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SINGLE VARIABLE

CALCULUS:

An Introduction

To

Integration 2nd ed.

Delia D. Samuel, Ph.D.

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Table of Contents

PREFACE ..................................................................................................................................................... i

INTRODUCTION ....................................................................................................................................... ii

CHAPTER 1: AN INTRODUCTION TO INTEGRATION .................................................................. 1

CHAPTER 2: THE DEFINITE INTEGRAL ......................................................................................... 10

CHAPTER 3: THE FUNDAMENTAL THEOREM OF CALCULUS ................................................ 26

CHAPTER 4: THE INDEFINITE INTEGRAL .................................................................................... 35

CHAPTER 5: TECHNIQUES OF INTEGRATION I .......................................................................... 43

CHAPTER 6: TECHNIQUES OF INTEGRATION II ......................................................................... 56

CHAPTER 7: TECHNIQUES OF INTEGRATION III ....................................................................... 66

CHAPTER 8: TECHNIQUES OF INTEGRATION IV ....................................................................... 85

CHAPTER 9: REVIEW OF TECHNIQUES OF INTEGRATION ................................................... 102

CHAPTER 10: APPROXIMATING DEFINITE INTEGRALS ........................................................ 109

CHAPTER 11: IMPROPER INTEGRALS ......................................................................................... 117

CHAPTER 12: AREA ............................................................................................................................ 125

CHAPTER 13: VOLUME ...................................................................................................................... 142

CHAPTER 14: ARC LENGTH ............................................................................................................. 157

CHAPTER 15: HYDROSTATIC PRESSURE AND FORCE ............................................................ 163

CHAPTER 16: MOMENTS AND CENTER OF MASS ..................................................................... 172

CHAPTER 17: POLAR COORDINATES ........................................................................................... 185

CHAPTER 18: SEQUENCES ............................................................................................................... 209

CHAPTER 19: SERIES I – GEOMETRIC SERIES ........................................................................... 221

CHAPTER 20: SERIES II – CONVERGENCE OF SERIES ............................................................ 230

CHAPTER 21: SERIES III – CONVERGENCE OF SERIES ........................................................... 237

CHAPTER 22: SERIES IV – CONVERGENCE OF SERIES ........................................................... 245

CHAPTER 23: POWER SERIES ......................................................................................................... 258

CHAPTER 24: SERIES EXPANSIONS ............................................................................................... 268

CHAPTER 25: AN INTRODUCTION TO DIFFERENTIAL EQUATIONS .................................. 280

CHAPTER 26: SOLVING DIFFERENTIAL EQUATIONS ............................................................. 291

REFERENCES ........................................................................................................................................ 304

i

PREFACE

Welcome to Calculus II. It promises to be an exciting time for you. You will learn much, work

hard, but also have fun.

For some of you, this may be the first class that you are taking at college, having just graduated

from high school. Some of you are looking forward to this class and can’t wait to start learning

about integration, what you can do with it, and how you can apply it to the discipline you are

currently studying. Yet, others may have a fear for Calculus. You may have heard some horror

stories about how difficult Calculus is, especially Calculus II. However, I would like to put your

fears to rest. Indeed, Calculus II is more challenging than Calculus I (Differential Calculus) but

you will enjoy it, and if you apply certain principles to learning and studying this course, you

will excel, and in hindsight, you will wonder what you were afraid of.

This book explains Calculus II concepts adequately, comprehensively and concisely. It is

student-friendly and will cater to your individual needs of the students. It also makes use of

illustrations like graphs and tables which will increase your comprehension of concepts.

The concepts taught in this book gives an adequate picture of Calculus II and prepares you for

other disciplines like Engineering and Physics, as well as higher-level Mathematics courses.

This is second edition of the textbook which was published in 2017. Some edits were made and

a new chapter added (Ch. 23 – Power Series).

Enjoy this book! Have fun learning a very interesting area of Calculus.

ii

INTRODUCTION

In this preliminary chapter, I will show you how you can excel at Calculus. You will learn how

to eliminate math anxiety, what professors expect of Calculus students, and tips for becoming a

Calculus expert. Read these tips throughout this course and practice them consistently, and you

will be amazed at how much you will enjoy this course and how well you will do.

Overcoming Math Anxiety

Math anxiety is characterized by an overwhelming feeling of fear of mathematics. An individual

experiencing such anxiety feels frustrated and helpless and shudders at the thought of being

tested in mathematics. He/she feels incapable of understanding mathematics.

Students with math anxiety invariably perform poorly in mathematics. A terrified student taking

a math exam will develop a mental block and will perform below his potential on the exam.

When he/she obtains the exam results, he/she will feel justified for the fear of mathematics and

will again declare emphatically, “I just cannot do mathematics. I hate it.” This negative attitude

will lower the student’s self-esteem and increase math anxiety, which will in turn produce low

test scores. It is a self-fulfilling prophecy.

Math anxiety

Poor performance

Lower self esteem

iii

Thus, it is very important to control such anxiety in all students. Here are some tips for

overcoming math anxiety:

1. Be aware of your math anxiety and find the reasons or sources for this anxiety. Some

reasons include:

• An unpleasant experience in mathematics

• A lack of encouragement from your role models

2. Work on having a positive attitude towards mathematics.

• Speak positively to yourself reminding yourself again and again that you can

succeed, in spite of possibly having a history of failure or lower-than-your-

potential scores in mathematics.

3. Develop a rapport with your instructor.

• Ask questions no matter how trivial you may find them to be.

• Visit your instructor during office hours.

4. Seek help – from instructors, peers, tutors, online.

5. Have a solid foundation in basic mathematics.

• Review all material that are pre-requisites for learning Calculus II. Some of this

includes algebra, differential calculus, and trigonometry.

6. Do homework every night.

• You must spend 2 – 3 hours on homework for every hour spent in class.

• Read your textbook actively. Take notes. Highlight important points.

7. Practice anxiety reduction techniques.

• Find your happy place and visit that place often.

• Meditate.

• Practice deep breathing when you find yourself getting anxious.

iv

• Relax and eliminate tension.

8. Do not compare yourself with others.

• You are unique with your own talents. There will always be students who are

better at mathematics than you and there will always be students who are not as

good as you. Instead of comparing yourself with others, work on your

mathematical deficiencies and focus on your strengths.

9. Practice proper examination techniques.

• Study regularly.

• Do not cram right before an exam.

• Get a good night’s sleep the night before an exam.

• Do something fun before an exam to release any tension.

10. Never give up!

Instructors’ expectations of Calculus II students

All students should:

• Have a solid background in algebra, trigonometry and differential calculus

• Have a positive attitude and be motivated and willing to learn

• Be an active reader able to analyze and dissect information

• Have problem-solving skills

• Be able to communicate mathematically, think critically and logically

• Write clearly, legibly and sequentially

• Practice effective time-management skills

• Complete daily homework

• Purchase required materials – textbook, calculator, notebook for notetaking

• Take notes

• Attend class and be on time

v

More Tips for learning Calculus II

1. Find a study group made up of peers you can work with and who can be of support.

2. Always ask questions when you do not understand concepts taught.

3. Learn the vocabulary of Calculus and use it.

4. Practice problems often.

5. Make note cards of important formulas and concepts.

6. Do not wait until test time to study. Study regularly.

7. Seek help.

8. Do not procrastinate and get behind with work.

9. Use textbook effectively. Read the section being taught before coming to class and after

class. Review material needed to understand concepts being learned.

10. Develop an effective schedule and plan for study and homework

11. Read the syllabus. It spells out exactly what is expected of you.

1

CHAPTER 1: AN INTRODUCTION TO INTEGRATION

The problem of finding the line tangent to a given curve was solved through differential

calculus. The problem of finding the area enclosed by a given curve is solved through integral

calculus.

We can easily find the area under a line graph. This bounded region would either be a triangle or

trapezoid which we can find the area of. Recall:

The area of a triangle = 𝟏

𝟐 × 𝒃𝒂𝒔𝒆 × 𝒉𝒆𝒊𝒈𝒉𝒕

The area of a trapezoid = 𝟏

𝟐 × (𝒔𝒖𝒎 𝒐𝒇 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒔𝒊𝒅𝒆𝒔) × 𝒉𝒆𝒊𝒈𝒉𝒕

Let us look at some examples:

Example 1.1

1. Find the area under the line 𝑦 = 3 − 𝑥 from 𝑥 = 0 to 𝑥 = 3.

2

The shaded region under the graph is a triangle with

Height = 3

Base = 3

Area under line = Area of shaded triangle = 1

2 × 3 × 3 = 4.5

2. Find the area under the line 𝑦 = 𝑥 + 2 from 𝑥 = 0 to 𝑥 = 3.

Area of region = Area of trapezoid = 1

2 × (2 + 5) × 3 = 10.5

3

AREA UNDER A CURVE

How do we find the area under a curve? First, we shall approximate the area by dividing the

region into rectangles and finding the sum of the areas of the rectangles. We can form the

rectangles in a few ways: by taking the right end points of each interval, the left end points, or

the midpoints.

Example 1.2

1. Estimate the area under the parabola 𝑦 = 𝑥2 from 1 to 3 by dividing the area into 4

approximating rectangles and using:

a) The right end points

b) The left end points

c) The midpoints of each interval

a) We first form rectangles by using the right end points of each interval.

4

Note that the width of each rectangle is 1

2 .

The length of each rectangle is the value of the y-coordinate at the right end points.

For example, the length of the first rectangle = 𝑓 (3

2) = (

3

2)

2.

The length of the second rectangle = 𝑓(2) = (2)2, and so on….

Thus the area under the curve is approximately equal to the total area of the rectangles,

which is

1

2(

3

2)

2

+1

2(2)2 +

1

2(

5

2)

2

+1

2(3)2 = 10

3

4

Note that this is an overestimation of the exact area under the curve.

b) We form rectangles by using the left end points of each interval.

5

Again, the width of each rectangle is 1

2 .

The length of each rectangle is the value of the y-coordinate at the left end points.

For example, the length of the first rectangle = 𝑓(1) = (1)2.

The length of the second rectangle = 𝑓 (3

2) = (

3

2)

2, and so on….


which is

1

2(1)2 +

1

2(

3

2)

2

+1

2(2)2 +

1

2(

5

2)

2

= 63

4

Note that this is an underestimation of the exact area under the curve.

c) We now form rectangles by using the mid-points of each interval.

6

The width of each rectangle is 1

2 .

The length of each rectangle is the value of the y-coordinate at the mid-points.

For example, the length of the first rectangle = 𝑓 (5

4) = (

5

4)

2.

The length of the second rectangle = 𝑓 (7

4) = (

7

4)

2, and so on….


which is

1

2[(

5

4)

2

+ (7

4)

2

+ (9

4)

2

+ (11

4)

2

] =69

8

This estimation is closer to the exact area under the curve than the previous two

estimations.




b) The left end points of each interval

a)

7

Area ≈

0.25[(1.25)2 + (1.5)2 + (1.75)2 + (2)2 + (2.25)2 + (2.5)2 + (2.75)2 + (3)2] = 9.6875

b)

Area ≈

0.25[(1)2 + (1.25)2 + (1.5)2 + (1.75)2 + (2)2 + (2.25)2 + (2.5)2 + (2.75)2] = 7.6875

8

HOMEWORK ON CHAPTER 1




b) The left end points of each interval.

Show a sketch of the graph and the rectangles in each case.

Classify each case as either an overestimation or an underestimation of the actual

area.

2. Estimate the area under the curve 𝑦 = sin 𝑥 from 0 to 𝜋

2 by dividing the area into 6




c) The midpoints of each interval.


3. Estimate the area under the line 𝑦 = 4 − 𝑥 from 0 to 4 by dividing the area into 8




c) The midpoints of each interval.


Find the exact area under the line and then classify each case as either an

overestimation or an underestimation of the actual area.

9

4. Estimate the area under the graph of 𝑦 =1

𝑥 from 1 to 6 by dividing the area into 5


a) The left end points

b) The right end points

c) The mid points of each interval.

10

CHAPTER 2: THE DEFINITE INTEGRAL

Review of limits at infinity

In this chapter we will make use of your previous knowledge of finding limits of rational

functions at infinity. Let us review it here:

For rational functions, if substitution results in indeterminate form ∞

∞, divide each term by the

highest power of x in the expression, then find the limit.

This simply means that we can first find the highest power of x in the rational function. The limit

is the quotient of the coefficients of this power of x in the numerator and denominator.

Example 2.1

1. lim𝑥→∞

𝑥2−2𝑥+1

3𝑥2+4𝑥−5=

1

3

2. lim𝑥→∞

4𝑥3−2𝑥+1

5𝑥3+4𝑥−5=

4

5

3. lim𝑥→∞

7𝑥6−2𝑥+1

4𝑥6+4𝑥2−5=

7

4

4. lim𝑥→∞

2𝑥+1

3𝑥2+4𝑥−5=

0

3= 0

Riemann Sums

Recall that the problem of finding the line tangent to a given curve was solved through

differential calculus. The problem of finding the area enclosed by a given curve was solved

through integral calculus. The method of Riemann Sums was used to calculate such an area. It

led to the definition of the definite integral.

11

In the previous chapter, we estimated the area under the parabola 𝑦 = 𝑥2 from 1 to 3 by dividing

the area into 4 and 8 approximating rectangles and using left and right end points and midpoints

of each interval. We can increase the number of rectangles and find even more approximations.

Let us examine approximations by using the left and right end points of the intervals. Denote the

area obtained by using n rectangles and their left end points by 𝐿𝑛, and 𝑅𝑛, the area obtained by

using n rectangles and their right end points. The table below shows the approximate areas

obtained based on a specified number of rectangles:

n 𝑳𝒏 𝑹𝒏

10 7.88 9.48

20 8.27 9.07

30 8.40148148148148 8.93481481481481

40 8.4675 8.8675

50 8.5072 8.8272

100 8.5868 8.7468

200 8.6267 8.7067

10000 8.66586668 8.66746668

100000 8.6665866668 8.6667466668

Notice that as the number of rectangles increase, the estimates get better, and 𝐿𝑛 gets closer to

𝑅𝑛. This suggests that as8

𝑛 → ∞, 𝐿𝑛 → 𝑅𝑛 → 𝑒𝑥𝑎𝑐𝑡 𝑎𝑟𝑒𝑎

So, 𝑅𝑛 → ? as 𝑛 → ∞

It appears that as 𝑛 → ∞, 𝑅𝑛 → 8.667 or 82

3

We can show that

12

lim𝑛→∞

𝑅𝑛 = 82

3

Recall when we divided the area into 4 rectangles, we found that:

𝑅4 =1

2(

3

2)

2

+1

2(2)2 +

1

2(

5

2)

2

+1

2(3)2

=2

4(1 +

2

4)

2

+2

4(1 +

4

4)

2

+2

4(1 +

6

4)

2

+2

4(1 +

8

4)

2

Similarly, for 8 rectangles, we obtained:

𝑅8 = 0.25[(1.25)2 + (1.5)2 + (1.75)2 + (2)2 + (2.25)2 + (2.5)2 + (2.75)2 + (3)2]

=2

8[(1 +

2

8)

2

+ (1 +4

8)

2

+ (1 +6

8)

2

+ (1 +8

8)

2

+ (1 +10

8)

2

+ (1 +12

8)

2

+ (1 +14

8)

2

+ (1 +16

8)

2

]

Observing the pattern in both formulas, we can write a general formula when we divide the area

into n rectangles, where n is any positive integer. Thus,

𝑅𝑛 =2

𝑛[(1 +

2

𝑛)

2

+ (1 +4

𝑛)

2

+ (1 +6

𝑛)

2

+ ⋯ + (1 +2𝑛

8)

2

]

It can then be shown that

𝑅𝑛 = 82

3

We have just demonstrated the method of Riemann Sums:

1. Approximate the area under the curve by drawing n rectangles which can be formed by

choosing an arbitrary point within each rectangle.

2. Find the total area of the rectangles.

3. Find the limit as 𝑛 → ∞ of this total area. This gives the exact area of the region.

13

Riemann Sum

Let 𝑦 = 𝑓(𝑥) be a function which is positive and continuous on the interval [a, b]. This is how

the area bounded by the curve 𝑦 = 𝑓(𝑥), the x-axis and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏 can be

approximated.

Divide [a, b] into n equal intervals with end points 𝑥0, 𝑥1, … , 𝑥𝑛. Each interval has length ∆𝑥,

where ∆𝑥 =𝑏−𝑎

𝑛.

Approximate the area of the region by drawing rectangles whose total area is close to the actual

area. We will choose an arbitrary point 𝑥𝑖 in each interval, which can be the left end point,

midpoint or right end point of each interval. For our purposes we will use the right end point of

each interval.

Since the arbitrary point (say, the right end point) chosen to form the rectangles is denoted by 𝑥𝑖,

then the length of each rectangle is 𝑓(𝑥𝑖). As mentioned previously, the width of each rectangle

is ∆𝑥𝑖 . So the area of each rectangle is 𝑓(𝑥𝑖)∆𝑥𝑖. The area under the curve is approximately equal

to the sum of the areas of each of the n rectangles, which is:

∑ 𝑓(𝑥𝑖)

𝑛

𝑖=1

∆𝑥

This is called the Riemann Sum.

14

We obtain the exact area of the region by taking the limit of this sum as 𝑛 → ∞. This leads to the

definition of the definite integral.

The definite integral

The definite integral of a function f from 𝑥 = 𝑎 to 𝑥 = 𝑏, denoted by

∫ 𝑓(𝑥) 𝑑𝑥

𝑏

𝑎

finds the area of the region under the graph 𝑦 = 𝑓(𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏.

𝑓(𝑥) is called the integrand.

The limits of integration are a and b.

dx is called the differential of the variable x. It indicates that we are integrating with respect to x.

15

Example 2.2

Find the area under the line 𝑦 = 2 − 𝑥 from 𝑥 = 0 to 𝑥 = 2.

Let us first sketch the region:

So

∫ (2 − 𝑥) 𝑑𝑥 =1

2× 2 × 2 = 2

2

0

Definition

Suppose f is continuous on [a, b]. Then the definite integral of f from a to b is defined as:

∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎

= lim𝑛→∞

(∑ 𝑓(𝑥𝑖) ∆𝑥

𝑛

1=1

)

16

where ∆𝑥 =𝑏−𝑎

𝑛. 𝑥𝑖 is an arbitrary point within each interval.

For uniformity, we will use 𝑥𝑖 as the right end point of each interval. So

𝑥𝑖 = 𝑎 + ∆𝑥 . 𝑖

You may find these formulas useful:

∑ 𝑖 = 1 + 2 + 3 + 4 + ⋯ + 𝑛 =𝑛(𝑛 + 1)

2

𝑛

𝑖=1

∑ 𝑖2 = 12 + 22 + 32 + 42 + ⋯ + 𝑛2 =𝑛(𝑛 + 1)(2𝑛 + 1)

6

𝑛

𝑖=1

∑ 𝑖3 = 13 + 23 + 33 + 43 + ⋯ + 𝑛3 =𝑛2(𝑛 + 1)2

4

𝑛

𝑖=1

Examples 2.3

Use Riemann sums to find:

1. ∫ 𝑥 𝑑𝑥4

0

Recall:


𝑎

= lim𝑛→∞

(∑ 𝑓(𝑥𝑖) ∆𝑥

𝑛

1=1

)

𝑓(𝑥) = 𝑥, ∆𝑥 =𝑏 − 𝑎

𝑛=

4 − 0

𝑛=

4

𝑛, 𝑥𝑖 = 𝑎 + ∆𝑥. 𝑖 = 0 +

4

𝑛𝑖 =

4𝑖

𝑛

So

∫ 𝑥 𝑑𝑥 = lim𝑛→∞

(∑ 𝑓 (4𝑖

𝑛)

4

𝑛

𝑛

1=1

)4

0

17

= lim𝑛→∞

(∑4𝑖

𝑛.4

𝑛

𝑛

1=1

)

= lim𝑛→∞

(∑16𝑖

𝑛2

𝑛

1=1

)

= lim𝑛→∞

16

𝑛2(∑ 𝑖

𝑛

1=1

)

= lim𝑛→∞

16

𝑛2 .

𝑛(𝑛 + 1)

2

=16

2

= 8

2. ∫ 𝑥2 𝑑𝑥3

0

Recall:


𝑎

= lim𝑛→∞

(∑ 𝑓(𝑥𝑖) ∆𝑥

𝑛

1=1

)

𝑓(𝑥) = 𝑥2, ∆𝑥 =𝑏 − 𝑎

𝑛=

3 − 0

𝑛=

3

𝑛, 𝑥𝑖 = 𝑎 + ∆𝑥. 𝑖 = 0 +

3

𝑛𝑖 =

3𝑖

𝑛

So

∫ 𝑥2 𝑑𝑥 = lim𝑛→∞

(∑ 𝑓 (3𝑖

𝑛)

3

𝑛

𝑛

1=1

)3

0

18

= lim𝑛→∞

(∑9𝑖2

𝑛2.3

𝑛

𝑛

1=1

)

= lim𝑛→∞

(∑27𝑖2

𝑛3

𝑛

1=1

)

= lim𝑛→∞

27

𝑛3(∑ 𝑖2

𝑛

1=1

)

= lim𝑛→∞

27

𝑛3 .

𝑛(𝑛 + 1)(2𝑛 + 1)

6

=27 × 2

6

= 9

3. ∫ 𝑥3 𝑑𝑥2

0


𝑎

= lim𝑛→∞

(∑ 𝑓(𝑥𝑖) ∆𝑥

𝑛

1=1

)

𝑓(𝑥) = 𝑥3, ∆𝑥 =𝑏 − 𝑎

𝑛=

2 − 0

𝑛=

2

𝑛, 𝑥𝑖 = 𝑎 + ∆𝑥. 𝑖 = 0 +

2

𝑛𝑖 =

2𝑖

𝑛

So

∫ 𝑥3 𝑑𝑥 = lim𝑛→∞

(∑ 𝑓 (2𝑖

𝑛)

2

𝑛

𝑛

1=1

)2

0

= lim𝑛→∞

(∑8𝑖3

𝑛3.2

𝑛

𝑛

1=1

)

19

= lim𝑛→∞

(∑16𝑖3

𝑛4

𝑛

1=1

)

= lim𝑛→∞

16

𝑛4(∑ 𝑖3

𝑛

1=1

)

= lim𝑛→∞

16

𝑛4 .

𝑛2(𝑛 + 1)2

4

=16

4

= 4

4. ∫ (3 − 𝑥) 𝑑𝑥4

1


𝑎

= lim𝑛→∞

(∑ 𝑓(𝑥𝑖) ∆𝑥

𝑛

1=1

)

𝑓(𝑥) = 3 − 𝑥, ∆𝑥 =𝑏 − 𝑎

𝑛=

4 − 1

𝑛=

3

𝑛, 𝑥𝑖 = 𝑎 + ∆𝑥. 𝑖 = 1 +

3

𝑛𝑖 = 1 +

3𝑖

𝑛

So

∫ (3 − 𝑥) 𝑑𝑥 = lim𝑛→∞

(∑ 𝑓 (1 +3𝑖

𝑛)

3

𝑛

𝑛

1=1

)4

1

= lim𝑛→∞

(∑ [3 − (1 +3𝑖

𝑛)] .

3

𝑛

𝑛

1=1

)

= lim𝑛→∞

(∑ [2 −3𝑖

𝑛]

𝑛

1=1

.3

𝑛)

20

= lim𝑛→∞

(∑ [6

𝑛−

9𝑖

𝑛2]

𝑛

1=1

)

= lim𝑛→∞

(∑6

𝑛− ∑

9𝑖

𝑛2

𝑛

𝑖=1

𝑛

1=1

)

= lim𝑛→∞

[6

𝑛∑ 1 −

9

𝑛2∑ 𝑖

𝑛

𝑖=1

𝑛

𝑖=1

]

= lim𝑛→∞

[6

𝑛. 𝑛 −

9

𝑛2.𝑛(𝑛 + 1)

2]

= 6 −9

2

=3

2

Interpreting Areas of regions

Consider the graph of 𝑦 = 𝑓(𝑥) which lies above the x-axis. So 𝑓(𝑥) is positive.

Consider the area of the region bounded by 𝑦 = 𝑓(𝑥), the x-axis, and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏.

21

Since 𝑓(𝑥) is positive, then ∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎 is positive.

Now consider the area of this region S bounded by 𝑦 = 𝑓(𝑥), the x-axis, and the lines 𝑥 = 𝑎 and

𝑥 = 𝑏, but this time, 𝑓(𝑥) and the region lie below the 𝑥-axis. So 𝑓(𝑥) is negative. Thus,


𝑎 is negative. But area is positive. Thus, the area of this region is

|∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎

|

Example 2.4

Consider the region below bounded by 𝑦 = 𝑓(𝑥) and the x-axis.

Suppose ∫ 𝑓(𝑥) 𝑑𝑥 = −81

4

3

0. Find:

a) ∫ 𝑓(𝑥) 𝑑𝑥0

−3

Note that since the region corresponding to ∫ 𝑓(𝑥) 𝑑𝑥3

0 lies below the x-axis, then its area is

81

4.

Since this graph is symmetrical about the origin, then

∫ 𝑓(𝑥) 𝑑𝑥0

−3

=81

4

22

b) ∫ 𝑓(𝑥) 𝑑𝑥3

−3

∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 +0

−3

3

−3

∫ 𝑑𝑥3

0

= −81

4+

81

4= 0

c) the area of the shaded region.

The area of the shaded region is:

81

4+

81

4=

81

2

Examples 2.5

Find each definite integral by interpreting it in terms of the area of the region:

1. ∫ √9 − 𝑥2 𝑑𝑥3

0

We would first need to sketch the region to determine whether it is a recognizable region that we

easily find the area of.

The function is

𝑦 = √9 − 𝑥2

Squaring both sides and simplifying, we get

𝑦2 = 9 − 𝑥2

𝑥2 + 𝑦2 = 9

This is a circle center the origin, radius 3.

But 𝑦 = √9 − 𝑥2 is positive so it represents the part of the circle lying above the x-axis.

23

Moreover, from the limits of integration, we see that x goes from 0 to 3. This means that the

region is a quarter of a circle as seem in the graph below:

So ∫ √9 − 𝑥2 𝑑𝑥3

0 is the area of a quarter circle with radius 3 which is

1

4 . 𝜋 . 32 =

9

4 𝜋

2. ∫ |𝑥| 𝑑𝑥1

−3

Again, we first sketch the region:

24

The area of the region is the sum of the areas of the two triangles:

(1

2 × 3 × 3) + (

1

2× 1 × 1) = 5

Some properties of the definite integral

1. ∫ 𝑘 𝑓(𝑥) 𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎

𝑏

𝑎 where k is a constant.

2. ∫ [𝑓(𝑥) ± 𝑔(𝑥)] 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ 𝑔(𝑥) 𝑑𝑥𝑏

𝑎

𝑏

𝑎

𝑏

𝑎

3. If 𝑎 < 𝑐 < 𝑏, then

∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑐

𝑐

𝑎

𝑏

𝑎

4. ∫ 𝑓(𝑥) 𝑑𝑥 = − ∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎

𝑎

𝑏

25


1. Use Riemann Sums to find:

a) ∫ 𝑥21

0 𝑑𝑥

b) ∫ 2𝑥 𝑑𝑥5

0

c) ∫ 4𝑥32

0 𝑑𝑥

d) ∫ (1 − 2𝑥) 𝑑𝑥5

2

2. Find the following integrals by first sketching the region and interpreting the area as the

area of a known region:

a) ∫ √4 − 𝑥2 𝑑𝑥2

0

b) ∫ √9 − 𝑥2 𝑑𝑥3

−3

c) ∫ √25 − 𝑥2 𝑑𝑥0

−5

d) ∫ |𝑥 − 3| 𝑑𝑥5

0

e) ∫ (2 − 𝑥) 𝑑𝑥3

−1

26

CHAPTER 3: THE FUNDAMENTAL THEOREM OF CALCULUS

THE FUNDAMENTAL THEOREM OF CALCULUS

The fundamental theorem of Calculus is a very important theorem. It shows us how the processes

of differentiation and integration are related. There are two parts to this theorem.

You will need to recall some of the basic rules of differentiation below:

y 𝒅𝒚

𝒅𝒙

𝑥𝑛, n is a constant

𝑛𝑥𝑛−1

𝑒𝑥 𝑒𝑥

ln 𝑥

1

𝑥

sin 𝑥 cos 𝑥

cos 𝑥 − sin 𝑥

tan 𝑥 𝑠𝑒𝑐2𝑥

sec 𝑥 sec 𝑥 tan 𝑥

csc 𝑥 − csc 𝑥 cot 𝑥

cot 𝑥 −𝑐𝑠𝑐2𝑥

27

y 𝒅𝒚

𝒅𝒙

sin−1 𝑥 1

√1 − 𝑥2

cos−1 𝑥 −

1

√1 − 𝑥2

tan−1 𝑥 1

1 + 𝑥2

csc−1 𝑥 −

1

𝑥√𝑥2 − 1

sec−1 𝑥 1

𝑥√𝑥2 − 1

cot−1 𝑥 −

1

1 + 𝑥2

Part 1 of the Fundamental theorem of Calculus

The first part enables us to compute the derivative of a definite integral.

Theorem

Suppose f is continuous on an interval. Let

𝐹(𝑥) = ∫ 𝑓(𝑡) 𝑑𝑡𝑥

𝑎

where a is a number in that interval. Then

𝐹′(𝑥) =𝑑

𝑑𝑥(∫ 𝑓(𝑡) 𝑑𝑡

𝑥

𝑎

) = 𝑓(𝑥)

Recall that ∫ 𝑓(𝑡) 𝑑𝑡𝑥

𝑎 is the area under the graph of 𝑦 = 𝑓(𝑡) from a to x. Thus this area is

denoted by 𝐹(𝑥). This theorem enables us to find the derivative of 𝐹(𝑥), that is, the rate of

28

change of the area, which is equal to the value of the function at x. Thus, we are able to evaluate

the derivative of a definite integral without evaluating that integral.

Note that the lower limit a can be any constant as long as f is continuous at a.

Example 3.1

Find the derivatives of the following integrals:

1. ∫ sin 𝑡 𝑑𝑡𝑥

𝑎 where a is an arbitrary constant.

𝐹(𝑥) = ∫ sin 𝑡 𝑑𝑡𝑥

𝑎

We note that the sine function is continuous everywhere. So

𝐹′(𝑥) = sin 𝑥

2. ∫ (1 + 𝑡3)−4 𝑑𝑡𝑥

0

𝐹(𝑥) = ∫ (1 + 𝑡3)−4 𝑑𝑡𝑥

0

𝐹′(𝑥) = (1 + 𝑥3)−4

3. ∫ 2𝑡 sin 𝑡 𝑑𝑡1

𝑥

We first note that the limits of integration are reversed in that the constant 1 is now the

upper limit. So we would need to change the order of the limits so that 1 is now the lower

limit. We will make use of this theorem:

∫ 𝑓(𝑥) 𝑑𝑥 = − ∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎

𝑎

𝑏

Thus

29

𝐹(𝑥) = ∫ 2𝑡 sin 𝑡 𝑑𝑡 = − ∫ 2𝑡 sin 𝑡 𝑑𝑡𝑥

1

1

𝑥

𝐹′(𝑥) = −2𝑥 sin 𝑥

4. ∫ (1 + 𝑡3)−3 𝑑𝑡𝑥2

1

Notice that in this case the upper limit is a function of x. It is not simply x. We would

therefore need to replace it with a single variable like u. Thus,

𝐹(𝑥) = ∫ (1 + 𝑡3)−3 𝑑𝑡 = ∫ (1 + 𝑡3)−3 𝑑𝑡𝑢

1

𝑥2

1

𝐹′(𝑢) = (1 + 𝑢3)−3

However, we need to find 𝐹′(𝑥). Further work is necessary.

Recall the chain rule from differentiation:

If 𝒚 = 𝒇(𝒖) and 𝒖 = 𝒈(𝒙) are both differentiable functions, then

𝒅𝒚

𝒅𝒙=

𝒅𝒚

𝒅𝒖 .

𝒅𝒖

𝒅𝒙

So from above, let 𝑦 = 𝐹(𝑥). We want to find 𝑑𝑦

𝑑𝑥= 𝐹′(𝑥).

Also 𝑑𝑦

𝑑𝑢= 𝐹′(𝑢)

And since 𝑢 = 𝑥2,𝑑𝑢

𝑑𝑥= 2𝑥

So

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢 .

𝑑𝑢

𝑑𝑥

𝐹′(𝑥) = 𝐹′(𝑢) . 2𝑥

30

𝐹′(𝑥) = (1 + 𝑢3)−3 . 2𝑥

We then rewrite this result in terms of x by substituting 𝑢 = 𝑥2.

𝐹′(𝑥) = (1 + (𝑥2)3)−3 . 2𝑥 = 2𝑥(1 + 𝑥6)−3

This brings us to the following rule for using the first part of the fundamental theorem of

calculus when the upper limit is a function of x.

Let

𝑭(𝒙) = ∫ 𝒇(𝒕) 𝒅𝒕𝒈(𝒙)

𝒂

where 𝒈(𝒙) is a function of x. Then

𝑭′(𝒙) = 𝒈′(𝒙) . 𝒇(𝒈(𝒙))

5. ∫ cos(𝑡5) 𝑑𝑡𝑥3

1

𝐹(𝑥) = ∫ cos(𝑡5) 𝑑𝑡𝑥3

1

𝐹′(𝑥) =𝑑

𝑑𝑥(𝑥3) . cos((𝑥3)5)

𝐹′(𝑥) = 3𝑥2 cos(𝑥15)

Part 2 of the Fundamental theorem of Calculus

This theorem shows that integration is the inverse operation of differentiation and allows us to

evaluate definite integrals without using Riemann Sums.

31

Theorem

Let f be continuous on [a, b]. If 𝐹′(𝑡) = 𝑓(𝑡), then

∫ 𝑓(𝑡) 𝑑𝑡 = 𝐹(𝑏) − 𝐹(𝑎)𝑏

𝑎

The function F is called the antiderivative of the function f.

Thus, to evaluate a definite integral of f we would first need to find the function F whose

derivative is the function f.

Example 3.2

Evaluate the definite integrals by using the fundamental theorem of calculus:

1. ∫ 2𝑥 𝑑𝑥3

1

The derivative of 𝑥2 is 2𝑥. This means that the antiderivative of 2𝑥 is 𝑥2. So

𝐹(𝑥) = 𝑥2

Thus

∫ 2𝑥 𝑑𝑥3

1

= 𝐹(3) − 𝐹(1)

= 9 − 1

= 8

2. ∫ 3 sin 𝑥 𝑑𝑥𝜋

20

The antiderivative of 3 sin 𝑥 is −3 cos 𝑥. So 𝐹(𝑥) = −3 cos 𝑥. So

32

∫ 3 sin 𝑥 𝑑𝑥

𝜋2

0

= 𝐹 (𝜋

2) − 𝐹(0)

= 0 − −3

= 3

3. ∫ (1 + 4𝑥3) 𝑑𝑥3

1

The antiderivative of 1 + 4𝑥3 is 𝑥 + 𝑥4. So 𝐹(𝑥) = 𝑥 + 𝑥4. Thus

∫ (1 + 4𝑥3)𝑑𝑥3

1

= 𝐹(3) − 𝐹(1)

= (3 + 81) − (1 + 1)

= 82

4. ∫ (𝑥2 − 4𝑒𝑥)𝑑𝑥3

0

The antiderivative of 𝑥2 − 4𝑒𝑥 is 1

3𝑥3 − 4𝑒𝑥. So 𝐹(𝑥) =

1

3𝑥3 − 4𝑒𝑥. Thus

∫ (𝑥2 − 4𝑒𝑥)𝑑𝑥3

1

= 𝐹(3) − 𝐹(1)

= (9 − 4𝑒3) − (1

3− 4𝑒)

=26

3− 4𝑒3 + 4𝑒

33


1. Find the derivatives of the following integrals:

a) ∫ (1 + 3𝑡5)−4𝑥

0𝑑𝑡

b) ∫ √𝑡 − 𝑡33

𝑥𝑑𝑡

c) ∫ sin(𝑡2) 𝑑𝑥𝑥4

1

d) ∫ √𝑡0

𝑥2 tan(𝑡3) 𝑑𝑥

e) ∫ (1 + 𝑡4)20 𝑑𝑡𝑥

2

f) ∫ cos−1(2𝑡) 𝑑𝑡0

𝑥

g) ∫ cos(√𝑡) 𝑑𝑡𝑥4

3

2. Evaluate the definite integrals by using the fundamental theorem of calculus:

a) ∫ (6𝑥5 + 2𝑒𝑥)1

0 𝑑𝑥

b) ∫ (1 − 3𝑥 + 2𝑥2)2

1 𝑑𝑥

c) ∫ cos 𝑥 𝑑𝑥𝜋

20

d) ∫ (2𝑥 + 2𝑥3 − 3) 𝑑𝑥2

1

e) ∫1

1+𝑥2 𝑑𝑥1

0

34

f) ∫ 5𝑥4 𝑑𝑥1

0

g) ∫ (2𝑥 − 𝑥3) 𝑑𝑥2

1

h) ∫ cos 𝑥 𝑑𝑥𝜋

0

i) ∫ 3𝑥5 + 𝑥2

1 𝑑𝑥

j) ∫ (2𝑒𝑥 + 3𝑥2 − 1) 𝑑𝑥2

0

k) ∫ 𝑠𝑒𝑐2𝑥 𝑑𝑥𝜋

4𝜋

6

35

CHAPTER 4: THE INDEFINITE INTEGRAL

In the previous chapters, you learned how to evaluate definite integrals by using Riemann sums

and the fundamental theorem of calculus. Recall that definite integrals have limits of integration.

Integrals without such limits are called indefinite integrals. In this chapter, you will learn how to

evaluate them.

Observe:

The derivative of 𝑥2 is 2𝑥.

The derivative of 𝑥2 + 1 is 2𝑥.

The derivative of 𝑥2 − 500 is 2𝑥.

The derivative of 𝑥2 + 0.009 is also 2𝑥.

So the derivative of 𝑥2 + 𝑎𝑛𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 is 2𝑥

In the last chapter we saw that the 𝑥2 is the antiderivative of 2𝑥. Another way of expressing the

antiderivative is by using the indefinite integral.

Since the derivative of 𝑥2 + 𝑎𝑛𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 is 2𝑥, we say that the indefinite integral of 2𝑥 is

𝑥2 + 𝑎𝑛𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. This leads us to the following definition:

Definition

If 𝐹′(𝑥) = 𝑓(𝑥) for all x on an interval [a, b], then F is called an antiderivative or indefinite

integral of f :

∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶

where C is an arbitrary constant.

36

To find the indefinite integral, we will need to apply some basic integration formulas. This will

also enable us to evaluate the definite integral easily.

Some basic integration formulas

1. ∫ 𝑘 𝑑𝑥 = 𝑘𝑥 + 𝐶, where k is a constant.

For example:

∫ 2 𝑑𝑥 = 2𝑥 + 𝑐

2. ∫ 𝑥𝑛𝑑𝑥 =𝑥𝑛+1

𝑛+1+ 𝐶, 𝑛 ≠ −1

For example:

∫ 𝑥4 𝑑𝑥 = 𝑥4+1

4 + 1+ 𝐶 =

𝑥5

5+ 𝐶

∫ 3𝑥2 𝑑𝑥 = 3𝑥3

3+ 𝐶 = 𝑥3 + 𝐶

∫ √𝑥 𝑑𝑥 = ∫ 𝑥12 𝑑𝑥 =

𝑥32

32

+ 𝐶 =2

3𝑥

32 + 𝐶

3. ∫1

𝑥𝑑𝑥 = ln |𝑥| + 𝐶

The absolute value is necessary since by using the chain rule:

𝑑

𝑑𝑥(ln 𝑥) =

1

𝑥

𝑑

𝑑𝑥(ln(−𝑥)) = −

1

𝑥 . −1 =

1

𝑥

37

For example:

∫3

𝑥𝑑𝑥 = 3 ln|𝑥| + 𝐶

∫1

5𝑥 𝑑𝑥 = ∫

1

5 .

1

𝑥 𝑑𝑥 =

1

5 ∫

1

𝑥 𝑑𝑥 =

1

5ln|𝑥| + 𝐶

4. ∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝐶

For example:

∫ 2𝑒𝑥 𝑑𝑥 = 2𝑒𝑥 + 𝐶

5. ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶

For example:

∫ −3 sin 𝑥 𝑑𝑥 = 3 cos 𝑥 + 𝐶

6. ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶

For example:

∫1

2cos 𝑥 𝑑𝑥 =

1

2sin 𝑥 + 𝐶

7. ∫ 𝑠𝑒𝑐2𝑥 𝑑𝑥 = tan 𝑥 + 𝐶

For example:

38

∫ 5 𝑠𝑒𝑐2𝑥 𝑑𝑥 = 5 tan 𝑥 + 𝐶

8. ∫ 𝑐𝑠𝑐2𝑥 𝑑𝑥 = −cot 𝑥 + 𝐶

For example:

∫ 4 𝑐𝑠𝑐2𝑥 𝑑𝑥 = −4 cot 𝑥 + 𝐶

9. ∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶

For example:

∫1

3 sec 𝑥 tan 𝑥 𝑑𝑥 =

1

3 sec 𝑥 + 𝐶

10. ∫ csc 𝑥 cot 𝑥 𝑑𝑥 = −csc 𝑥 + 𝐶

For example:

∫ 0.6 csc 𝑥 cot 𝑥 𝑑𝑥 = − 0.6 csc 𝑥 + 𝐶

11. ∫1

1+𝑥2 𝑑𝑥 = tan−1 𝑥 + 𝐶

For example:

∫7

1 + 𝑥2𝑑𝑥 = 7 ∫

1

1 + 𝑥2 𝑑𝑥 = 7 tan−1 𝑥 + 𝐶

12. ∫1

√1−𝑥2𝑑𝑥 = sin−1 𝑥 + 𝐶

For example:

39

∫3

√1 − 𝑥2𝑑𝑥 = 3 ∫

1

√1 − 𝑥2𝑑𝑥 = 3 sin−1 𝑥 + 𝐶

Please note that it is important to add the constant C after evaluating indefinite integrals.

Now, let us try some examples. In each example, note that we integrate term by term.

Example 4.1

Find:

1. ∫(𝑥3 − 5𝑥 + 2) 𝑑𝑥 = 𝑥4

4−

5𝑥2

2+ 2𝑥 + 𝐶 =

1

4𝑥4 −

5

2𝑥2 + 2𝑥 + 𝐶

2. ∫ (5𝑡4 +𝑡

2+

5

𝑡− 3 cos 𝑡) 𝑑𝑡 =

5𝑡5

5+

𝑡2

2 . 2+ 5 ln|𝑡| − 3 sin 𝑡 + 𝐶

= 𝑡5 +1

4𝑡2 + 5 ln|𝑡| − 3 sin 𝑡 + 𝐶

3. ∫(4 sin 𝑥 −1

7 𝑠𝑒𝑐2𝑥 −

2

√𝑥 + 5𝑒𝑥 + √𝑥

3) 𝑑𝑥

= ∫ (4 sin 𝑥 −1

7 𝑠𝑒𝑐2𝑥 − 2𝑥−

12 + 5𝑒𝑥 + 𝑥

13) 𝑑𝑥

= −4 cos 𝑥 −1

7tan 𝑥 −

2𝑥12

12

+ 5𝑒𝑥 +𝑥

43

43

+ 𝐶

= −4 cos 𝑥 − 1

7tan 𝑥 − 4𝑥

12 + 5𝑒𝑥 +

3

4𝑥

43 + 𝐶

4. ∫(𝑥 − 1)(𝑥2 + 5) 𝑑𝑥 = ∫(𝑥3 + 5𝑥 − 𝑥2 − 5) 𝑑𝑥

= 1

4 𝑥4 +

5

2𝑥2 −

1

3𝑥3 − 5𝑥 + 𝐶

40

5. ∫ 𝑥−3(2𝑥 + 7) 𝑑𝑥 = ∫(2𝑥−2 + 7𝑥−3) 𝑑𝑥

=2𝑥−1

−1+

7𝑥−2

−2+ 𝐶

= −2𝑥−1 −7

2𝑥−2 + 𝐶

6. ∫2+√𝑥+4𝑥2

𝑥3 𝑑𝑥 = ∫ 𝑥−3 (2 + 𝑥1

2 + 4𝑥2) 𝑑𝑥

= ∫ (2𝑥−3 + 𝑥−52 + 4𝑥−1) 𝑑𝑥

=2𝑥−2

−2+

𝑥−32

−32

+ 4 ln|𝑥| + 𝐶

= −𝑥−2 −2

3𝑥−

32 + 4 ln|𝑥| + 𝐶

Evaluating definite integrals

We can now use the basic rules of integration to evaluate definite integrals. Pay attention to the

set-up of these solutions.

Example 4.2

1. ∫ (𝑥2 − 2𝑥 − 1) 𝑑𝑥2

1

∫ (𝑥2 − 2𝑥 − 1) 𝑑𝑥 = [1

3𝑥3 − 𝑥2 − 𝑥]

1

22

1

= [8

3− 4 − 2] − [

1

3− 1 − 1]

= −5

3

41

2. ∫ (𝑒𝑥 − cos 𝑥) 𝑑𝑥𝜋

0

∫ (𝑒𝑥 − sin 𝑥) 𝑑𝑥

𝜋2

0

= [𝑒𝑥 + cos 𝑥]0

𝜋2

= [𝑒𝜋2 + cos

𝜋

2] − [𝑒0 + cos 0]

= 𝑒𝜋2 − 2

42


Evaluate the following indefinite integrals:

1. ∫ (7𝑥11 − 1000 +55

𝑥− 11𝑒𝑥) 𝑑𝑥

2. ∫ (8 csc 𝑥 cot 𝑥 + 1

9sin 𝑥 − √𝑥) 𝑑𝑥

3. ∫ (2

𝑥4 + 5√𝑥3

+6

7𝑥3) 𝑑𝑥

4. ∫(5𝑥2 − 2𝑥 + 1)(𝑥 − 3) 𝑑𝑥

5. ∫(𝑥4 + 7)(2𝑥3 − 5𝑥) 𝑑𝑥

6. ∫ 4√𝑥3 (𝑥2 − 3𝑥 + 1) 𝑑𝑥

7. ∫ (2 +√𝑥

5) (√𝑥 −

4

√𝑥) 𝑑𝑥

8. ∫2𝑥+3𝑥2−5

√𝑥 𝑑𝑥

9. ∫5𝑥18+ √𝑥34

−4

2𝑥3 𝑑𝑥

10. ∫(𝑡2+4𝑡)(𝑡−1)

2𝑡5 𝑑𝑡

11. ∫ (5𝑥3 +7

𝑥− 9) 𝑑𝑥

12. ∫(−2 𝑐𝑠𝑐2𝑥 + 3𝑒𝑥 − √𝑥4

) 𝑑𝑥

13. ∫ (𝑥2 −1

𝑥) (2𝑥3 + 4) 𝑑𝑥

14. ∫5𝑥7+6𝑥−2√𝑥+4

𝑥2 𝑑𝑥

43

CHAPTER 5: TECHNIQUES OF INTEGRATION I

So far, you have learned how to integrate simple functions like 𝑥2,5

𝑥, etc. by using the basic rules

of integration seen in the last chapter. Now, we will begin to learn techniques of integration that

we can use for integrating more intricate functions like √4𝑥 + 7, 𝑥𝑒𝑥2, etc. In order to employ

such techniques, you will need to learn the basic rules of integration.

In this chapter, you will learn the techniques for integrating a function of a linear function, and

the technique of integration by substitution.

Integrating a function of a linear function

A function of linear function is a composite function 𝑓(𝑔(𝑥)) whose inner function 𝑔(𝑥) is a

linear function of the form 𝑚𝑥 + 𝑏, where m and b are constants. Examples of such functions

are: (3𝑥 + 5)7, 𝑒5𝑥−9, 6

5−4𝑥 , sin (2𝑡 − 𝜋), etc.

To integrate a function of a linear function: Apply the basic rules of integration and divide

by the coefficient of x, the variable within the linear function.

Recall the basic rules of integration:

1. ∫ 𝑘 𝑑𝑥 = 𝑘𝑥 + 𝐶


𝑛+1+ 𝐶, 𝑛 ≠ −1

3. ∫1


4. ∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝐶






44


11. ∫1

1+𝑥2 𝑑𝑥 = tan−1 𝑥 + 𝐶

12. ∫1

√1−𝑥2𝑑𝑥 = sin−1 𝑥 + 𝐶

Example 5.1

1. ∫(3𝑥 − 7)4 𝑑𝑥

We use the basic rule #1 above and divide by 3, the coefficient of x, to get:

∫(3𝑥 − 7)4 𝑑𝑥 = 1

3 .

1

5 (3𝑥 − 7)5

= 1

15 (3𝑥 − 7)5 + 𝐶

2. ∫1

√1−5𝑥3 𝑑𝑥

We would first need to write the integrand in exponential form, then apply basic rule #1,

and divide by the coefficient of x.

∫1

√1 − 5𝑥3 𝑑𝑥 = ∫(1 − 5𝑥)−

13 𝑑𝑥

= −1

5 .

3

2 (1 − 5𝑥)

23 + 𝐶

= −3

10 (1 − 5𝑥)

23 + 𝐶

3. ∫ cos 9𝑥 𝑑𝑥

Here we use basic rule #6:

∫ cos 9𝑥 𝑑𝑥 = 1

9sin 9𝑥 + 𝐶

45

4. ∫ 𝑒−2𝑥+3 𝑑𝑥

We use basic rule #4:

∫ 𝑒−2𝑥+3 𝑑𝑥 = −1

2 𝑒−2𝑥+3 + 𝐶

5. ∫4

7𝑥+11 𝑑𝑥


∫4

7𝑥 + 11 𝑑𝑥 =

4

7ln|7𝑥 + 11| + 𝐶

6. ∫1

1+9𝑥2 𝑑𝑥


∫1

1 + 9𝑥2 𝑑𝑥 = ∫

1

1 + (3𝑥)2 𝑑𝑥

=1

3 𝑡𝑎𝑛−1(3𝑥) + 𝐶

Integration by Substitution

This technique of integration is used for integrating some products and quotients, and also

functions of a linear function. We cannot use the method of integration by substitution to

integrate all products and quotients. Thus, we would first need to decide when such a technique

is appropriate.

Under what conditions do we use this technique for integrating products and quotients?

We examine the inner function. If the derivative of the inner function is a multiple of the other

function, then we use this method. For example:

46

Consider the product 3𝑥(1 − 6𝑥2)5. The inner function is (1 − 6𝑥2). Its derivative is −12𝑥

which is a multiple of 3𝑥. So we can use integration by substitution to integrate this product.

The goal of this technique is to rewrite the integrand in a simpler form so that it is easier to

integrate. This is accomplished through a change of variables.

The method of integration by substitution

1. If the integrand is a product or quotient, let u be the inner function. If the integrand

is a function of a linear function, let u be the linear function.

2. Find du by differentiating both sides and solve for dx.

3. Using substitution, rewrite the integrand in terms of u and du.

4. Integrate the now-simpler integrand.

5. Rewrite the answer in terms of x.

Example 5.2

1. ∫ 3𝑥2(4 + 𝑥3)5 𝑑𝑥

The derivative of the inner function (4 + 𝑥3) is 3𝑥2 which is the other function 3𝑥2. So

we can use integration by substitution.

𝑢 = 4 + 𝑥3

𝑑𝑢 = 3𝑥2 𝑑𝑥

𝑑𝑥 =𝑑𝑢

3𝑥2

We now substitute into the integral so that the variables are changed from x to u.

∫ 3𝑥2(4 + 𝑥3)5 𝑑𝑥 = ∫ 3𝑥2 . 𝑢5 .𝑑𝑢

3𝑥2

47

= ∫ 𝑢5 𝑑𝑢

= 1

6𝑢6 + 𝐶

= 1

6(4 + 𝑥3)6 + 𝐶

2. ∫3𝑥

√𝑥2−1 𝑑𝑥

This integral can be written as ∫ 3𝑥(𝑥2 − 1)−1

2 𝑑𝑥.

The inner function is (𝑥2 − 1) and its derivative is 2𝑥 which is a multiple of 3𝑥. So again

we use integration by substitution.

𝑢 = 𝑥2 − 1

𝑑𝑢 = 2𝑥 𝑑𝑥

𝑑𝑥 =𝑑𝑢

2𝑥

We substitute into the integral:

∫ 3𝑥(𝑥2 − 1)−12 𝑑𝑥 = ∫ 3𝑥 . 𝑢−

12 .

𝑑𝑢

2𝑥

= ∫3

2 𝑢−

12 𝑑𝑢

= 2 .3

2 . 𝑢

12 + 𝐶

= 3√𝑥2 − 1 + 𝐶

3. ∫ 𝑥𝑒5𝑥2−3 𝑑𝑥

The derivative of the inner function 5𝑥2 − 3 is 10𝑥 which is a multiple of 𝑥.

𝑢 = 5𝑥2 − 3


48

𝑑𝑥 =𝑑𝑢

10𝑥

Substitute:

∫ 𝑥𝑒5𝑥2−3 𝑑𝑥 = ∫ 𝑥 . 𝑒𝑢 . 𝑑𝑢

10𝑥

= ∫1

10𝑒𝑢 𝑑𝑢

=1

10𝑒𝑢 + 𝐶

=1

10𝑒5𝑥2−3 + 𝐶

4. ∫ 7𝑥4 cos(2𝑥5) 𝑑𝑥

The derivative of the inner function 2𝑥5 is 10𝑥4 which is a multiple of 7𝑥4.

𝑢 = 2𝑥5


𝑑𝑥 =𝑑𝑢

10𝑥

So

∫ 7𝑥4 cos(2𝑥5) 𝑑𝑥 = ∫ 7𝑥4 cos 𝑢 . 𝑑𝑢

10𝑥4

= ∫7

10cos 𝑢 𝑑𝑢

=7

10sin 𝑢 + 𝐶

=7

10sin(2𝑥5) + 𝐶

49

5. ∫ 𝑠𝑖𝑛6𝑥 cos 𝑥 𝑑𝑥

The integral can be written as ∫(sin 𝑥)6 cos 𝑥 𝑑𝑥.

Again we use integration by substitution with our inner function sin 𝑥.

𝑢 = sin 𝑥

𝑑𝑢 = cos 𝑥 𝑑𝑥

𝑑𝑥 =𝑑𝑢

cos 𝑥

So

∫ 𝑠𝑖𝑛6𝑥 cos 𝑥 𝑑𝑥 = ∫ 𝑢6 . cos 𝑥 . 𝑑𝑢

cos 𝑥

= ∫ 𝑢6 𝑑𝑢

=1

7𝑢7 + 𝐶

=1

7𝑠𝑖𝑛7𝑥 + 𝐶

6. ∫ 𝑡𝑎𝑛53𝑥 𝑠𝑒𝑐23𝑥 𝑑𝑥

In this case, what would be our inner function? Here we rely on our knowledge of

derivatives. Since the derivative of tan 𝑥 is 𝑠𝑒𝑐2𝑥 we would need to choose the tangent

function as our inner function. In this example, the inner function is tan 3𝑥. Its derivative

(by the chain rule) is 3 𝑠𝑒𝑐2 3𝑥. Recall the chain rule: we apply the basic rules of

differentiation and then multiply by the derivative of the inner function 3𝑥.

Again, we use integration by substitution.

𝑢 = tan 3𝑥

𝑑𝑢 = 3 𝑠𝑒𝑐2 3𝑥 𝑑𝑥

𝑑𝑥 =𝑑𝑢

3 𝑠𝑒𝑐2 3𝑥

50

So

∫ 𝑡𝑎𝑛53𝑥 𝑠𝑒𝑐23𝑥 𝑑𝑥 = ∫ 𝑢5 . 𝑠𝑒𝑐23𝑥 .𝑑𝑢

3 𝑠𝑒𝑐2 3𝑥

= ∫1

3 𝑢5 𝑑𝑢

=1

18𝑢6 + 𝐶

=1

18(tan 3𝑥)6 + 𝐶

Integration by substitution for definite integrals

We have seen how to apply integration by substitution to indefinite integrals. How do we apply it

to definite integrals where there are upper and lower limits of integration?

We follow the same procedure we used before but before we make the substitution we would

need to first change the limits to limits with respect to the new variable of integration. That is,

suppose that we are integrating with respect to x, then the limits are the values of x. We then

change the limits from x-limits to u-limits by finding the value of u at these x-values. Only then,

do we make the substitution. Thus, the definite integral will have a new variable of integration u

and new limits for u.

The Method of integration by substitution for definite integrals

1. If the integrand is a product or quotient, let u be the inner function. If the integrand

is a function of a linear function, let u be the linear function.

2. Find du by differentiating both sides and solve for dx.

3. Change limits.

51

4. Using substitution, rewrite the integrand in terms of u and du.

5. Integrate and evaluate.

Example 5.3

1. ∫3𝑥3

√𝑥4+9 𝑑𝑥

2

0

We rewrite the integral as ∫ 2𝑥3(𝑥4 + 9)−1

22

0 𝑑𝑥

𝑢 = 𝑥4 + 9


𝑑𝑥 =𝑑𝑢

4𝑥3

Now change limits:

When 𝑥 = 0, 𝑢 = 0 + 9 = 9

When 𝑥 = 2, 𝑢 = 16 + 9 = 25

Now we substitute:

∫ 2𝑥3(𝑥4 + 9)−12

2

0

𝑑𝑥 = ∫ 2𝑥3 . 𝑢−12 .

𝑑𝑢

4𝑥3

25

9

=1

2 ∫ 𝑢−

12

25

9

𝑑𝑢

=1

2 . 2 [𝑢

12]

9

25

= 2512 − 9

12

= 5 − 3

= 2

52

2. ∫𝑥

1+𝑥4 𝑑𝑥1

0

In order to use integration by substitution, we rewrite the integral as ∫𝑥

1+(𝑥2)2 𝑑𝑥1

0

𝑢 = 𝑥2


𝑑𝑥 =𝑑𝑢

2𝑥

Change limits:

When 𝑥 = 0, 𝑢 = 0

When 𝑥 = 1, 𝑢 = 1

∫𝑥

1 + (𝑥2)2 𝑑𝑥

1

0

= ∫𝑥

1 + 𝑢2 .

𝑑𝑢

2𝑥

1

0

=1

2∫

1

1 + 𝑢2 𝑑𝑢

1

0

=1

2[tan−1 𝑢]0

1

=1

2[tan−1 1 − tan−1 0]

=1

2[𝜋

4− 0]

=𝜋

8

Harder Integration by Substitution

Sometimes we are presented with an integral and we are unable to find the technique for

integration because the established rules for using the technique do not apply. In such cases, it is

advisable to apply the technique of integration by substitution. However, we would need to

rewrite all terms in the integrand in terms of u since cancelling will not occur as in previous

examples.

53

Example 5.4

1. ∫ 𝑥2√𝑥 + 3 𝑑𝑥

We rewrite the integral as ∫ 𝑥2(𝑥 + 3)1

2 𝑑𝑥

𝑢 = 𝑥 + 3

𝑑𝑢 = 𝑑𝑥

We now rewrite 𝑥2 in terms of u.

𝑢 = 𝑥 + 3

𝑥 = 𝑢 − 3

𝑥2 = (𝑢 − 3)2

So

∫ 𝑥2(𝑥 + 3)12 𝑑𝑥 = ∫(𝑢 − 3)2 . 𝑢

12 𝑑𝑥

= ∫ 𝑢12(𝑢2 − 6𝑢 + 9) 𝑑𝑢

= ∫ (𝑢52 − 6𝑢

32 + 9𝑢

12) 𝑑𝑢

=2

7 𝑢

72 −

12

5𝑢

52 + 6𝑢

32 + 𝐶

=2

7 (𝑥 + 3)

72 −

12

5(𝑥 + 3)

52 + 6(𝑥 + 3)

32 + 𝐶

54


Evaluate the following integrals:

1. ∫(8𝑥 + 3)5 𝑑𝑥

2. ∫(1 + 11𝑥)6 𝑑𝑥

3. ∫ √2 − 3𝑥 𝑑𝑥

4. ∫ 𝑒4𝑥 𝑑𝑥

5. ∫ sin(7𝑥 + 3𝜋) 𝑑𝑥

6. ∫ 2 𝑠𝑒𝑐2(10 𝑡) 𝑑𝑡

7. ∫1

1−9𝑥 𝑑𝑥

8. ∫ √4 − 5𝑥3

𝑑𝑥

9. ∫ 𝑠𝑒𝑐2 (11𝑥) 𝑑𝑥

10. ∫7

3𝑥+1 𝑑𝑥

11. ∫ −2𝑒3𝑥+2 𝑑𝑥

12. ∫ sec 5𝑥 tan 5𝑥 𝑑𝑥

13. ∫1

√1−4𝑥2 𝑑𝑥

14. ∫ 4 sin(2𝜋 − 9𝑥) 𝑑𝑥

15. ∫3𝑥3

(1+𝑥4)2 𝑑𝑥

16. ∫ 5𝑥2√2𝑥3 + 7 𝑑𝑥

17. ∫ 𝑐𝑜𝑠3𝑥 sin 𝑥 𝑑𝑥

18. ∫𝑥

1+3𝑥2

1

0 𝑑𝑥

19. ∫ 𝑥3𝑒𝑥4 𝑑𝑥

20. ∫2𝑥4

𝑥5−2 𝑑𝑥

21. ∫ 3𝑥(7𝑥2 + 2)4 𝑑𝑥

22. ∫ 𝑐𝑜𝑠35𝑥 sin 5𝑥 𝑑𝑥

55

23. ∫1

√𝑥𝑒√𝑥 𝑑𝑥

24. ∫ 𝑠𝑒𝑐32𝑥 tan 2𝑥 𝑑𝑥

25. ∫ 𝑥(1 + 𝑥2)32

1

26. ∫ 2𝑥 (𝑥2 + 1)1

3 𝑑𝑥√7

0

27. ∫ 𝑡𝑎𝑛3𝑥 𝑠𝑒𝑐2𝑥 𝑑𝑥𝜋

30

28. ∫ 2𝑥√𝑥 − 1 𝑑𝑥

29. ∫ 𝑥√𝑥 + 1 𝑑𝑥

56

CHAPTER 6: TECHNIQUES OF INTEGRATION II

Integration by Parts

Integration by parts is another technique of integration that can be used to integrate products

when other techniques (like substitution and distribution) cannot be used. For example, to find

∫(𝑥2 − 4)(2𝑥3 + 𝑥) 𝑑𝑥

we first distribute and then integrate. Also, to find

∫ 𝑥(7𝑥2 + 5)11 𝑑𝑥

we use integration by substitution. However, consider the following integrals:

∫ 𝑥2𝑒𝑥 𝑑𝑥, ∫ 𝑥3 ln 𝑥 𝑑𝑥, ∫ 𝑥 sin 𝑥 𝑑𝑥, ∫ sin−1 𝑥 𝑑𝑥

Integration by substitution cannot be used in these cases, neither will distribution work. So use

the technique of integration by parts.

Formula for integration by parts

∫ 𝒖 𝒗′ 𝒅𝒙 = 𝒖𝒗 − ∫ 𝒖′𝒗 𝒅𝒙

The left hand side of this formula represents the product to be integrated and the right hand side

is the result upon application of integration by parts. Note that the right hand side also contains

an integral which, in most cases, will be a simpler integral to evaluate.

In examining the left hand side, notice that one of the functions in the product to be integrated

should be u and the other 𝑣′. We would need to wisely choose the function representing u. How

do we choose u?

57

In most cases, u is the simpler function, that is, the function which evaluates to 0 when

differentiated repeatedly. However, if the integrand contains the natural logarithmic or

inverse trigonometric function, then u = natural logarithmic function or the inverse

trigonometric function.

Example 6.1

1. ∫ 𝑥 cos 𝑥 𝑑𝑥

In this case the simpler function is x. So 𝑢 = 𝑥 and 𝑣′ = cos 𝑥. From the formula


we would need to find 𝑢′ by differentiating u, and v by integrating 𝑣′. We set up in this

way:

𝑢 = 𝑥 𝑣′ = cos 𝑥

𝑢′ = 1 𝑣 = sin 𝑥

We now substitute these terms into the formula to get:

∫ 𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 − ∫ 1 . sin 𝑥 𝑑𝑥

Simplify the integrand on the right hand side and then integrate to get:

∫ 𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 − ∫ sin 𝑥 𝑑𝑥

∫ 𝑥 cos 𝑥 = 𝑥 sin 𝑥 − (− cos 𝑥) + 𝐶

∫ 𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 + cos 𝑥 + 𝐶

58

2. ∫ 2𝑥 𝑒5𝑥 𝑑𝑥

𝑢 = 2𝑥 𝑣′ = 𝑒5𝑥

𝑢′ = 2 𝑣 =1

5 𝑒5𝑥


So

∫ 2𝑥 𝑒5𝑥 𝑑𝑥 = 2𝑥 .1

5 𝑒5𝑥 − ∫ 2 .

1

5 𝑒5𝑥 𝑑𝑥

= 2

5 𝑥 𝑒5𝑥 − ∫

2

5 𝑒5𝑥 𝑑𝑥

= 2

5 𝑥 𝑒5𝑥 −

2

25 𝑒5𝑥 + 𝐶

3. ∫ 𝑥2 ln 𝑥 𝑑𝑥

In this example, since the integrand contains the natural logarithm function, then

𝑢 = ln 𝑥.

𝑢 = ln 𝑥 𝑣′ = 𝑥2

𝑢′ =1

𝑥 𝑣 =

1

3 𝑥3

So

∫ 𝑥2 ln 𝑥 𝑑𝑥 = ln 𝑥 . 1

3 𝑥3 − ∫

1

𝑥 .

1

3 𝑥3 𝑑𝑥

= 1

3 𝑥3 ln 𝑥 − ∫

1

3 𝑥2 𝑑𝑥

= 1

3 𝑥3 ln 𝑥 −

1

9 𝑥3 + 𝐶

59

4. ∫ tan−1 𝑥 𝑑𝑥

Any integral containing an inverse trigonometric function should be integrated using

integration by parts and u must be the inverse trigonometric function. In this example, the

integrand can be written as a product of tan−1 𝑥 . 1. Thus

𝑢 = tan−1 𝑥 𝑣′ = 1

𝑢′ =1

1 + 𝑥2 𝑣 = 𝑥

So

∫ tan−1 𝑥 𝑑𝑥 = tan−1 𝑥 . 𝑥 − ∫1

1 + 𝑥2 . 𝑥 𝑑𝑥

= 𝑥 tan−1 𝑥 − ∫𝑥

1 + 𝑥2 𝑑𝑥

Note that to integrate the remaining integral we will need to use integration by

substitution. So to find

∫𝑥

1 + 𝑥2 𝑑𝑥

𝑢 = 1 + 𝑥2


𝑑𝑥 =𝑑𝑢

2𝑥

∫𝑥

1 + 𝑥2 𝑑𝑥 = ∫

𝑥

𝑢 .

𝑑𝑢

2𝑥

= ∫1

2 .

1

𝑢 𝑑𝑢

= 1

2 ln|𝑢|

= 1

2ln |1 + 𝑥2|

60

Thus

∫ tan−1 𝑥 𝑑𝑥 = 𝑥 tan−1 𝑥 − 1

2ln|1 + 𝑥2| + 𝐶

Repeated integration by parts

In the last example, after applying integration by parts, we saw that we had to apply integration

by substitution to the integral in the result. Sometimes, we may have to apply integration by parts

again to the integral in the result. This technique is called repeated integration by parts.

Example 6.2

1. ∫ 𝑥2 𝑒2𝑥 𝑑𝑥

𝑢 = 𝑥2 𝑣′ = 𝑒2𝑥

𝑢′ = 2𝑥 𝑣 =1

2𝑒2𝑥

So

∫ 𝑥2 𝑒2𝑥 𝑑𝑥 = 1

2𝑥2𝑒2𝑥 − ∫ 𝑥𝑒2𝑥 𝑑𝑥 … … … … … … … . . (1)

Notice that the only technique that we can apply to ∫ 𝑥𝑒2𝑥 𝑑𝑥 is integration by parts.

∫ 𝑥𝑒2𝑥 𝑑𝑥

𝑢 = 𝑥 𝑣′ = 𝑒2𝑥

𝑢′ = 1 𝑣 = 1

2𝑒2𝑥

So

∫ 𝑥𝑒2𝑥 𝑑𝑥 = 1

2𝑥𝑒2𝑥 − ∫

1

2𝑒2𝑥 𝑑𝑥

61

=1

2𝑥𝑒2𝑥 −

1

4 𝑒2𝑥

Now we substitute this into (1) above:

∫ 𝑥2 𝑒2𝑥 𝑑𝑥 = 1

2𝑥2𝑒2𝑥 − (

1

2𝑥𝑒2𝑥 −

1

4 𝑒2𝑥) + 𝐶

=1

2𝑥2𝑒2𝑥 −

1

2𝑥𝑒2𝑥 +

1

4 𝑒2𝑥 + 𝐶

2. ∫ 𝑥3𝑒𝑥 𝑑𝑥

In this example, we apply integration by parts 3 times.

𝑢 = 𝑥3 𝑣′ = 𝑒𝑥

𝑢′ = 3𝑥2 𝑣 = 𝑒𝑥

So

∫ 𝑥3𝑒𝑥 𝑑𝑥 = 𝑥3𝑒𝑥 − ∫ 3𝑥2𝑒𝑥 𝑑𝑥 … … … … … … … … (1)

We apply integration by parts to ∫ 3𝑥2𝑒𝑥 𝑑𝑥:

∫ 3𝑥2𝑒𝑥 𝑑𝑥

𝑢 = 3𝑥2 𝑣′ = 𝑒𝑥

𝑢′ = 6𝑥 𝑣 = 𝑒𝑥

So

∫ 3𝑥2𝑒𝑥 𝑑𝑥 = 3𝑥2𝑒𝑥 − ∫ 6𝑥𝑒𝑥 𝑑𝑥 … … … … … … … … … (2)

We apply integration by parts to ∫ 6𝑥𝑒𝑥 𝑑𝑥:

∫ 6𝑥𝑒𝑥 𝑑𝑥

𝑢 = 6𝑥 𝑣′ = 𝑒𝑥

62

𝑢′ = 6 𝑣 = 𝑒𝑥

So

∫ 6𝑥𝑒𝑥 𝑑𝑥 = 6𝑥𝑒𝑥 − ∫ 6𝑒𝑥 𝑑𝑥

= 6𝑥𝑒𝑥 − 6𝑒𝑥

Substitute in (2):

∫ 3𝑥2𝑒𝑥 𝑑𝑥 = 3𝑥2𝑒𝑥 − (6𝑥𝑒𝑥 − 6𝑒𝑥)

= 3𝑥2𝑒𝑥 − 6𝑥𝑒𝑥 + 6𝑒𝑥

Now substitute in (1):

∫ 𝑥3𝑒𝑥 𝑑𝑥 = 𝑥3𝑒𝑥 − (3𝑥2𝑒𝑥 − 6𝑥𝑒𝑥 + 6𝑒𝑥) + 𝐶

= 𝑥3𝑒𝑥 − 3𝑥2𝑒𝑥 + 6𝑥𝑒𝑥 − 6𝑒𝑥 + 𝐶

3. ∫ 𝑒𝑥 cos 𝑥 𝑑𝑥

In this example, it appears that we may have to apply integration by parts multiple times,

but this is not so. Here, you will learn a useful trick.

In this case, none of the functions obey our definition of what a simple function is. So we

can choose any function as u.

𝑢 = 𝑒𝑥 𝑣′ = cos 𝑥

𝑢′ = 𝑒𝑥 𝑣 = sin 𝑥

∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 = 𝑒𝑥 sin 𝑥 − ∫ 𝑒𝑥 sin 𝑥 𝑑𝑥 … … … … … … . . (1)

63

We again apply integration by parts and choose u to be the same type of function we

chose it to be previously:

∫ 𝑒𝑥 sin 𝑥 𝑑𝑥

𝑢 = 𝑒𝑥 𝑣′ = sin 𝑥

𝑢′ = 𝑒𝑥 𝑣 = − cos 𝑥

∫ 𝑒𝑥 sin 𝑥 𝑑𝑥 = −𝑒𝑥 cos 𝑥 + ∫ 𝑒𝑥 cos 𝑥 𝑑𝑥

It would be futile to apply integration by parts again since ∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 is the integral

we started with in the first place. So instead we substitute our result into (1):

∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 = 𝑒𝑥 sin 𝑥 − (−𝑒𝑥 cos 𝑥 + ∫ 𝑒𝑥 cos 𝑥 𝑑𝑥)

∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 = 𝑒𝑥 sin 𝑥 + 𝑒𝑥 cos 𝑥 − ∫ 𝑒𝑥 cos 𝑥 𝑑𝑥

Notice that the same integral appears on both sides of the integration. We can therefore

collect like terms:

∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 + ∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 = 𝑒𝑥 sin 𝑥 + 𝑒𝑥 cos 𝑥

2 ∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 = 𝑒𝑥 sin 𝑥 + 𝑒𝑥 cos 𝑥

So

∫ 𝑒𝑥 cos 𝑥 𝑑𝑥 = 1

2(𝑒𝑥 sin 𝑥 + 𝑒𝑥 cos 𝑥 ) + 𝐶

Definite integration by parts

Here is one example of definite integration by parts. We use the same procedure for integration

by parts, but after integrating we substitute the limits of integration as we did before.

64

Example 6.3

1. ∫ 𝑥 𝑒−𝑥 𝑑𝑥3

0

𝑢 = 𝑥 𝑣′ = 𝑒−𝑥

𝑢′ = 1 𝑣 = −𝑒−𝑥

∫ 𝑥 𝑒−𝑥 𝑑𝑥 = [−𝑥𝑒−𝑥]03 + ∫ 𝑒−𝑥 𝑑𝑥

3

0

3

0

= [−𝑥𝑒−𝑥 − 𝑒−𝑥]03

= [−3𝑒−3 − 𝑒−3] − [−1]

= −3𝑒−3 − 𝑒−3 + 1

65



1. ∫ 𝑥𝑒𝑥 𝑑𝑥

2. ∫ 𝑥2 cos 𝑥 𝑑𝑥

3. ∫ ln 𝑥 𝑑𝑥

4. ∫ 3𝑥 sin 𝑥 𝑑𝑥

5. ∫ 7𝑥 ln 𝑥 𝑑𝑥

6. ∫ sin−1 𝑥 𝑑𝑥

7. ∫ 𝑥5 ln 𝑥 𝑑𝑥

8. ∫ 5𝑥 cos 2𝑥 𝑑𝑥𝜋

0

9. ∫(𝑥2 − 3𝑥)𝑒2𝑥 𝑑𝑥

10. ∫ 𝑥2 sin 𝑥 𝑑𝑥

11. ∫ 𝑡3𝑒−𝑡 𝑑𝑡1

0

12. ∫ 𝑒3𝑥 sin 5𝑥 𝑑𝑥

13. ∫ 𝑥 . √3 − 𝑥3

𝑑𝑥

14. ∫ 7𝑥 tan−1(𝑥2) 𝑑𝑥

15. ∫ 𝑥5𝑒𝑥3 𝑑𝑥 (Hint: express the integral as a different product: 𝑥3 . 𝑥2𝑒𝑥3

66

CHAPTER 7: TECHNIQUES OF INTEGRATION III

INTEGRATING RATIONAL FUNCTIONS

In this chapter you will learn how to integrate a rational function using other methods not

previously taught. You already know how to integrate some rational functions like 𝑥2−5

𝑥3 or

𝑥

(4𝑥2−5)4. In this chapter we will examine other rational functions and learn other techniques to

integrate them.

Integrating Improper Rational functions

A rational function 𝑓(𝑥)

𝑔(𝑥) is said to be improper if the degree of 𝑓(𝑥) is greater than or equal to the

degree of 𝑔(𝑥). Examples of improper rational functions are 𝑥−1

𝑥+3 and

𝑥2

𝑥−5.

To integrate an improper rational function 𝑓(𝑥)

𝑔(𝑥) , we first divide the numerator by the

denominator to obtain a result of the form 𝑝(𝒙) +𝒒(𝒙)

𝒈(𝒙). Then, integrate the result.

Example 7.1

1. ∫𝑥+1

𝑥−5 𝑑𝑥

We divide 𝑥 + 1 by 𝑥 − 5 to get:

1 +6

𝑥 − 5

So

∫𝑥 + 1

𝑥 − 5 𝑑𝑥 = ∫ (1 +

6

𝑥 − 5) 𝑑𝑥 = 𝑥 + 6 ln|𝑥 − 5| + 𝐶

67

2. ∫𝑥2+4𝑥−2

𝑥2+1 𝑑𝑥

Divide 𝑥2 + 4𝑥 − 2 by 𝑥2 + 1 to get 1 +4𝑥−3

𝑥2+1 . So

∫𝑥2 + 4𝑥 − 2

𝑥2 + 1 𝑑𝑥 = ∫ (1 +

4𝑥 − 3

𝑥2 + 1) 𝑑𝑥

= ∫ (1 +4𝑥

𝑥2 + 1−

3

𝑥2 + 1 ) 𝑑𝑥

= 𝑥 + 2 ln|𝑥2 + 1| − 3 tan−1 𝑥 + 𝐶

Note that integration by substitution was used to integrate the second term.

Integrating Rational functions whose denominators contain irreducible quadratic factors

An irreducible quadratic factor is a quadratic expression that cannot be factored. For example,

𝑥2 + 3, 𝑥2 − 𝑥 − 1 are irreducible quadratic factors.

To integrate such rational functions, we first complete the square of the denominator of the

rational function, and then integrate the resulting rational function by using the integration by

substitution technique.

Review of completing the square

To complete the square of a quadratic expression 𝑎𝑥2 + 𝑏𝑥 + 𝑐, first factor out the coefficient of

𝑥2 so that the coefficient of 𝑥2 is 1. Then add and subtract from the expression, the square of half

the coefficient of x. Factor the trinomial. That is,

𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 𝑎 (𝑥2 +𝑏

𝑎𝑥 +

𝑐

𝑎)

= 𝑎 (𝑥2 +𝑏

𝑎𝑥 + (

𝑏

2𝑎)

2

+𝑐

𝑎− (

𝑏

2𝑎)

2

)

= 𝑎 [(𝑥 +𝑏

2𝑎)

2

+𝑐

𝑎−

𝑏2

4𝑎2]

68

= 𝑎 [(𝑥 +𝑏

2𝑎)

2

−𝑏2 − 4𝑎𝑐

4𝑎2]

= 𝑎 (𝑥 +𝑏

2𝑎)

2

−𝑏2 − 4𝑎𝑐

4𝑎

When integrating the resulting expression, you will find these formulas useful:

∫𝟏

𝟏 + 𝒙𝟐 𝒅𝒙 = 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 + 𝑪

∫𝟏

𝒂𝟐 + 𝒙𝟐 𝒅𝒙 =

𝟏

𝒂𝐚𝐫𝐜𝐭𝐚𝐧

𝒙

𝒂+ 𝑪

∫𝟏

√𝟏 − 𝒙𝟐 𝒅𝒙 = 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙 + 𝑪

∫𝟏

√𝒂𝟐 − 𝒙𝟐 𝒅𝒙 = 𝐚𝐫𝐜𝐬𝐢𝐧

𝒙

𝒂+ 𝑪

Example 7.2

1. ∫1

𝑥2+2𝑥+5 𝑑𝑥

Since 𝑥2 + 2𝑥 + 5 is irreducible, first complete the square.

𝑥2 + 2𝑥 + 5 = 𝑥2 + 2𝑥 + 1 + 5 − 1

= (𝑥 + 1)2 + 4

So

∫1

𝑥2 + 2𝑥 + 5 𝑑𝑥 = ∫

1

(𝑥 + 1)2 + 4 𝑑𝑥

Now use integration by substitution:

𝑢 = 𝑥 + 1

𝑑𝑢 = 𝑑𝑥

69

∫1

𝑥2 + 2𝑥 + 5 𝑑𝑥 = ∫

1

(𝑥 + 1)2 + 4 𝑑𝑥

= ∫1

𝑢2 + 4 𝑑𝑢

To integrate this expression, use the formula:

∫𝟏

𝒂𝟐 + 𝒙𝟐 𝒅𝒙 =

𝟏


𝒙

𝒂+ 𝑪

In this case 𝑎2 is 4. So

∫1

𝑥2 + 2𝑥 + 5 𝑑𝑥 = ∫

1

(𝑥 + 1)2 + 4 𝑑𝑥

= ∫1

𝑢2 + 4 𝑑𝑢

= ∫1

𝑢2 + 22 𝑑𝑢

=1

2tan−1

𝑢

2+ 𝐶

=1

2tan−1

𝑥 + 1

2+ 𝐶

2. ∫1

4𝑥2+4𝑥+5 𝑑𝑥

Since 4𝑥2 + 4𝑥 + 5 is irreducible, first complete the square.

4𝑥2 + 4𝑥 + 5 = 4 (𝑥2 + 𝑥 +5

4)

= 4 (𝑥2 + 𝑥 +1

4+

5

4−

1

4)

= 4 (𝑥2 + 𝑥 +1

4+ 1)

= 4 [(𝑥 +1

2)

2

+ 1]

70

= 4 (𝑥 +1

2)

2

+ 4

= 22 (𝑥 +1

2)

2

+ 4

= (2𝑥 + 1)2 + 4

So

∫1

4𝑥2 + 4𝑥 + 5 𝑑𝑥 = ∫

1

(2𝑥 + 1)2 + 4 𝑑𝑥

Now use integration by substitution:

𝑢 = 2𝑥 + 1

𝑑𝑢 = 2 𝑑𝑥

𝑑𝑥 =𝑑𝑢

2

∫1

4𝑥2 + 4𝑥 + 5 𝑑𝑥 = ∫

1

(2𝑥 + 1)2 + 4 𝑑𝑥

=1

2∫

1

𝑢2 + 4 𝑑𝑢

To integrate this expression, use the formula:

∫𝟏

𝒂𝟐 + 𝒙𝟐 𝒅𝒙 =

𝟏


𝒙

𝒂+ 𝑪

In this case 𝑎2 is 4. So

∫1

4𝑥2 + 4𝑥 + 5 𝑑𝑥 = ∫

1

(2𝑥 + 1)2 + 4 𝑑𝑥

=1

2∫

1

𝑢2 + 4 𝑑𝑢

71

=1

2∫

1

𝑢2 + 22 𝑑𝑢

=1

2 .

1

2tan−1

𝑢

2+ 𝐶

=1

4tan−1

2𝑥 + 1

2+ 𝐶

So far, you have learned how to integrate rational expressions whose denominators are

irreducible quadratic factors and numerators are constants. You will now learn how to integrate

rational expressions whose denominators are irreducible quadratic factors and numerators are

functions of 𝑥. To integrate such rational functions, you will again complete the square of the

denominator of the rational function. Then express the numerator in terms of the term which is

squared in the denominator. Finally, integrate the resulting rational function by using the

integration by substitution technique.

You will find this formula useful:

∫𝒇′(𝒙)

𝒇(𝒙) 𝒅𝒙 = 𝐥𝐧|𝒇(𝒙)| + 𝑪

For example, since the derivative of 𝑥2 + 5 is 2𝑥, then

∫2𝑥

𝑥2 + 5 𝑑𝑥 = ln|𝑥2 + 5| + 𝐶

Also

∫6𝑥

𝑥2 + 5 𝑑𝑥 = 3 ∫

2𝑥

𝑥2 + 5 𝑑𝑥 = 3 ln|𝑥2 + 5| + 𝐶

∫𝑥

𝑥2 + 5 𝑑𝑥 =

1

2∫

2𝑥

𝑥2 + 5 𝑑𝑥 =

1

2ln|𝑥2 + 5| + 𝐶

72

Example 7.3

1. ∫3𝑥+1


Since 𝑥2 + 4𝑥 + 5 is irreducible, first complete the square.

𝑥2 + 4𝑥 + 5 = 𝑥2 + 4𝑥 + 4 + 5 − 4

= (𝑥 + 2)2 + 1

We now express 3𝑥 + 1 in terms of 𝑥 + 2. We replace x with 𝑥 + 2 and write an expression in

terms of 𝑥 + 2 that is equal to 3𝑥 + 1.

3𝑥 + 1 = 3(𝑥 + 2) − 5

∫3𝑥 + 1

𝑥2 + 4𝑥 + 5 𝑑𝑥 = ∫

3(𝑥 + 2) − 5

(𝑥 + 2)2 + 1 𝑑𝑥

We now use integration by substitution:

𝑢 = 𝑥 + 2

𝑑𝑢 = 𝑑𝑥

∫3𝑥 + 1

𝑥2 + 4𝑥 + 5 𝑑𝑥 = ∫

3(𝑥 + 2) − 5

(𝑥 + 2)2 + 1 𝑑𝑥

= ∫3𝑢 − 5

𝑢2 + 1 𝑑𝑢

= ∫3𝑢

𝑢2 + 1 𝑑𝑢 − ∫

5

𝑢2 + 1 𝑑𝑢

=3

2∫

2𝑢

𝑢2 + 1 𝑑𝑢 − ∫

5

𝑢2 + 1 𝑑𝑢

=3

2ln|𝑢2 + 1| − 5 tan−1 𝑢 + 𝐶

=3

2ln|(𝑥 + 2)2 + 1| − 5 tan−1(𝑥 + 2) + 𝐶

73

2. ∫2𝑥+5

𝑥2−6𝑥+25 𝑑𝑥

𝑥2 − 6𝑥 + 25 = 𝑥2 − 6𝑥 + 9 + 25 − 9

= (𝑥 − 3)2 + 16

We now express 2𝑥 + 5 in terms of 𝑥 − 3. We replace x with 𝑥 − 3 and write an expression in

terms of 𝑥 − 3 that is equal to 2𝑥 + 5.

2𝑥 + 5 = 2(𝑥 − 3) + 11

∫2𝑥 + 5

𝑥2 − 6𝑥 + 25 𝑑𝑥 = ∫

2(𝑥 − 3) + 11

(𝑥 − 3)2 + 16 𝑑𝑥

Integrate by substitution:

𝑢 = 𝑥 − 3

𝑑𝑢 = 𝑑𝑥

∫2𝑥 + 5

𝑥2 − 6𝑥 + 25 𝑑𝑥 = ∫

2(𝑥 − 3) + 11

(𝑥 − 3)2 + 16 𝑑𝑥

= ∫2𝑢 + 11

𝑢2 + 16 𝑑𝑢

= ∫2𝑢

𝑢2 + 16 𝑑𝑢 + ∫

11

𝑢2 + 16 𝑑𝑢

= ln|𝑢2 + 16| + 11

4 tan−1

𝑢

4+ 𝐶

= ln|(𝑥 − 3)2 + 16| + 11

4 tan−1

𝑥 − 3

4+ 𝐶

Integrating Rational Functions whose denominators contain reducible quadratic factors

If the denominator 𝑄(𝑥) of the rational function 𝑃(𝑥)

𝑄(𝑥) can be factored, we first decompose the

rational function into partial fractions, and then integrate. You will now learn how to decompose

a rational function into partial fractions.

74

Partial Fractions

We know how to simplify 1

𝑥−1+

1

𝑥+1 to get

2𝑥

(𝑥−1)(𝑥+1).

Now given 2𝑥

(𝑥−1)(𝑥+1) and expressing it in the form:

2𝑥

(𝑥 − 1)(𝑥 + 1)=

1

𝑥 − 1+

1

𝑥 + 1

is known as decomposing 2𝑥

(𝑥−1)(𝑥+1) into partial fractions.

There are various rules for decomposing a rational function into partial fractions. This depends

on the factors of the denominator of the rational function.

Rule 1

Consider the rational function 𝑃(𝑥)

𝑄(𝑥).

Suppose 𝑄(𝑥) can be factored into linear factors, then the partial fraction decomposition is

𝑃(𝑥)

𝑄(𝑥)=

𝐴

𝑎𝑥 + 𝑏+

𝐵

𝑐𝑥 + 𝑑+ ⋯

where A and B are constants.

How to find the values of the constants

1. Rewrite the R.H.S. (right hand side) as one fraction by finding the LCD – least common

denominator.

2. Compare the numerators.

3. Find the constants by choosing appropriate values for x.

75

Example 7.4

Decompose into partial fractions:

1. 4

(𝑥+3)(𝑥−2)

Since the denominator is made up of linear factors we decompose as follows, then

simplify the R.H.S:

4

(𝑥 + 3)(𝑥 − 2)=

𝐴

𝑥 + 3+

𝐵

𝑥 − 2

4

(𝑥 + 3)(𝑥 − 2)=

𝐴(𝑥 − 2) + 𝐵(𝑥 + 3)

(𝑥 + 3)(𝑥 − 2)

Equate numerators:

4 = 𝐴(𝑥 − 2) + 𝐵(𝑥 + 3)

To find the values of the constants, we choose values for x which will make each of the

terms on the left equal to 0.

When 𝑥 = 2:

4 = 5𝐵

𝐵 =4

5

When 𝑥 = −3:

4 = −5𝐴

𝐴 = −4

5

So

4

(𝑥 + 3)(𝑥 − 2)=

−45

𝑥 + 3+

45

𝑥 − 2

2. 3𝑥+1

𝑥2−6𝑥+5

3𝑥 + 1

𝑥2 − 6𝑥 + 5=

3𝑥 + 1

(𝑥 − 5)(𝑥 − 1)=

𝐴

𝑥 − 5+

𝐵

𝑥 − 1

76

3𝑥 + 1

(𝑥 − 5)(𝑥 − 1)=

𝐴(𝑥 − 1) + 𝐵(𝑥 − 5)

(𝑥 − 5)(𝑥 − 1)

3𝑥 + 1 = 𝐴(𝑥 − 1) + 𝐵(𝑥 − 5)

When 𝑥 = 1:

3 + 1 = −4𝐵

𝐵 = −1

When 𝑥 = −5:

−15 + 1 = −6𝐴

𝐴 =7

3

So

3𝑥 + 1

𝑥2 − 6𝑥 + 5=

73

𝑥 − 5−

1

𝑥 − 1

Rule 2

Suppose Q(x) has repeated linear factors of the form (𝑎𝑥 + 𝑏)𝑛, use the decomposition:

𝑃(𝑥)

𝑄(𝑥)=

𝐴1

𝑎𝑥 + 𝑏+

𝐴2

(𝑎𝑥 + 𝑏)2+ ⋯ +

𝐴𝑛

(𝑎𝑥 + 𝑏)𝑛

where the 𝐴𝑖′𝑠 are constants.


1. Rewrite the R.H.S. as one fraction by finding the LCD.

77



4. Equate coefficients of 𝑥2 or any other variable to find the last constant.

Example 7.5

1. 3𝑥

(𝑥+2)(𝑥−1)2

3𝑥

(𝑥 + 2)(𝑥 − 1)2=

𝐴

𝑥 + 2+

𝐵

𝑥 − 1+

𝐶

(𝑥 − 1)2

3𝑥

(𝑥 + 2)(𝑥 − 1)2=

𝐴(𝑥 − 1)2 + 𝐵(𝑥 + 2)(𝑥 − 1) + 𝐶(𝑥 + 2)

(𝑥 + 2)(𝑥 − 1)2

3𝑥 = 𝐴(𝑥 − 1)2 + 𝐵(𝑥 + 2)(𝑥 − 1) + 𝐶(𝑥 + 2)

When 𝑥 = 1:

3 = 3𝐶

𝐶 = 1

When 𝑥 = −2:

−6 = 9𝐴

𝐴 = −2

3

We now equate coefficients of 𝑥2 on both sides of the equation to find B:

0 = 𝐴 + 𝐵

0 = −2

3+ 𝐵

𝐵 =2

3

3𝑥

(𝑥 + 2)(𝑥 − 1)2=

−23

𝑥 + 2+

23

𝑥 − 1+

1

(𝑥 − 1)2

78

2. 𝑥2+4𝑥+3

2𝑥3−𝑥2

𝑥2 + 4𝑥 + 3

2𝑥3 − 𝑥2=

𝑥2 + 4𝑥 + 3

𝑥2(2𝑥 − 1)=

𝐴

𝑥+

𝐵

𝑥2+

𝐶

2𝑥 − 1

𝑥2 + 4𝑥 + 3

𝑥2(2𝑥 − 1)=

𝐴𝑥(2𝑥 − 1) + 𝐵(2𝑥 − 1) + 𝐶𝑥2

𝑥2(2𝑥 − 1)

𝑥2 + 4𝑥 + 3 = 𝐴𝑥(2𝑥 − 1) + 𝐵(2𝑥 − 1) + 𝐶𝑥2

When 𝑥 = 0:

3 = −𝐵

𝐵 = −3

When 𝑥 =1

2:

1

4+ 2 + 3 =

1

4𝐶

𝐶 = 21

Equate coefficients of 𝑥2:

1 = 2𝐴 + 𝐶

1 = 2𝐴 + 21

−20 = 2𝐴

𝐴 = −10

So

𝑥2 + 4𝑥 + 3

2𝑥3 − 𝑥2=

−10

𝑥−

3

𝑥2+

21

2𝑥 − 1

Rule 3

Suppose Q(x) has an irreducible quadratic factor 𝑎𝑥2 + 𝑏𝑥 + 𝑐, then use the decomposition

𝐴𝑥 + 𝐵

𝑎𝑥2 + 𝑏𝑥 + 𝑐

79


1. Rewrite the R.H.S. as one fraction by finding the LCD.



4. Substitute 𝑥 = 0.

5. Equate coefficients of 𝑥2 to find the last constant.

Example 7.6

1. 4

(𝑥+1)(𝑥2+3)

4

(𝑥 + 1)(𝑥2 + 3)=

𝐴

𝑥 + 1+

𝐵𝑥 + 𝐶

𝑥2 + 3

4

(𝑥 + 1)(𝑥2 + 3)=

𝐴(𝑥2 + 3) + (𝐵𝑥 + 𝐶)(𝑥 + 1)

(𝑥 + 1)(𝑥2 + 3)

4 = 𝐴(𝑥2 + 3) + (𝐵𝑥 + 𝐶)(𝑥 + 1)

When 𝑥 = −1:

4 = 4𝐴

𝐴 = 1

When 𝑥 = 0:

4 = 3𝐴 + 𝐶

4 = 3 + 𝐶

𝐶 = 1


80

0 = 𝐴 + 𝐵

0 = 1 + 𝐵

𝐵 = −1

So

4

(𝑥 + 1)(𝑥2 + 3)=

1

𝑥 + 1+

1 − 𝑥

𝑥2 + 3

2. 𝑥2

(𝑥−2)(𝑥2+𝑥+1)

𝑥2

(𝑥 − 2)(𝑥2 + 𝑥 + 1)=

𝐴

𝑥 − 2+

𝐵𝑥 + 𝐶

𝑥2 + 𝑥 + 1

𝑥2

(𝑥 − 2)(𝑥2 + 𝑥 + 1)=

𝐴(𝑥2 + 𝑥 + 1) + (𝐵𝑥 + 𝐶)(𝑥 − 2)

(𝑥 − 2)(𝑥2 + 𝑥 + 1)

𝑥2 = 𝐴(𝑥2 + 𝑥 + 1) + (𝐵𝑥 + 𝐶)(𝑥 − 2)

When 𝑥 = 2:

4 = 7𝐴

𝐴 =4

7

When 𝑥 = 0:

0 = 𝐴 − 2𝐶

0 =4

7− 2𝐶

81

𝐶 =2

7


1 = 𝐴 + 𝐵

1 =4

7+ 𝐵

𝐵 =3

7

So

𝑥2

(𝑥 − 2)(𝑥2 + 𝑥 + 1)=

47

𝑥 − 2+

37 𝑥 +

27

𝑥2 + 𝑥 + 1

Once you have resolved a rational function into its partial fractions, it is now easy to

integrate. You will find these formulas useful:

∫1

1 + 𝑥2 𝑑𝑥 = arctan 𝑥 + 𝐶

∫1

𝑎2 + 𝑥2 𝑑𝑥 =

1

𝑎arctan

𝑥

𝑎+ 𝐶

∫𝑓′(𝑥)

𝑓(𝑥) 𝑑𝑥 = ln|𝑓(𝑥)| + 𝐶

Example 7.7

1. ∫4

(𝑥+3)(𝑥−2) 𝑑𝑥

Earlier on, we decomposed this rational function and found that

82

4

(𝑥 + 3)(𝑥 − 2)=

−45

𝑥 + 3+

45

𝑥 − 2

So

∫4

(𝑥 + 3)(𝑥 − 2) 𝑑𝑥 = ∫ (

−45

𝑥 + 3+

45

𝑥 − 2) 𝑑𝑥

= −4

5ln|𝑥 + 3| +

4

5ln|𝑥 − 2| + 𝐶

2. ∫3𝑥+1

𝑥2−6𝑥+5 𝑑𝑥

We found that

3𝑥 + 1

𝑥2 − 6𝑥 + 5=

73

𝑥 − 5−

1

𝑥 − 1

So

∫3𝑥 + 1

𝑥2 − 6𝑥 + 5 𝑑𝑥 = ∫ (

73

𝑥 − 5−

1

𝑥 − 1) 𝑑𝑥

=7

3ln|𝑥 − 5| − ln|𝑥 − 1| + 𝐶

Example 7.7

1. ∫3𝑥

(𝑥+2)(𝑥−1)2 𝑑𝑥

3𝑥

(𝑥 + 2)(𝑥 − 1)2=

−23

𝑥 + 2+

23

𝑥 − 1+

1

(𝑥 − 1)2

So

∫3𝑥

(𝑥 + 2)(𝑥 − 1)2 𝑑𝑥 = ∫ (

−23

𝑥 + 2+

23

𝑥 − 1+

1

(𝑥 − 1)2) 𝑑𝑥

83

= ∫ (−

23

𝑥 + 2+

23

𝑥 − 1+ (𝑥 − 1)−2) 𝑑𝑥

= −2

3ln|𝑥 + 2| +

2

3ln|𝑥 − 1| −

1

𝑥 − 1+ 𝐶

2. ∫𝑥2+4𝑥+3

2𝑥3−𝑥2 𝑑𝑥

𝑥2 + 4𝑥 + 3

2𝑥3 − 𝑥2=

−10

𝑥−

3

𝑥2+

21

2𝑥 − 1

So

∫𝑥2 + 4𝑥 + 3

2𝑥3 − 𝑥2 𝑑𝑥 = ∫ (

−10

𝑥−

3

𝑥2+

21

2𝑥 − 1) 𝑑𝑥

= −10 ln|𝑥| +3

𝑥+

21

2ln|2𝑥 − 1| + 𝐶

Example 7.8

1. ∫4

(𝑥+1)(𝑥2+3) 𝑑𝑥

4

(𝑥 + 1)(𝑥2 + 3)=

1

𝑥 + 1+

1 − 𝑥

𝑥2 + 3

So

∫4

(𝑥 + 1)(𝑥2 + 3) 𝑑𝑥 = ∫ (

1

𝑥 + 1+

1 − 𝑥

𝑥2 + 3) 𝑑𝑥

= ∫ (1

𝑥 + 1+

1

𝑥2 + 3−

𝑥

𝑥2 + 3) 𝑑𝑥

= ln|𝑥 + 1| + 1

√3tan−1

𝑥

√3−

1

2ln|𝑥2 + 3| + 𝐶

84


Evaluate the following rational integrals:

1. ∫𝑥

𝑥−3 𝑑𝑥

2. ∫𝑥

𝑥+2 𝑑𝑥

3. ∫𝑥2+5

𝑥−1 𝑑𝑥

4. ∫𝑥3+𝑥+1

𝑥2+4 𝑑𝑥

5. ∫1

𝑥2−6𝑥+13 𝑑𝑥

6. ∫1

𝑥2+4𝑥+13 𝑑𝑥

7. ∫5

9𝑥2+12𝑥+5 𝑑𝑥

8. ∫3𝑥−1

𝑥2+4𝑥+13 𝑑𝑥

9. ∫𝑥+3

𝑥2+2𝑥+10 𝑑𝑥

10. ∫2𝑥−5

𝑥2−8𝑥+41 𝑑𝑥

11. ∫5𝑥

(𝑥+4)(𝑥+2) 𝑑𝑥

12. ∫6

3𝑥2+2𝑥−1 𝑑𝑥

13. ∫𝑥2

𝑥2−1 𝑑𝑥

14. ∫1

(𝑥+2)(𝑥−1)2 𝑑𝑥

15. ∫2𝑥2+25

𝑥3+5𝑥 𝑑𝑥

16. ∫1

(𝑥−2)(𝑥2+1) 𝑑𝑥

17. ∫𝑥2+5𝑥+3

(𝑥−1)(𝑥2+2) 𝑑𝑥

18. ∫2𝑥−1


19. ∫2𝑥−1

𝑥 (𝑥+3)2 𝑑𝑥

20. ∫10𝑥

(𝑥+3)(𝑥2+1) 𝑑𝑥

85

CHAPTER 8: TECHNIQUES OF INTEGRATION IV

TRIGONOMETRIC INTEGRALS AND TRIGONOMETRIC SUBSTITUTIONS

In this chapter, you will learn how to solve more complicated trigonometric integrals, as well as

integrate expressions using trigonometric substitutions.

Trigonometric Integrals

To evaluate trigonometric integrals, you will need to use some of the trigonometric identities you

learned in a trigonometry course. Recall these trigonometric identities:

𝒄𝒐𝒔𝟐𝒙 + 𝒔𝒊𝒏𝟐𝒙 = 𝟏 … … … … … . (𝟏)

Dividing each term in equation (1) by 𝑐𝑜𝑠2𝑥 we get:

𝟏 + 𝒕𝒂𝒏𝟐𝒙 = 𝒔𝒆𝒄𝟐𝒙 … … … … . (𝟐)

The double angle formula for cosine is:

𝒄𝒐𝒔 𝟐𝒙 = 𝒄𝒐𝒔𝟐𝒙 − 𝒔𝒊𝒏𝟐𝒙 … … . . (𝟑)

Replacing 𝑐𝑜𝑠2𝑥 with 1 − 𝑠𝑖𝑛2𝑥 and solving for 𝑠𝑖𝑛2𝑥 we get:

𝒔𝒊𝒏𝟐𝒙 =𝟏

𝟐(𝟏 − 𝐜𝐨𝐬 𝟐𝒙) … … … … (𝟒)

Similarly, replacing 𝑠𝑖𝑛2𝑥 with 1 − 𝑐𝑜𝑠2𝑥 and solving for 𝑐𝑜𝑠2𝑥 we get:

𝒄𝒐𝒔𝟐𝒙 =𝟏

𝟐(𝟏 + 𝐜𝐨𝐬 𝟐𝒙) … … … … (𝟓)

Equations (4) and (5) are called half-angle identities.

Note that equations (4) and (5) enable us to write any squared cosine and sine function in terms

of its double angle. For example,

86

𝑠𝑖𝑛23𝑥 =1

2(1 − cos 6𝑥)

𝑐𝑜𝑠2𝑥

2=

1

2(1 + cos 𝑥)

Integrating even powers of sine and cosine

To integrate even powers of sine and cosine, replace the even power with its half-angle identity.

Then simplify the result and integrate. For even powers higher than 2, you would need to use the

half angle identities more than once.

Example 8.1

1. ∫ 𝑠𝑖𝑛22𝑥 𝑑𝑥

∫ 𝑠𝑖𝑛22𝑥 𝑑𝑥 = ∫1

2(1 − cos 4𝑥) 𝑑𝑥

= ∫ (1

2−

1

2cos 4𝑥) 𝑑𝑥

=1

2𝑥 −

1

8sin 4𝑥 + 𝐶

2. ∫ 𝑐𝑜𝑠43𝑥 𝑑𝑥

∫ 𝑐𝑜𝑠43𝑥 𝑑𝑥 = ∫(𝑐𝑜𝑠23𝑥)2 𝑑𝑥

= ∫ [1

2(1 + cos 6𝑥)]

2

𝑑𝑥

=1

4∫(1 + 2 cos 6𝑥 + 𝑐𝑜𝑠26𝑥) 𝑑𝑥

=1

4∫ [1 + 2 cos 6𝑥 +

1

2(1 + cos 12𝑥)] 𝑑𝑥

=1

4∫ (1 + 2 cos 6𝑥 +

1

2+

1


87

= ∫ (3

8+ 2 cos 6𝑥 +

1


=3

8𝑥 +

1

3sin 6𝑥 +

1

24sin 12𝑥 + 𝐶

Integrating odd powers of sine and cosine

To integrate odd powers of sine and cosine:

• Factor out 𝑐𝑜𝑠2𝑥 or 𝑠𝑖𝑛2𝑥

• Substitute for 𝑐𝑜𝑠2𝑥 or 𝑠𝑖𝑛2𝑥 by using the identity 𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 = 1

• Use integration by substitution to then evaluate the integral.

Example 8.2


∫ 𝑐𝑜𝑠32𝑥 𝑑𝑥 = ∫ cos 2𝑥 . 𝑐𝑜𝑠22𝑥 𝑑𝑥

= ∫ cos 2𝑥(1 − 𝑠𝑖𝑛22𝑥) 𝑑𝑥

𝑢 = sin 2𝑥

𝑑𝑢 = 2 cos 2𝑥 𝑑𝑥

𝑑𝑥 =𝑑𝑢

2 cos 2𝑥

Substituting:

∫ cos 2𝑥 (1 − 𝑢2) . 𝑑𝑢

2 cos 2𝑥= ∫

1

2(1 − 𝑢2) 𝑑𝑢

=1

2(𝑢 −

1

3𝑢3) + 𝐶

=1

2sin 2𝑥 −

1

6 𝑠𝑖𝑛32𝑥 + 𝐶

88

Integrating 𝑠𝑖𝑛𝑚 𝑥 𝑐𝑜𝑠𝑛 𝑥 where m and n are positive integers

Follow the rules above for odd and even powers of sine and cosine. However, it is wise to first

manipulate the odd power of sine or cosine, if m or n is odd.

Example 8.3

1. ∫ 𝑠𝑖𝑛4𝑥 𝑐𝑜𝑠3𝑥 𝑑𝑥

Since the power of cosine is odd we will use the rule for odd powers of cosine.

∫ 𝑠𝑖𝑛4𝑥 𝑐𝑜𝑠3𝑥 𝑑𝑥 = ∫ 𝑠𝑖𝑛4𝑥 cos 𝑥 . 𝑐𝑜𝑠2𝑥 𝑑𝑥

= ∫ 𝑠𝑖𝑛4𝑥 cos 𝑥 (1 − 𝑠𝑖𝑛2𝑥) 𝑑𝑥

= ∫ cos 𝑥 (𝑠𝑖𝑛4𝑥 − 𝑠𝑖𝑛6𝑥) 𝑑𝑥

𝑢 = sin 𝑥

𝑑𝑢 = cos 𝑥 𝑑𝑥

Substituting:

∫(𝑢4 − 𝑢6) 𝑑𝑢 = 1

5𝑢5 −

1

7𝑢7 + 𝐶

=1

5𝑠𝑖𝑛5𝑥 −

1

7𝑠𝑖𝑛7𝑥 + 𝐶


∫ 𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥 = ∫1

2(1 − cos 2𝑥) .

1

2 (1 + cos 2𝑥) 𝑑𝑥

=1

4∫(1 − 𝑐𝑜𝑠22𝑥) 𝑑𝑥

=1

4∫ [1 −

1

2(1 + cos 4𝑥)] 𝑑𝑥

= ∫ (1

8−

1


89

=1

8𝑥 −

1

32sin 4𝑥 + 𝐶

Integrating powers of tangent

Use the identity 1 + 𝑡𝑎𝑛2𝑥 = 𝑠𝑒𝑐2𝑥 and substitute 𝑡𝑎𝑛2𝑥 with 𝑠𝑒𝑐2𝑥 − 1.

Example 8.4

1. ∫ 𝑡𝑎𝑛25𝑥 𝑑𝑥

∫ 𝑡𝑎𝑛25𝑥 𝑑𝑥 = ∫(𝑠𝑒𝑐25𝑥 − 1) 𝑑𝑥

=1

5tan 5𝑥 − 𝑥 + 𝐶


∫ 𝑡𝑎𝑛4𝑥 𝑑𝑥 = ∫ 𝑡𝑎𝑛2𝑥 . 𝑡𝑎𝑛2𝑥 𝑑𝑥

= ∫ 𝑡𝑎𝑛2𝑥 (𝑠𝑒𝑐2𝑥 − 1) 𝑑𝑥

= ∫(𝑡𝑎𝑛2𝑥 𝑠𝑒𝑐2𝑥 − 𝑡𝑎𝑛2𝑥) 𝑑𝑥

= ∫(𝑡𝑎𝑛2𝑥 𝑠𝑒𝑐2𝑥 − 𝑠𝑒𝑐2𝑥 + 1) 𝑑𝑥

=1

3𝑡𝑎𝑛3𝑥 − tan 𝑥 + 𝑥 + 𝐶

Note that we used integration by substitution to integrate the first term.

Integrating 𝑡𝑎𝑛𝑚 𝑥 𝑠𝑒𝑐𝑛 𝑥 where m and n are positive integers

In this case it would be better to manipulate the power of the secant function.

For even powers of secant:

90

• Factor out 𝑠𝑒𝑐2𝑥

• Use the substitution 𝑠𝑒𝑐2𝑥 = 1 + 𝑡𝑎𝑛2𝑥

• To integrate, use integration by substitution by letting 𝑢 = tan 𝑥

For odd powers of secant:

• Factor out sec 𝑥 tan 𝑥

• Use the substitution 𝑡𝑎𝑛2𝑥 = 𝑠𝑒𝑐2𝑥 − 1

• To integrate, use integration by substitution by letting 𝑢 = sec 𝑥

Example 8.5

1. ∫ 𝑡𝑎𝑛5𝑥 𝑠𝑒𝑐3 𝑥 𝑑𝑥

∫ 𝑡𝑎𝑛5𝑥 𝑠𝑒𝑐3 𝑥 𝑑𝑥 = ∫ 𝑡𝑎𝑛4𝑥 𝑠𝑒𝑐2 𝑥 . sec 𝑥 tan 𝑥 𝑑𝑥

= ∫(𝑡𝑎𝑛2𝑥)2 𝑠𝑒𝑐2𝑥 . sec 𝑥 tan 𝑥 𝑑𝑥

= ∫(𝑠𝑒𝑐2𝑥 − 1)2 𝑠𝑒𝑐2𝑥 . sec 𝑥 tan 𝑥 𝑑𝑥

𝑢 = sec 𝑥

𝑑𝑢 = sec 𝑥 tan 𝑥 𝑑𝑥

Substituting:

∫(𝑢2 − 1)2. 𝑢2 𝑑𝑢 = ∫(𝑢4 − 2𝑢2 + 1)𝑢2 𝑑𝑢

= ∫(𝑢6 − 2𝑢4 + 𝑢2) 𝑑𝑢

=1

7𝑢7 −

2

5𝑢5 +

1

3𝑢3 + 𝐶

=1

7𝑠𝑒𝑐7𝑥 −

2

5𝑠𝑒𝑐5𝑥 +

1

3𝑠𝑒𝑐3𝑥 + 𝐶

91


∫ 𝑡𝑎𝑛5𝑥 𝑠𝑒𝑐4𝑥 𝑑𝑥 = ∫ 𝑡𝑎𝑛5𝑥 𝑠𝑒𝑐2𝑥 𝑠𝑒𝑐2𝑥 𝑑𝑥

= ∫ 𝑡𝑎𝑛5𝑥 (𝑡𝑎𝑛2𝑥 + 1 )𝑠𝑒𝑐2𝑥 𝑑𝑥

𝑢 = tan 𝑥

𝑑𝑢 = 𝑠𝑒𝑐2𝑥 𝑑𝑥

Substituting:

∫ 𝑢5(𝑢2 + 1) 𝑑𝑢 =1

8𝑢8 +

1

6𝑢6 + 𝐶

=1

8𝑡𝑎𝑛8𝑥 +

1

6𝑡𝑎𝑛6𝑥 + 𝐶

Trigonometric substitutions

Earlier one, we used these integration formulas:

∫1

1 + 𝑥2 𝑑𝑥 = arctan 𝑥 + 𝐶

∫1

𝑎2 + 𝑥2 𝑑𝑥 =

1

𝑎arctan

𝑥

𝑎+ 𝐶

The results of these integrals were found by making a trigonometric substitution in the integral.

We use the same method for integration by substitution to evaluate such integrals. Then to

simplify the resulting integral we use trigonometric identities, then integrate.

Example 8.6

1. Use the substitution 𝑥 = 3 tan 𝜃 to find

∫𝑑𝑥

(𝑥2 + 9)

𝑥 = 3 tan 𝜃

92

𝑑𝑥 = 3 𝑠𝑒𝑐2𝜃 𝑑𝜃

Substituting:

∫𝑑𝑥

(𝑥2 + 9) = ∫

3 𝑠𝑒𝑐2𝜃

9 𝑡𝑎𝑛2𝜃 + 9 𝑑𝜃

= ∫3 𝑠𝑒𝑐2𝜃

9 (𝑡𝑎𝑛2𝜃 + 1) 𝑑𝜃

= ∫3 𝑠𝑒𝑐2𝜃

9 𝑠𝑒𝑐2𝜃 𝑑𝜃

= ∫1

3 𝑑𝜃

=1

3𝜃 + 𝐶

Now

𝑥 = 3 tan 𝜃

𝑥

3= tan 𝜃

𝜃 = tan−1𝑥

3

So

∫𝑑𝑥

(𝑥2 + 9) =

1

3tan−1

𝑥

3+ 𝐶

2. Use the substitution 2𝑥 = 3 sin 𝜃 to find

∫𝑥2

√9 − 4𝑥2 𝑑𝑥

2𝑥 = 3 sin 𝜃

93

𝑥 =3

2sin 𝜃

𝑑𝑥 =3

2cos 𝜃 𝑑𝜃

Substituting:

∫𝑥2

√9 − 4𝑥2 𝑑𝑥 = ∫

(32

sin 𝜃)2

√9 − 4 (32

sin 𝜃)2

. 3

2cos 𝜃 𝑑𝜃

= ∫

94

𝑠𝑖𝑛2𝜃

√9 − 4 .94

𝑠𝑖𝑛2𝜃

. 3

2cos 𝜃 𝑑𝜃

= ∫

94

𝑠𝑖𝑛2𝜃

√9 − 9𝑠𝑖𝑛2𝜃 .

3

2cos 𝜃 𝑑𝜃

= ∫

94

𝑠𝑖𝑛2𝜃

√9(1 − 𝑠𝑖𝑛2𝜃) .

3

2cos 𝜃 𝑑𝜃

=9

4 .

1

3.

3

2∫

𝑠𝑖𝑛2𝜃

√(1 − 𝑠𝑖𝑛2𝜃) . cos 𝜃 𝑑𝜃

=9

8 ∫

𝑠𝑖𝑛2𝜃

cos 𝜃 . cos 𝜃 𝑑𝜃

=9

8∫ 𝑠𝑖𝑛2𝜃 𝑑𝜃

=9

8∫

1

2(1 − cos 2𝜃) 𝑑𝜃

=9

16(𝜃 −

1

2sin 2𝜃) + 𝐶

Now

2𝑥 = 3 sin 𝜃

94

2𝑥

3= sin 𝜃

θ = sin−12𝑥

3

How do we write sin 2𝜃 in terms of x. We use the double angle formula:

sin 2𝜃 = 2 sin 𝜃 cos 𝜃

We then rewrite sin 𝜃 and cos 𝜃 in terms of x. We already know this for sin 𝜃 but we will need to

calculate this for cos 𝜃. We can do this in several ways. One way is by using the identity

𝑐𝑜𝑠2𝑥 + 𝑠𝑖𝑛2𝑥 = 1

Then substitute 2𝑥

3 for sin 𝜃 thus enabling us to find cos 𝜃. Another way is to use a right-angled

triangle which I will illustrate here.

We know that

sin 𝜃 = 2𝑥

3=

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

Now draw a right-angled triangle whose opposite is 2x and hypotenuse 3.

3

2x

We use the Pythagorean theorem to find the base which is

√9 − 4𝑥2

So from the triangle

cos 𝜃 =√9 − 4𝑥2

3

Thus

95

9

16(𝜃 −

1

2sin 2𝜃) + 𝐶 =

9

16(𝜃 − sin 𝜃 cos 𝜃) + 𝐶

= 9

16(sin−1

2𝑥

3−

2𝑥

3 .

√9 − 4𝑥2

3) + 𝐶

∫𝑥2

√9 − 4𝑥2 𝑑𝑥 =

9

16(sin−1

2𝑥

3−

2𝑥√9 − 4𝑥2

9) + 𝐶

Evaluating definite integrals using trigonometric substitution

We use the same method we learned in Chapter 5 for evaluating definite integrals by integration

by substitution. That is, follow the same procedure above with one difference: change the limits

of integration.

Example 8.7

1. Use the substitution 3𝑥 = 4 sin 𝜃 to evaluate

∫𝑑𝑥

√16 − 9𝑥2

43

0

3𝑥 = 4 sin 𝜃

𝑥 =4

3sin 𝜃

𝑑𝑥 =4

3cos 𝜃 𝑑𝜃

Change limits:

𝑥 =4

3sin 𝜃

When 𝑥 =4

3:

96

4

3=

4

3sin 𝜃

sin 𝜃 = 1

𝜃 =𝜋

2

When 𝑥 = 0:

0 =4

3sin 𝜃

𝜃 = 0

Thus

∫𝑑𝑥

√16 − 9𝑥2

43

0

= ∫

43 cos 𝜃

√16 − 9(43

sin 𝜃)2

𝜋2

0

𝑑𝜃

= ∫

43 cos 𝜃

√16 − 9 . 169

𝑠𝑖𝑛2𝜃

𝜋2

0

𝑑𝜃

= ∫

43 cos 𝜃

√16(1 − 𝑠𝑖𝑛2𝜃)

𝜋2

0

𝑑𝜃

=4

3 .

1

4∫

cos 𝜃

cos 𝜃

𝜋2

0

𝑑𝜃

=1

3∫ 𝑑𝜃

𝜋2

0

=1

3[𝜃]

0

𝜋2

=1

3 .

𝜋

2

=𝜋

6

97

Trigonometric substitutions with no given substitutions

How do you know that you would need to use a trigonometric substitution to evaluate an

integral, and how would you know what substitution to make?

Previously, you learned how to evaluate an integral using a given trigonometric substitution. In

the following examples you will have to determine what trigonometric substitution to make by

following the following rules:

Expression contains Trigonometric substitution Use Identity

𝑎2 − 𝑥2

𝑥 = a sin 𝜃 1 − 𝑠𝑖𝑛2𝜃 = 𝑐𝑜𝑠2𝜃

𝑎2 + 𝑥2

𝑥 = 𝑎 tan 𝜃 1 + 𝑡𝑎𝑛2𝜃 = 𝑠𝑒𝑐2𝜃

𝑥2 − 𝑎2

𝑥 = 𝑎 sec 𝜃 𝑠𝑒𝑐2𝜃 − 1 = 𝑡𝑎𝑛2𝜃

For example, suppose the integral contains (16 − 𝑥2)3

2, then the trigonometric substitution

should be 𝑥 = 4 sin 𝜃.

If the integral contains √16 + 25𝑥2, then the trigonometric substitution should be 5𝑥 = 4 tan 𝜃.

Example 8.8

1. ∫𝑥

√4−25𝑥2 𝑑𝑥

Use the substitution:

5𝑥 = 2 sin 𝜃

𝑥 =2

5sin 𝜃

98

𝑑𝑥 =2

5cos 𝜃 𝑑𝜃

∫𝑥

√4 − 25𝑥2 𝑑𝑥 = ∫

25

sin 𝜃

√4 − 25 (25

sin 𝜃)2

. 2

5cos 𝜃 𝑑𝜃

= ∫

25

sin 𝜃

√4 − 25.4

25𝑠𝑖𝑛2𝜃

. 2

5cos 𝜃 𝑑𝜃

= ∫

25

sin 𝜃

√4(1 − 𝑠𝑖𝑛2𝜃) .

2

5cos 𝜃 𝑑𝜃

=2

5.1

2.

2

5∫

sin 𝜃

cos 𝜃 . cos 𝜃 𝑑𝜃

=4

25∫ sin 𝜃 𝑑𝜃

= −4

25cos 𝜃 + 𝐶

Now

sin 𝜃 =5𝑥

2=

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒


𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = √4 − 25𝑥2

5x 2

So

cos 𝜃 =√4 − 25𝑥2

2

Thus

99

∫𝑥

√4 − 25𝑥2 𝑑𝑥 = −

4

25 .

√4 − 25𝑥2

2+ 𝐶

= − 2√4 − 25𝑥2

25+ 𝐶

2. ∫1

√9+𝑥2 𝑑𝑥

3

√3

Use the substitution

𝑥 = 3 tan 𝜃

𝑑𝑥 = 3 𝑠𝑒𝑐2𝜃 𝑑𝜃

Change limits:

When 𝑥 = 3:

3 = 3 tan 𝜃

tan 𝜃 = 1

𝜃 =𝜋

4

When 𝑥 = √3:

√3 = 3 tan 𝜃

tan 𝜃 = 1

√3

𝜃 =𝜋

6

∫1

√9 + 𝑥2 𝑑𝑥 = ∫

1

√9 + 9 𝑡𝑎𝑛2𝜃

𝜋4

𝜋6

3

√3

. 3 𝑠𝑒𝑐2𝜃 𝑑𝜃

100

= ∫1

3 sec 𝜃 . 3 𝑠𝑒𝑐2𝜃 𝑑𝜃

𝜋4

𝜋6

= ∫ sec 𝜃 𝑑𝜃

𝜋4

𝜋6

= [ln | sec 𝜃 + tan 𝜃|]𝜋6

𝜋4

= ln | sec𝜋

4+ tan

𝜋

4| − ln | sec

𝜋

6+ tan

𝜋

6|

= ln |√2 + 1| − ln |2

√3+

1

√3|

= ln|√2 + 1| − ln √3

= ln (√2 + 1

√3)

101













12. Use the substitution 𝑥 = 5 sin 𝜃 to evaluate ∫𝑥3

√25−𝑥2 𝑑𝑥.

13. Use the substitution 3𝑥 = 7 tan 𝜃 to evaluate ∫𝑑𝑥

9𝑥2+49

7

30

.

14. Use an appropriate substitution to evaluate:

a) ∫1

√16−25𝑥2 𝑑𝑥

b) ∫1

(4+𝑥2)32

2

0 𝑑𝑥

102

CHAPTER 9: REVIEW OF TECHNIQUES OF INTEGRATION

In this chapter we will summarize the techniques of integration and review when each technique

should be used to evaluate an integral.

First, you must know the basic integration formulas:

1. ∫ 𝑘 𝑑𝑥 = 𝑘𝑥 + 𝐶, where k is a constant.


𝑛+1+ 𝐶, 𝑛 ≠ −1

3. ∫1


4. ∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝐶







11. ∫ sec 𝑥 𝑑𝑥 = ln | sec 𝑥 + tan 𝑥|

12. ∫1

1+𝑥2 𝑑𝑥 = tan−1 𝑥 + 𝐶

13. ∫1

√1−𝑥2𝑑𝑥 = sin−1 𝑥 + 𝐶

14. ∫1

𝑎2+𝑥2 𝑑𝑥 =1

𝑎tan−1 𝑥

𝑎+ 𝐶

15. ∫1

√𝑎2−𝑥2𝑑𝑥 = sin−1 𝑥

𝑎+ 𝐶

16. ∫𝑓′(𝑥)

𝑓(𝑥) 𝑑𝑥 = ln|𝑓(𝑥)| 𝑑𝑥

Here is a step by step approach which should enable you to determine when a particular

technique should be used.

A) Check whether the basic integration formulas can be applied without using any

technique. It may be necessary to first simplify the integrand. For example, consider

103

∫(𝑥2 + 4)(𝑥 − 1) 𝑑𝑥

We can easily distribute the function and then use the basic integral formulas to evaluate

the integral.

B) If the integrand is a function of a linear function, then apply the basic integration

formulas and divide the result by the coefficient of 𝑥. For example,

∫ 𝑒3𝑥+2 𝑑𝑥

C) If the integrand is a quotient or product, look for an obvious substitution and use the

technique of integration by substitution. For example, consider

∫𝑥

𝑥2 − 1 𝑑𝑥, ∫ 4𝑥3(5𝑥4 + 10)5 𝑑𝑥

We use integration by substitution to evaluate both integrals: in the first one, since x is a

multiple of the derivative of 𝑥2 − 1, and in the second, since 4𝑥3 is a multiple of the

derivative of 5𝑥4 + 10.

Note also, that we could have used another technique to integrate the first. Since 𝑥2 − 1

can be factored, we could have first resolved the rational function into partial fractions

and then integrate. However, this latter technique may take a longer time so even though

the denominator of a rational function can be factored, it is advisable to first check

whether integration by substitution can be used.

D) If the integrand is a product and integration by substitution does not work, use integration

by parts. For example, consider

∫ 𝑥 𝑒𝑥 𝑑𝑥, ∫ ln 𝑥 𝑑𝑥, ∫ 𝑥2 sin−1 𝑥 𝑑𝑥

104

Each of these integrals can be evaluated using integration by parts. Recall that the

formula is:


Remember the general rule is to let the simpler function be u except when the integrand

contains the natural logarithm or inverse trigonometric function, in which case u would

be that function.

E) Suppose the integrand is a rational function:

• If the integrand is an improper rational function, then perform long division and

apply the techniques for integrating the result. For example,

∫2𝑥 − 1

5𝑥 + 3 𝑑𝑥

• If the rational function is proper and the denominator can be factored, decompose

it into partial fractions, then integrate. For example,

∫𝑥

𝑥2 + 4𝑥 + 3 𝑑𝑥

• If the rational function is proper and the quadratic denominator cannot be

factored, use the concept of completing the square on the denominator, then

integrate by using integration by substitution. For example,

∫3

𝑥2 + 2𝑥 + 10 𝑑𝑥, ∫

2𝑥 + 5

𝑥2 + 6𝑥 + 34 𝑑𝑥

F) Check for and apply other methods:

• Trigonometric integrals

• Trigonometric substitution

105

Example 9.1

1. ∫2𝑥+1

𝑥2+𝑥−30 𝑑𝑥

∫2𝑥 + 1

𝑥2 + 𝑥 − 30 𝑑𝑥 = ∫

2𝑥 + 1

(𝑥 + 6)(𝑥 − 5) 𝑑𝑥

We first decompose the rational function into partial fractions:

2𝑥 + 1

(𝑥 + 6)(𝑥 − 5)=

𝐴

𝑥 + 6+

𝐵

𝑥 − 5

2𝑥 + 1

(𝑥 + 6)(𝑥 − 5)=

𝐴(𝑥 − 5) + 𝐵(𝑥 + 6)

(𝑥 + 6)(𝑥 − 5)

2𝑥 + 1 = 𝐴(𝑥 − 5) + 𝐵(𝑥 + 6)

When 𝑥 = 5:

11 = 11𝐵

𝐵 = 1

When 𝑥 = −6:

−11 = −11𝐴

𝐴 = 1

2𝑥 + 1

(𝑥 + 6)(𝑥 − 5)=

1

𝑥 + 6+

1

𝑥 − 5

∫1

𝑥 + 6+

1

𝑥 − 5 𝑑𝑥 = ln|𝑥 + 6| + ln|𝑥 − 5| + 𝐶

= ln |𝑥 + 6

𝑥 − 5| + 𝐶

106

2. ∫𝑥−1

𝑥2−8𝑥+41 𝑑𝑥

Since the denominator of the rational function is irreducible, we complete the square:

𝑥2 − 8𝑥 + 41 = 𝑥2 − 8𝑥 + 16 + 41 − 16

= (𝑥 − 4)2 + 25

∫𝑥 + 1

𝑥2 − 8𝑥 + 41 𝑑𝑥 = ∫

(𝑥 − 4) + 5

(𝑥 − 4)2 + 25 𝑑𝑥

𝑢 = 𝑥 − 4

𝑑𝑢 = 𝑑𝑥

∫𝑢 + 5

𝑢2 + 25 𝑑𝑢 = ∫

𝑢

𝑢2 + 25 𝑑𝑢 + ∫

5

𝑢2 + 25 𝑑𝑢

=1

2ln|𝑢2 + 25| +

5

5 tan−1

𝑢

5+ 𝐶

=1

2ln|(𝑥 − 4)2 + 25| + tan−1 (

𝑥 − 4

5) + 𝐶

3. ∫(𝑥3 − 5)(𝑥2 + 1) 𝑑𝑥

We apply the basic integration formulas after distributing:

∫(𝑥3 − 5)(𝑥2 + 1) 𝑑𝑥 = ∫(𝑥5 + 𝑥3 − 5𝑥2 − 5) 𝑑𝑥

=1

5𝑥6 +

1

4𝑥4 −

5

3𝑥3 − 5𝑥 + 𝐶

4. ∫(3𝑥 + 7)11 𝑑𝑥

The integrand is a function of a linear function.

∫(3𝑥 + 7)11 𝑑𝑥 =1

3 .

1

12 (3𝑥 + 7)12 + 𝐶

107

=1

36 (3𝑥 + 7)12 + 𝐶

5. ∫ 3𝑥2𝑒4𝑥3−5 𝑑𝑥

We use integration by substitution here.

𝑢 = 4𝑥3 − 5


𝑑𝑥 =𝑑𝑢

12𝑥2

∫ 3𝑥2𝑒4𝑥3−5 𝑑𝑥 = ∫ 3𝑥2𝑒𝑢 . 𝑑𝑢

12𝑥2

=1

4∫ 𝑒𝑢 𝑑𝑢

=1

4𝑒𝑢 + 𝐶

=1

4𝑒4𝑥3−5 + 𝐶

6. ∫ 𝑥5 ln 3𝑥 𝑑𝑥

We use integration by parts here.

𝑢 = ln 3𝑥 𝑑𝑣 = 𝑥5

𝑢′ =3

3𝑥=

1

𝑥 𝑣 =

1

6𝑥6

∫ 𝑥5 ln 3𝑥 𝑑𝑥 =1

6𝑥6 ln 3𝑥 − ∫

1

𝑥 .

1

6𝑥6 𝑑𝑥

=1

6𝑥6 ln 3𝑥 − ∫

1

6𝑥5 𝑑𝑥

=1

6𝑥6 ln 3𝑥 −

1

36𝑥6 + 𝐶

108



1. ∫ 𝑥5𝑒𝑥6+2 𝑑𝑥

2. ∫3𝑥


3. ∫𝑥2−4𝑥+1

𝑥2+9 𝑑𝑥

4. ∫(𝑥4 − 2𝑥2 + 5)(𝑥3 − 1) 𝑑𝑥

5. ∫2

𝑥2+4𝑥+40 𝑑𝑥

6. ∫ 3𝑥𝑒2𝑥+1 𝑑𝑥

7. ∫5𝑥+1

𝑥2+2𝑥+82 𝑑𝑥

8. ∫(ln 𝑥)2 𝑑𝑥

9. ∫ √7 − 4𝑥 𝑑𝑥

10. ∫11𝑥

(𝑥2−9)2 𝑑𝑥

109

CHAPTER 10: APPROXIMATING DEFINITE INTEGRALS

Sometimes it is difficult or impossible to find the exact value of a definite integral. For example,

∫ 𝑒𝑥2 𝑑𝑥

1

0 cannot be integrated by conventional means. In these cases, we use numerical

methods to find approximate values for the definite integrals. We approximate these values by

finding an approximation for the required area under the curve. We will examine the following

methods:

1. Midpoint Rule

2. Trapezoidal Rule

3. Simpson’s Rule

Midpoint Rule

Here, we divide the required area into intervals. Then, approximate the area of the region by

drawing rectangles whose total area is close to the actual area. We will choose the midpoint of

each interval to draw the rectangle.

Let’s derive the formula. We will choose 6 rectangles to illustrate. Denote the midpoints of the

intervals 𝑥1∗, 𝑥2

∗, 𝑥3∗, 𝑥4

∗, 𝑥5∗, 𝑥6

∗ . Let the width of each rectangle be ∆𝑥. Then the area of the

region is approximately equal to

110

∆𝑥[𝑓(𝑥1∗) + 𝑓(𝑥2

∗) + 𝑓(𝑥3∗) + 𝑓(𝑥4

∗) + 𝑓(𝑥5∗) + 𝑓(𝑥6

∗)]

= ∆𝑥 [𝑓 (𝑥0 + 𝑥1

2) + 𝑓 (

𝑥1 + 𝑥2

2) + 𝑓 (

𝑥2 + 𝑥3

2) + 𝑓 (

𝑥3 + 𝑥4

2) + 𝑓 (

𝑥4 + 𝑥5

2)

+ 𝑓 (𝑥5 + 𝑥6

2)]

Extend this to n rectangles and we obtain the midpoint rule.

Midpoint rule (formula)

∫ 𝑓(𝑥) 𝑑𝑥 ≈ 𝑀𝑛 = ∆𝑥[𝑓(𝑥1∗) + 𝑓(𝑥2

∗) + 𝑓(𝑥3∗) + 𝑓(𝑥4

∗) + 𝑓(𝑥5∗) + 𝑓(𝑥6

∗)]𝑏

𝑎

where 𝑥𝑖∗ is the midpoint of the 𝑖𝑡ℎ interval.

∆𝑥 =𝑏 − 𝑎

𝑛

𝑥𝑖∗ =

𝑥𝑖−1 + 𝑥𝑖

2

Example 10.1

1. Use the midpoint rule with 𝑛 = 4 to approximate ∫1

𝑥 𝑑𝑥

3

1. Then find the exact value of

the definite integral and the error in using the approximation.

𝑓(𝑥) =1

𝑥

∆𝑥 =3 − 1

4= 0.5

The end-points of the intervals are: 1, 1.5, 2, 2.5, 3. So the midpoints of the intervals are:

1.25, 1.75, 2.25, 2.75

∫1

𝑥 𝑑𝑥 ≈ 0.5[𝑓(1.25) + 𝑓(1.75) + 𝑓(2.25) + 𝑓(2.75)]

3

1

111

= 0.5 [1

1.25+

1

1.75+

1

2.25+

1

2.75]

= 1.08975469

The exact value of the integral is:

∫1

𝑥 𝑑𝑥

3

1

= [ln|𝑥|]13

= ln 3 − ln 1

= 1.098612289

Error = 1.098612289 – 1.08975469 = 0.00886

Trapezoidal Rule

We approximate the area under the curve by using trapezoids.

Let’s derive the formula.

Suppose that we have again divided the area of the region into 6 intervals. We find the area of

the region by summing the areas of the trapezoids. Thus the area of the region is approximately

equal to

112

∆𝑥 [(𝑓(𝑥0) + 𝑓(𝑥1)

2) + (

𝑓(𝑥1) + 𝑓(𝑥2)

2) + (

𝑓(𝑥2) + 𝑓(𝑥3)

2) + (

𝑓(𝑥3) + 𝑓(𝑥4)

2)

+ (𝑓(𝑥4) + 𝑓(𝑥5)

2) + (

𝑓(𝑥5) + 𝑓(𝑥6)

2)]

= ∆𝑥

2[𝑓(𝑥0) + 𝑓(𝑥1) + 𝑓(𝑥1) + 𝑓(𝑥2) + 𝑓(𝑥2) + 𝑓(𝑥3) + 𝑓(𝑥3) + 𝑓(𝑥4) + 𝑓(𝑥4) + 𝑓(𝑥5)

+ 𝑓(𝑥5) + 𝑓(𝑥6)]

= ∆𝑥

2[𝑓(𝑥0) + 2{𝑓(𝑥1) + 𝑓(𝑥2) + 𝑓(𝑥3) + 𝑓(𝑥4) + 𝑓(𝑥5)} + 𝑓(𝑥6)]

Extend this to n rectangles and we obtain the trapezoidal rule:

∫ 𝑓(𝑥) 𝑑𝑥 ≈ 𝑇𝑛 = ∆𝑥

2[𝑓(𝑥0) + 2{𝑓(𝑥1) + 𝑓(𝑥2) + ⋯ + 𝑓(𝑥𝑛−1)} + 𝑓(𝑥𝑛)]

𝑏

𝑎

∆𝑥 =𝑏 − 𝑎

𝑛

Example 10.2

1. Use the trapezoidal rule with 𝑛 = 4 to approximate ∫1

𝑥 𝑑𝑥

3

1. Then find the exact value of

the definite integral and the error in using the approximation.

(𝑥) =1

𝑥

∆𝑥 =3 − 1

4= 0.5

∫1

𝑥 𝑑𝑥 ≈

0.5

2[𝑓(1) + 2{𝑓(1.5) + 𝑓(2) + 𝑓(2.5)} + 𝑓(3)]

3

1

=0.5

2[1 + 2 (

1

1.5+

1

2+

1

2.5) +

1

3]

= 1.116666667

113

From the last example,

∫1

𝑥 𝑑𝑥 = 1.098612289

3

1

Error = 1.098612289 − 1.116666667 = −0.01805437767

Some observations

1. Approximations may overestimate or underestimate the exact area. This depends on the

concavity of the curve.

For a concave down curve, 𝑀𝑛, the mid-point rule formula, gives an overestimation and

𝑇𝑛, the trapezoidal rule, gives an underestimation.

For a concave up curve, 𝑀𝑛, the mid-point rule formula, gives an underestimation and

𝑇𝑛, the trapezoidal rule, gives an overestimation.

Note that in the example above, the defined region corresponds to a concave up curve.

2. We get more accurate approximations when we increase the value of n.

3. The errors in 𝑇𝑛 and 𝑀𝑛 are opposite in sign.

Simpson’s Rule

This is a more accurate approximation. We approximate the definite integral by using parabolas

instead of line segments.

114

The formula is:

∫ 𝑓(𝑥) 𝑑𝑥 ≈ 𝑆𝑛 = ∆𝑥

3[𝑓(𝑥0) + 4𝑓(𝑥1) + 2𝑓(𝑥2) + ⋯ + 2𝑓(𝑥𝑛−2) + 4𝑓(𝑥𝑛−1) + 𝑓(𝑥𝑛)]

𝑏

𝑎

One difference with this approximation is that n must be even.

The approximation can also found by taking a weighted average of 𝑇𝑛 and 𝑀𝑛.

∫ 𝑓(𝑥) 𝑑𝑥 ≈ 𝑆𝑛 = 2𝑀𝑛 + 𝑇𝑛

3

𝑏

𝑎

Example 10.3

1. Use Simpson’s rule with 𝑛 = 4 to approximate ∫1

𝑥 𝑑𝑥

3

1. Then find the exact value of the

definite integral and the error in using the approximation.

∫1

𝑥 𝑑𝑥 ≈

0.5

3[𝑓(1) + 4𝑓(1.5) + 2𝑓(2) + 4𝑓(2.5) + 𝑓(3)]

3

1

=0.5

3[1 +

4

1.5+

2

2+

4

2.5+

1

3]

= 1.1

From the last example,

∫1

𝑥 𝑑𝑥 = 1.098612289

3

1

Error = 1.098612289 − 1.1 = −0.001387711

Note that using the weighted average formula gives:

∫ 𝑓(𝑥) 𝑑𝑥 ≈ 𝑆𝑛 = 2𝑀𝑛 + 𝑇𝑛

3

𝑏

𝑎

115

∫1

𝑥 𝑑𝑥

3

1

=2(1.08975469) + 1.116666667

3

≈ 1.1

116


Approximate the integral

∫ 𝑥3 𝑑𝑥3

0

with 𝑛 = 6 by using:

1. The midpoint rule

2. The trapezoidal rule

3. Simpsons rule

Then find the exact value of the definite integral and in each case, the error in using the

approximation.

117

CHAPTER 11: IMPROPER INTEGRALS

The integral ∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎 is said to be an improper integral if a or b is infinite or if 𝑓(𝑥) becomes

infinite within or at an extremity of the range of integration. For example,

∫ 𝑒−3𝑥 𝑑𝑥∞

2 is an improper integral because its upper limit is infinite.

∫1

𝑥4 𝑑𝑥5

0 is an improper integral because 𝑓(𝑥) =

1

𝑥4 is infinite when 𝑥 = 0.

∫1

𝑥 𝑑𝑥

5

−2 is an improper integral because 𝑓(𝑥) =

1

𝑥 is infinite when 𝑥 = 0 and −2 ≤ 0 ≤ 5.

To evaluate such integrals, please review finding limits at infinity which can be found in Chapter

2.

We will first learn how to evaluate improper integrals in which the upper and /or lower limits of

integration are infinite.

Improper integrals with infinite limits

∫ 𝑓(𝑥) 𝑑𝑥 = lim𝑡 →∞

∫ 𝑓(𝑥) 𝑑𝑥𝑡

𝑎

∞

𝑎

∫ 𝑓(𝑥) 𝑑𝑥 = lim𝑡 → −∞


𝑡

𝑏

−∞

∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥∞

𝑏

𝑏

−∞

∞

−∞

where b is any real number at which 𝑓(𝑥) is defined.

So, to evaluate such integrals, we write them in the appropriate format above. We then evaluate

the definite integral and then find the limit at infinity.

118

An improper integral is said to converge if the limit exists, otherwise it diverges.

Example 11.1

Evaluate the following improper integrals, that is, investigate their convergence.

1. ∫ 𝑒−2𝑥 𝑑𝑥∞

1

∫ 𝑒−2𝑥 𝑑𝑥∞

1

= lim𝑡→∞

∫ 𝑒−2𝑥 𝑑𝑥𝑡

1

= lim𝑡→∞

[−1

2 𝑒−2𝑥]

1

𝑡

= lim𝑡→∞

[−1

2 𝑒−2𝑡 +

1

2𝑒−2]

= lim𝑡→∞

[−1

2𝑒2𝑡 +

1

2𝑒−2]

= 0 +1

2𝑒−2

=1

2𝑒−2

This improper integral converges.

2. ∫𝑑𝑥

𝑥3

∞

2

∫𝑑𝑥

𝑥3=

∞

2

∫ 𝑥−3 𝑑𝑥∞

2

= lim𝑡→∞

∫ 𝑥−3 𝑑𝑥𝑡

2

= lim𝑡→∞

[−2𝑥−2]2𝑡

= lim𝑡→∞

[−2

𝑥2]

2

𝑡

= lim𝑡→∞

[−2

𝑡2+

2

4]

119

=1

2


3. ∫ ln 𝑥 𝑑𝑥1

−∞

∫ ln 𝑥 𝑑𝑥 = lim𝑡→−∞

∫ ln 𝑥 𝑑𝑥1

𝑡

1

−∞

We use integration by parts to find ∫ ln 𝑥 𝑑𝑥1

𝑡

𝑢 = ln 𝑥 𝑣′ = 1

𝑢′ =1

𝑥 𝑣 = 𝑥

∫ ln 𝑥 𝑑𝑥1

𝑡

= [𝑥 ln 𝑥]𝑡1 − ∫ 𝑑𝑥

1

𝑡

= [𝑥 ln 𝑥 − 𝑥]𝑡1

= (0 − 1) − (𝑡 ln 𝑡 − 𝑡)

= −1 − 𝑡 ln 𝑡 + 𝑡

So

lim𝑡→−∞

(−1 − 𝑡 ln 𝑡 + 𝑡)

= ∞

This improper integral diverges.

120

4. ∫𝑑𝑥

1+𝑥2 ∞

−∞

∫𝑑𝑥

1 + 𝑥2

∞

−∞

= ∫𝑑𝑥

1 + 𝑥2+ ∫

𝑑𝑥

1 + 𝑥2

∞

0

0

−∞

∫𝑑𝑥

1 + 𝑥2= lim

𝑡→−∞∫

𝑑𝑥

1 + 𝑥2

0

𝑡

0

−∞

= lim𝑡→−∞

[tan−1 𝑥]𝑡0

= lim𝑡→−∞

(tan−1 0 − tan−1 𝑡)

= 0 − (−𝜋

2)

=𝜋

2

∫𝑑𝑥

1 + 𝑥2= lim

𝑡→∞∫

𝑑𝑥

1 + 𝑥2

𝑡

0

∞

0

= lim𝑡→∞

[tan−1 𝑥]0𝑡

= lim𝑡→∞

(tan−1 𝑡 − tan−1 0)

=𝜋

2

So

∫𝑑𝑥

1 + 𝑥2=

𝜋

2+

𝜋

2

∞

−∞

= 𝜋


121

We now evaluate improper integrals whose limit(s) of integration make the integrand infinite.

These are called infinite integrands.

Infinite integrands

Suppose f is continuous and positive on [𝑎, 𝑏). If lim𝑥 →𝑏

𝑓(𝑥) is infinite, then

∫ 𝑓(𝑥) 𝑑𝑥 = lim𝑡 →𝑏−

∫ 𝑓(𝑥) 𝑑𝑥𝑡

𝑎

𝑏

𝑎

Similarly, if lim𝑥 →𝑎

𝑓(𝑥) is infinite, then

∫ 𝑓(𝑥) 𝑑𝑥 = lim𝑡 →𝑎+


𝑡

𝑏

𝑎

Example 11.2

1. ∫1

√𝑥−1 𝑑𝑥

5

1

Note that the lower limit of integration 1 makes the integrand infinite. So

∫1

√𝑥 − 1 𝑑𝑥 = lim

𝑡→1+∫ (𝑥 − 1)−

12 𝑑𝑥

5

𝑡

5

1

= lim𝑡→1+

[2(𝑥 − 1)12]

𝑡

5

= 2 lim𝑡→1+

[(5 − 1)12 − (𝑡 − 1)

12]

= 2(2 − 0)

= 4


122

2. ∫𝑥

𝑥2−9 𝑑𝑥

3

0

The upper limit of 3 makes the integrand infinite. So

∫𝑥

𝑥2 − 9 𝑑𝑥 = lim

𝑡→3−∫

𝑥

𝑥2 − 9 𝑑𝑥

𝑡

0

3

0

= lim𝑡→3−

[1

2ln |𝑥2 − 9|]

0

𝑡

= lim𝑡→3−

1

2 [ln|𝑡2 − 9| − ln |0 − 9|]

which does not exist since the natural logarithm of a negative number is undefined.

This improper integral diverges.

3. ∫1

𝑥13

𝑑𝑥8

−1

We note that 0 makes the integrand infinite. Since 0 is within the range of integration we

must split up the integral as follows:

∫1

𝑥13

𝑑𝑥8

−1

= ∫ 𝑥−13 𝑑𝑥 + ∫ 𝑥−

13 𝑑𝑥

8

0

0

−1

∫ 𝑥−13 𝑑𝑥 = lim

𝑡→0−∫ 𝑥−

13 𝑑𝑥

𝑡

−1

0

−1

= lim𝑡→0−

[3

2𝑥

23]

−1

𝑡

=3

2lim

𝑡→0− [𝑡

23 − 1]

=3

2

123

∫ 𝑥−13 𝑑𝑥 = lim

𝑡→0+∫ 𝑥−

13 𝑑𝑥

8

𝑡

8

0

= lim𝑡→0+

[3

2𝑥

23]

𝑡

8

=3

2lim

𝑡→0− [8

23 − 𝑡

23]

=3

2 . 4

= 6

∫1

𝑥13

𝑑𝑥 = 3

2+ 6

8

−1

= 15

2


124


Evaluate the following improper integrals. State whether the integral converges or diverges.

1. ∫𝑑𝑥

√𝑥

∞

9

2. ∫ 𝑥𝑒−𝑥 𝑑𝑥∞

1

3. ∫𝑑𝑥

9+𝑥2

√3

−∞

4. ∫𝑑𝑥

√1−𝑥2

0

−1

5. ∫𝑑𝑥

𝑥−3

5

2

6. ∫5

𝑥 𝑑𝑥

∞

−∞

125

CHAPTER 12: AREA

Recall:

The definite integral

∫ 𝒇(𝒙) 𝒅𝒙𝒃

𝒂

gives the area of the region bounded by the graph of 𝑦 = 𝑓(𝑥), the x-axis and the lines 𝑥 = 𝑎

and 𝑥 = 𝑏.

Let us now apply this to finding the area of regions. When finding the area of such a region, it is

advisable to make a sketch of that region.

Example 12.1

1. Find the area bounded by 𝑦 = 𝑥2, the x-axis, and the line 𝑥 = 3.

126

Area =

∫ 𝑥2 𝑑𝑥 = [1

3𝑥3]

0

33

0

= 9

2. Find the area of the region bounded by the curve 𝑦 = 9 − 𝑥2 and the x-axis.

127

In this example, we would need to find the limits of integration which are the x-intercepts

of the graph. We therefore solve

9 − 𝑥2 = 0

𝑥2 = 9

𝑥 = ±3

Area =

∫ (9 − 𝑥2) 𝑑𝑥 = [9𝑥 −1

3𝑥3]

−3

33

−3

= (27 − 9) − (−27 + 9)

= 36

Note that since 𝑦 = 𝑥2 is an even function, the region is symmetrical about the y-axis.

Thus, we could have found the area of the region in this way:

Area =

2 ∫ (9 − 𝑥2) 𝑑𝑥 = 2 [9𝑥 −1

3𝑥3]

0

33

0

= 2(27 − 9)

= 36

3. Find the area of the region bounded by the curve 𝑦 = 𝑥(𝑥2 − 4) and the x-axis.

128

We find the limits of integration:

𝑥(𝑥2 − 4) = 0

𝑥(𝑥 − 2)(𝑥 + 2) = 0

𝑥 = 0, 𝑥 = 2, 𝑥 = −2

We will evaluate the area of each region separately.

∫ 𝑥(𝑥2 − 4) 𝑑𝑥0

−2

= ∫ (𝑥3 − 4𝑥) 𝑑𝑥 0

−2

= [1

4𝑥4 − 2𝑥2]

−2

0

= [0 − (4 − 8)]

= 4

And

∫ 𝑥(𝑥2 − 4) 𝑑𝑥 = ∫ (𝑥3 − 4𝑥) 𝑑𝑥2

0

2

0

129

= [1

4𝑥4 − 2𝑥2]

0

2

= (4 − 8)

= −4

Recall that this definite integral is negative since the region lies below the x-axis. We

therefore take the absolute value to find the area. Thus

The area of the required region = 4 + 4 = 8

Again, note that since the graph of 𝑦 = 𝑥(𝑥2 − 4) is an odd function (symmetric about

the origin), we could have found the area in this way.

Area =

2 ∫ 𝑥(𝑥2 − 4) 𝑑𝑥0

−2

= 2 ∫ (𝑥3 − 4𝑥) 𝑑𝑥 0

−2

= 2 [1

4𝑥4 − 2𝑥2]

−2

0

= 2[0 − (4 − 8)]

= 8

Now consider the following region:

y x=f(y)

d

R

c

0 x

130

The area of the region R bounded by the curve 𝑥 = 𝑓(𝑦), the y-axis and the lines 𝑦 = 𝑐 and

𝑦 = 𝑑 is given by:

∫ 𝒇(𝒚) 𝒅𝒚𝒅

𝒄

Example 12.2

1. Find the area between the curve 𝑥 = 9 − 𝑦2 and the y-axis.

We first find the limits of integration by calculating the y-intercepts.

0 = 9 − 𝑦2

𝑦2 = 9

𝑦 = ±3

Area =

∫ (9 − 𝑦2) 𝑑𝑦 = [9𝑦 −1

3𝑦3]

−3

33

−3

= (27 − 9) − (−27 + 9)

131

= 36

2. Find the area of the region between the curve 𝑦 = 𝑥2, the y-axis, and the lines 𝑦 = 1 and

𝑦 = 9.

We would first need to rewrite the equation for the curve 𝑦 = 𝑥2 as:

𝑥 = √𝑦

Area =

∫ √𝑦9

1

𝑑𝑦 = ∫ 𝑦12

9

1

𝑑𝑦

= [2

3𝑦

32]

1

9

= 2

3(9

32 − 1

32)

132

= 2

3(27 − 1)

= 52

3

Area between two curves

Let us find the area enclosed between two curves.

y y = g(x)

R

y = f(x)

x

Let the functions f and g be continuous with 𝑓(𝑥) ≥ 𝑔(𝑥) for all x in [𝑎, 𝑏]. Let 𝑥 = 𝑎 and 𝑥 = 𝑏

be the x-intercepts of the points of intersection of the graphs of 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥). Then

the area of the region R bounded by the curves 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥) is:

∫ [𝑓(𝑥) − 𝑔(𝑥)] 𝑑𝑥𝑏

𝑎

Note that in the integrand, 𝑓(𝑥) appears first since 𝑓(𝑥) ≥ 𝑔(𝑥). That is, the graph of 𝑓(𝑥) lies

above the graph of 𝑔(𝑥).

Note

• It does not matter whether the curves are below or above the x-axis. The same formula

applies.

133

• Always sketch the region.

• Sometimes the area of the region comprises of the sum of more than one integral. In such

a case the region is not a vertically simple region. Picture a vertically simple region in

this way: take a vertical strip and move it throughout the region. If the top of that strip

always touches 𝑦 = 𝑓(𝑥) and the bottom always touches 𝑦 = 𝑔(𝑥) as the strip moves

throughout the region, then the region is called vertically simple. Otherwise, the region is

not vertically simple and you would have to divide it into more than one simple region

before using the formula above. An example of this is illustrated in Example 3 below.

Example 12.3

1. Find the area bounded by the curves 𝑦 = 𝑥2 and 𝑦 = 𝑥 .

Note that this region is vertically simple.

We first find the x-coordinates of the points of intersection.

𝑥2 = 𝑥

𝑥2 − 𝑥 = 0

134

𝑥(𝑥 − 1) = 0

𝑥 = 0, 𝑥 = 1

Area of the region =

∫ (𝑥 − 𝑥2) 𝑑𝑥 = [1

2𝑥2 −

1

3𝑥3]

0

11

0

= 1

2−

1

3

= 1

6

2. Find the area of the region enclosed by the curves 𝑦 = 𝑥2 and 𝑦 = √𝑥.

This is another vertically simple region.

First find the points of intersection:

𝑥2 = √𝑥

135

𝑥4 = 𝑥

𝑥4 − 𝑥 = 0

𝑥(𝑥3 − 1) = 0

𝑥 = 0, 𝑥 = 1

Area =

∫ (√𝑥 − 𝑥2) 𝑑𝑥 = [2

3𝑥

32 −

1

3𝑥3]

0

11

0

= 2

3−

1

3

= 1

3

3. Find the area of the region enclosed by the parabola 𝑥 = 𝑦2 and the line 𝑦 = 𝑥 − 2.

This is not a vertically simple region. We would therefore need to divide the region into

two. We then find the points of intersection:

136

𝑥 = (𝑥 − 2)2

𝑥 = 𝑥2 − 4𝑥 + 4

𝑥2 − 5𝑥 + 4 = 0

(𝑥 − 4)(𝑥 − 1) = 0

𝑥 = 1, 𝑥 = 4


∫ (√𝑥 − (−√𝑥)) 𝑑𝑥 + ∫ (√𝑥 − (𝑥 − 2)) 𝑑𝑥4

1

1

0

= ∫ 2√𝑥 𝑑𝑥 + ∫ (√𝑥 − 𝑥 + 2) 𝑑𝑥4

1

1

0

= [4

3𝑥

32]

0

1

+ [2

3𝑥

32 −

1

2𝑥2 + 2𝑥]

1

4

= 4

3+ [(

16

3− 8 + 8) − (

2

3−

1

2+ 2)]

= 9

2

This could have been solved in a simpler manner by making it a horizontally simple region. This

would necessitate the use of the following formula:

137

y

d

R x = f(y)

x = g(y)

c

0 x

Let f and g be continuous with 𝑓(𝑦) ≥ 𝑔(𝑦) for all y in [𝑐, 𝑑]. Let 𝑦 = 𝑐 and 𝑦 = 𝑑 be the y-

intercepts of the points of intersection of the graphs of 𝑥 = 𝑓(𝑦) and 𝑥 = 𝑔(𝑦), or simply the

bounds of the region. Then the area of the region bounded by the curves 𝑥 = 𝑔(𝑦) and 𝑥 = 𝑓(𝑦)

is:

∫ [𝑓(𝑦) − 𝑔(𝑦)] 𝑑𝑦𝑑

𝑐

Note that in the integrand, 𝑓(𝑦) appears first since 𝑓(𝑦) ≥ 𝑔(𝑦). That is, the graph of 𝑓(𝑦) lies

to the right of the graph of 𝑔(𝑦).

Again, sometimes the area of the region comprises of the sum of more than one integral. In such

a case the region is not a horizontally simple region. Picture a horizontally simple region in this

way: take a horizontal strip and move it throughout the region. If the right of that strip always

touches 𝑥 = 𝑓(𝑦) and the left always touches 𝑥 = 𝑔(𝑦) as the strip moves throughout the region,

then the region is called horizontally simple. Otherwise, the region is not horizontally simple and

you would have to divide it into more than one simple region before using the formula above.

138

Example 12.4

1. Find the area bounded by the curves 𝑦 = 𝑥2 and 𝑦 = 𝑥 .

Note that this region is horizontally simple.

The equations of both functions must be expressed in the form 𝑥 = 𝑓(𝑦). 𝑦 = 𝑥 is

already in this form. So we now do the same to 𝑦 = 𝑥2. Note that the portion of this

curve being considered corresponds to positive values of x. So

𝑥 = √𝑦

We now find the y-coordinates of the points of intersection.

√𝑦 = 𝑦

𝑦 = 𝑦2

𝑦 − 𝑦2 = 0

𝑦(1 − 𝑦) = 0

𝑦 = 0, 𝑦 = 1


139

∫ (√𝑦 − 𝑦) 𝑑𝑦 = [2

3𝑦

32 −

1

2𝑦2]

0

11

0

= 2

3−

1

2

= 1

6

2. Find the area of the region enclosed by the parabola 𝑥 = 𝑦2 and the line 𝑦 = 𝑥 − 2.

Earlier on we saw that this region was not vertically simple. However, it is horizontally

simple.

𝑦 = 𝑥 − 2 ⇒ 𝑥 = 𝑦 + 2

We find the y-coordinates of the points of intersection.

𝑦2 = 𝑦 + 2

𝑦2 − 𝑦 − 2 = 0

(𝑦 − 2)(𝑦 + 1) = 0

𝑦 = 2, 𝑦 = −1

140


∫ (𝑦 + 2 − 𝑦2) 𝑑𝑦 = [1

2𝑦2 + 2𝑦 −

1

3𝑦3]

−1

22

−1

= (2 + 4 −8

3) − (

1

2− 2 +

1

3)

= 9

2

141


1. Find the area of the region bounded by the curve 𝑦 = 𝑥3, the x-axis, and 𝑥 = 0 and

𝑥 = 2.

2. Find the area of the region bounded by the curve 𝑦 = 𝑥3, the y-axis, and 𝑦 = 1 and

𝑦 = 8.

3. Find the area of the region enclosed by the parabola 𝑦 = 6 − 𝑥2 and the line 𝑦 = −𝑥.

4. Find the area of the region enclosed by the curves 𝑦 = 𝑥3 and 𝑥 = 𝑦2.

5. Find the area of the region enclosed by the curves 𝑦 = 𝑥2 − 16 and 𝑦 = −𝑥2 − 4𝑥.

6. Find the area of the region enclosed by the curve 𝑦2 − 4𝑥 = 4 and 4𝑥 − 𝑦 = 16.

7. Find the area of the region between the curve 𝑦 = 𝑥2 + 5 and the line 𝑦 = 3𝑥 + 9.

8. Find the area of the region enclosed by the curves 𝑦 = 𝑥2 and 𝑦 = −𝑥2 + 6𝑥.

142

CHAPTER 13: VOLUME

We will examine two methods for finding the volume of a solid:

• By slicing

• By rotation about an axis

Volume by slicing

Here we define volumes of solids whose cross sections are plane regions. A cross section of a solid

S is a plane region formed by intersecting S with a plane.

The idea behind finding the volume of the solid by slicing is to cut the solid into slices, find the

volume of each slice and add up these volumes.

Consider the solid lying on an interval [𝑎, 𝑏]

A x b

Partition [𝑎, 𝑏] into subintervals. Slice the solid. Consider the 𝑘𝑡ℎ slice with base area 𝐴(𝑥𝑘) and

height ∆𝑘.

Volume of the 𝑘𝑡ℎ slice = 𝑉𝑘 = 𝐴(𝑥𝑘)∆𝑘

143

So the volume of the solid V =

lim𝑛 → ∞

∑ 𝐴(𝑥𝑘)∆𝑘 = ∫ 𝐴(𝑥) 𝑑𝑥𝑏

𝑎

𝑛

𝑘=1

The volume of a solid of known integrable cross-sectional area 𝑨(𝒙) from 𝒙 = 𝒂 to 𝒙 = 𝒃

is:

𝑽 = ∫ 𝑨(𝒙) 𝒅𝒙𝒃

𝒂

Steps for calculating volume by slicing

1. Sketch the solid and a typical cross section

2. Find a formula for 𝐴(𝑥), the area of a typical cross section

3. Find the limits of integration

4. Integrate

Example 13.1

1. Find the volume of a sphere of radius r.

We make a sketch of the sphere with radius r and center it at the origin. We now draw a cross-

section (slice) on the y-axis, positioned y units from the origin. This cross-section is a circle nor

disk. Let the radius of the circle be 𝑥.

144

We need to find an expression for the radius 𝑥 of the cross-section. From the diagram, knowing

that the radius of the sphere is r, we use the Pythagorean theorem to find 𝑥. Thus

𝑥2 + 𝑦2 = 𝑟2

𝑥 = √𝑟2 − 𝑦2

Therefore, the area of the slice is given by:

𝐴(𝑦) = 𝜋𝑥2 = 𝜋(𝑟2 − 𝑦2)

Since the radius of the sphere is r, then the limits of integration will go from -r to r. So

Volume =

∫ 𝜋(𝑟2 − 𝑦2) 𝑑𝑦 = 𝜋 [𝑟2𝑦 −1

3𝑦3]

−𝑟

𝑟𝑟

−𝑟

= 𝜋 [(𝑟3 −1

3𝑟3) − (−𝑟3 +

1

3𝑟3)]

= 4

3𝜋𝑟3

2. Find the volume of a right circular cone of base radius 2 and height 6.

y

145

We sketch the solid with its height on the y-axis and the origin as the center of the base.

At a particular value of y on the y-axis, we sketch a slice with radius x on the y-axis,

positioned y units from the origin. Note that the slice is a circle. We need to find the area

of that slice. To do so we will use the concept of similar triangles

6 − 𝑦

𝑥=

6

2

6 − 𝑦 = 3𝑥

𝑥 = 2 −𝑦

3

So, the area of the slice =

𝜋 (2 −𝑦

3)

2

Since the height of the cone is 6, the limits of integration are from 0 to 6.

Volume =

∫ 𝜋 (2 −𝑦

3)

2

𝑑𝑦 = 𝜋 [−3 . 1

3 (2 −

𝑦

3)

3

]0

66

0

= 𝜋(0 + 23) = 8𝜋

146

Volume of solids of revolution

Another way of finding the volume of solids is to rotate some plane region about an axis. The

solid generated is called a solid of revolution.

The shaded regions below are rotated about the x-axis to form the solids.

We can find the volume of these solids by slicing. This will enable us to arrive at a formula to

find the volume of solids of revolution.

Suppose these solids are bounded on the x-axis by the lines 𝑥 = 𝑎 and 𝑥 = 𝑏. We slice the solid

at some point x. This cross-section is a circle with radius [𝑓(𝑥)]2. So the area of the cross-section

is:

𝜋[𝑓(𝑥)]2

The volume of the solid is therefore:

𝜋 ∫ [𝑓(𝑥)]2 𝑑𝑥𝑏

𝑎

Thus, we can state the formula as follows:

147

Consider the region bounded by the curve 𝑦 = 𝑓(𝑥), the x-axis, and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏.

We rotate that region about the x-axis to obtain the following solid of revolution:

The volume of the solid obtained by rotating the shaded region about the x-axis is:

𝝅 ∫ 𝒚𝟐 𝒅𝒙𝒃

𝒂

= 𝝅 ∫ [𝒇(𝒙)]𝟐 𝒅𝒙𝒃

𝒂

Similarly, consider the region bounded by the curve 𝑥 = 𝑔(𝑦), the y-axis, and the lines 𝑦 = 𝑐

and 𝑦 = 𝑑

The volume of the solid obtained by rotating this region about the y-axis is:

𝝅 ∫ 𝒙𝟐 𝒅𝒚𝒅

𝒄

= 𝝅 ∫ [𝒈(𝒚)]𝟐 𝒅𝒚𝒅

𝒄

Example 13.2

1. The region bounded by 𝑦 = 𝑥3, the x-axis and the line 𝑥 = 1 is rotated about the x-axis.

Calculate the volume of the solid generated.

148

Volume =

𝜋 ∫ (𝑥3)2 𝑑𝑥 = 1

0

𝜋 ∫ 𝑥6 𝑑𝑥1

0

= [1

7𝑥7]

0

1

= 1

7𝜋

2. Find the volume of the solid obtained by rotating about the y-axis the region bounded by

𝑦 = 𝑥2 in the first quadrant, the line 𝑦 = 4 and y-axis.

𝑦 = 𝑥2

𝑥 = √𝑦

Volume =

𝜋 ∫ (√𝑦)2

𝑑𝑦 = 𝜋 ∫ 𝑦 𝑑𝑦4

0

4

0

= 𝜋 [1

2𝑦2]

0

4

= 8𝜋

149

Rotation of a plane region about other horizontal and vertical lines

Example 13.3

1. Find the volume of the solid generated by revolving the region bounded by 𝑦 = 𝑥3 and the

lines 𝑦 = 1 and 𝑥 = 2 about the line 𝑦 = 1.

When we rotate the solid and find a cross-section, observe that the cross-section is a

circle with radius 𝑥3 − 1. So the area of the cross-section is

𝜋(𝑥3 − 1)2

Volume of the generated solid =

𝜋 ∫ (𝑥3 − 1)2 𝑑𝑥 = 𝜋 ∫ (𝑥6 − 2𝑥3 + 1) 𝑑𝑥2

1

2

1

= 𝜋 [1

7𝑥7 −

1

2𝑥4 + 𝑥]

1

2

= 𝜋 [(128

7− 8 + 2) − (

1

7−

1

2+ 1)]

= 163

14𝜋

150

2. Find the volume of the solid generated by revolving the region between the parabola

𝑥 = 𝑦2 and the line 𝑥 = 4 about the line 𝑥 = 4.

The radius of the cross-section (a circle) is 4 − 𝑦2.

Area of the cross-section is

𝜋(4 − 𝑦2)2

To find the limits of integration we solve

𝑦2 = 4

𝑦 = ±2

Volume of the solid =

𝜋 ∫ (4 − 𝑦2)2 𝑑𝑦 = 𝜋 ∫ (16 − 8𝑦2 + 𝑦4) 𝑑𝑦2

−2

2

−2

= 𝜋 [16𝑦 −8

3𝑦3 +

1

5𝑦5]

−2

2

151

= 𝜋 [(32 −64

3+

32

5) − (−32 +

64

3−

32

5)]

= 512

15𝜋

The washer method

If the region we revolve to generate a solid does not border on or cross the axis of revolution, the

solid has a hole in it.

Consider the region lying between the curves 𝑓(𝑥) =𝑥2

8+ 1 and 𝑔(𝑥) =

𝑥2

16+

𝑥

8+

1

2. The solid

formed when this region is rotated about the x-axis, denoted by S, can be seen in the following

diagram.

𝑆1 and 𝑆2 are the solids formed when the regions under the curves 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥)

respectively are rotated about the x-axis, as seen in the following diagrams:

152

𝑆1:

𝑆2:

The cross-section of solid S is the region between two concentric circles and is called an annulus

or washer. The area of the annulus is the difference between the areas of the circles, which are the

cross-sections of solids 𝑆1 and 𝑆2. So the area of the annulus is:

153

𝜋[𝑓(𝑥)]2 − 𝜋[𝑔(𝑥)]2

Our limits of integration are a and b. Thus,

The volume of the solid obtained by rotating the shaded region R about the x-axis is:

𝝅 ∫ [𝒇(𝒙)]𝟐 − [𝒈(𝒙)]𝟐 𝒅𝒙𝒃

𝒂

Note that 𝑦 = 𝑓(𝑥) is the curve lying above the region, and 𝑦 = 𝑔(𝑥) is the curve below the

region.

Example 13.4

1. Find the volume of the solid generated when the region bounded by 𝑦 = 4 − 𝑥2 and

𝑦 = 𝑥 + 2 is rotated about the x-axis.

Let us first find the limits of integration by finding the points of intersection of the

graphs.

𝑥 + 2 = 4 − 𝑥2

154

𝑥2 + 𝑥 − 2 = 0

(𝑥 + 2)(𝑥 − 1) = 0

𝑥 = −2, 𝑥 = 1

Volume =

𝜋 ∫ [(4 − 𝑥2)2 − (𝑥 + 2)2] 𝑑𝑥 = 𝜋 ∫ (16 − 8𝑥2 + 𝑥4 − 𝑥2 − 4𝑥 − 4) 𝑑𝑥1

−2

1

−2

= 𝜋 ∫ (𝑥4 − 9𝑥2 − 4𝑥 + 12) 𝑑𝑥1

−2

= 𝜋 [1

5𝑥5 − 3𝑥3 − 2𝑥2 + 12𝑥]

−2

1

= 𝜋 [(1

5− 3 − 2 + 12) − (−

32

5+ 24 − 8 − 24)]

= 108

5𝜋

2. The plane region bounded by the curves 𝑦 = 𝑥3 and 𝑥 = 𝑦2 is rotated about the y-axis.

Calculate the volume of the solid generated.

155

We find the limits of integration:

𝑦 = (𝑦2)3

𝑦 = 𝑦6

𝑦6 − 𝑦 = 0

𝑦(𝑦5 − 1) = 0

𝑦 = 0, 𝑦 = 1

Note that we are rotating the region about the y-axis. So the formula for volume would be

something like this:

𝜋 ∫ (𝑟𝑖𝑔ℎ𝑡 𝑐𝑢𝑟𝑣𝑒)2 − (𝑙𝑒𝑓𝑡 𝑐𝑢𝑟𝑣𝑒)2 𝑑𝑦1

0

Volume =

𝜋 ∫ [(𝑦13)

2

− (𝑦2)2 ] 𝑑𝑦 = 𝜋 ∫ (𝑦23 − 𝑦4)

1

0

1

0

𝑑𝑦

= 𝜋 [3

5𝑦

53 −

1

5𝑦5]

0

1

= 2

5𝜋

156


1. Find the volume (by slicing) of a pyramid with height 5 cm and whose base is an

equilateral triangle with side 2 cm.

2. Find the volume of a right circular cone of base radius 6 and height 8.

3. Find the volume of the solid obtained by rotating the region bounded by the curve

𝑦 = 9 − 𝑥2 and the x-axis about the x-axis.

4. Consider the curve 𝑦 = √𝑥 − 4. Find the volume of the solid obtained by rotating about

the x-axis, the region bounded by this curve, the line 𝑥 = 5 and the x-axis.


𝑥 = √𝑦, the lines 𝑦 = 1 and 𝑦 = 3, and the y-axis about the y-axis.


𝑦 = 𝑒𝑥, the lines 𝑦 = 1 and 𝑦 = 2, and the y-axis about the y-axis.

7. The plane region in the first quadrant bounded by the curve 𝑦 = 𝑥3 and the line 𝑦 = 4𝑥

is rotated about the x-axis. Calculate the volume of the solid generated.

8. The plane region bounded by the curve 𝑥 = 𝑦2 and the line 𝑦 = 3𝑥 is rotated about the y-

axis. Calculate the volume of the solid generated.

157

CHAPTER 14: ARC LENGTH

In this chapter we will learn how to find the length of a curve using calculus.

Consider the curve 𝑦 = 𝑓(𝑥). We want to find the length of the curve between 𝑥 = 𝑎 and 𝑥 = 𝑏.

Just as we did when we used Riemann sums to find the area under a curve, we will subdivide the

curve into sub-curves as follows:

a ______________________________________________________________b

We now find the length of each line segment connecting each of the points 𝑃𝑖. So denote the

length of the curve by L. Then

𝐿 ≈ |𝑃0𝑃1| + |𝑃1𝑃2| + |𝑃2𝑃3| + |𝑃3𝑃4|

This can be written as

𝐿 ≈ ∑|𝑃𝑖−1𝑃𝑖|

4

𝑖=1

Recall that the greater the number of intervals, the more accurate the approximation. So as the

number of intervals approach infinity the sum above approaches the exact length of the curve.

Thus

𝐿 = lim𝑛→∞

∑|𝑃𝑖−1𝑃𝑖|

𝑛

𝑖=1

Let us now express |𝑃𝑖−1𝑃𝑖| in another way by using the Pythagorean theorem.

158

a___________________________________________________________ b

Consider the line segment 𝑃𝑖−1𝑃𝑖 which is the hypotenuse of a right-angled triangle whose base

is ∆𝑥 and height ∆𝑦.

𝑃𝑖−1

∆𝑦

∆𝑥 𝑃𝑖

∆𝑥 = 𝑥𝑖 − 𝑥𝑖−1

∆𝑦 = 𝑦𝑖 − 𝑦𝑖−1 = 𝑓(𝑥𝑖) − 𝑓(𝑥𝑖−1)

So by the Pythagorean theorem:

|𝑃𝑖−1𝑃𝑖| = √(∆𝑥)2 + (∆𝑦)2

Now recall the Mean Value Theorem from Calculus I – Differential Calculus:

Let f be a function that satisfies the following hypotheses:

1. f is continuous on the closed interval [a, b].

2. f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

159

𝑓′(𝑐) =𝑓(𝑏) − 𝑓(𝑎)

𝑏 − 𝑎

So choose a number 𝑥𝑖∗ within the interval [𝑥𝑖−1, 𝑥𝑖]. Then from the Mean Value Theorem

𝑓′(𝑥𝑖∗ ) =

𝑓(𝑥𝑖) − 𝑓(𝑥𝑖−1)

𝑥𝑖 − 𝑥𝑖−1=

∆𝑦

∆𝑥

∆𝑦 = 𝑓′(𝑥𝑖∗ )∆𝑥

So

|𝑃𝑖−1𝑃𝑖| = √(∆𝑥)2 + (∆𝑦)2

= √(∆𝑥)2 + (𝑓′(𝑥𝑖∗ )∆𝑥)2

= √(∆𝑥)2[1 + (𝑓′(𝑥𝑖∗ ))2]

= √[1 + (𝑓′(𝑥𝑖∗ ))2] . ∆𝑥

So

𝐿 = lim𝑛→∞

(∑ √[1 + (𝑓′(𝑥𝑖∗ ))2] . ∆𝑥

𝑛

𝑖=1

)

Recall the definition of the definite integral from Chapter 2:


𝑎

= lim𝑛→∞

(∑ 𝑓(𝑥𝑖) ∆𝑥

𝑛

1=1

)

This enables us to write the formula for the arc length or length of a curve:

The length of the curve 𝒚 = 𝒇(𝒙) from 𝒙 = 𝒂 to 𝒙 = 𝒃 is given by:

𝑳 = ∫ √𝟏 + (𝒇′(𝒙))𝟐 𝒃

𝒂

𝒅𝒙 = ∫ √𝟏 + (𝒅𝒚

𝒅𝒙)

𝟐

𝒃

𝒂

𝒅𝒙

160

Similarly, the length of the curve 𝒙 = 𝒈(𝒚) from 𝒚 = 𝒄 to 𝒚 = 𝒅 is given by:

𝑳 = ∫ √𝟏 + (𝒈′(𝒚))𝟐 𝒅

𝒄

𝒅𝒚 = ∫ √𝟏 + (𝒅𝒙

𝒅𝒚)

𝟐

𝒃

𝒂

𝒅𝒚

Example 14.1

1. Find the arc length of 𝑦 = 2𝑥 + 1 from 𝑥 = −1 to 𝑥 = 0.

𝐿 = ∫ √1 + (𝑓′(𝑥))2 𝑏

𝑎

𝑑𝑥 = ∫ √1 + (𝑑𝑦

𝑑𝑥)

2

𝑏

𝑎

𝑑𝑥

𝑦 = 2𝑥 + 1

𝑑𝑦

𝑑𝑥= 2

𝐿 = ∫ √1 + 22 𝑑𝑥0

−1

= √5 ∫ 𝑑𝑥0

−1

= √5(0 − (−1))

= √5

2. Find the length of the curve 𝑦 =1

12𝑥3 +

1

𝑥 from 𝑥 = 2 to 𝑥 = 4.

𝑑𝑦

𝑑𝑥=

𝑥2

4−

1

𝑥2

(𝑑𝑦

𝑑𝑥)

2

= (𝑥2

4−

1

𝑥2)

2

= 𝑥4

16−

1

2+

1

𝑥4

𝐿 = ∫ √1 +𝑥4

16−

1

2+

1

𝑥4

4

2

𝑑𝑥

161

= ∫ √𝑥4

16+

1

2+

1

𝑥4

4

2

𝑑𝑥

= ∫ √(𝑥2

4+

1

𝑥2)

2

4

2

𝑑𝑥

= ∫ (𝑥2

4+

1

𝑥2) 𝑑𝑥

4

2

= [1

12𝑥3 −

1

𝑥]

2

4

= (16

3−

1

4) − (

2

3−

1

2)

= 59

12

3. Find the length of the curve 𝑥 = 2𝑦3

2 from 𝑥 = 0 to 𝑥 = 3.

𝐿 = ∫ √1 + (𝑔′(𝑦))2 𝑑

𝑐

𝑑𝑦 = ∫ √1 + (𝑑𝑥

𝑑𝑦)

2

𝑏

𝑎

𝑑𝑦

𝑑𝑥

𝑑𝑦= 3𝑦

12

(𝑑𝑥

𝑑𝑦)

2

= 9𝑦

𝐿 = ∫ √1 + 9𝑦 𝑑𝑦3

0

= ∫ (1 + 9𝑦)12 𝑑𝑦

3

0

= [2

3 .

1

9 (1 + 9𝑦)

32]

0

3

=2

27(28)

32

162


1. Find the arc length of 𝑦 = 2𝑥 + 1 from 𝑥 = −3 to 𝑥 = 2.

2. Find the length of the curve 𝑥 =𝑦4

8+

1

4𝑦2 from 𝑦 = 1 to 𝑦 = 3.

3. Find the length of the curve 𝑦 =1

3𝑥

3

2 − 𝑥1

2 from 𝑥 = 4 to 𝑥 = 9.

4. Find the length of the curve 𝑦 = ln (cos 𝑥) from 𝑥 = 0 to 𝑥 =𝜋

6.

163

CHAPTER 15: HYDROSTATIC PRESSURE AND FORCE

In this chapter you will learn how to find hydrostatic force, the force due to some fluid pressure.

For the most part, we will consider water pressure. This is yet another application of integral

calculus.

The basic strategy for finding hydrostatic force is to break physical quantity into smaller parts,

approximate each small part, add results, take limits and evaluate the resulting integrals. This is

again similar to the technique of Riemann sums. To understand the idea behind the method for

finding hydrostatic force, we should note that

1. Water pressure increases with depth.

2. At any point in a liquid, the pressure is the same in all directions

Suppose that a thin horizontal plate with area A square meters is submerged in a fluid of mass

density 𝜌 𝑘𝑔/𝑚3 at a depth of d meters below the surface of the fluid.

The volume of the fluid directly above the plate is:

𝑉 = 𝐴𝑑

The mass of the fluid directly above the plate is:

𝑚 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 × 𝑣𝑜𝑙𝑢𝑚𝑒

= 𝜌𝐴𝑑

where 𝜌 is the Greek letter rho used here to denote the mass density of the fluid.

The force exerted by the fluid on the plate is hydrostatic force and it is given by:

𝑭 = 𝒎𝒈 = 𝝆𝒈𝑨𝒅

where 𝑔 = 9.8 𝑚/𝑠2 is acceleration due to gravity.

The hydrostatic pressure on the plate is the force per unit area and is given by:

164

𝑃 =𝐹

𝐴

𝑷 = 𝝆𝒈𝒅

This formula for hydrostatic pressure can also be written as:

𝑷 = 𝜹𝒅

Similarly, the formula for hydrostatic force can be written as:

𝑭 = 𝜹𝑨𝒅

where 𝛿 = 𝜌𝑔 is the Greek letter delta used to denote the weight density of the fluid.

The mass density of water is: 𝜌 = 1000 𝑘𝑔/𝑚3

The weight density of water is: 𝛿 = 62.5 𝑙𝑏/𝑓𝑡3

Steps for finding hydrostatic force

1. Divide submerged region into horizontal strips.

2. Find area A and depth d of the 𝑖𝑡ℎ horizontal strip.

3. Find the force F acting on the 𝑖𝑡ℎ horizontal strip:

F = pressure x Area

= 𝜌𝑔𝐴𝑑

= 𝛿𝐴𝑑

4. Integrate the force expression over the interval of depths occupied by the object.

Example 15.1

1. The end of a certain tank is in the shape of an isosceles triangle. The top of the triangle is

24 feet across. The tank is 8 feet deep at the tip of the triangle and 30 feet long. If the

165

tank is full of water, find the hydrostatic force on the end of the tank. (Assume water

weighs 62.5 𝑙𝑏𝑠/𝑓𝑡3).

We need to set up an axis system. Let 𝑦 = 0 correspond to the water surface and 𝑦 = 8

the depth.

Now break up the triangle into horizontal strips of width ∆𝑦. We assume that ∆𝑦 is small

enough so that the hydrostatic pressure on each strip is constant. Place the 𝑖𝑡ℎ strip at a

depth of y. So the remaining depth is 8 − 𝑦.

We will use similar triangles to find the length of the strip which will allow us to find its

area. Thus we would need to draw a vertical line midway through the triangle.

Let a be half the length of the strip. Using similar triangles we get

𝑎

8 − 𝑦=

12

8

8𝑎 = 12(8 − 𝑦)

𝑎 =3

2(8 − 𝑦)

So the length of the strip is

2𝑎 = 3(8 − 𝑦) = 24 − 3𝑦

166

Since the width of the strip is ∆𝑦, then the area of the strip is

(24 − 3𝑦)∆𝑦

Hydrostatic force on the strip is:

𝐹 = 𝛿𝐴𝑑

𝐹 = 62.5(24 − 3𝑦)𝑦 ∆𝑦

Hydrostatic force on the end of the tank =

∫ 62.5(24 − 3𝑦)𝑦 𝑑𝑦 = 62.5 ∫ (24𝑦 − 3𝑦2) 𝑑𝑥8

0

8

0

= 62.5 [12𝑦2 − 𝑦3]08

= 62.5 (768 − 512)

= 16000 𝑙𝑏𝑠

2. A dam has the shape of the trapezoid shown below. The height is 10 meters and the width

is 30 meters at the top and 20 meters at the bottom. Find the force on the dam due to the

hydrostatic pressure if

a) the dam is full

b) the water level is 2 meters from the top of the dam.

167

a) Again, we need to set up an axis system. Let 𝑦 = 0 correspond to the water surface and

𝑦 = 10 the depth.

Break up the trapezoid into horizontal strips of width ∆𝑦. We assume that ∆𝑦 is small

enough so that the hydrostatic pressure on each strip is constant. Place the 𝑖𝑡ℎ strip at a

depth of y. So the remaining depth is 10 − 𝑦.

We will use similar triangles to find the length of the strip which will allow us to find its

area. Thus we would need to draw a vertical line as shown in the diagram below:

Using similar triangles, we get

𝑎

10 − 𝑦=

5

10=

1

2

2𝑎 = 10 − 𝑦

𝑎 =1

2(10 − 𝑦)


20 + 2𝑎 = 20 + 10 − 𝑦 = 30 − 𝑦

168


(30 − 𝑦)∆𝑦


𝐹 = 𝜌𝑔𝐴𝑑

where 𝜌 = 1000 𝑘𝑔/𝑚3 and 𝑔 = 9.8 𝑚/𝑠2. So

𝐹 = 1000 . 9.8 (30 − 𝑦)𝑦 ∆𝑦


∫ 980(30 − 𝑦)𝑦 𝑑𝑦 = 980 ∫ (30𝑦 − 𝑦2) 𝑑𝑦10

0

10

0

= 980 [15𝑦2 −1

3𝑦3]

0

10

= 980 (1500 −1000

3)

=3430000

3 𝑁

a) We do this again with the water level 2 meters from the top of the dam.

169

Again, using similar triangles we get

𝑎

8 − 𝑦=

5

10=

1

2

2𝑎 = 8 − 𝑦

𝑎 =1

2(8 − 𝑦)


20 + 2𝑎 = 20 + 8 − 𝑦 = 28 − 𝑦


(28 − 𝑦)∆𝑦


𝐹 = 𝜌𝑔𝐴𝑑

where 𝜌 = 1000 𝑘𝑔/𝑚3 and 𝑔 = 9.8 𝑚/𝑠2. So

𝐹 = 1000 . 9.8 (28 − 𝑦)𝑦 ∆𝑦

Note that the upper limit of integration is now 8.


∫ 980(28 − 𝑦)𝑦 𝑑𝑦 = 980 ∫ (28𝑦 − 𝑦2) 𝑑𝑦8

0

8

0

= 980 [14𝑦2 −1

3𝑦3]

0

8

= 980 (896 −512

3)

=2132480

3 𝑁

170


1. A trough whose cross-section is a trapezoid, measures 6 meters across the bottom and 8

meters across the top, and is 3 meters deep. If the trough is filled with a liquid of mass

density 𝜌, what is the force due to hydrostatic pressure on one end of the trough? Write

your answer in terms of 𝜌.

2. A trough whose cross section is a trapezoid is 4 meters across the bottom, 10 meters

across the top, and 4 m deep. If the trough is filled to 3 meters with water, what is the

force due to water pressure on one end of the trough?

171

3. A trough whose cross section is an equilateral triangle with side 6 meters long is filled

with water. What is the force due to water pressure on one end of the trough?

172

CHAPTER 16: MOMENTS AND CENTER OF MASS

In this chapter, you will learn yet another application of integral calculus – the calculation of

moments and centers of mass. Our goal is to find the point P on which a thin plate of any shape

balances horizontally.

P is called the plate’s centroid or center of mass.

Moment is the tendency of an object to rotate about an axis.

Simple example

2 masses 𝑚1 and 𝑚2 are attached to a rod of negligible mass at distances 𝑑1 and 𝑑2 from the

fulcrum.

Under what conditions will the rod balance?

𝑑1 𝑑2

𝑚1 𝑚2

Moment of each mass is the product of the mass and its distance from the fulcrum, that is, 𝑚1𝑑1

and 𝑚2𝑑2.

The rod will balance when the moments are equal, that is, when

𝑚1𝑑1 = 𝑚2𝑑2

This is the center of mass condition for such a system.

173

Now suppose that the rod lies along the x-axis with 𝑚1 at 𝑥1 and 𝑚2 at 𝑥2. Denote the center of

mass by �̅�.

y

𝑥1 �̅� 𝑥2 x

𝑚1 𝑚2

Then

𝑚1(�̅� − 𝑥1) = 𝑚2(𝑥2 − �̅�)

𝑚1�̅� − 𝑚1𝑥1 = 𝑚2𝑥2 − 𝑚2�̅�

𝑚1�̅� + 𝑚2�̅� = 𝑚1𝑥1 + 𝑚2𝑥2

(𝑚1 + 𝑚2)�̅� = 𝑚1𝑥1 + 𝑚2𝑥2

�̅� =𝑚1𝑥1 + 𝑚2𝑥2

𝑚1 + 𝑚2

Thus the center of mass �̅� is given by

𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠𝑒𝑠

𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠𝑒𝑠

In general, if there is a system of n masses 𝑚1, 𝑚2, … , 𝑚𝑛 located at the points 𝑥1, 𝑥2, … , 𝑥𝑛 on

the x-axis, the center of mass is:

�̅� =𝑚1𝑥1 + 𝑚2𝑥2 + ⋯ + 𝑚𝑛𝑥𝑛

𝑚1 + 𝑚2 + ⋯ + 𝑚𝑛=

∑ 𝑚𝑖𝑥𝑖𝑛𝑖=1

∑ 𝑚𝑖𝑛𝑖=1

= ∑ 𝑚𝑖𝑥𝑖

𝑛𝑖=1

𝑚

where m is the sum of the masses and 𝑚𝑖𝑥𝑖 denotes the moment of respective mass.

174

𝑀 = ∑ 𝑚𝑖𝑥𝑖𝑛𝑖=1 is called the moment of the system about the origin.

Now consider a system of n particles with masses 𝑚1, 𝑚2, … , 𝑚𝑛 located at points

(𝑥1, 𝑦1), … , (𝑥𝑛, 𝑦𝑛).

y

∗ 𝑚1 𝑦1

∗ 𝑚2 𝑦2

𝑦𝑛 ∗ 𝑚𝑛

𝑥1 𝑥2 𝑥𝑛 x

Let us find a formula for the moment of the system about the y-axis. To do this, we multiply each

mass by its distance from the y-axis and add up the products. Denote the moment about the y-

axis by 𝑀𝑦. We define 𝑀𝑦 to be

𝑴𝒚 = 𝒎𝟏𝒙𝟏 + 𝒎𝟐𝒙𝟐 + ⋯ + 𝒎𝒏𝒙𝒏 = ∑ 𝒎𝒊𝒙𝒊

𝒏

𝒊=𝟏

Similarly, we define 𝑀𝑥, the moment of the system about the x-axis to be:

𝑴𝒙 = 𝒎𝟏𝒚𝟏 + 𝒎𝟐𝒚𝟐 + ⋯ + 𝒎𝒏𝒚𝒏 = ∑ 𝒎𝒊𝒚𝒊

𝒏

𝒊=𝟏

Denote the center of mass to be the point (�̅�, �̅�). Then

175

�̅� = 𝒎𝟏𝒙𝟏 + 𝒎𝟐𝒙𝟐 + ⋯ + 𝒎𝒏𝒙𝒏

𝒎𝟏 + 𝒎𝟐 + ⋯ + 𝒎𝒏=

∑ 𝒎𝒊𝒙𝒊𝒏𝒊=𝟏

𝒎 =

𝑴𝒚

𝒎

�̅� =𝒎𝟏𝒚𝟏 + 𝒎𝟐𝒚𝟐 + ⋯ + 𝒎𝒏𝒚𝒏

𝒎𝟏 + 𝒎𝟐 + ⋯ + 𝒎𝒏=

∑ 𝒎𝒊𝒚𝒊𝒏𝒊=𝟏

𝒎=

𝑴𝒙

𝒎

Example 16.1

Find the moments and center of mass of the system of objects that have masses 2, 5 and 6 at the

points (−1,1), (2,0) and (3, −1).

𝑀𝑦 = ∑ 𝑚𝑖𝑥𝑖

𝑛

𝑖=1

So the moment of the system about the y-axis is:

𝑀𝑦 = 2(−1) + 5(2) + 6(3) = 26

𝑀𝑥 = ∑ 𝑚𝑖𝑦𝑖

𝑛

𝑖=1

So the moment of the system about the x-axis is:

𝑀𝑥 = 2(1) + 5(0) + 6(−1) = −4

Sum of the masses = 2 + 5 + 6 = 13

�̅� = 𝑀𝑦

𝑚

�̅� =26

13= 2

�̅� = 𝑀𝑥

𝑚

�̅� = −4

13

176

So the center of mass is (2, −4

13)

Now consider a lamina with uniform density 𝜌 that occupies a region R in the plane. We want to

locate the center of mass (or centroid) of the plate.

If the region R is symmetric about a line, then we can use the symmetry principle to find the

centroid:

Symmetry principle: If R is symmetric about a line L, then the centroid of R lies on L.

If the region R has 2 lines of symmetry, then the centroid is the point of intersection of those

lines. For example, the centroid of a circle is its center.

Now suppose that R is not a symmetric region. Let us find a formula for the centroid.

We divide R into rectangles like we did when finding Riemann sums:

Divide [𝑎, 𝑏] into n sub-intervals 𝑥1, 𝑥2, … , 𝑥𝑛 with width ∆𝑥 where

∆𝑥 =𝑏 − 𝑎

𝑛

Let 𝑥�̅� be the midpoint of each interval.

Recall that the centroid of a rectangle is the point of intersection of the two lines of symmetry.

So the centroid of the 𝑖𝑡ℎ rectangle 𝑅𝑖 is (𝑥�̅� ,1

2𝑓(𝑥�̅� )).

To find moments, we would need to find the mass of the system. Let us find the mass of the 𝑖𝑡ℎ

rectangle.

177

𝑀𝑎𝑠𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 × 𝑎𝑟𝑒𝑎

Area of the 𝑖𝑡ℎ rectangle = 𝑓(𝑥�̅�)∆𝑥

Mass of the 𝑖𝑡ℎ rectangle = 𝜌𝑓(𝑥�̅�)∆𝑥

Moment of 𝑅𝑖 about y-axis = Mass × Distance of centroid to y-axis

That is, 𝑀𝑦(𝑅𝑖) = [ 𝜌𝑓(𝑥�̅�)∆𝑥]𝑥�̅�

So

𝑴𝒚 = 𝐥𝐢𝐦𝒏→∞

∑ 𝝆𝒙�̅�𝒇(𝒙�̅�)∆𝒙 = 𝝆 ∫ 𝒙𝒇(𝒙) 𝒅𝒙𝒃

𝒂

𝒏

𝒊=𝟏

Similarly

𝑴𝒙 = 𝝆 ∫𝟏

𝟐[𝒇(𝒙)]𝟐

𝒃

𝒂

𝒅𝒙

Mass is given by:

𝒎 = 𝝆 ∫ 𝒇(𝒙) 𝒅𝒙𝒃

𝒂

So

�̅� = 𝑴𝒚

𝒎=

𝝆 ∫ 𝒙𝒇(𝒙) 𝒅𝒙𝒃

𝒂

𝝆 ∫ 𝒇(𝒙) 𝒅𝒙𝒃

𝒂

= ∫ 𝒙𝒇(𝒙) 𝒅𝒙

𝒃

𝒂


𝒂

�̅� = 𝑴𝒙

𝒎=

𝝆 ∫𝟏𝟐

[𝒇(𝒙)]𝟐𝒃

𝒂 𝒅𝒙

𝝆 ∫ 𝒇(𝒙) 𝒅𝒙𝒃

𝒂

= ∫

𝟏𝟐

[𝒇(𝒙)]𝟐𝒃

𝒂 𝒅𝒙


𝒂

Note that the centroid is not dependent on the density 𝜌.

178

Example 16.2

1. Find the center of mass of a semicircular plate of radius 1 cm.

The plate is symmetrical about the y-axis so �̅� = 0.

Let us now find �̅�.

Equation of circle is 𝑥2 + 𝑦2 = 1.

Solving for y we get the equation of the semi-circle:

𝑦 = √1 − 𝑥2

This is our function 𝑓(𝑥).

�̅� = 𝑀𝑥

𝑚=

∫12

[𝑓(𝑥)]2𝑏

𝑎 𝑑𝑥


𝑎

For the semi-circle

∫1

2[𝑓(𝑥)]2

𝑏

𝑎

𝑑𝑥 = ∫1

2

1

−1

(1 − 𝑥2) 𝑑𝑥

=1

2[𝑥 −

1

3𝑥3]

−1

1

179

=1

2[(1 −

1

3) − (−1 +

1

3)]

=2

3

Now ∫ 𝑓(𝑥) 𝑑𝑥𝑏

𝑎 is the area of the semi-circle which can be easily found without

integrating.

Area of the semicircle =

1

2 . 𝜋 . 12 =

𝜋

2

So

�̅� = 2

3⁄𝜋

2⁄=

4

3𝜋

Center of mass = (0,4

3𝜋).

2. Find the centroid of the lamina occupying the region bounded by 𝑦 = 𝑥2, 𝑦 = 0, 𝑥 = 0,

and 𝑥 = 1.

180

�̅� = 𝑀𝑦

𝑚=

∫ 𝑥𝑓(𝑥) 𝑑𝑥𝑏

𝑎


𝑎

�̅� = 𝑀𝑥

𝑚=

∫12

[𝑓(𝑥)]2𝑏

𝑎 𝑑𝑥


𝑎

𝑚 = ∫ 𝑥2 𝑑𝑥1

0

= [1

3𝑥3]

0

1

= 1

3

𝑀𝑦 = ∫ 𝑥 . 𝑥2 𝑑𝑥1

0

= ∫ 𝑥3 𝑑𝑥1

0

= [1

4𝑥4]

0

1

= 1

4

�̅� = 1

4⁄

13⁄

= 3

4

𝑀𝑥 = ∫1

2 (𝑥2)2 𝑑𝑥

1

0

181

= ∫1

2𝑥4 𝑑𝑥

1

0

= [1

10𝑥5]

0

1

= 1

10

�̅� = 1

10⁄

13⁄

= 3

10

Centroid = (3

4,

3

10)

3. Find the centroid of the region bounded by the line 𝑦 = −𝑥 and 𝑦 = 𝑥2.

To find the centroid of regions bounded by two curves we apply the same principles

when we found area between two curves. Thus

182

�̅� = 𝑀𝑦

𝑚=

∫ 𝑥[𝑓(𝑥) − 𝑔(𝑥)] 𝑑𝑥𝑏

𝑎

∫ 𝑓(𝑥) − 𝑔(𝑥) 𝑑𝑥𝑏

𝑎

�̅� = 𝑀𝑥

𝑚=

∫12

{[𝑓(𝑥)]2 − [𝑔(𝑥)]2}𝑏

𝑎 𝑑𝑥

∫ 𝑓(𝑥) − 𝑔(𝑥) 𝑑𝑥𝑏

𝑎

where 𝑓(𝑥) is the function above the region and 𝑔(𝑥) is the function above the region.

We first find the limits of integration:

𝑥2 = −𝑥

𝑥2 + 𝑥 = 0

𝑥(1 + 𝑥) = 0

𝑥 = 0, 𝑥 = −1

So

𝑀𝑦 = ∫ 𝑥(−𝑥 − 𝑥2)0

−1

𝑑𝑥

= ∫ (−𝑥2 − 𝑥3)0

−1

𝑑𝑥

= [−1

3𝑥3 −

1

4𝑥4]

−1

0

= 0 − (1

3−

1

4)

= −1

12

𝑚 = ∫ (−𝑥 − 𝑥2)0

−1

𝑑𝑥

183

= [−1

2𝑥2 −

1

3𝑥3]

−1

0

= 0 − (−1

2+

1

3)

= 1

6

�̅� = − 1

12⁄

16⁄

= − 1

2

𝑀𝑥 = ∫1

2[(−𝑥)2 − (𝑥2)2] 𝑑𝑥

0

−1

= ∫1

2(𝑥2 − 𝑥4) 𝑑𝑥

0

−1

= 1

2[1

3𝑥3 −

1

5𝑥5]

−1

0

= 0 −1

2(−

1

3+

1

5)

= 1

15

�̅� = 1

15⁄

16⁄

= 2

5

Centroid = (−1

2,

2

5)

184


1. Find the centroid of the region bounded by 𝑦 = 9 − 𝑥2 and the x-axis.

2. Find the centroid of the lamina occupying the region bounded by 𝑦 = 𝑥3, 𝑦 = 0,

𝑥 = 0, and 𝑥 = 1.

3. Find the centroid of the region bounded by 𝑦 = 6 − 𝑥2 and 𝑦 = 𝑥.

4. Find the centroid of the region bounded by 𝑦 = 𝑥2 and 𝑦 = √𝑥.

185

CHAPTER 17: POLAR COORDINATES

To understand the concepts in this chapter, you will need to review the unit circle.

The Unit Circle

From this unit circle, you can easily find the sine and cosine of some angles given in radians and

degrees. For example,

sin 210° = −1

2

186

cos 210° = −√3

2

sin7𝜋

4= −

√2

2

cos7𝜋

4=

√2

2

Recall that

tan 𝑥 =sin 𝑥

cos 𝑥

So, for example

tan 210° =− 1

2⁄

− √32

⁄=

1

√3

Polar Coordinates

So far we have dealt with points in the Cartesian plane (that is, x-y plane) which are specified

using Cartesian or rectangular coordinates (𝑥, 𝑦). However, in some situations, it would be better

to specify a point in another way for ease of calculation. In this chapter, you will learn about

writing a point using polar coordinates (𝑟, 𝜃).

r is the length of the line from the origin to the point (𝑥, 𝑦)

𝜃 is the angle the line makes with the positive x-axis. When we move in an counterclockwise

direction, 𝜃 is positive and 𝜃 is negative when we move in a clockwise direction.

187

𝑟 can be negative. To plot a point with a negative r-coordinate, we plot the point with the

positive r-coordinate and then extend it in the opposite direction to get the negative r-coordinate.

Thus if (𝑟, 𝜃), 𝑟 > 0 is in the third quadrant, then (−𝑟, 𝜃) is in the first quadrant.

Example 17.1

Plot the following points:

1. (3,2𝜋

3)

188

2. (5, −𝜋

4)

2. (−2,5𝜋

4)

189

Converting from polar to rectangular coordinates

Consider the diagram:

From the right angled triangle and basic trigonometric formulas, recall that

sin 𝜃 =𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒


cos 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡


So

sin 𝜃 =𝑦

𝑟

cos 𝜃 = 𝑥

𝑟

Thus to convert a point from polar to rectangular coordinates, use the formulas:

𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃

Example 17.2

Convert to rectangular coordinates:

1. (2,𝜋

2)

𝑥 = 2 cos𝜋

2= 0

190

𝑦 = 2 sin𝜋

2= 2

(2,𝜋

2) = (0,2)

2. (−2,7𝜋

6)

𝑥 = −2 cos7𝜋

6= −2 . −

√3

2= √3

𝑦 = −2 sin7𝜋

6= −2 . −

1

2= 1

(−2,7𝜋

6) = (√3, 1)

Converting from rectangular to polar coordinates

Again, from the diagram:

𝑟2 = 𝑥2 + 𝑦2 from the Pythagorean theorem, and tan 𝜃 = 𝑦

𝑥.

Thus to convert a point from rectangular coordinates to polar coordinates, first illustrate the point

and to find 𝑟 and 𝜃, use the formulas

𝑟2 = 𝑥2 + 𝑦2

tan 𝜃 = 𝑦

𝑥

191

But exercise caution! Consider which quadrant the point lies in. It is important to refer to the unit

circle. Also recall that tangent is positive in the first and third quadrants, and negative in the

second and fourth quadrants.

Example 17.3

Convert to polar coordinates. Find 4 different sets of points such that the angle satisfies

−2𝜋 ≤ 𝜃 ≤ 2𝜋

1. (1, −√3)

We first plot the point:

𝑟 = √1 + 3 = 2

tan 𝜃 = −√3

192

But 𝜃 is in the 4th quadrant. From the unit circle,

𝜃 =5𝜋

3

But recall 𝜃 can be negative when moving in a clockwise direction. Thus

𝜃 =5𝜋

3− 2𝜋 = −

𝜋

3

This gives us these 2 points:

(2,5𝜋

3) , (2, −

𝜋

3)

Now 𝑟 can be negative. So extend the line in the opposite direction so that we are now in the 2nd

quadrant.

From the unit circle

𝜃 =2𝜋

3

That is,

𝜃 = 𝜋 −𝜋

3=

2𝜋

3

Again, 𝜃 can be negative. So

193

𝜃 =2𝜋

3− 2𝜋 = −

4𝜋

3

We now have 2 more points:

(−2,2𝜋

3) , (−2, −

4𝜋

3)

2. (2, 2)

We first plot the point:

The point is in the first quadrant.

𝑟 = √4 + 4 = √8 = 2√2

tan 𝜃 = 1

𝜃 =𝜋

4

194

We calculate a negative value for 𝜃:

𝜃 =𝜋

4− 2𝜋 = −

7𝜋

4

This gives us these 2 points:

(2√2,𝜋

4) , (2√2, −

7𝜋

4)

𝑟 can be negative. So extend the line in the opposite direction so that we are now in the

3rd quadrant.

From the unit circle,

𝜃 =5𝜋

4

That is,

𝜃 = 𝜋 +𝜋

4=

5𝜋

4

For negative 𝜃:

195

𝜃 =5𝜋

4− 2𝜋 = −

3𝜋

4

We now have 2 more points:

(2√2,5𝜋

4) , (2√2, −

3𝜋

4)

Graphs of polar equations

So far, you know how to draw graphs in the Cartesian plane. Now, you will learn how to draw

polar curves.

Simple example

1. 𝑟 = 1 is the unit circle.

2. 𝑟 = 3 is a circle centered at the origin with radius 3.

Now, you will learn to sketch more complicated curves by hand. Again, refer to the position of

the angles on the unit circle to help you draw the graphs. You first construct a table of values.

Example 17.4

Graph

1. 𝑟 = 2 cos 𝜃

Here we will use intervals of 𝜋

2 for 𝜃.

𝜃 0 𝜋

2 𝜋

𝑟 2 0 -2

196

Now draw the graph.

2. 𝑟 = 2 sin 𝜃

𝜃 0 𝜋

2 𝜋

𝑟 0 2 0

197

3. 𝑟 = −3 sin 𝜃 for 0 ≤ 𝜃 ≤𝜋

2

𝜃 0 𝜋

4

𝜋

2

𝑟 0 −

3

√2

-3

4. 𝑟 = cos 2𝜃

𝜃 0 𝜋

4

𝜋

2

3𝜋

4

𝜋

𝑟 1 0 -1 0 1

198

Points of intersection of polar curves

We find points of intersection the usual way by solving the two equations simultaneously. This is

useful when finding the area between two polar curves.

Example 17.5

Find the point(s) of intersection the curves 𝑟 = sin 𝜃 and 𝑟 = cos 𝜃

We graph the two functions:

199

Solving the two equations we get:

sin 𝜃 = cos 𝜃

Dividing by cos 𝜃, cos 𝜃 ≠ 0:

tan 𝜃 = 1

𝜃 =𝜋

4

Exercise caution when using the graphs of the function to find the points of intersection.

From the graph it appears that (0,0) is indeed a point of intersection. But this is false

since 𝜃 = 0 does not satisfy

sin 𝜃 = cos 𝜃

Area in polar coordinates

Consider the shaded region R below bounded by the curve 𝑟 = 𝑓(𝜃) and the lines representing

𝜃 = 𝛼 and 𝜃 = 𝛽.

200

The area of R is given by:

1

2∫ [𝑟(𝜑)]2 𝑑𝜑

𝑏

𝑎

Let us look at some simple examples. Note that you should always make a sketch of the curve

and shade the region. You can use your calculator to graph the curves.

How to graph polar curves on the TI-84

• Press the MODE key.

• Use the ARROW keys and the ENTER key to select Pol.

• Press the Y= key to enter the function. If there are two functions, then enter the second

one below the first.

• Use the WINDOW key to change the domain and range of the function.

• Press GRAPH.

201

Example 17.6

In the following examples, only find the integrals, but do not evaluate them.

1. Find the integral which gives the area inside 𝑟 = 2 cos 𝜃.

Recall: we had graphed this function as follows:

𝜃 0 𝜋

2 𝜋

𝑟 2 0 -2

Notice that 0 ≤ 𝜃 ≤ 𝜋 maps out the entire circle. So these would be the limits of

integration.

Thus, to find the area, evaluate:

1

2∫ (2 cos 𝜃)2 𝑑𝜃 = ∫ 2 𝑐𝑜𝑠2𝜃 𝑑𝜃

𝜋

0

𝜋

0

202

2. Identify the integral which gives area inside 𝑟 = 1 + sin 𝜃

Here this region is defined for 0 ≤ 𝜃 ≤ 2𝜋. So to find the area, evaluate:

1

2∫ (1 + sin 𝜃)2 𝑑𝜃

2𝜋

0

Now let us examine some other examples in which we will find the area between curves.

Steps to finding the area between two curves

1. Graph the curves

2. Label the curves

3. Shade the region

4. Find 𝜃 from the point of intersection

5. Find the limits of integration

6. It may be necessary to draw a line from the origin to the point of intersection. The

importance of this will be revealed shortly.

203

7. Use the concept of symmetry when finding area of symmetrical regions.

Example 17.7

Set up, but do not evaluate the integrals:

1. Find the area of the region outside the circle 𝑟 = 3 cos 𝜃 but inside the cardioid

𝑟 = 1 + cos 𝜃.

We graph, label the curves and shade the region.

Notice that the region is symmetrical so it suffices to take either the lower or upper region

and multiply its area by 2 to find the area of the entire region. Let us take the upper

region.

We now find the points of intersection:

1 + cos 𝜃 = 3 cos 𝜃

2 cos 𝜃 = 1

204

cos 𝜃 =1

2

𝜃 =𝜋

3

Now draw a line from 𝜋

3 to the origin. Notice that in order to find the required area we

will have to subtract the area for the region bounded by the circle from the region

bounded by the cardioid. Let us set up these integrals separately:

Circle: The curve for that particular region is defined for 𝜋

3≤ 𝜃 ≤

𝜋

2. So the integral to be

evaluated is:

1

2∫ (3 cos 𝜃)2 𝑑𝜃

𝜋2

𝜋3

Cardioid: The curve for that particular region is defined for 𝜋

3≤ 𝜃 ≤ 𝜋. So the integral to

be evaluated is:

1

2∫ (1 + cos 𝜃)2 𝑑𝜃

𝜋

𝜋3

Thus the integral for evaluating the required region is:

2 [1

2∫ (1 + cos 𝜃)2 𝑑𝜃 −

1

2

𝜋

𝜋3

∫ (3 cos 𝜃)2 𝑑𝜃

𝜋2

𝜋3

]

= ∫ (1 + cos 𝜃)2 𝑑𝜃 − 𝜋

𝜋3

∫ (3 cos 𝜃)2 𝑑𝜃

𝜋2

𝜋3

2. Find the area of the region outside the circle 𝑟 = cos 𝜃 but inside the circle

𝑟 = sin 𝜃.


205


sin 𝜃 = cos 𝜃

tan 𝜃 = 1

𝜃 =𝜋

4

Now draw a line from 𝜋


will have to subtract the area for the region bounded by 𝑟 = cos 𝜃 from the region

bounded by 𝑟 = sin 𝜃. Let us set up these integrals separately:

𝑟 = sin 𝜃: The curve for that particular region is defined for 𝜋

4≤ 𝜃 ≤ 𝜋. So the integral to

be evaluated is:

1

2∫ (sin 𝜃)2 𝑑𝜃

𝜋

𝜋4

𝑟 = cos 𝜃: The curve for that particular region is defined for 𝜋

4≤ 𝜃 ≤

𝜋

2. So the integral to

be evaluated is:

206

1

2∫ (cos 𝜃)2 𝑑𝜃

𝜋2

𝜋4


=1

2∫ (sin 𝜃)2 𝑑𝜃

𝜋

𝜋4

−1

2∫ (cos 𝜃)2 𝑑𝜃

𝜋

𝜋3

3. Find the area of the region inside 𝑟 = 2 + sin 𝜃 and inside 𝑟 = −3 sin 𝜃.


Notice that the region is symmetrical so it suffices to take either the left or right region

and multiply its area by 2 to find the area of the entire region. Let us take the left region.


2 + sin 𝜃 = −3 sin 𝜃

207

−4 sin 𝜃 = 2

sin 𝜃 = −1

2

𝜃 =7𝜋

6

But this value of 𝜃 applies to 𝑟 = 2 + sin 𝜃 and not to 𝑟 = −3 sin 𝜃 since the latter is

negative.

For 𝑟 = −3 sin 𝜃, 𝜃 =𝜋

6.

Now draw a line from 7𝜋


will have to add the area for the region bounded by 𝑟 = 2 + sin 𝜃 to the region bounded

by 𝑟 = −3 sin 𝜃. Let us set up these integrals separately:

𝑟 = −3 sin 𝜃: The curve for that particular region is defined for 0 ≤ 𝜃 ≤𝜋

6. So the

integral to be evaluated is:

1

2∫ (−3 sin 𝜃)2 𝑑𝜃

𝜋6

0

𝑟 = 2 + sin 𝜃: The curve for that particular region is defined for 7𝜋

6≤ 𝜃 ≤

3𝜋

2. So the

integral to be evaluated is:

1

2∫ (2 + sin 𝜃)2 𝑑𝜃

3𝜋2

7𝜋6


2 [1

2∫ (−3 sin 𝜃)2 𝑑𝜃

𝜋6

0

−1

2∫ (2 + sin 𝜃)2 𝑑𝜃

3𝜋2

7𝜋6

]

= ∫ (−3 sin 𝜃)2 𝑑𝜃

𝜋6

0

− ∫ (2 + sin 𝜃)2 𝑑𝜃

3𝜋2

7𝜋6

208


Set up, but do not evaluate the integrals:

1. Find the area of the region inside the circle 𝑟 = cos 𝜃 but outside the circle

𝑟 = sin 𝜃.

2. Find the area of the region inside the circle 𝑟 = 2 but outside 𝑟 = 3 + 2 sin 𝜃.

3. Find the area of the region enclosed by the curve 𝑟 = 2 sin 𝜃.

4. Find the area of the region that lies inside the curves 𝑟 = 2 cos 𝜃 and 𝑟 = 1.

5. Find the area of the region that lies inside 𝑟 = 2 + sin 𝜃 and outside 𝑟 = 3 sin 𝜃.

6. Find the area of the region inside the cardioid 𝑟 = 3 − 3 sin 𝜃 and outside the

cardioid 𝑟 = 1 + sin 𝜃.

209

CHAPTER 18: SEQUENCES

In this chapter, you will learn about sequences which is an important pre-requisite for learning

about series, which will follow subsequently.

Definition

A sequence is a list of numbers written in a definite order.

e.g. {1, 3, 5, 7, …}

The numbers 1, 3, 5, 7 are called the terms of the sequence. For example, 1 is the first term of the

sequence.

Notation

The sequence {𝑎1, 𝑎2, … , 𝑎𝑛, … } is denoted by {𝑎𝑛}

𝑎1 is the first term

𝑎2 is the second term

𝑎𝑛 is the 𝑛𝑡ℎ term. It is also called the general term or general formula for the sequence.

Listing the terms of a sequence

To list the terms of a sequence, we substitute appropriate values for n. For example, to find the

2nd term of a sequence, evaluate the 𝑛𝑡ℎ term at 𝑛 = 2.

Example 18.1

List the first 4 terms in the sequences:

1. 𝑎𝑛 = 2𝑛 + 1

210

𝑎1 = 2(1) + 1 = 3

𝑎2 = 2(2) + 1 = 5

𝑎3 = 2(3) + 1 = 7

𝑎4 = 2(4) + 1 = 9

So

{𝑎𝑛} = {3, 5, 7, 9}

2. 𝑎𝑛 = (−1)𝑛 𝑛+3

2𝑛

𝑎1 = (−1)11 + 3

21= −2

𝑎2 = (−1)22 + 3

22=

5

4

𝑎3 = (−1)33 + 3

23= −

3

4

𝑎4 = (−1)44 + 3

24=

7

16

So

{ 𝑎𝑛} = {−2,5

4, −

3

4,

7

16}

Notice that when the 𝑛𝑡ℎ term of a sequence contains (−1)𝑛 or similar, the terms of the sequence

alternate in sign.

211

Finding the formula for 𝒂𝒏, the general term of a sequence

Example 18.2

1. {𝑎𝑛} = {1

3,

1

6,

1

9,

1

12, … }

What general formula can be used to define the sequence? Let us look at the terms to see

whether we can find a pattern:

𝑎1 =1

3=

1

3(1)

𝑎2 =1

6=

1

3(2)

𝑎3 =1

9=

1

3(3)

𝑎4 =1

12=

1

3(4)

So

𝑎𝑛 =1

3𝑛

2. {𝑎𝑛} = {1

2,

1

4,

1

8,

1

16, … }

𝑎1 =1

2=

1

21

𝑎2 =1

4=

1

22

𝑎3 =1

8=

1

23

𝑎4 =1

16=

1

24

So

212

𝑎𝑛 =1

2𝑛

3. {𝑎𝑛} = [2, 5, 8, 11, … }

Note that terms increase by 3. This means that 𝑎𝑛 will be of the form:

𝑎𝑛 = 3𝑛 + 𝑥

We find x by evaluating the terms in the sequence. For example:

𝑎1 = 2 = 3(1) − 1

𝑎2 = 5 = 3(2) − 1

So

𝑎𝑛 = 3𝑛 − 1

4. {𝑎𝑛} = {−7

5,

11

25, −

15

125,

19

625, −

23

3125, … }}

Note the terms have alternating signs which means that 𝑎𝑛 contains a power of (-1).

Since the first term is negative and every odd numbered term is negative, then 𝑎𝑛

contains (−1)𝑛.

The denominators of the terms are powers of 5 so the denominator of 𝑎𝑛 is 5𝑛.

Now consider the numerators 7, 11, 19, 19, 23.

The terms increase by 4. This is similar to the previous example. So the numerator of 𝑎𝑛

is 4𝑛 + 3.

Thus

𝑎𝑛 = (−1)𝑛 4𝑛 + 3

5𝑛

213

Recursively defined sequences

Consider the sequence

{𝑎𝑛} = [2, 5, 8, 11, … }

We found that the general term of this sequence is:

𝑎𝑛 = 3𝑛 − 1

However, this sequence could also have been defined as follows:

𝑎1 = 2

𝑎𝑛+1 = 𝑎𝑛 + 3, 𝑛 ≥ 1

This is a recursive definition of the sequence where one term is given (usually the first term), and

subsequent terms are defined in terms of the preceding ones.

Note that 𝑎𝑛+1 is the term following 𝑎𝑛.

So in the example above, since we are given 𝑎1, we can now find 𝑎2. We are not able to find 𝑎3

unless we know 𝑎2. Similarly, we are not able to find 𝑎31 unless we know 𝑎30. This is how a

recursively defined sequence is evaluated, which is unlike what we have done before. Thus

𝑎2 = 𝑎1 + 3 = 2 + 3 = 5

𝑎3 = 𝑎2 + 3 = 5 + 3 = 8

𝑎4 = 𝑎3 + 3 = 8 + 3 = 11

A famous recursively defined sequence is the Fibonacci sequence:

{1,1,2,3,5,8,13, … }

It is defined recursively as follows:

𝑎1 = 1, 𝑎2 = 1

𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2, 𝑛 ≥ 3

214

Example 18.3

Find the first 5 terms of the following sequences:

1. 𝑎1 = 1, 𝑎𝑛+1 = 3𝑎𝑛 − 1

𝑎1 = 1

𝑎2 = 3𝑎1 − 1 = 3(1) − 1 = 2

𝑎3 = 3𝑎2 − 1 = 3(2) − 1 = 5

𝑎4 = 3𝑎3 − 1 = 3(5) − 1 = 14

𝑎5 = 3𝑎4 − 1 = 3(14) − 1 = 41

2. 𝑎1 = 2, 𝑎𝑛+1 = 𝑎𝑛 + 3𝑛

𝑎1 = 2

𝑎2 = 𝑎1 + 3(1) = 2 + 3 = 5

𝑎3 = 𝑎2 + 3(2) = 5 + 6 = 11

𝑎4 = 𝑎3 + 3(3) = 11 + 9 = 20

𝑎5 = 𝑎4 + 3(4) = 20 + 12 = 32

3. 𝑎𝑛 = 𝑎𝑛−1 − 𝑎𝑛−2, 𝑎1 = 5, 𝑎2 = 3

𝑎1 = 5

𝑎2 = 3

𝑎3 = 𝑎2 − 𝑎1 = 3 − 5 = −2

𝑎4 = 𝑎3 − 𝑎2 = −2 − 3 = −5

215

𝑎5 = 𝑎4 − 𝑎3 = −5 − (−2) = −5 = −3

Convergence of sequences

A sequence {𝑎𝑛} has a limit L, if lim𝑛→∞

𝑎𝑛 = 𝐿.

If L exists, then the sequence converges; otherwise it diverges.

Note that you will need to review finding limits at infinity. Recall, this was reviewed in Chapter

2. Please return to that chapter and review that topic if necessary.

Example 18.4

1. Consider the sequence whose 𝑛𝑡ℎ term is given by:

𝑎𝑛 =1

3𝑛

lim𝑛→∞

1

3𝑛= 0

.

So the sequence

{𝑎𝑛} = {1

3,1

6,1

9,

1

12, … }

converges to 0.

2. Consider the sequence whose 𝑛𝑡ℎ term is given by:

𝑎𝑛 = 𝑛

lim𝑛→∞

𝑛 = ∞

.

So the sequence {𝑎𝑛} = 𝑛 diverges.

216

Helpful hints for testing the convergence of sequences

1. Consider a sequence {𝑎𝑛} whose terms alternate in sign. As we have seen before, these

are sequences whose 𝑛𝑡ℎ term has a form containing (−1)𝑛.

If lim𝑛→∞

|𝑎𝑛| = 0, then lim𝑛→∞

𝑎𝑛 = 0; otherwise the sequence diverges.

For example, the sequence given by

𝑎𝑛 = (−1)𝑛1

3𝑛

converges to 0 since

lim𝑛→∞

1

3𝑛= 0

But the sequence given by

𝑎𝑛 = (−1)𝑛𝑛

3𝑛 + 1

diverges since

lim𝑛→∞

𝑛

3𝑛 + 1= −

1

3

The following points give other techniques for evaluating limits at infinity which are useful when

testing for convergence of sequences.

2. lim𝑛→∞

𝑟𝑛 = {0, − 1 < 𝑟 < 11, 𝑟 = 1∞, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

3. lim𝑛→∞

1

𝑛𝑝 = 0 if p > 0.

4. The laws of limits apply to sequences.

217

5. Use the squeeze theorem where applicable. That is:

If 𝑎𝑛 ≤ 𝑏𝑛 ≤ 𝑐𝑛 for all n and lim𝑛→∞

𝑎𝑛 = lim𝑛→∞

𝑐𝑛 = 𝐿, then lim𝑛→∞

𝑏𝑛 = 𝐿.

6. L’H�̂�pital’s rule may need to be used.

Review of L’H�̂�pital’s rule

Suppose f and g are differentiable functions and 𝑔′(𝑥) ≠ 0 on an open interval containing c

except possibly at c. Suppose that

lim𝑥→𝑐

𝑓(𝑥) = 0 and lim𝑥→𝑐

𝑔(𝑥) = 0

or that

lim𝑥→𝑐

𝑓(𝑥) = ±∞ and lim𝑥→𝑐

𝑔(𝑥) = ±∞

Then

lim𝑥→𝑐

𝑓(𝑥)

𝑔(𝑥)= lim

𝑥→𝑐

𝑓′(𝑥)

𝑔′(𝑥)

A second application of L’H�̂�pital’s rule is sometimes necessary if upon the first application, the

result is still in indeterminate form.

For example:

lim𝑛→∞

ln 𝑛

𝑛= lim

𝑛 →∞

1𝑛⁄

1= 0

lim𝑛→∞

𝑒𝑛

𝑛2= lim

𝑛 →∞

𝑒𝑛

2𝑛= lim

𝑛 →∞

𝑒𝑛

2= ∞

Example 18.5

Determine whether the following sequences converge. If a sequence converges, what is its limit?

218

1. 𝑎𝑛 = 3𝑛 − 1

lim𝑛→∞

(3𝑛 − 1) = ∞

The sequence diverges.

2. 𝑎𝑛 =3𝑛3+5𝑛2−1

4𝑛3+7

lim𝑛→∞

3𝑛3 + 5𝑛2 − 1

4𝑛3 + 7=

3

4

The sequence converges to 3

4.

3. 𝑎𝑛 = (1

2)

𝑛

Since −1 <1

2< 1,

lim𝑛→∞

(1

2)

𝑛

= 0

The sequence converges to 0.

4. 𝑎𝑛 = (−1)𝑛 3𝑛

𝑛5+2

lim𝑛→∞

3𝑛

𝑛5 + 2= 0


219

5. 𝑎𝑛 = (7

3)

𝑛

Since 7

3> 1,

lim𝑛→∞

(7

3)

𝑛

= ∞

The sequence diverges.

6. 𝑎𝑛 =3𝑛2

𝑒2𝑛

lim𝑛→∞

3𝑛2

𝑒2𝑛= lim

𝑛→∞

6𝑛

2𝑒2𝑛= lim

𝑛→∞

6

4𝑒2𝑛= 0


220


1. List the first 5 terms of the sequence:

a) 𝑎𝑛 =2𝑛

3𝑛+1

b) 𝑎𝑛 = (−1)𝑛+1 𝑛+3

𝑛!

c) 𝑎1 = 2, 𝑎𝑛+1 = 3 − 5𝑎𝑛

d) 𝑎1 = 1, 𝑎2 = 3, 𝑎𝑛 = 2𝑎𝑛−1 − 𝑎𝑛−2

2. Find a formula for the general term 𝑎𝑛 of the sequence:

a) {5, 7, 9,11, … . }

b) {1, 4,16, 64, 256, … }

c) {−5

3,

10

9, −

15

27,

20

81, … }

d) {1, −1, 1, −1, 1, −1 … }

3. Determine whether the sequence converges or diverges. If it converges, find its limit:

a) 𝑎𝑛 = 5 −1

𝑛2

b) 𝑎𝑛 = 𝑛4

2𝑛3−1

c) 𝑎𝑛 = (0.4)𝑛

d) 𝑎𝑛 = 3𝑛5−5

11𝑛5−4𝑛3−1

e) 𝑎𝑛 = 𝑛(𝑛 − 1)

f) 𝑎𝑛 = (9

8)

𝑛

g) 𝑎𝑛 = 𝑛3𝑒−𝑛

h) 𝑎𝑛 = ln 𝑛

ln 2𝑛

i) 𝑎𝑛 = (−1)𝑛 1

2𝑛3−1

j) 𝑎𝑛 = (−1)𝑛+1 2𝑛3

3𝑛3−5

k) 𝑎𝑛 = 𝑛 sin1

𝑛

221

CHAPTER 19: SERIES I – GEOMETRIC SERIES

Review of summation notation

The summation symbol, denoted by ∑ , tells us to add up its elements.

Here are some important terms associated with the summation symbol:

Example 19.1

1. ∑ 𝑛 = 1 + 2 + 3 + 4 + 5 …5𝑛=1

2. ∑ 𝑛2 = 12 + 22 + 32 + 42 + ⋯∞𝑛=1

3. ∑ 3𝑛 − 1 = 5 + 8 + 11 + 14 + ⋯∞𝑛=2

Definition of a series

An expression of the form 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 + … is called an infinite series (or series) and is

denoted by ∑ 𝑎𝑛∞𝑛=1 or simply ∑ 𝑎𝑛.

The 𝑎𝑖′𝑠 are the terms of the series.

For example:

∑ 2𝑛 = 2 + 4 + 6 + 8 + ⋯

∞

𝑛=1

An important example of an infinite series is the geometric series which can take the following

forms:

∑ 𝑎𝑟𝑛−1 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ .

∞

𝑛=1

222

∑ 𝑎𝑟𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ .

∞

𝑛=0

In the geometric series, each term is obtained from the preceding one by multiplying it by a

constant called the common ratio. The common ratio is denoted by r and the first term of the

geometric series is denoted by a.

For example, consider the geometric series:

∑ 5 (1

2)

𝑛∞

𝑛=1

=5

2+

5

4+

5

8+

5

16+ ⋯

𝑎 =5

2, 𝑟 =

1

2

Example 19.2

Decide which of the following are geometric series. For those which are, find a and r.

1. 2 + 10 + 50 + 250 + ⋯

This is a geometric series with 𝑟 = 5, 𝑎 = 2

2. 3 + 12 + 16 + 64 + 256 + ⋯

This is not a geometric series because there is no common ratio.

3. 1 + 4 + 7 + 10 + 13 + ⋯

This is not a geometric series.

223

4. 3 +3

2+

3

3+

3

4+ ⋯.

This is not a geometric series.

5. 5 −10

3+

20

9−

40

27+ ⋯

This is a geometric series with 𝑟 = −2

3, 𝑎 = 5

Sum of a finite geometric series

We want a formula for finding 𝑆𝑛, the sum of the first n terms of the finite geometric series:

𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ 𝑎𝑟𝑛−1 … … … … … … … … . (1)

Multiply this equation by r to get (2):

𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ 𝑎𝑟𝑛−1 … … … … … … … … . (1)

𝑟𝑆𝑛 = 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ 𝑎𝑟𝑛−1 + 𝑎𝑟𝑛 … … … … … … … … . (2)

Subtracting the 2 equations we get:

𝑆𝑛 − 𝑟𝑆𝑛 = 𝑎 − 𝑎𝑟𝑛

𝑆𝑛(1 − 𝑟) = 𝑎(1 − 𝑟𝑛)

So

𝑺𝒏 =𝒂(𝟏 − 𝒓𝒏)

𝟏 − 𝒓, 𝒓 ≠ 𝟏

Thus to find the sum of a finite geometric series, you would need the common ration r, the first

term a, and the number of terms n.

224

Example 19.3

Find the sum of the finite geometric series:

1. 2 + 1 +1

2+

1

4+

1

8+ ⋯ +

1

26

𝑎 = 2, 𝑟 =1

2, 𝑛 = 8

𝑆𝑛 =𝑎(1 − 𝑟𝑛)

1 − 𝑟

So

𝑆8 =2(1 − (0.5)8)

1 − 0.5

= 3.984375

2. 3 + 3(0.2) + 3(0.2)2 + ⋯ + 3(0.2)9

𝑎 = 3, 𝑟 = 0.2, 𝑛 = 10

𝑆20 =3(1 − (0.2)10)

1 − 0.2

≈ 3.75

Sum of an infinite geometric series

Using the same approach for finding 𝑆𝑛:

𝑆∞ = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ … … … … … … … … . (1)

𝑟𝑆𝑛 = 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ … … … … … … … … . (2)

Subtracting the 2 equations we get:

𝑆𝑛 − 𝑟𝑆𝑛 = 𝑎

𝑆𝑛(1 − 𝑟) = 𝑎

So

225

𝑆𝑛 =𝑎

1 − 𝑟

We can find this formula in another way:

Denote the partial sum, the sum of the first n terms of an infinite geometric series, by 𝑆𝑛 .

Denote the sum of an infinite series by S. Then

𝑆 = lim𝑛→∞

𝑆𝑛

So, to find the sum of an infinite geometric series, we find:


𝑆𝑛 = lim𝑛→∞

𝑎(1 − 𝑟𝑛)

1 − 𝑟

which depends on the value of r.

Recall: lim𝑛→∞

𝑟𝑛 = {0, − 1 < 𝑟 < 11, 𝑟 = 1∞, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

So


𝑎(1 − 𝑟𝑛)

1 − 𝑟=

𝑎

1 − 𝑟

for −1 < 𝑟 < 1.

That is, the sum of an infinite geometric series is given by

𝑺 =𝒂

𝟏 − 𝒓

for |𝒓| < 𝟏.

We say that the infinite geometric series ∑ 𝑎𝑛 converges to 𝑎

1−𝑟 for |𝑟| < 1; otherwise it diverges.

226

Example 19.4

1. Find the sum of the geometric series:

a) 5 −5

4+

5

16−

5

64+ ⋯

𝑎 = 5, 𝑟 = −1

4

𝑆 =𝑎

1 − 𝑟

So

𝑆 =5

1 − (−14

)

= 4

The series converges to 4.

b) ∑ (2

5)

𝑛∞𝑛=2

𝑎 = (2

5)

2

=4

25

𝑟 =2

5

So

𝑆 =

425

1 −25

=4

15

The series converges to 4

15.

227

Applications

Example 19.5

1. Write the number 1.345̅̅̅̅ = 1.3454545 … as a ratio of integers.

1.345̅̅̅̅ = 1.3454545 …

= 1.3 + 0.045 + 0.00045 + 0.0000045 + ⋯.

=13

10+

45

103+

45

105+

45

107+

Observe that if the first term is excluded then, the infinite series is a geometric series with

𝑎 =45

103=

45

1000

𝑟 =1

102=

1

100

The sum of this infinite geometric series is:

451000

1 −1

100

=45

990

So

1.345̅̅̅̅ = 1.3454545 …

=13

10+

45

990

=1332

990

228

2. How much money will you have at the end of 25 years if you deposit $100 at 5%

compounded annually?

To find the amount at the end of each year we would add the interest accrued to the

principal.

Interest = 5% of principal or 0.05 × principal

Principal +Interest = 1.05 × principal

Amount at the end of year 1 = $1.05(100) = $105

Amount at the end of year 2 = $1.05(105) = 1.05(1.05)(100) = 1.052(100)

Amount at the end of year 3 = $1.05(1.05)(105) = 1.05(1.05)(1.05)(100) = 1.053(100)

So, the amount at the end of 25 years is:

$1.0525(100) = $338.64

229


1. Find the sum of the finite geometric series:

a) ∑ (0.8)𝑛10𝑛=2

b) 1 +1

5+

1

52 + ⋯ +1

513

2. Determine whether the infinite geometric series converges or diverges. If it converges,

find its sum.

a) ∑ 3(0.4)𝑛∞𝑛=0

b) ∑ (1.6)𝑛∞𝑛=2

c) 2 −4

5+

8

25−

16

125+ ⋯.

3. Write the number 1.72̅ as a ratio of integers.

4. Sam takes 200 mg of a drug at 6:00 pm every day. 3% of the drug remains in the body

after the drug is taken.

a) How much of the drug is in the body after the 2nd day?

b) How much of the drug is in the body after the 5th day?

c) How much of the drug is in the body after the nth day?

d) How much of the drug remains in the body in the long run?

230

CHAPTER 20: SERIES II – CONVERGENCE OF SERIES

So far you have learned how to test the geometric series for convergence. In this chapter, you

will learn some techniques for testing series for convergence

The partial sum technique

Recall that the 𝑛𝑡ℎ partial sum 𝑆𝑛 is the sum of the first n terms of an infinite series. So {𝑆𝑛} is

the sequence of partial sums. Consider the infinite series

∑ 𝑎𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯

∞

𝑛=1

Then

𝑆1 = 𝑎1

𝑆2 = 𝑎1 + 𝑎2

𝑆3 = 𝑎1 + 𝑎2 + 𝑎3

𝑆𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ 𝑎𝑛

The sequence of partial sums is then given by:

{𝑆𝑛} = {𝑆1, 𝑆2, 𝑆3, … }

If the sequence {𝑆𝑛} converges, then the series ∑ 𝑎𝑛 converges. Recall, that to test a sequence

{𝑆𝑛} for convergence, we find lim𝑛→∞

𝑆𝑛. Thus,

If 𝐥𝐢𝐦𝒏→∞

𝑺𝒏 exists, then ∑ 𝒂𝒏 converges. Moreover, the sum of the series is 𝐥𝐢𝐦𝒏→∞

𝑺𝒏.

This technique for testing convergence of a series and finding its sum is called the partial sum

technique.

231

Example 20.1

1. Show that the series ∑1

(𝑛+1)(𝑛+2)∞𝑛=1 converges, and find its sum.

We write out 𝑆𝑛 and try to find an expression for the sum so that we can find the limit at

infinity.

𝑆𝑛 =1

6+

1

12+

1

20+ ⋯ +

1

(𝑛 + 1)(𝑛 + 2)

This expression is not helpful to us so we try another technique.

We first decompose 1

(𝑛+1)(𝑛+2) into partial fractions:

1

(𝑛 + 1)(𝑛 + 2)=

𝐴

𝑛 + 1+

𝐵

𝑛 + 2

1 = 𝐴(𝑛 + 2) + 𝐵(𝑛 + 1)

When 𝑛 = −2:

1 = −𝐵

𝐵 = −1

When 𝑛 = −1:

1 = 𝐴

So

1

(𝑛 + 1)(𝑛 + 2)=

1

𝑛 + 1−

1

𝑛 + 2

We again write out 𝑆𝑛 and try to find an expression for the sum:

𝑆𝑛 = (1

2−

1

3) + (

1

3−

1

4) + (

1

4−

1

5) + ⋯ (

1

𝑛−

1

𝑛 + 1) + (

1

𝑛 + 1−

1

𝑛 + 2)

232

Notice that the middle terms cancel out. Thus, we can simplify this sum to:

𝑆𝑛 =1

2−

1

𝑛 + 2

lim𝑛→∞


1

2−

1

𝑛 + 2

=1

2

So, the series converges to 1

2. The sum of the series is

1

2.

2. Investigate the convergence of the series ∑ 𝑙𝑛 (𝑛+4

𝑛+3)∞

𝑛=1

We write out 𝑆𝑛 and try to find an expression for the sum so that we can find the limit at

infinity.

𝑆𝑛 = ln (5

4) + ln (

6

5) + ln (

7

6) + ln (

8

7) + … + ln (

𝑛 + 3

𝑛 + 2) + 𝑙𝑛 (

𝑛 + 4

𝑛 + 3)

= ln (5

4 .

6

5.

7

6.

8

7. ….

𝑛 + 3

𝑛 + 2 .

𝑛 + 4

𝑛 + 3)

= ln (𝑛 + 4

4)

lim𝑛→∞


ln (𝑛 + 4

4)

= ∞

The series diverges.

The 𝒏𝒕𝒉 term test for divergence

The following test only proves that the series ∑ 𝑎𝑛 diverges.

If 𝐥𝐢𝐦𝒏→∞

𝒂𝒏 ≠ 𝟎, then ∑ 𝒂𝒏 diverges

233

Note: If 𝐥𝐢𝐦𝒏→∞

𝒂𝒏 = 𝟎, it does not necessarily mean that the series converges. In this case,

further testing is necessary.

Example 20.2

Test for convergence:

1. ∑3𝑛2+1

2𝑛2−5∞𝑛=1

lim𝑛→∞

3𝑛2 + 1

2𝑛2 − 5=

3

2≠ 0


2. ∑ ln 3𝑛∞𝑛=1

lim𝑛→∞

ln 3𝑛 = ∞


3. ∑𝑛2

𝑛3+1∞𝑛=1

lim𝑛→∞

𝑛2

𝑛3 + 1= 0

This test fails. We will then need to use another test for convergence which you will learn

about in the next chapter.

234

Popular series test

There are some well-known series whose convergence is known (one we have already learned

about from the last chapter). Thus, another technique used for testing an infinite series for

convergence is to determine whether it is a popular series whose convergence is known. Here are

some of these known series:

(i) The Geometric series

∑ 𝑎𝑟𝑛−1 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ .

∞

𝑛=1

The geometric series converges for |𝒓| < 𝟏 with a sum of 𝒂

𝟏−𝒓.

(ii) The Harmonic series

∑1

𝑛= 1 +

1

2+

1

3+ ⋯

∞

𝑛=1

The harmonic series diverges.

For example,

∑2

𝑛∞𝑛=1 is a harmonic series and so diverges.

(iii) The p-series

∑1

𝑛𝑝=

1

1𝑝+

1

2𝑝+

1

3𝑝+ ⋯

∞

𝑛=1

The p-series converges for 𝒑 > 𝟏 and diverges for 𝒑 ≤ 𝟏.

For example,

235

a) ∑1

𝑛4∞𝑛=1 converges since it is a p-series with p = 4 >1.

b) ∑2

𝑛1

2⁄∞𝑛=1 diverges since it is a p-series with 𝑝 =

1

2< 1

Example 20.3


1. ∑3

𝑛7∞𝑛=1

The series converges because it is a p-series with 𝑝 = 7 > 1.

2. ∑4

𝑛∞𝑛=1

The series diverges because it is a harmonic series.

3. ∑ (1

5)

𝑛∞𝑛=1

The series converges because it is a geometric series with 𝑟 =1

5.

236


1. Find the 𝑛𝑡ℎ partial sum of the series and determine whether the series converges or

diverges. If the series converges, find its sum.

a) ∑1

(𝑛+3)(𝑛+4)∞𝑛=1

b) ∑4

𝑛2−1∞𝑛=2

c) ∑ ln (𝑛+1

𝑛+2)∞

𝑛=1

2. Test the series for convergence:

a) ∑𝑛−1

3𝑛+1∞𝑛=1

b) ∑ (1.3)𝑛∞𝑛=1

c) ∑1

3𝑛∞𝑛=1

d) ∑1

2𝑛5∞𝑛=1

e) ∑1

√𝑛∞𝑛=1

f) ∑ 5 (1

3)

𝑛∞𝑛=1

g) ∑11

𝑛∞𝑛=1

237

CHAPTER 21: SERIES III – CONVERGENCE OF SERIES

So far, we have examined the following tests for convergence of series:

1. The 𝑛𝑡ℎ partial sum test

2. The 𝑛𝑡ℎ term test for divergence

3. The popular series test.

Let us examine more tests for convergence of series.

The Integral Test

To test ∑ 𝑎𝑛∞𝑛=1 for convergence we apply the following theorem:

Suppose 𝑎𝑛 = 𝑓(𝑛), where 𝑓(𝑛) is decreasing and positive.

• If ∫ 𝑓(𝑛) 𝑑𝑛∞

1 converges, then ∑ 𝑎𝑛 converges


1 diverges, then ∑ 𝑎𝑛 diverges

To use the integral test, we must first show that 𝑎𝑛 = 𝑓(𝑛) is a decreasing, positive function.

Only then can we apply the integral test. Note that to show convergence or divergence, you will

need to evaluate an improper integral. You can review this in Chapter 11.

Let us now use the integral test to show why the p-series converges for 𝑝 > 1 and diverges for

𝑝 ≤ 1.

Example 21.1

We will show that the p-series ∑1

𝑛𝑝∞𝑛=1 is convergent for 𝑝 > 1 and divergent for 𝑝 ≤ 1 by using

the integral test.

𝑓(𝑛) =1

𝑛𝑝= 𝑛−𝑝

238

Now 𝑓(𝑛) > 0 for 𝑛 ≥ 1.

For 𝑝 > 0:

𝑓′(𝑛) = −𝑝𝑛−𝑝−1 < 0

So 𝑓(𝑛) is decreasing.

We can now use the integral test since both conditions hold.

∫1

𝑛𝑝 𝑑𝑛 = lim

𝑡→∞∫ 𝑛−𝑝 𝑑𝑛

𝑡

1

∞

1

= lim𝑡→∞

[1

1 − 𝑝 𝑛1−𝑝]

1

𝑡

=1

1 − 𝑝 lim𝑡 →∞

[𝑡1−𝑝 − 1]

=1

1 − 𝑝 lim𝑡 →∞

[1

𝑡𝑝−1− 1]

If 𝑝 = 1, the integral diverges, so the p-series diverges.

If 𝑝 < 1, the integral diverges, so the p-series diverges.

If 𝑝 > 1, the integral converges to 1

𝑝−1, so the p-series converges.

Example 21.2

Use the integral test to test the following series for convergence:

1. The harmonic series ∑1

𝑛∞𝑛=1

𝑓(𝑛) =1

𝑛

𝑓(𝑛) > 0

Also

239

𝑓′(𝑛) = −1

𝑛2< 0


∫1

𝑛 𝑑𝑛 = lim

𝑡→∞∫

1

𝑛 𝑑𝑛

𝑡

1

∞

1

= lim𝑡→∞

[ln 𝑛]1𝑡

= lim𝑡 →∞

[ln 𝑡]

= ∞

The integral diverges so the harmonic series diverges.

2. ∑𝑛

𝑛2+1∞𝑛=1

𝑓(𝑛) =𝑛

𝑛2 + 1

𝑓(𝑛) > 0

𝑓′(𝑛) =1. (𝑛2 + 1) − 2𝑛 . 𝑛

(𝑛2 + 1)2

=−𝑛2 + 1

(𝑛2 + 1)2< 0


∫𝑛

𝑛2 + 1 𝑑𝑛 = lim

𝑡→∞∫

𝑛

𝑛2 + 1 𝑑𝑛

𝑡

1

∞

1

= lim𝑡→∞

[1

2ln|𝑛2 + 1|]

1

𝑡

=1

2 lim𝑡 →∞

[ln(𝑡2 + 1) − ln 2]

= ∞

240

The integral diverges, so the series diverges.

The following test allow us to test positive term series for convergence by finding other series to

compare them to. There are two comparison tests. Please be careful in using both tests. Ensure

that the conditions hold for convergence and divergence.

The Comparison Test

Let ∑ 𝑎𝑛 and ∑ 𝑏𝑛 be positive term series. Then

1. If 𝑎𝑛 ≤ 𝑏𝑛 for all n and ∑ 𝑏𝑛 converges, then ∑ 𝑎𝑛 converges.

2. If 𝑎𝑛 ≥ 𝑏𝑛 for all n and ∑ 𝑏𝑛 diverges, then ∑ 𝑎𝑛 diverges.

The Limit Comparison Test

If lim𝑛→∞

𝑎𝑛

𝑏𝑛 exists and is positive, then either both series converge or both diverge.

So, to test ∑ 𝑎𝑛 for convergence, we will first need to find a known series ∑ 𝑏𝑛 for comparison.

To find such a series, we usually find the quotient of the dominant terms in the numerator and

denominator of the rational function represented by 𝑎𝑛.

Example 21.3

Use the comparison test, to test the series for convergence:

1. ∑1

𝑛(𝑛+1)∞𝑛=1

𝑎𝑛 =1

𝑛(𝑛 + 1)

241

Our comparing series is:

𝑏𝑛 =1

𝑛. 𝑛=

1

𝑛2

This comparing series converges because it is a p-series with 𝑝 = 2 > 1. Thus we will

use the first condition in the comparison test and check if it holds.

1

𝑛(𝑛 + 1)<

1

𝑛2

since they both have the same numerator, and the denominator of 1

𝑛(𝑛+1) is greater than

that of 1

𝑛2.

So by the comparison test, ∑1

𝑛(𝑛+1)∞𝑛=1 also converges.

2. ∑1

𝑛 3𝑛∞𝑛=1

𝑎𝑛 =1

𝑛 3𝑛


𝑏𝑛 =1

3𝑛

This comparing series converges because it is a geometric series with 𝑟 =1

3. Thus we will

again use the first condition in the comparison test and check if it holds.

1

𝑛 3𝑛<

1

3𝑛


𝑛 3𝑛∞𝑛=1 also converges.

242

3. ∑1

5𝑛−1∞𝑛=1

𝑎𝑛 =1

5𝑛 − 1


𝑏𝑛 =1

5𝑛

This comparing series diverges because it is a harmonic series. Thus we will use the

second condition in the comparison test and check if it holds.

1

5𝑛 − 1>

1

5𝑛


5𝑛−1∞𝑛=1 diverges.

4. ∑𝑛2

𝑛5−𝑛−2∞𝑛=1

𝑎𝑛 =𝑛2

𝑛5 − 𝑛 − 2


𝑏𝑛 =𝑛2

𝑛5=

1

𝑛3

This comparing series converges because it is a p-series with 𝑝 = 3 > 1. Thus we will

use the first condition in the comparison test and check if it holds.

However, observe that

𝑛2

𝑛5 − 𝑛 − 2>

𝑛2

𝑛5=

1

𝑛3

243

This means that in this case, the comparison test fails. We will therefore use the limit

comparison test.

lim𝑛→∞

𝑎𝑛

𝑏𝑛= lim

𝑛→∞

𝑛2

𝑛5 − 𝑛 − 2 . 𝑛3

lim𝑛→∞

𝑛5

𝑛5 − 𝑛 − 2

= 1

This limit exists and is positive. So by the limit comparison test, both series converge.

That is, ∑𝑛2

𝑛5−𝑛−2∞𝑛=1 converges.

244


1. Use the integral test to test the series for convergence:

a) ∑1

𝑛2∞𝑛=1

b) ∑1

𝑛+5∞𝑛=1

c) ∑ln 𝑛

𝑛2∞𝑛=1

2. Use the comparison test or the limit comparison test to test the series for convergence:

a) ∑1

√𝑛2𝑛∞𝑛=1

b) ∑𝑛2

𝑛5+3𝑛+5∞𝑛=1

c) ∑1

√𝑛−1∞𝑛=2

d) ∑𝑛3

𝑛5−3𝑛−5∞𝑛=1

245

CHAPTER 22: SERIES IV – CONVERGENCE OF SERIES

In the last three chapters, you learned how to test the convergence of positive term series. In this

chapter, you will learn how to determine the convergence of general series, that is, series with

positive and/or negative terms.

Alternating Series

An alternating series is an infinite series of the form

∑(−1)𝑛+1𝑎𝑛 = 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + ⋯

∞

𝑛=1

Alternating series Test

If 𝑎𝑛 > 𝑎𝑛+1 > 0 for all n and lim𝑛→∞

𝑎𝑛 = 0 then the alternating series

∑(−1)𝑛+1𝑎𝑛 = 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + ⋯

∞

𝑛=1

converges.

Thus, to test an alternating series for convergence, we must show that:

1. 𝑎𝑛 is decreasing

2. lim𝑛→∞

𝑎𝑛 = 0

Example 22.1

Test the alternating series for convergence:

1. ∑(−1)𝑛

√𝑛3

∞𝑛=1

246

𝑎𝑛 =1

√𝑛3

𝑎𝑛+1 =1

√𝑛 + 13

1

√𝑛3 >

1

√𝑛 + 13 > 0

lim𝑛→∞

1

√𝑛3 = 0

So by the alternating series test, ∑(−1)𝑛

√𝑛3

∞𝑛=1 converges.

2. ∑(−1)𝑛+1

𝑛2∞𝑛=1

𝑎𝑛 =1

𝑛2

𝑎𝑛+1 =1

(𝑛 + 1)2

1

𝑛2>

1

(𝑛 + 1)2> 0

lim𝑛→∞

1

𝑛2= 0

So by the alternating series test, ∑(−1)𝑛+1

𝑛2∞𝑛=1 converges.

Absolute Convergence

The series ∑ 𝑎𝑛 is said to converge absolutely if the series ∑ |𝑎𝑛 | converges, where

∑ |𝑎𝑛 | = |𝑎1| + |𝑎2| + ⋯

247

Theorem

If ∑ 𝑎𝑛 converges absolutely, that is, if ∑ |𝑎𝑛 | converges, then so does the series ∑ 𝑎𝑛.

However, if ∑ |𝑎𝑛 | diverges, it does not necessarily mean that ∑ 𝑎𝑛 diverges.

Conditional Convergence

Earlier on, we saw that the series ∑(−1)𝑛

√𝑛3


But ∑ |(−1)𝑛

√𝑛3

∞𝑛=1 | = ∑

1

√𝑛3

∞𝑛=1 diverges since it is a p-series with 𝑝 =

1

3< 1.

The series ∑(−1)𝑛

√𝑛3

∞𝑛=1 is called conditionally convergent – it converges but it is not absolutely

convergent.

Example 22.2

Show that the series ∑(−1)𝑛+1

𝑛∞𝑛=1 is conditionally convergent.

𝑎𝑛 =1

𝑛

𝑎𝑛+1 =1

𝑛 + 1

1

𝑛>

1

𝑛 + 1> 0

lim𝑛→∞

1

𝑛= 0

So by the alternating series test, ∑(−1)𝑛+1

𝑛∞𝑛=1 converges.

But ∑ |(−1)𝑛+1

𝑛|∞

𝑛=1 = ∑1

𝑛∞𝑛=1 which diverges since it is a harmonic series.

248

The Ratio Test

For the series ∑ 𝑎𝑛, suppose lim𝑛→∞

|𝑎𝑛+1

𝑎𝑛| = 𝐿.

1. If L < 1, then ∑ 𝑎𝑛 converges.

2. If L > 1 or 𝐿 = ±∞, then ∑ 𝑎𝑛 diverges.

3. If L = 1, this test is inconclusive. Use a different test.

This section uses the concept of factorial notation.

Review of factorial notation

𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) … 3 . 2 . 1

So

5! = 5 . 4 . 3 . 2 . 1

Factorials can also be defined recursively. For example,

(𝑛 + 1)! = (𝑛 + 1) 𝑛!

Example 22.3

Test the following series for convergence by using the ratio test:

1. ∑1

2𝑛∞𝑛=1

𝑎𝑛 =1

2𝑛 𝑎𝑛+1 =

1

2𝑛+1

𝑎𝑛+1

𝑎𝑛=

1

2𝑛+1 ÷

1

2𝑛

=2𝑛

2𝑛+1

249

=2𝑛

2𝑛. 2

=1

2

lim𝑛→∞

1

2=

1

2< 1

By the ratio test, ∑1

2𝑛∞𝑛=1 converges.

2. ∑(−5)𝑛

𝑛!∞𝑛=1

We use absolute value here because this series contains negative terms.

|𝑎𝑛| =5𝑛

𝑛! |𝑎𝑛+1| =

5𝑛+1

(𝑛 + 1)!

|𝑎𝑛+1

𝑎𝑛| =

5𝑛+1

(𝑛 + 1)! ÷

5𝑛

𝑛!

=5𝑛+1

(𝑛 + 1)! .

𝑛!

5𝑛

=5𝑛. 5

(𝑛 + 1)𝑛! .

𝑛!

5𝑛

= 5

𝑛 + 1

lim𝑛→∞

5

𝑛 + 1= 0 < 1

By the ratio test, ∑(−5)𝑛

𝑛!∞𝑛=1 converges.

250

3. ∑2𝑛

𝑛∞𝑛=1

𝑎𝑛 =2𝑛

𝑛 𝑎𝑛+1 =

2𝑛+1

𝑛 + 1

𝑎𝑛+1

𝑎𝑛=

2𝑛+1

𝑛 + 1 ÷

2𝑛

𝑛

=2𝑛+1

𝑛 + 1 .

𝑛

2𝑛

=2𝑛

𝑛 + 1

lim𝑛→∞

2𝑛

𝑛 + 1= 2 > 1

By the ratio test, ∑2𝑛

𝑛∞𝑛=1 diverges.

The Root Test

For the series ∑ 𝑎𝑛, suppose lim𝑛→∞

√|𝑎𝑛|𝑛= 𝐿.

1. If L < 1, then ∑ 𝑎𝑛 converges.

2. If L > 1 or 𝐿 = ±∞, then ∑ 𝑎𝑛 diverges.


Example 22.4


1. ∑ (−1)𝑛+1 (2𝑛3−5

3𝑛3+1)

𝑛∞𝑛=1

|𝑎𝑛| = (2𝑛3 − 5

3𝑛3 + 1)

𝑛

251

√|𝑎𝑛|𝑛

= √(2𝑛3 − 5

3𝑛3 + 1)

𝑛𝑛

= (2𝑛3 − 5

3𝑛3 + 1)

lim𝑛→∞

2𝑛3 − 5

3𝑛3 + 1=

2

3< 1

By the root test, ∑ (−1)𝑛+1 (2𝑛3−5

3𝑛3+1)

𝑛∞𝑛=1 converges.

2. ∑ (3𝑛+1

5𝑛−4)

−2𝑛∞𝑛=1

𝑎𝑛 = (3𝑛 + 1

5𝑛 − 4)

−2𝑛

√𝑎𝑛𝑛 = √(

3𝑛 + 1

5𝑛 − 4)

−2𝑛𝑛

= (3𝑛 + 1

5𝑛 − 4)

−2

lim𝑛→∞

(3𝑛 + 1

5𝑛 − 4)

−2

= (3

5)

−2

= 25

9> 1

By the root test, ∑ (3𝑛+1

5𝑛−4)

−2𝑛 ∞

𝑛=1 diverges.

Strategies for testing series for convergence

You have learned various tests for investigating the convergence of the series ∑ 𝑎𝑛, however,

you will need to know when to apply each test. In this section we review each test and examine

ways of deciding which test to use.

1. The 𝑛𝑡ℎ term test for divergence:

If lim𝑛→∞

𝑎𝑛 ≠ 0, then ∑ 𝑎𝑛 diverges

252

Try this test first but remember it only proves divergence and not convergence.

2. The popular series test:

(i) The Geometric series: ∑ 𝑎𝑟𝑛−1 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ .∞𝑛=1

The geometric series converges for |𝑟| < 1 with a sum of 𝑎

1−𝑟.

(ii) The Harmonic series ∑1

𝑛= 1 +

1

2+

1

3+ ⋯∞

𝑛=1

The harmonic series diverges.

(iii) The p-series ∑1

𝑛𝑝 =1

1𝑝 +1

2𝑝 +1

3𝑝 + ⋯∞𝑛=1

The p-series converges for 𝑝 > 1 and diverges for 𝑝 ≤ 1.

This is a relatively straightforward test and should be used if the series is recognizable.

3. If the series is an alternating series, use the Alternating series theorem. The ratio test may

also work here.

Alternating series Test:

If 𝑎𝑛 > 𝑎𝑛+1 > 0 for all n and lim𝑛→∞

𝑎𝑛 = 0 then the alternating series

∑(−1)𝑛+1𝑎𝑛 = 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + ⋯

∞

𝑛=1

converges.

4. The Root test:

Suppose lim𝑛→∞

√|𝑎𝑛|𝑛= 𝐿.

(i) If L < 1, then ∑ 𝑎𝑛 converges.

(ii) If L > 1 or 𝐿 = ±∞, then ∑ 𝑎𝑛 diverges.

(iii) If L = 1, this test is inconclusive. Use a different test.

253

If 𝑎𝑛 is of the form (𝑓(𝑛))𝑔(𝑛), use the root test.

5. The Ratio test

Suppose lim𝑛→∞

|𝑎𝑛+1

𝑎𝑛| = 𝐿.

(i) If L < 1, then ∑ 𝑎𝑛 converges.

(ii) If L > 1 or 𝐿 = ±∞, then ∑ 𝑎𝑛 diverges.

(iii)If L = 1, this test is inconclusive. Use a different test.

For series containing factorials or terms with constants raised to some 𝑛𝑡ℎ power, use the

ratio test.

6. The Comparison test and Limit Comparison test.

The Comparison Test:

Let ∑ 𝑎𝑛 and ∑ 𝑏𝑛 be positive term series. Then

3. If 𝑎𝑛 ≤ 𝑏𝑛 for all n and ∑ 𝑏𝑛 converges, then ∑ 𝑎𝑛 converges.

4. If 𝑎𝑛 ≥ 𝑏𝑛 for all n and ∑ 𝑏𝑛 diverges, then ∑ 𝑎𝑛 diverges.

The Limit Comparison Test:

If lim𝑛→∞

𝑎𝑛

𝑏𝑛 exists and is positive, then either both series converge or both diverge.

Note that these tests are only valid for positive term series. Use this test when the general

term of the series is a rational function and all other tests fail for proving convergence.

7. The Integral Test:

Suppose 𝑎𝑛 = 𝑓(𝑛), where 𝑓(𝑛) is decreasing and positive.


1 converges, then ∑ 𝑎𝑛 converges

254


1 diverges, then ∑ 𝑎𝑛 diverges

Again, this test is valid only for positive term series. Use this test when all other tests fail

or when the improper integral can be easily evaluated.

Example 22.5

Test the series for convergence:

1. ∑1

𝑛5+𝑒𝑛∞𝑛=1

Use the comparison test:

𝑎𝑛 =1

𝑛5 + 𝑒𝑛


𝑏𝑛 =1

𝑛5

This comparing series converges because it is a p-series with 𝑝 = 5 > 1.

1

𝑛5 + 𝑒𝑛<

1

𝑛5


𝑛5+𝑒𝑛∞𝑛=1 also converges.

2. ∑ (𝑛3+1

3𝑛3+5)

3𝑛∞𝑛=1

Use the root test:

𝑎𝑛 = (𝑛3 + 1

3𝑛3 + 5)

3𝑛

255

√𝑎𝑛𝑛 = √(

𝑛3 + 1

3𝑛3 + 5)

3𝑛𝑛

= (𝑛3 + 1

3𝑛3 + 5)

3

lim𝑛→∞

(𝑛3 + 1

3𝑛3 + 5)

3

= (1

3)

3

= 1

27< 1

By the root test, ∑ (𝑛3+1

3𝑛3+5)

3𝑛


3. ∑ (−1)𝑛 3𝑛 . 𝑛

(2𝑛)!∞𝑛=1

Use the ratio test:

|𝑎𝑛| =3𝑛 . 𝑛

(2𝑛)! |𝑎𝑛+1| =

3𝑛+1 . (𝑛 + 1)

[2(𝑛 + 1)]!=

3𝑛+1 . (𝑛 + 1)

(2𝑛 + 2)!

|𝑎𝑛+1

𝑎𝑛| =

3𝑛+1 . (𝑛 + 1)

(2𝑛 + 2)! ÷

3𝑛 . 𝑛

(2𝑛)!

=3𝑛+1 . (𝑛 + 1)

(2𝑛 + 2)! .

(2𝑛)!

3𝑛 . 𝑛

=3𝑛 . 3 . (𝑛 + 1)

(2𝑛 + 2)(2𝑛 + 1)(2𝑛)! .

(2𝑛)!

3𝑛 . 𝑛

= 3(𝑛 + 1)

(2𝑛 + 2)(2𝑛 + 1)𝑛

lim𝑛→∞

3(𝑛 + 1)

(2𝑛 + 2)(2𝑛 + 1)𝑛= 0 < 1

By the ratio test, ∑ (−1)𝑛 3𝑛 . 𝑛

(2𝑛)!∞𝑛=1 converges.

256

4. ∑ (−1)𝑛 3𝑛+1

𝑛+2∞𝑛=1

Use the 𝑛𝑡ℎ test for divergence:

lim𝑛→∞

3𝑛 + 1

𝑛 + 2= 3 ≠ 0

By the 𝑛𝑡ℎ test for divergence, ∑ (−1)𝑛 3𝑛+1

𝑛+2∞𝑛=1 diverges.

5. ∑5

11𝑛∞𝑛=1

Use the popular series test:

This is a harmonic series, so ∑5

11𝑛∞𝑛=1 diverges.

257


1. Use the ratio test or the root test to investigate the convergence of the series

a) ∑ (−1)𝑛 𝑛2

5𝑛∞𝑛=1

b) ∑ 1

𝑛𝑛∞𝑛=1

c) ∑ (−1)𝑛 (4𝑛3+𝑛−1

3𝑛3+2)

𝑛∞𝑛=1

d) ∑ 𝑛!

7𝑛∞𝑛=1

e) ∑ (1 +1

𝑛)

𝑛2

∞𝑛=1

2. Determine whether the series converges:

a) ∑ 𝑛3

𝑛!∞𝑛=1

b) ∑ 3𝑛5+2𝑛+7

5𝑛5+1∞𝑛=1

c) ∑ 1

𝑛2𝑛∞𝑛=1

d) ∑ (𝑛

2𝑛+5)

−𝑛∞𝑛=1

e) ∑ 3𝑛

𝑛5∞𝑛=1

f) ∑ 1

2𝑛+𝑛∞𝑛=1

g) ∑ 𝑛2−2𝑛−1

𝑛3+5𝑛+4∞𝑛=1

h) ∑ 1

𝑛32

∞𝑛=1

i) ∑ 1

𝑛 ln 𝑛∞𝑛=2

258

CHAPTER 23: POWER SERIES

In this chapter you will learn about a special type of series called a power series and how to

determine the conditions under which such a series would converge.

Definition

A power series is a series of the form

∑ 𝑐𝑛𝑥𝑛 = 𝑐0 + 𝑐1𝑥 + 𝑐2𝑥2 + 𝑐3𝑥3 + ⋯

∞

𝑛=0

where the 𝑐𝑛′𝑠 are constants called the coefficients of the series.

A power series about 𝑥 = 𝑎 is a series of the form:

∑ 𝑐𝑛(𝑥 − 𝑎)𝑛 = 𝑐0 + 𝑐1(𝑥 − 𝑎) + 𝑐2(𝑥 − 𝑎)2 + 𝑐3(𝑥 − 𝑎)3 + ⋯

∞

𝑛=0

For example,

∑(−1)𝑛𝑥𝑛 = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯

∞

𝑛=0

is a power series that is centered at 0.

An important question is: “When does a power series converge?” When we test a power series

for convergence, we determine the values of x for which the series converges.

In previous chapters, we learned various tests for determining the convergence of series. Any of

these can be used to test power series for convergence, but the most popular test to use in these

cases is the ratio test.

259

The following examples makes use of another test, other than the ratio test.

Example

1. ∑ (2𝑥)𝑛 = 1 + 2𝑥 + 4𝑥2 + 8𝑥3 + ⋯∞𝑛=0

This power series is a geometric series with common ratio 𝑟 = 2𝑥.

Recall that a geometric series converges for |𝑟| < 1. So the series converges for

|2𝑥| < 1

−1 < 2𝑥 < 1

−1

2< 𝑥 <

1

2

2. 1 −1

3(𝑥 − 1) +

1

9(𝑥 − 1)2 −

1

27(𝑥 − 1)3 + ⋯.

This is another geometric series with common ratio −1

3(𝑥 − 1).

The series converges for

|−1

3(𝑥 − 1)| < 1

−1 <1

3(𝑥 − 1) < 1

−3 < 𝑥 − 1 < 1

−2 < 𝑥 < 2

260

Not all power series are popular series like the geometric series. Generally, to test a power series

for convergence, use the ratio test. Recall the ratio test:

For the series ∑ 𝒂𝒏, suppose 𝐥𝐢𝐦𝒏→∞

|𝒂𝒏+𝟏

𝒂𝒏| = 𝑳.

4. If L < 1, then ∑ 𝒂𝒏 converges.

5. If L > 1 or 𝑳 = ±∞, then ∑ 𝒂𝒏 diverges.


Note that if 𝐿 = 1, the test is inconclusive. This means that we would need to do some further

testing at these x-values.

Example

For what values of x do the following power series converge?

1. ∑ 𝑛𝑥𝑛∞𝑛=1

|𝑎𝑛| = | 𝑛𝑥𝑛| |𝑎𝑛+1| = |(𝑛 + 1)𝑥𝑛+1|

|𝑎𝑛+1

𝑎𝑛| = |

(𝑛 + 1)𝑥𝑛+1

𝑛𝑥𝑛|

=(𝑛 + 1)|𝑥|

𝑛

lim𝑛 →∞

|𝑎𝑛+1

𝑎𝑛| = lim

𝑛 →∞

(𝑛 + 1)|𝑥|

𝑛

= |𝑥| lim𝑛 →∞

𝑛 + 1

𝑛

= |𝑥|. 1

= |𝑥|

By the ratio test, this series converges for |𝑥| < 1, that is for −1 < 𝑥 < 1.

But because the test is inconclusive for |𝑥| = 1, we would need to do further testing at

𝑥 = 1 and at 𝑥 = −1, that is, at the end-points of the interval. We check these, by

substituting these values of x, into the actual series and then test for convergence.

261

Check at 𝑥 = 1:

∑ 𝑛𝑥𝑛

∞

𝑛=1

= ∑ 𝑛(1)𝑛

∞

𝑛=1

= ∑ 𝑛

∞

𝑛=1

= 1 + 2 + 3 + ⋯

This series clearly diverges. But we can test it by using the 𝑛𝑡ℎ term test for divergence:

lim𝑛 →∞

𝑛 = ∞ ≠ 0

Check at 𝑥 = −1:

∑ 𝑛𝑥𝑛

∞

𝑛=1

= ∑ 𝑛(−1)𝑛

∞

𝑛=1

= −1 + 2 − 3 + ⋯

Again, this series clearly diverges. We can test it by using the 𝑛𝑡ℎ term test for

divergence:

lim𝑛 →∞

|𝑛(−1)𝑛| = lim𝑛 →∞

𝑛 = ∞ ≠ 0

Thus, the power series ∑ 𝑛𝑥𝑛∞𝑛=1 converges for −1 < 𝑥 < 1.

2. ∑𝑥2𝑛

𝑛!∞𝑛=1

|𝑎𝑛| = | 𝑥2𝑛

𝑛!| |𝑎𝑛+1| = |

𝑥2(𝑛+1)

(𝑛 + 1!| = |

𝑥2𝑛+2

(𝑛 + 1!|

|𝑎𝑛+1

𝑎𝑛| = |

𝑥2𝑛+2

(𝑛 + 1! .

𝑛!

𝑥2𝑛|

=𝑥2

𝑛 + 1

lim𝑛 →∞

|𝑎𝑛+1

𝑎𝑛| = lim

𝑛 →∞

𝑥2

𝑛 + 1

= 𝑥2 lim𝑛 →∞

1

𝑛 + 1

= 𝑥2 . 0

262

= 0 < 1

By the ratio test, the series ∑𝑥2𝑛

𝑛!∞𝑛=1 converges for all values of 𝑥.

3. ∑𝑥𝑛

3𝑛𝑛2∞𝑛=1

|𝑎𝑛| = | 𝑥𝑛

3𝑛𝑛2| |𝑎𝑛+1| = |

𝑥𝑛+1

3𝑛+1(𝑛 + 1)2|

|𝑎𝑛+1

𝑎𝑛| = |

𝑥𝑛+1

3𝑛+1(𝑛 + 1)2 .

3𝑛𝑛2

𝑥𝑛|

=|𝑥|𝑛2

3(𝑛 + 1)2

lim𝑛 →∞

|𝑎𝑛+1

𝑎𝑛| = lim

𝑛 →∞

|𝑥|𝑛2

3(𝑛 + 1)2

= |𝑥| lim𝑛 →∞

|𝑛2

3(𝑛 + 1)2

=|𝑥|

3

By the ratio test, this series converges for |𝑥|

3< 1, that is for −3 < 𝑥 < 3.

We would need to do further testing at the end-points 𝑥 = 3 and at 𝑥 = −3.

Check at 𝑥 = 3:

∑𝑥𝑛

3𝑛𝑛2

∞

𝑛=1

= ∑3𝑛

3𝑛𝑛2

∞

𝑛=1

= ∑1

𝑛2

∞

𝑛=1

This is a p-series with 𝑝 = 2. So the series converges at 𝑥 = 3.

Check at 𝑥 = −3:

263

∑𝑥𝑛

3𝑛𝑛2

∞

𝑛=1

= ∑(−3)𝑛

3𝑛𝑛2

∞

𝑛=1

= ∑(−1)𝑛

𝑛2

∞

𝑛=1

Now

∑ |(−1)𝑛

𝑛2|

∞

𝑛=1

= ∑1

𝑛2

∞

𝑛=1

which converges. So the series converges at 𝑥 = −3.

Thus, the power series 𝑥𝑛

3𝑛𝑛2 converges for −3 ≤ 𝑥 ≤ 3.

We will now define two concepts: the radius of convergence and the interval of convergence.

Theorem

For a given power series

∑ 𝑐𝑛(𝑥 − 𝑎)𝑛 = 𝑐0 + 𝑐1(𝑥 − 𝑎) + 𝑐2(𝑥 − 𝑎)2 + 𝑐3(𝑥 − 𝑎)3 + ⋯

∞

𝑛=0

There are only 3 possibilities:

1. The series converges only when 𝑥 = 𝑎

2. The series converges for all x

3. There is a positive number R such that the series converges if |𝑥 − 𝑎| < 𝑅

R is called the radius of convergence. In the first two possibilities above:

1. 𝑅 = 0

2. 𝑅 = ∞

Thus in the examples above:

Example 1: R = 1

Example 2: 𝑅 = ∞

264

Example 3: 𝑅 = 3

The interval of convergence

The interval of convergence of a power series is the interval that consists of all values of x for

which the series converges. Again in the possibilities above:

1. Interval of convergence is a single point a.

2. Interval of convergence = (−∞, ∞)

3. Interval of convergence can be one of the following:

a) (a – R, a + R)

b) [a – R, a + R]

c) (a – R, a + R]

d) [a – R, a + R)

Example

Find the radius and interval of convergence of the series:

1. ∑(−1)𝑛(𝑥−3)𝑛

𝑛∞𝑛=0

|𝑎𝑛| = | (−1)𝑛(𝑥 − 3)𝑛

𝑛| |𝑎𝑛+1| = |

(−1)𝑛+1(𝑥 − 3)𝑛+1

𝑛 + 1|

|𝑎𝑛+1

𝑎𝑛| = |

(𝑥 − 3)𝑛+1

𝑛 + 1 .

𝑛

(𝑥 − 3)𝑛|

=|𝑥 − 3|𝑛

𝑛 + 1

lim𝑛 →∞

|𝑎𝑛+1

𝑎𝑛| = lim

𝑛 →∞

|𝑥 − 3|𝑛

𝑛 + 1

= |𝑥 − 3| lim𝑛 →∞

𝑛

𝑛 + 1

265

= |𝑥 − 3|

By the ratio test, this series converges for |𝑥 − 3| < 1, that is for

−1 < 𝑥 − 3 < 1

2 < 𝑥 < 4

The radius of convergence is 1.

.

We would need to do further testing at the end-points 𝑥 = 2 and at 𝑥 = 4.

Check at 𝑥 = 2:

∑(−1)𝑛(𝑥 − 3)𝑛

𝑛

∞

𝑛=1

= ∑(−1)𝑛. (−1)𝑛

𝑛

∞

𝑛=1

= ∑1

𝑛

∞

𝑛=1

This is the harmonic series which diverges. So the series diverges at 𝑥 = 2.

Check at 𝑥 = 4:

∑(−1)𝑛(𝑥 − 3)𝑛

𝑛

∞

𝑛=1

= ∑(−1)𝑛. (1)𝑛

𝑛

∞

𝑛=1

= ∑(−1)𝑛

𝑛

∞

𝑛=1

We use the alternating series test to test this series for convergence.

𝑎𝑛 =1

𝑛

𝑎𝑛+1 =1

𝑛 + 1

1

𝑛>

1

𝑛 + 1> 0

lim𝑛→∞

1

𝑛= 0

266

So by the alternating series test, ∑(−1)𝑛

𝑛∞𝑛=1 converges. So the series converges for

𝑥 = 4.

Thus, the power series ∑(−1)𝑛(𝑥−3)𝑛

𝑛∞𝑛=0 converges for 2 < 𝑥 ≤ 4.

The interval of convergence is 𝟐 < 𝒙 ≤ 𝟒.

267


Find the radius of convergence and interval of convergence of the series:

1. ∑𝑥2𝑛

(2𝑛)!∞𝑛=1

2. ∑2𝑛𝑥𝑛

𝑛!∞𝑛=1

3. ∑ 𝑛3(𝑥 − 1)𝑛∞𝑛=1

4. ∑𝑥𝑛3𝑛

𝑛𝑛∞𝑛=1

5. ∑(−1)𝑛(𝑥+1)𝑛

5𝑛√𝑛∞𝑛=1

6. ∑ 5𝑛𝑥𝑛∞𝑛=1

7. ∑𝑛!𝑥𝑛

7𝑛∞𝑛=1

268

CHAPTER 24: SERIES EXPANSIONS

This chapter teaches you a general method for writing a function as a series. You will learn how

to approximate functions by polynomials. For example,

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯

Taylor Polynomials

Taylor’s polynomial gives the expansion of 𝑓(𝑥) in powers of 𝑥 − 𝑎, with 𝑎 being a constant.

Let’s derive the Taylor polynomial expansion:

Suppose

𝑓(𝑥) = 𝑏0 + 𝑏1(𝑥 − 𝑎) + 𝑏2(𝑥 − 𝑎)2 + 𝑏3(𝑥 − 𝑎)3 + ⋯ + 𝑏𝑛(𝑥 − 𝑎)𝑛 … . . (1)

Substitute 𝑥 = 𝑎. We get:

𝒃𝟎 = 𝒇(𝒂)

We differentiate (1):

𝑓′(𝑥) = 𝑏1 + 2𝑏2(𝑥 − 𝑎) + 3𝑏3(𝑥 − 𝑎)2 + ⋯ + 𝑛𝑏𝑛(𝑥 − 𝑎)𝑛−1


𝒃𝟏 = 𝒇′(𝒂)

Differentiate again:

𝑓′′(𝑥) = 2𝑏2 + 3 ∙ 2𝑏3(𝑥 − 𝑎) + ⋯ + 𝑛(𝑛 − 1)𝑏𝑛(𝑥 − 𝑎)𝑛−2


𝒃𝟐 =𝒇′′(𝒂)

𝟐!

269

Continuing we get:

𝒃𝟑 =𝒇′′′(𝒂)

𝟑!

𝒃𝒌 =𝒇(𝒌)(𝒂)

𝒌!

Substituting these values for the 𝑏𝑖′𝑠 in 𝑓(𝑥) gives us:

Taylor polynomial of degree n approximating 𝒇(𝒙) near 𝒙 = 𝒂 denoted by 𝑷𝒏(𝒙)

𝒇(𝒙) ≈ 𝑷𝒏(𝒙)

where

𝑷𝒏(𝒙) = 𝒇(𝒂) + 𝒇′(𝒂)(𝒙 − 𝒂) +𝒇′′(𝒂)

𝟐!(𝒙 − 𝒂)𝟐 +

𝒇′′′(𝒂)

𝟑!(𝒙 − 𝒂)𝟑 + ⋯ +

𝒇(𝒏)(𝒂)

𝒏!(𝒙 − 𝒂)𝒏

𝒏𝒕𝒉 degree Taylor polynomial about 𝒙 = 𝟎 is:

𝒇(𝒙) ≈ 𝒇(𝟎) + 𝒇′(𝟎)𝒙 +𝒇′′(𝟎)

𝟐!𝒙𝟐 +

𝒇′′′(𝟎)

𝟑!𝒙𝟑 + ⋯ +

𝒇(𝒏)(𝟎)

𝒏!𝒙𝒏

Example 24.1

1. Find the 5𝑡ℎ degree Taylor polynomial of 𝑓(𝑥) = 𝑒2𝑥 at 𝑥 = 0.

We find up to the 5𝑡ℎ derivative of 𝑓(𝑥) at 𝑥 = 0:

𝑓(𝑥) = 𝑒2𝑥 𝑓(0) = 1

𝑓′(𝑥) = 2𝑒2𝑥 𝑓′(0) = 2

𝑓′′(𝑥) = 4𝑒2𝑥 𝑓′′(0) = 4

270

𝑓′′′(𝑥) = 8𝑒2𝑥 𝑓′′′(0) = 8

𝑓(4)(𝑥) = 16𝑒2𝑥 𝑓(4)(0) = 16

𝑓(5)(𝑥) = 32𝑒2𝑥 𝑓(5)(0) = 32

𝑓(𝑥) ≈ 𝑓(0) + 𝑓′(0)𝑥 +𝑓′′(0)

2!𝑥2 +

𝑓′′′(0)

3!𝑥3 +

𝑓(4)(0)

4!𝑥4 +

𝑓(5)(0)

5!𝑥5

So

𝑒2𝑥 ≈ 1 + 2𝑥 +4

2!𝑥2 +

8

3!𝑥3 +

16

4!𝑥4 +

32

5!𝑥5

𝑒2𝑥 ≈ 1 + 2𝑥 + 2𝑥2 +4

3𝑥3 +

2

3𝑥4 +

4

15𝑥5

2. Find the 3𝑟𝑑 degree Taylor polynomial of 𝑓(𝑥) = cos 𝑥 at 𝑥 =𝜋

3.

We find up to the 3𝑟𝑑 derivative of 𝑓(𝑥) at 𝑥 =𝜋

3:

𝑓(𝑥) = cos 𝑥 𝑓 (𝜋

3) =

1

2

𝑓′(𝑥) = −sin 𝑥 𝑓′ (𝜋

3) = −

√3

2

𝑓′′(𝑥) = −cos 𝑥 𝑓′′ (𝜋

3) = −

1

2

𝑓′′′(𝑥) = sin 𝑥 𝑓′′′ (𝜋

3) =

√3

2

𝑓(𝑥) ≈ 𝑓(𝑎) + 𝑓′(𝑎)(𝑥 − 𝑎) +𝑓′′(𝑎)

2!(𝑥 − 𝑎)2 +

𝑓′′′(𝑎)

3!(𝑥 − 𝑎)3

So

271

cos 𝑥 ≈ 𝑓 (𝜋

3) + 𝑓′ (

𝜋

3) (𝑥 −

𝜋

3) +

𝑓′′ (𝜋3

)

2!(𝑥 −

𝜋

3)

2

+ 𝑓′′′ (

𝜋3

)

3!(𝑥 −

𝜋

3)

3

cos 𝑥 ≈ 1

2−

√3

2(𝑥 −

𝜋

3) −

1

2(𝑥 −

𝜋

3)

2

+√3

2(𝑥 −

𝜋

3)

3

3. Find the 𝑛𝑡ℎ degree Taylor polynomial of 𝑓(𝑥) = ln(1 + 𝑥) at 𝑎 = 0.

𝑓(𝑥) = ln(1 + 𝑥) 𝑓(0) = 0

𝑓′(𝑥) =1

1 + 𝑥 𝑓′(0) = 1

𝑓′′(𝑥) = −1

(1 + 𝑥)2 𝑓′′(0) = −1

𝑓′′′(𝑥) =2

(1 + 𝑥)3 𝑓′′′(0) = 2

𝑓(4)(𝑥) = −6

(1 + 𝑥)4 𝑓(4)(0) = −6

𝑓(5)(𝑥) =24

(1 + 𝑥)5 𝑓(5)(0) = 24

𝑓(𝑥) ≈ 𝑓(0) + 𝑓′(0)𝑥 +𝑓′′(0)

2!𝑥2 +

𝑓′′′(0)

3!𝑥3 + ⋯ +

𝑓(𝑛)(0)

𝑛!𝑥𝑛

So

ln(1 + 𝑥) ≈ 𝑥 −1

2!𝑥2 +

2

3!𝑥3 −

6

4!𝑥4 +

24

5!𝑥5 … +

𝑓(𝑛)(0)

𝑛!𝑥𝑛

ln(1 + 𝑥) ≈ 𝑥 −𝑥2

2+

𝑥3

3−

𝑥4

4+

𝑥5

5− ⋯ + (−1)𝑛+1

𝑥𝑛

𝑛

272

Taylor series

The Taylor series for 𝑓(𝑥) about 𝑥 = 𝑎 is an infinite series given by:

𝒇(𝒙) = 𝒇(𝒂) + 𝒇′(𝒂)(𝒙 − 𝒂) +𝒇′′(𝒂)

𝟐!(𝒙 − 𝒂)𝟐 +

𝒇′′′(𝒂)

𝟑!(𝒙 − 𝒂)𝟑 + ⋯ +

𝒇(𝒏)(𝒂)

𝒏!(𝒙 − 𝒂)𝒏+..

The Taylor series for 𝑓(𝑥) about 𝑥 = 0 is called the Maclaurin series and is given by:

𝒇(𝒙) = 𝒇(𝟎) + 𝒇′(𝟎)𝒙 +𝒇′′(𝟎)

𝟐!𝒙𝟐 +

𝒇′′′(𝟎)

𝟑!𝒙𝟑 + ⋯ +

𝒇(𝒏)(𝟎)

𝒏!𝒙𝒏 + ⋯

The following are common Maclaurin series:

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯ +

𝑥𝑛

𝑛!+ ⋯

sin 𝑥 = 𝑥 −𝑥3

3!+

𝑥5

5!−

𝑥7

7!+ ⋯ + ⋯ (−1)𝑛

𝑥2𝑛+1

(2𝑛 + 1)!+ ⋯

cos 𝑥 = 1 −𝑥2

2!+

𝑥4

4!−

𝑥6

6!+ ⋯ + ⋯ (−1)𝑛

𝑥2𝑛

(2𝑛)!+ ⋯

ln(1 + 𝑥) ≈ 𝑥 −𝑥2

2+

𝑥3

3−

𝑥4

4+

𝑥5

5− ⋯ + (−1)𝑛+1

𝑥𝑛

𝑛+ ⋯

arctan 𝑥 = 𝑥 −𝑥3

3+

𝑥5

5−

𝑥7

7+ ⋯ + ⋯ (−1)𝑛

𝑥2𝑛+1

(2𝑛 + 1)+ ⋯

1

1 − 𝑥= 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ + 𝑥𝑛 + ⋯

We can use these common series to write out series expansions for related functions without

using the Taylor formula.

273

Example 24.2

Find the Taylor series for:

1. 𝑒3𝑥

We know that

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯ +

𝑥𝑛

𝑛!+ ⋯

So to find the series expansion for 𝑒3𝑥, we replace 𝑥 with 3𝑥 to get:

𝑒3𝑥 = 1 + 3𝑥 +(3𝑥)2

2!+

(3𝑥)3

3!+ ⋯ +

(3𝑥)𝑛

𝑛!+ ⋯

𝑒3𝑥 = 1 + 3𝑥 +9𝑥2

2!+

27𝑥3

3!+ ⋯ +

3𝑛𝑥𝑛

𝑛!+ ⋯

𝑒3𝑥 = 1 + 3𝑥 +9𝑥2

2+

9𝑥3

2+ ⋯ +

3𝑛𝑥𝑛

𝑛!+ ⋯

2. cos (𝑥2)

We know that

cos 𝑥 = 1 −𝑥2

2!+

𝑥4

4!−

𝑥6

6!+ ⋯ + ⋯ (−1)𝑛

𝑥2𝑛

(2𝑛)!+ ⋯

So

cos 𝑥2 = 1 −(𝑥2)2

2!+

(𝑥2)4

4!−

(𝑥2)6

6!+ ⋯ + ⋯ (−1)𝑛

(𝑥2)2𝑛

(2𝑛)!+ ⋯

cos 𝑥2 = 1 −𝑥4

2!+

𝑥8

4!−

𝑥12

6!+ ⋯ + ⋯ (−1)𝑛

𝑥4𝑛

(2𝑛)!+ ⋯

274

3. 𝑒𝑥 sin 𝑥

Find the series expansion up to the term in 𝑥3.

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!= 1 + 𝑥 +

𝑥2

2+

𝑥3

3

sin 𝑥 = 𝑥 −𝑥3

3!= 𝑥 −

𝑥3

6

So

𝑒𝑥 sin 𝑥 = (1 + 𝑥 +𝑥2

2+

𝑥3

3) (𝑥 −

𝑥3

6)

= 𝑥 −𝑥3

6+ 𝑥2 +

𝑥3

2

= 𝑥 + 𝑥2 +𝑥3

3

Binomial Series

An important series which often appears in applications is the binomial series. This series is the

Maclaurin series expansion of (1 + 𝑥)𝑛 and is given by:

(𝟏 + 𝒙)𝒏 = 𝟏 + 𝒏𝒙 +𝒏(𝒏 − 𝟏)

𝟐!𝒙𝟐 +

𝒏(𝒏 − 𝟏)(𝒏 − 𝟐)

𝟑!𝒙𝟑 + ⋯

This expansion is valid for −𝟏 < 𝒙 < 𝟏 or |𝒙| < 𝟏. That is, the series converges for |𝒙| < 𝟏.

Example 24.3

1. Expand (1 + 𝑥)4 as a power series up to the 5th term.

(1 + 𝑥)4 = 1 + 𝑛𝑥 +𝑛(𝑛 − 1)

2!𝑥2 +

𝑛(𝑛 − 1)(𝑛 − 2)

3!𝑥3 +

𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)

4!𝑥4

275

= 1 + 4𝑥 +4(4 − 1)

2!𝑥2 +

4(4 − 1)(4 − 2)

3!𝑥3 +

4(4 − 1)(4 − 2)(4 − 3)

4!𝑥4

= 1 + 4𝑥 + 6𝑥2 + 4𝑥3 + 𝑥4

2. Expand (1 − 2𝑥)1

2⁄ as a power series up to the 4th term.

In the binomial expansion formula, we would need to let 𝑛 =1

2 and replace 𝑥 with −2𝑥.

(1 − 2𝑥)1

2⁄ = 1 +1

2(−2𝑥) +

12

(12

− 1)

2!(−2𝑥)2 +

12

(12

− 1) (12

− 2)

3!(−2𝑥)3

= 1 − 𝑥 +

12

. −12

2!4𝑥2 +

12

. −12

. −32

3!. −8𝑥3

= 1 − 𝑥 −1

2𝑥2 −

1

2𝑥3

3. Find the series expansion up to the 4th term for 1

3+𝑥.

1

3 + 𝑥= (3 + 𝑥)−1

In order to use the binomial series expansion, we would need to first factor out 3. So

(3 + 𝑥)−1 = [3 (1 +𝑥

3)]

−1

= 3−1 (1 +𝑥

3)

−1

=1

3[1 + (−1) (

𝑥

3) +

−1 . (−1 − 1)

2!(

𝑥

3)

2

+−1 . (−1 − 1)(−1 − 2)

3!(

𝑥

3)

3

]

=1

3[1 −

𝑥

3+

𝑥2

9−

𝑥3

27]

276

=1

3−

𝑥

9+

𝑥2

27−

𝑥3

81

We can use the binomial series expansion for estimation.

Example 24.4

1. Expand √1 + 𝑥 as a power series. Hence estimate √1.01 to 3 decimal places.

√1 + 𝑥 = (1 + 𝑥)12

= 1 +1

2𝑥 +

12

(12

− 1)

2!𝑥2 +

12

(12

− 1) (12

− 2)

3!𝑥3 + ⋯

= 1 +1

2𝑥 +

12

. −12

2!𝑥2 +

12

. −12

. −32

3!. 𝑥3 + ⋯

= 1 +1

2𝑥 −

1

8𝑥2 +

1

16𝑥3 + ⋯

It was necessary to expand only up to the term in 𝑥3 since the estimation must be

rounded to 3 decimal places.

To find √1.01 we express it as √1 + 0.01 and substitute 0.01 for 𝑥 in the series

expansion. So

√1.01 = √1 + 0.01

= 1 +1

2(0.01) −

1

8(0.01)2 +

1

16(0.01)3 + ⋯

= 1 + 0.005 − 0

= 1.005

277

Finding series expansions through differentiation and integration of other series

Example 24.5

1. Find the Taylor series expansion for 1

(1−𝑥)2 from the series expansion for 1

1−𝑥.

1

1 − 𝑥= 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ + 𝑥𝑛 + ⋯

Observe that

(1

1 − 𝑥)

′

=1

(1 − 𝑥)2

So to find the series expansion of 1

(1−𝑥)2 we differentiate the series expansion of 1

1−𝑥. So

1

(1 − 𝑥)2=

𝑑

𝑑𝑥(1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ + 𝑥𝑛 + ⋯ )

= 1 + 2𝑥 + 3𝑥2 + 4𝑥3 + ⋯ + 𝑛𝑥𝑛−1 + ⋯

2. Find the Taylor series expansion for tan−1 𝑥 from the series expansion for 1

1+𝑥2.

1

1 + 𝑥2= (1 + 𝑥2)−1 = 1 − 𝑥2 +

−1 . −2

2! (𝑥2)2 +

−1 . −2 . −3

3! (𝑥2)3 +

= 1 − 𝑥2 + 𝑥4 − 𝑥6 + 𝑥8 − ⋯

Observe that

∫1

1 + 𝑥2 𝑑𝑥 = tan−1 𝑥 + 𝐶

So to find the series expansion for tan−1 𝑥 we integrate the series expansion for 1

1+𝑥2. So

tan−1 𝑥 = ∫(1 − 𝑥2 + 𝑥4 − 𝑥6 + 𝑥8 − ⋯ ) 𝑑𝑥

278

𝑥 −1

3𝑥3 +

1

5𝑥5 −

1

7𝑥7 +

1

9𝑥9 − ⋯ + 𝐶

To find C, substitute 𝑥 = 0 on both sides of the equation to get

𝐶 = 0

So

tan−1 𝑥 = 𝑥 −1

3𝑥3 +

1

5𝑥5 −

1

7𝑥7 +

1

9𝑥9 − ⋯

279


1. Use the formula for the binomial series expansion to find the series expansion for the

following functions:

a) 𝑓(𝑥) =1

√1+3𝑥

b) 𝑓(𝑥) =2

4−𝑥

c) 𝑓(𝑥) = (1 + 5𝑥)4

2. Find the Taylor series expansion for the function about the given value of 𝑎 up to the

term in 𝑥4:

a) 𝑓(𝑥) = 𝑥−3 about 𝑎 = 2

b) 𝑓(𝑥) = 𝑥4 + 3𝑥3 − 1 about 𝑎 = −1

3. Find the Maclaurin series expansion up to the term in 𝑥4 for:

a) 𝑒−2𝑥

b) cos(𝑥3)

c) 𝑒−𝑥 sin 2𝑥

4. Expand √1 − 𝑥. Hence estimate √0.999 up to 4 decimal places.

5. Find the Taylor series expansion for 1

(1+𝑥)2 from the series expansion for 1

1+𝑥.

6. Find the Taylor series expansion for ln(1 + 𝑥) from the series expansion for 1

1+𝑥.

280

CHAPTER 25: AN INTRODUCTION TO DIFFERENTIAL EQUATIONS

Suppose that we need to determine the rate of change of some quantity with respect to time (for

example, the growth or decay of a population, the cooling temperature of a hot cup of tea), we

would study differential equations. The importance of differential equations can be observed in

disciplines such as pure and applied mathematics, physics, engineering, biology, and economics.

A differential equation is an equation that contains an unknown function and one or more of its

derivatives. The order of a differential equation is the order of the highest derivative that occurs

in the equation. For example,

𝑑𝑦

𝑑𝑥= 𝑥2 is a first order differential equation

𝑥𝑑2𝑦

𝑑𝑥2 =1

𝑦 is a second order differential equation.

The solution of a differential equation

𝑦 = 𝑓(𝑥) is a solution of a differential equation if it satisfies the differential equation.

For example, consider the differential equation 𝑑𝑦

𝑑𝑥= 3𝑥2 + 2𝑥. The following are solutions for

this differential equation:

𝑦 = 𝑥3 + 𝑥2

𝑦 = 𝑥3 + 𝑥2 − 1

𝑦 = 𝑥3 + 𝑥2 + 5

𝑦 = 𝑥3 + 𝑥2 + 𝐶

Observe that the derivative of each of these solutions is the differential equation. When we solve

a differential equation, we are finding all possible solutions of the equation. We usually solve

differential equations by integration.

281

Example 25.1

Solve the differential equations:

1. 𝑑𝑦

𝑑𝑥= 4𝑥3 − 2𝑥 + 5

𝑦 = ∫(4𝑥3 − 2𝑥 + 5) 𝑑𝑥

= 𝑥4 − 𝑥2 + 5𝑥 + 𝐶

2. . 𝑑𝑦

𝑑𝑥= 𝑥2 +

1

𝑥

𝑦 =1

3𝑥3 + ln 𝑥 + 𝐶

This solution of a differential equation is called the general solution because it contains C, a

constant.

When we graph the general solution we obtain a family of solutions.

Consider the differential equation

𝑑𝑦

𝑑𝑥= 2𝑥

The general solution is:

𝑦 = 𝑥2 + 𝐶

Substituting values for C would give a family of solutions. The following graph represents such a

family:

282

Usually, when we solve a differential equation we are interested in finding a particular solution,

i.e. a solution that satisfies some requirement.

A differential equation is usually presented with an initial condition, that is, the value of the

dependent variable at some initial value of the independent variable. For example,

𝑑𝑦

𝑑𝑥= 𝑥2, 𝑦(0) = 1

The initial condition 𝑦(0) = 1 means that when 𝑥 = 0, 𝑦 = 1. This problem is called an initial

value problem and its solution will be a particular solution for the differential equation.

Some differential equation models

1. Population growth:

Model: The rate of change of a population is proportional to the size of the population

𝒅𝑷

𝒅𝒕= 𝒌𝑷

283

where 𝑃 is the population size, 𝑡 is time, and k is a constant of proportionality

This is a simple model and it does not take into consideration factors that influence

population growth.

The solution of this differential equation is

𝑃 = 𝑃0𝑒𝑘𝑡

where 𝑃0 is the initial population.

If 𝑘 > 0, the population grows exponentially. Such a model does not take into account

that resources are limited.

A better model for population growth is:

𝒅𝑷

𝒅𝒕= 𝒌 (𝟏 −

𝑷

𝑵) 𝑷

where N is the carrying capacity of the population.

2. Newton’s law of cooling:

This law states that the rate at which the temperature of a body changes is proportional to

the difference between the temperature of the body and the temperature of the

surroundings. The differential equation modeling that law is:

𝒅𝑻

𝒅𝒕= 𝒌(𝑻 − 𝑻𝒔)

where 𝑇 is the temperature of the body, and 𝑇𝑠 is the temperature of the surroundings.

Slope fields

Slope fields enable us to answer qualitative questions about the solution of a differential

equation. For example, how does the solution behave near a certain point?


284

𝑑𝑦

𝑑𝑥= 𝑥 + 𝑦

How do we sketch the graph of its solution without solving the differential equation? We do so

by plotting a slope field.


𝑑𝑦

𝑑𝑥= 𝑓(𝑥, 𝑦)

A slope field is a collection of short straight line segments with slope 𝑓(𝑥, 𝑦) at several points.

For example, in the differential equation

𝑑𝑦

𝑑𝑥= 𝑥 + 𝑦

𝑑𝑦

𝑑𝑥|

(1,3)= 1 + 3 = 4

This means that the slope of the function (which is the solution of the differential equation) at the

point (1,3) is 4. Thus, the slope field would contain a short line segment at (1,3) with slope 4.

𝑑𝑦

𝑑𝑥|

(4,−1)= 4 − 1 = 3

This means that the slope of the function (which is the solution of the differential equation) at the

point (4, −1) is 3, and the slope field would contain a short line segment at (4, −1) with slope 3.

Example 25.1

The following is the slope field for

𝑑𝑦

𝑑𝑥= 𝑦 − 𝑥

285

Here is the slope field with solution curves:

Euler’s Method

This method enables us to find a numerical approximation to the solution of a differential

equation.

286

Euler used a slope field as a road map to finding an approximate solution. The idea is to start

with a first approximation (𝑥0, 𝑦0). To find the next approximation we draw the tangent line to

the curve 𝑦 = 𝑓(𝑥) at (𝑥0, 𝑦0), move a short distance away to (𝑥1, 𝑦1). Draw a tangent line to the

curve at this new approximation (𝑥1, 𝑦1) and find another approximation (𝑥2, 𝑦2). Continue this

procedure.

Let us find a formula for Euler’s method by using the first two approximations: (𝑥0, 𝑦0) and

(𝑥1, 𝑦1).

Finding Euler’s formula

Given

𝑑𝑦

𝑑𝑥= 𝑓(𝑥, 𝑦), 𝑦(𝑥0) = 𝑦0

Recall that 𝑑𝑦

𝑑𝑥= 𝑓(𝑥, 𝑦) represents the slope of the function. Also recall that we drew a tangent

line to the curve at (𝑥0, 𝑦0) to find (𝑥1, 𝑦1). The slope of that tangent line is therefore 𝑓(𝑥0, 𝑦0).

But the slope of the line joining (𝑥0, 𝑦0) to (𝑥1, 𝑦1) is also

𝑦1 − 𝑦0

𝑥1 − 𝑥0

So

𝑦1 − 𝑦0

𝑥1 − 𝑥0= 𝑓(𝑥0, 𝑦0)

Let the step size

𝑥1 − 𝑥0 = ℎ

Then

𝑦1 − 𝑦0 = ℎ𝑓(𝑥0, 𝑦0)

𝑦1 = 𝑦0 + ℎ𝑓(𝑥0, 𝑦0)

287

Extending this to arbitrary points (𝑥𝑛, 𝑦𝑛) and (𝑥𝑛+1, 𝑦𝑛+1), we obtain Euler’s formula:

Euler’s formula:

𝒚𝒏+𝟏 = 𝒚𝒏 + 𝒉𝒇(𝒙𝒏, 𝒚𝒏)

where h is the step size.

Example 25.2

1. Use Euler’s method to find 3 more approximate solutions to the initial value problem:

𝑑𝑦

𝑑𝑥= 2𝑥 + 𝑦, 𝑦(1) = 3

Use the step size ℎ = 0.1.

Since we want 3 more approximations, we will first write 4 approximations for 𝑥𝑛

starting with 𝑥0.

𝑥0 = 1 𝑥1 = 1.1 𝑥2 = 1.2 𝑥2 = 1.3

Also

𝑦0 = 3

Euler’s formula is:

𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑓(𝑥𝑛, 𝑦𝑛)

where

𝑓(𝑥𝑛, 𝑦𝑛) = 2𝑥 + 𝑦

So

𝑦1 = 𝑦0 + 0.2(2𝑥0 + 𝑦0)

= 3 + 0.2[2(1) + 3]

𝒚𝟏 = 𝟒

288

𝑦2 = 𝑦1 + 0.2(2𝑥1 + 𝑦1)

= 4 + 0.2[2(1.1) + 4]

𝒚𝟐 = 𝟓. 𝟐𝟒

𝑦3 = 𝑦2 + 0.2(2𝑥2 + 𝑦2)

= 5.24 + 0.2[2(1.2) + 5.24]

𝒚𝟑 = 𝟔. 𝟕𝟔𝟖

2. Use Euler’s method to find 3 more approximate solutions to the initial value problem:

𝑑𝑦

𝑑𝑡= 𝑦 − 𝑡2, 𝑦(0) = 2, ℎ = 0.5

Since we want 3 more approximations, we will first write 4 approximations for 𝑡𝑛

starting with 𝑡0.

𝑡0 = 0 𝑡1 = 0.5 𝑡2 = 1 𝑡3 = 1.5

Also

𝑦0 = 2

Euler’s formula is:

𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑓(𝑡𝑛, 𝑦𝑛)

where

𝑓(𝑥𝑛, 𝑦𝑛) = 𝑦 − 𝑡2

So

𝑦1 = 𝑦0 + 0.5(𝑦0 − 𝑡02)

= 2 + 0.5[2 − 0]

𝒚𝟏 = 𝟑

289

𝑦2 = 𝑦1 + 0.5(𝑦1 − 𝑡12)

= 3 + 0.5[3 − 0.52]

𝒚𝟐 = 𝟒. 𝟑𝟕𝟓

𝑦3 = 𝑦2 + 0.5(𝑦2 − 𝑡22)

= 4.375 + 0.5[4.375 − 12]

𝒚𝟏 = 𝟔. 𝟎𝟔𝟐𝟓

290


1. Solve the following differential equations:

a) 𝑑𝑦

𝑑𝑥= 2𝑥4 +

1

3𝑥− 1

b) 𝑑𝑦

𝑑𝑥= 𝑥7 + 3𝑥5 − 6𝑥 − 4

2. Use Euler’s method with step size 0.4 to compute the approximate values 𝑦0, 𝑦1, 𝑦2, and

𝑦3 of the solution to the initial value problem

𝑑𝑦

𝑑𝑡= 𝑦2 − 2𝑡, 𝑦(0) = 1

3. Use Euler’s method with step size 0.5 to compute the approximate values 𝑦0, 𝑦1, 𝑦2, and

𝑦3 of the solution to the initial value problem

𝑑𝑦

𝑑𝑥= 𝑥2𝑦 − 3𝑥, 𝑦(1) = 4

291

CHAPTER 26: SOLVING DIFFERENTIAL EQUATIONS

In this chapter we will learn to solve specific types of differential equations using the method of

separation of variables.

Separation of variables

The differential equation

𝑑𝑦

𝑑𝑥= 𝑔(𝑥)ℎ(𝑦)

is said to be separable because we can basically separate the variables by placing the dependent

variable 𝑦 with the differential 𝑑𝑦 on the left side of the equation, and placing the independent

variable 𝑥 with the differential 𝑑𝑥 on the right side of the equation. We achieve this by

multiplication and/or division.

Steps to solving a differential equation by separation of variables

1. Separate the variables in the differential equation 𝑑𝑦

𝑑𝑥= 𝑔(𝑥)ℎ(𝑦) to get

1

ℎ(𝑦) 𝑑𝑦 = 𝑔(𝑥) 𝑑𝑥

2. Integrate both sides

∫1

ℎ(𝑦) 𝑑𝑦 = ∫ 𝑔(𝑥) 𝑑𝑥

3. Solve by integrating both sides of the equation and putting the constant C on the right

hand side:

𝐻(𝑦) = 𝐺(𝑥) + 𝐶

4. If possible, solve for 𝑦.

292

Example 26.1

Solve:

1. 𝑑𝑦

𝑑𝑥= 2𝑥2𝑦

We separate variables by dividing by 𝑦 and multiplying by 𝑑𝑥.

1

𝑦 𝑑𝑦 = 2𝑥2 𝑑𝑥

Integrate:

∫1

𝑦 𝑑𝑦 = ∫ 2𝑥2 𝑑𝑥

ln|𝑦| = 2

3𝑥3 + 𝐾

We solve for 𝑦 by exponentiating both sides of the equation:

𝑒ln |𝑦| = 𝑒23𝑥3+𝐾

|𝑦| = 𝑒23𝑥3+𝐾

𝑦 = ±𝑒23𝑥3+𝐾

𝑦 = ±𝑒𝐾 . 𝑒23𝑥3

Now ±𝑒𝐾 is a constant which we can replace with the constant 𝐶. Thus the solution of

the differential equation is:

𝑦 = 𝐶𝑒23𝑥3

293

2. 𝑑𝑦

𝑑𝑥= 3√𝑦, 𝑦(2) = 4

This is an initial value problem so we will obtain a particular solution upon solving. Let

us first find the general solution.

𝑑𝑦

𝑑𝑥= 3√𝑦

Divide by √𝑦 and integrate:

∫1

√𝑦 𝑑𝑦 = ∫ 3 𝑑𝑥

∫ 𝑦−12 𝑑𝑦 = ∫ 3 𝑑𝑥

2𝑦12 = 3𝑥 + 𝐶

𝑦12 =

3

2𝑥 + 𝐶

Again C is a constant, so even though we have divided both sides of the equation by 2, it

will remain a constant. Instead of writing C, you may choose to write K instead and

replace 𝐾

2 with C.

Solving for 𝑦 we get:

𝑦 = (3

2𝑥 + 𝐶)

2

We find C by substituting the values in the initial condition:

4 = (3

2(2) + 𝐶)

2

2 = 3 + 𝐶

𝐶 = −1

So

𝑦 = (3

2𝑥 − 1)

2

294

3. 𝑑𝑥

𝑑𝑡=

𝑡𝑥

2+𝑥

∫2 + 𝑥

𝑥𝑑𝑥 = ∫ 𝑡 𝑑𝑡

∫ (2

𝑥+

𝑥

𝑥) 𝑑𝑥 =

1

2𝑡2 + 𝐶

∫ (2

𝑥+ 1) 𝑑𝑥 =

1

2𝑡2 + 𝐶

2 ln|𝑥| + 𝑥 =1

2𝑡2 + 𝐶

It would be non-trivial to solve for 𝑥 so we leave the solution as is.

4. 2𝑥 𝑑𝑦

𝑑𝑥− 𝑦 𝑑𝑥 = 0

We rewrite the equation as:

2𝑥 𝑑𝑦

𝑑𝑥= 𝑦 𝑑𝑥

We then separate variables:

2

𝑦 𝑑𝑦 =

1

𝑥 𝑑𝑥

2 ln|𝑦| = ln|𝑥| + ln 𝐶

When both sides of the solution contain a natural logarithm function, it would be better to

write ln 𝐶 instead of 𝐶.

Simplifying both sides by using the properties of logarithms we get:

ln 𝑦2 = ln 𝐶𝑥

295

So

𝑦2 = 𝐶𝑥

𝑦 = √𝐶𝑥

Applications of the method of separation of variables:

Population Growth

Example 26.2

1. The rate at which a herd of cattle grows is proportional to the number of animals present

at time t. At a certain instant the number of animals is 3000. After 10 days, the number of

animals is found to be 4000. How many animals are there after 20 days?

Let P be the number of animals at time t.

We first set up the differential equation

𝑑𝑃

𝑑𝑡= 𝑘𝑃

We solve the differential equation by separation of variables:

∫1

𝑃 𝑑𝑃 = ∫ 𝑘 𝑑𝑡

ln|𝑃| = 𝑘𝑡 + 𝐶

|𝑃| = 𝑒𝑘𝑡+𝐶

𝑃 = 𝐶𝑒𝑘𝑡

At time 𝑡 = 0, 𝑃 = 3000. So substituting in the general solution, we get

3000 = 𝐶𝑒0

𝐶 = 3000

296

So

𝑃 = 3000𝑒𝑘𝑡

When 𝑡 = 10, 𝑃 = 4000. Substituting we get

4000 = 3000𝑒10𝑘

We solve for 𝑘. To retain accuracy we will leave our solution in terms of 𝑒𝑘.

4000

3000= 𝑒10𝑘

𝑒𝑘 = (4

3)

110

Substituting this value into our formula for 𝑃 we get

𝑃 = 3000 (4

3)

𝑡10

We want to find 𝑃 when 𝑡 = 20.

𝑃 = 3000 (4

3)

2010

𝑃 ≅ 5333

2. A radioactive isotope decays at a rate proportional to the amount of the isotope present.

The half-life of the isotope is 8 days. After how much time will 10% of the isotope be

left?

Let A be the amount of the isotope present at time t. Then

𝑑𝐴

𝑑𝑡= −𝑘𝐴

Solving:

297

∫1

𝐴 𝑑𝐴 = ∫ −𝑘 𝑑𝑡

ln|𝐴| = −𝑘𝑡 + 𝐶

|𝐴| = 𝑒−𝑘𝑡+𝐶

𝐴 = 𝐶𝑒−𝑘𝑡

Suppose that at time 𝑡 = 0, 𝐴 = 𝐴0. So substituting in the general solution, we get

𝐴0 = 𝐶𝑒0

𝐶 = 𝐴0

So

𝐴 = 𝐴0𝑒−𝑘𝑡

The half-life of the isotope is 8 days. That is, when 𝑡 = 8, 𝐴 =𝐴0

2. Substituting:

𝐴0

2= 𝐴0𝑒−𝑘𝑡

1

2= 𝑒−8𝑘

𝑒−𝑘 = (1

2)

18

So

𝐴 = 𝐴0 (1

2)

𝑡8

When 𝐴 = 0.1𝐴0:

0.1𝐴0 = 𝐴0 (1

2)

𝑡8

0.1 = (0.5)𝑡8

298

ln 0.1 =𝑡

8ln 0.5

8 ln 0.1 = 𝑡 ln 0.5

𝑡 =8 ln 0.1

ln 0.5≅ 26.6 𝑑𝑎𝑦𝑠

Newton’s Law of Cooling

Recall:

Newton’s law if cooling states that the rate at which the temperature of a body changes is

proportional to the difference between the temperature of the body and the temperature of the

surroundings.

Example 26.3

1. Hot tea in a cup has an initial temperature of 170℉. The cup is placed on a table to cool

in a room maintained at a temperature of 70℉. After 20 minutes, the temperature of the

tea is 90℉. Find the temperature of the tea after 30 minutes.

Let T be the temperature of the tea at time t.

We first set up the differential equation for Newton’s law of cooling:

𝑑𝑇

𝑑𝑡= 𝑘(𝑇 − 70)

We solve the differential equation by separation of variables:

∫1

𝑇 − 70 𝑑𝑇 = ∫ 𝑘 𝑑𝑡

ln|𝑇 − 70| = 𝑘𝑡 + 𝐶

|𝑇 − 70| = 𝑒𝑘𝑡+𝐶

299

𝑇 = 70 + 𝐶𝑒𝑘𝑡

At time 𝑡 = 0, 𝑇 = 170. So substituting in the general solution, we get

170 = 70 + 𝐶𝑒0

𝐶 = 100

So

𝑇 = 70 + 100𝑒𝑘𝑡

When 𝑡 = 20, 𝑇 = 90. Substituting we get

90 = 70 + 100𝑒20𝑘

20 = 100𝑒20𝑘

We solve for 𝑘. To retain accuracy we will leave our solution in terms of 𝑒𝑘.

20

100= 𝑒20𝑘

𝑒𝑘 = (0.2)1

20

Substituting this value into our formula for 𝑇 we get

𝑇 = 70 + 100(0.2)𝑡

20

We want to find 𝑇 when 𝑡 = 30.

𝑇 = 70 + 100(0.2)3020

𝑇 ≅ 78.9 ℉

300

Mixing Problems

Example 26.3

1. A large tank contains 80 kg of sucrose dissolved in 500 liters of water. Water with a

sucrose concentration of 0.2 kg of sucrose per liter of water enters the tank at a rate of

50𝑙/𝑚𝑖𝑛. The water mixes and exits the tank at the same rate. Find an expression for the

amount of sucrose in that tank at time t.

Let S be the amount of sucrose in the tank at time t.

𝑑𝑆

𝑑𝑡= 𝑆𝑢𝑐𝑟𝑜𝑠𝑒 𝑟𝑎𝑡𝑒 𝑖𝑛 − 𝑆𝑢𝑐𝑟𝑜𝑠𝑒 𝑟𝑎𝑡𝑒 𝑜𝑢𝑡

Rate in is:

0.2 × 50 = 10

The amount of sucrose leaving the tank at any time depends on the concentration of the

sucrose in the tank at that time.

Concentration is:

𝑆

500

Rate out is:

𝑆

500 . 50 =

𝑆

10

So

𝑑𝑆

𝑑𝑡= 10 −

𝑆

10=

100 − 𝑆

10

Solving:

∫1

100 − 𝑆 𝑑𝑆 = ∫

1

10 𝑑𝑡

301

− ln|100 − 𝑆| =1

10𝑡 + 𝐶

ln|100 − 𝑆| = −1

10𝑡 + 𝐶

100 − 𝑆 = 𝐶𝑒−𝑡

10

𝑆 = 100 − 𝐶𝑒−𝑡

10

When 𝑡 = 0, 𝑆 = 80

80 = 100 − 𝐶𝑒0

20 = 𝐶

So

𝑆 = 100 − 20𝑒−𝑡

10

2. A tank contains 120 lbs. of salt dissolved in 300 gallons of water. Brine that contains 4

lbs. of salt per gallon of water enters the tank at the rate of 5 gallons per minute. The

mixture leaves the tank at the same rate. How much salt is in the tank after 60 minutes?

Let S be the amount of salt in the tank at time t.

𝑑𝑆

𝑑𝑡= 𝑟𝑎𝑡𝑒 𝑖𝑛 − 𝑟𝑎𝑡𝑒 𝑜𝑢𝑡

Rate in is:

4 × 5 = 20

The amount of salt leaving the tank at any time depends on the concentration of the salt

in the tank at that time.

Concentration is:

302

𝑆

300

Rate out is:

𝑆

300 . 5 =

𝑆

60

So

𝑑𝑆

𝑑𝑡= 20 −

𝑆

60=

1200 − 𝑆

60

Solving:

∫1

1200 − 𝑆 𝑑𝑆 = ∫

1

60 𝑑𝑡

− ln|1200 − 𝑆| =1

60𝑡 + 𝐶

ln|1200 − 𝑆| = −1

60𝑡 + 𝐶

1200 − 𝑆 = 𝐶𝑒−𝑡

60

𝑆 = 1200 − 𝐶𝑒−𝑡

60

When 𝑡 = 0, 𝑆 = 120

120 = 1200 − 𝐶𝑒0

1080 = 𝐶

So

𝑆 = 1200 − 1080𝑒−𝑡

60

When 𝑡 = 60:

𝑆 = 1200 − 1080𝑒−1 = 802.7 𝑙𝑏𝑠

303


1. A certain colony of bacteria grows at a rate proportional to the number of bacteria

present. At 𝑡 = 0, there are 10 bacteria. At 𝑡 = 4, there are 200 bacteria. At what time

will the colony have 3000 bacteria?

2. The amount of a drug present in the body decays at a rate proportional to the amount of

the drug present in the body. At time 𝑡 = 0, there is 25 mg of the drug in the body. The

half-life of the drug is 4 hours. How much of the drug is in the body after 10 hours?

3. A hard-boiled egg at 96℃ is put in a sink of 20℃ water. After 5 minutes, the egg’s

temperature is 36℃. Assuming that the water has not warmed appreciably, how much

longer will it take for the egg to reach 22℃?

4. A detective is investigating a murder. He finds the body at 6 p.m. in a room maintained at

a temperature of 68 ℉. At that time, the temperature of the body is 86 ℉. Thirty minutes

later, the temperature of the body is 85.4 ℉. Assuming that the body had a normal

temperature of 98.6 ℉, estimate the time of death.

5. A tank contains 30 kg of salt dissolved in 4000 liters of water. Brine that contains 0.02 kg

of salt per liter of water enters the tank at a rate of 25𝑙/𝑚𝑖𝑛. The solution is kept

thoroughly mixed and drains from the tank at the same rate. How much salt remains in

the tank after half an hour?

304

REFERENCES

Stewart, James. Single Variable Calculus: Early Transcendentals 7th ed. USA: Brooks/Cole,

2012.

Rogawski, Jon & Adams, Colin. Calculus 3rd ed. USA: W.H. Freeman and Company, 2015

Taalman, Laura & Kohn, Peter. Calculus. USA: W.H. Freeman and Company, 2014

Thomas, George B., Weir, Maurice & Haas, Joel. Thomas’ Calculus Early Transcendentals 11th

ed. USA: Pearson Education, Inc. 2008

www.tutorial.math.lamar.edu

www.columbia.edu/itc/sipa/math/summation.html

www.17calculus.com/infinite-series/radent-convergence

www.shmoop.com/differential-equations/slope-fields.html

www.calculuslab.deltacollege.edu/ODE/7-C-1/7-C-1-h-a.html

www.vias.org/calculus

www.cf.linnbenton.edu

www.csun.edu

http://www.tutorial.math.lamar.edu/

http://www.columbia.edu/itc/sipa/math/summation.html

http://www.17calculus.com/infinite-series/radent-convergence

http://www.shmoop.com/differential-equations/slope-fields.html

http://www.calculuslab.deltacollege.edu/ODE/7-C-1/7-C-1-h-a.html

http://www.vias.org/calculus

http://www.cf.linnbenton.edu/

http://www.csun.edu/

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