, Gottfried (inventor), 269 limit comparison test, 240–241 limits, evaluating, 37–39, 275 linear factors rational expressions with distinct, 154–155 repeated, 155–157
linear inputs, integrating compositions of functions with, 98–100 Lui Hui (mathematician), 270
M
Maclaurin series about, 255 answers and explanations, 262–266 example questions, 256–258 expressing functions as, 255–258 Madhava of Sangamagrama (mathematician), 271 mathematicians, 269–272 Mean Value Theorem, using for integration, 177–178 midpoint rectangles, estimating with, 57–58
N
negative exponents, 18–19 Newton, Isaac (inventor), 269 nth-term test, for divergence, 236–237
O
order of operation (PEMDAS), 38
P
parent functions about, 25–28 transformations, 28–31 partial fractions about, 151–154 answers and explanations, 161–166 distinct quadratic or higher-degree factors, 157–159 example questions, 152–155, 156–157, 158–159 integration with, 151–166 rational expressions with distinct linear factors, 154–155 repeated linear factors, 155–157 repeated quadratic factors, 160 Index
283
Power rule about, 41–42 for differentiation, 276 for integration, 86–87 powers of sines/cosines, integrating, 132–134 powers of tangents/secants, integrating, 134–137 pre-calculus about, 15–16 answers and explanations, 33–36 example questions, 16–18, 19–20, 21–25, 29–31, 32 factorials, 17–18 fractional exponents, 18–19 fractions, 16 negative exponents, 18–19 parent functions about, 25–28 transformations, 28–31 sigma notation for series, 31–32 simplifying rational functions, 20 skills from, 273–277 trigonometry, 20–25 Product rule about, 43–44 applying, 276 p-series, 230 Pythagorean identities, 96
Q
quadratic factors about, 157–159 repeated, 160 questions, example algebra, 16–18, 19–20, 21–25, 29–31, 32 area, 6–10, 11–12, 170–176 calculus, 38–43, 44–45 convergent series, 236–237, 238–245 differential equations, 212–215 divergent series, 236–237, 238–245
284
Calculus II Workbook For Dummies
Fundamental Theorem of Calculus (FTC2), 70–77 indefinite integrals, 70–77 integration, 84–86, 87, 92–94, 95–97, 118–124 Maclaurin series, 256–258 partial fractions, 152–155, 156–157, 158–159 pre-calculus, 16–18, 19–20, 21–25, 29–31, 32 Riemann sums, 54–62 sequences, 222–223, 225–226 series, 223–224 Taylor series, 259–261 3-D problems, 190–195, 196–198, 199–200, 201–203 trig substitution, 132–134, 135–137, 138–141 u-substitution, 104–111 Quotient rule, 43–44
R
radical functions, integrating with rational functions, 92–93 ratio test, 243–244 rational expressions, with distinct linear factors, 154–155 rational functions integrating with radical functions, 92–93 simplifying, 20 Remember icon, 2 repeated linear factors, 155–157 repeated quadratic factors, 160 revolutions finding volume of around horizontal axis, 190–195 around vertical axis, 195–198 surface area of, 198–200 Riemann sums about, 53 answers and explanations, 63–68 approximating area with, 53–68 calculating with left and right rectangles, 54–56
example questions, 54–62 finding estimates with midpoint rectangles, 57–58 improving estimates with Simpson’s rule, 61–62 with Trapezoid rule, 59–60 right rectangles, calculating Riemann sums with, 54–56 root test, 245 Ryan, Mark (author) Calculus For Dummies, 2
area problems, 169–188 separable differential equations, 214–215 3-D problems, 189–210 Sterling, Mary Jane (author) Algebra I For Dummies, 2 Algebra II For Dummies, 2 Trigonometry For Dummies, 2 Sum and Difference rule. See Sum rule Sum rule about, 42–43
S
for integration, 86–87 surface area, of revolutions, 198–200
secants, integrating powers of, 134–137 sequences about, 221–223 answers and explanations, 231–234 example questions, 222–223, 225–226 finding of partial sums for series, 225–226 series about, 221, 223–224 answers and explanations, 231–234 benchmark, 237 example questions, 223–224 finding sequences of partial sums for, 225–226 geometric, 227–229 Maclaurin, 255–258 p-series, 230 sigma notation for, 31–32 Taylor, 255, 259–261 using sigma notation for, 275 shell method, 200–203 sigma notation, for series, 31–32, 275 signed area, 75–77 simplifying rational functions, 20 Simpsons rule, estimating with, 61–62 sines, integrating powers of, 132–134 skills, for calculus II, 273–277 solving
T
tangents, integrating powers of, 134–137 Taylor series about, 255 answers and explanations, 262–266 example questions, 259–261 expressing functions as, 259–261 Technical Stuff icon, 2 3-D problems about, 189 answers and explanations, 204–210 example questions, 190–195, 196–198, 199–200, 201–203 finding volume of revolution around horizontal axis using disk/washer methods, 190–195 around vertical axis using inverse functions, 195–198 shell method, 200–203 solving, 189–210 surface area of revolutions, 198–200 Tip icon, 2 transformations of functions, 274–276 parent functions, 28–31
Index
285
Trapezoid rule, estimating with, 59–60 trig identities basic, 274 using to prepare functions for integration, 96–97 trig substitution about, 131 answers and explanations, 142–150 example questions, 132–134, 135–137, 138–141 integrating basic trig functions, 131–132 powers of sines and cosines, 132–134 powers of tangents and secants, 134–137 using, 137–141 trigonometry, 20–25 Trigonometry For Dummies (Sterling), 2 Type 1 (horizontally infinite) integral, 172–174 Type 2 (vertically infinite) integral, 172–174
U
u, assigning, 119–121 unsigned area, 75–77 u-substitution about, 103 answers and explanations, 112–114 example questions, 104–111 process of, 103–105
286
Calculus II Workbook For Dummies
using variable substitution with definite integrals, 109–111 when to use, 105–109
V
variable substitution, using with definite integrals, 109–111 vertical axis, finding volume of revolutions around, 195–198 vertically infinite (Type 2) integral, 172–174
W
Warning icon, 2 washer method, finding volume of revolution around horizontal axis using, 190–195
X
xy-graph calculating area defined by functions and curves on, 10–12 calculating area on, 5–7
Z
Zegarelli, Mark (author) Basic Math and Pre-Algebra For Dummies, 2 Calculus II For Dummies, 3rd Edition, 2 Zeno of Elea (mathematician), 270 Zu Chongzhi (mathematician), 271
About the Author Mark Zegarelli is the author of Calculus II For Dummies (Wiley), Basic Math and Pre-Algebra For Dummies (Wiley), SAT Math For Dummies (Wiley), and eight other books in the For Dummies series. He holds degrees in both English and math from Rutgers University. Most recently, he provides online lessons for kids as young as 4 years old, showing them easy ways to understand — and enjoy! — square numbers and square roots, factors, prime numbers, fractions, and even basic algebra concepts.
Dedication 这本书给光光, 我永远的朋友。
Author’s Acknowledgments Many thanks for the editorial guidance and wisdom of Vicki Adang, Saikarthick Kumarasamy, Lindsay Lefevere, Paul Levesque, Kristie Pyles, Scott Rosenstein, Ana Teodorescu, and Marylouise Wiack. Thanks also to my current students: Aaron, Brave, David, Gage, Greg, Iva, Jacob, Judah, Kevin, Leon, Maurice, Nevena, Nico, Orion, Ryal, Ryoma, Samantha, and Skylar. I have learned so much from each and every one of you!
Publisher’s Acknowledgments Acquisitions Editor: Lindsay Lefevere
Production Editor: Saikarthick Kumarasamy
Senior Project Editor: Paul Levesque
Cover Image: © oxygen/Getty Images
Copy Editor: Marylouise Wiack Tech Editor: Ana Teodorescu
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