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Last Updated: September 13, 2014

1

2

TEXTBOOK :

This book is strongly recommended for Calculus 102 as well as a reference

text for subsequent courses in mathematics. The pdf soft− copy of the three

chapters remain available for free download.

TYPESETTING :

The entire document was written in LaTeX, implemented for Windows us-

ing the MiKTeX 2.9 distribution. As for the text editor of my choice, I fancy

Notepad++ 6.6.8.

MORAL SUP PO RT:

My wife with my children Saleem and Sarah have had a very instrumental

role in providing moral support. Thank God for their patience, understand-

ing, encouragement, and prayers throughout the long process of writing

and editing.

Contents 3

1 Techniques of Integration 5

1.1 Basic Integration Rules . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Trigonometric Substitutions . . . . . . . . . . . . . . . . . . . . . 23

1.5 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.6 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.7 Strategy for Integration . . . . . . . . . . . . . . . . . . . . . . . . 37

2 Infinite Series 41

2.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.2 Series and Convergence . . . . . . . . . . . . . . . . . . . . . . . . 50

2.3 The Integral Test and p − Series . . . . . . . . . . . . . . . . . . . . 58

2.4 Comparisons of Series . . . . . . . . . . . . . . . . . . . . . . . . 61

2.5 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

2.6 The Ratio and Root Tests . . . . . . . . . . . . . . . . . . . . . . . 69

2.7 Strategies for Testing Series . . . . . . . . . . . . . . . . . . . . . . 72

2.8 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

2.9 Representation of Functions by Power Series . . . . . . . . . . . 80

2.10 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . 84

3 Conics, Parametric Equations, and Polar Coordinates 89

3

4CONTENTS

3.1 Introduction to Conics . . . . . . . . . . . . . . . . . . . . . . . . 89

3.2 Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3.3 Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

3.4 Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

3.5 Plane Curves and Parametric Equations . . . . . . . . . . . . . . 102

3.6 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

3.7 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

A Indeterminate Forms and L'H ˆ

ospital's Rule 113

A.1 Indeterminate Form 0 /0 , ∞/∞ . . . . . . . . . . . . . . . . . . . 113

A.2 Indeterminate Products 0 ·±∞ . . . . . . . . . . . . . . . . . . . 115

A.3 Indeterminate Differences ∞−∞ . . . . . . . . . . . . . . . . . . 116

A.4 Indeterminate Powers 00 , ∞0 , 1∞ . . . . . . . . . . . . . . . . . . 117

B Tables and Formulas 119

CHAPTER

In Calculus 1, you studied several basic techniques for evaluating simple inte-

grals. In this chapter, you will study other integration techniques, such as in-

tegration by parts, that are used to evaluate more complicated integrals. You

will also learn how to evaluate improper integrals.

1.1 Basic Integration Rules

In this chapter, you will study several integration techniques that greatly ex-

pand the set of integrals to which the basic integration rules can be applied.

A major step in solving any integration problem is recognizing which basic

integration rule to use. As shown in Example 1.1, slight differences in the in-

tegrand can lead to very different solution techniques.

Example 1.1. Evaluate each integral

a) Z 1

x2 +1 d x b) Z x

x2 +1 d x c) Z x 2

x2 +1 d x

Solution 1.1. a) Z 1

x2 +1 d x =tan−1 x +C.

b) Z x

x2 +1 d x =1

2Z 2x

x2 +1 d x =1

2ln ¡ 1+x 2 ¢ +C.

c) Z x 2

x2 +1 d x =Zµ 1 −1

x2 +1¶ d x =x −tan−1 x +C.

5

6CHAPTER 1. TECHNIQUES OF INTEGRATION

Some times you need to use two basic rules to solve a single integral as

shown in Example 1.2.

Example 1.2. Evaluate Z 1

0

x+3

p4 −x2 d x .

Solution 1.2. Begin by writing the integral as the sum of two integrals. Then

apply the Power (Substitution) Rule and the Arcsine Rule, as follows.

Z1

0

x+3

p4 −x2 d x =Z 1

0

x

p4 −x2 d x +Z 1

0

3

p4 −x2 d x

=− 1

2Z 1

0

2x

p4 −x2 d x +3Z 1

0

1

p22 −x2 d x

=h −p 4 −x2 + 3sin −1 x

2i 1

0

=2 −p 3 +π

2

Often you need your intelligence in the appropriate substitution to solve

the integration. Consider the following three examples.

Example 1.3. Evaluate Z 1

px − 3

pxd x .

Solution 1.3. Because two different radicals appear in the problem, the sub-

stitution x= u 6 , [6 = Least Common Multiple of 2 and 3] will eliminate both,

and you have

Z1

px − 3

pxd x =Z 6u5

u3 − u2 du =6Z u 3

u−1 du

=6Z· u 2 +u+1 +1

u−1¸ du

=2u 3 +4u 2 +6u +6ln | u −1 | +C

=2 p x +43

px +66

px +6ln ¯ ¯6

px −1¯ ¯+C

Example 1.4. Find Z x 2

16 +x 6 d x .

1.1. BASIC INTEGRATION RULES 7

Solution 1.4. Because the denominator can be written in the form 16 +x 6 =

42 +¡ x 3 ¢ 2 you can try the substitution u= x 3 . Then d u = 3x 2 d x and you have

Zx 2

16 +x 6 d x = 1

3Z 3x 2

42 +¡ x 3 ¢ 2 d x = 1

3Z 1

42 +u 2 du

=1

12 tan −1 u

4+C= 1

12 tan −1 x 3

4+C

Example 1.5. Evaluate Z 3ex + 5

2ex + 7d x .

Solution 1.5. One of the methods to solve this integral is by writing the inte-

gral as the sum of two integrals as in Example 1.2. To do this, we find constants

αand βsuch that

3ex + 5=α¡ 2ex + 7¢ +βd

d x ¡ 2 e x +7¢ = 2(α+ β )ex +7α

Comparing the coefficients in both sides of the above equation yields to solve

the two equations 2(α+ β )= 3 and 7α= 5 which gives us α= 5

7and β= 11

14 .

So,

Z3ex +5

2ex + 7d x = 5

7Z 



2ex + 7





2ex + 7d x + 11

14 Z 2ex

2ex + 7d x

=5

7x+ 11

14 ln ¡ 2ex +7¢ + C

Surprisingly, two of the most commonly overlooked integration rules are

the Log Rule and the Power (Substitution) Rule. Notice in the next two exam-

ples how these two integration rules can be disguised.

Example 1.6. Find Z 1

1+ex d x .

Solution 1.6. The integral does not appear to fit any of the basic rules. How-

ever, multiply both the numerator and the denominator by e−x and then the

quotient form suggests the Log Rule as follows.

Z1

1+ex d x = Z 1

1+ex × e −x

e−x d x = − Z − e −x

e−x +1 =ln ¡ e −x +1¢ +C

Example 1.7. Evaluate Z ( cot x )[ln ( sin x )] d x .

8CHAPTER 1. TECHNIQUES OF INTEGRATION

Solution 1.7. Again, the integral does not appear to fit any of the basic rules.

However, considering the two primary choices for u [u= cot x and u= ln ( sin x )]

you can see that the second choice is the appropriate one because

du = cos x

sin xd x = cot x d x

So, Z ( cot x )[ln (sin x )] d x =Z u du = 1

2u 2 +C= 1

2[ ln ( sin x )] 2 +C

Trigonometric identities can often be used to fit integrals to one of the ba-

sic integration rules.

Example 1.8. Find Z tan2(2x ) d x .

Solution 1.8. Note that tan2 tis not in the list of basic integration rules. How-

ever, sec2 t is in the list. This suggests the trigonometric identity tan2 t= sec2 t −

1. Z tan 2 (2x) d x =Z £ sec2(2 x )− 1¤ d x = 1

2tan(2x)− x+ C

Completing the square helps when quadratic functions are involved in the

integrand. For example, the quadratic ax 2 + b x + c can be written as the dif-

ference of two squares by adding and subtracting ( b /2)2 . If the leading coeffi-

cient is not 1, it helps to factor before completing the square.

Example 1.9. Find Z 1

x2 −4 x+7 d x .

Solution 1.9. You can write the denominator as the sum of two squares, as

follows.

x2 −4 x+7 =¡ x2 −4 x+4¢ −4+7 =( x−2)2 +3

Now, in this completed square form, we have

Z1

x2 −4 x+7 d x =Z 1

(x− 2)2 + 3 d x = 1

p3 tan −1 x − 2

p3 +C

1.1. BASIC INTEGRATION RULES 9

Exercise 1.1. Evaluate each of the following integrals.

1. Z 1

(5x− 3)4 d x .

2. Z 1

px ¡ 1 −2 px ¢ d x . Hint: Let u= 1− 2p x

3. Z ln ¡ x 2 ¢

xd x .Hint: Let u= ln ¡ x 2 ¢ = 2 ln x

4. Z 6

p10x− x2 d x .Hint: Complete the square of 10x− x 2

5. Z p ex − 1 d x . Hint: Let u 2 =ex − 1

6. Z 6

0

2x+ 5

p2x+4 d x

7. Z xe−1 + ex−1

xe + ex d x

8. Z e 2x − 1

e2x +1.

9. Z 1

x10 + xd x .Hint: Multiply by x −10

x−10

10. Z 1

px +1− p xd x .Hint: Multiply by p x+1 +p x

px +1+ p x

11. Zr x− 1

x+1 d x .Hint: Multiply by r x− 1

x−1

12. Z 4

2

pln(9 −x)

pln(9 −x)+ pln(x+ 3) d x .

Hint: As x goes from 2 to 4, 9 −x and x+ 3 go from 7 to 5, and from

5 to 7, respectively. This symmetry suggests the substitution x= 6− y

reversing the interval [2,4].

10 CHAPTER 1. TECHNIQUES OF INTEGRATION

1.2 Integration by Parts

In this section you will study an important integration technique called inte-

gration by parts. This technique can be applied to a wide variety of functions

and is particularly useful for integrands involving products of algebraic and

transcendental functions. For instance, integration by parts works well with

integrals such as

Zxn ln x d x , Zxn sin−1 x d x , Zxn eax d x , and Zeax sin(b x ) d x

Integration by parts is based on the formula for the derivative of a product of

two functions f (x ) and g (x ).

Theorem 1.2.1. If u and v are functions of x and have continuous derivatives,

then Z u d v = u v −Z v du.

This formula expresses the original integral in terms of another integral.

Depending on the choices of u and v it may be easier to evaluate the sec-

ond integral than the original one. However, some authors suggest a way for

selecting the first and second function. If we denote L ogarithmic, I nverse

trigonometric, A lgebraic, T rigonometric, and E xponential functions by their

first alphabet respectively, then the first function u is selected according to the

letters of the group LIATE.

Example 1.10. Evaluate Z xe x d x .

Solution 1.10. The LIATE suggests u= x as the first option and d v = ex d x.

So,

u= x→ du = d x and d v = ex d x → v= e x

Now, integration by parts produces

Zxe x d x = xe x − Zex d x = xe x − ex + C

Example 1.11. Find Z 2x 2 ln p x d x.

1.2. INTEGRATION BY PARTS 11

Solution 1.11. First notice that

Z2x2 ln p x d x = Z2x2 ln ³ x 1

2´d x =Z x 2 ln x d x

In this case, we let

u=ln x→ du = 1

xd x and d v = x2 d x → v= x 3

3

Integration by parts produces

Z2x2 ln p x d x = Zx 2 ln x d x

=1

3x 3 ln x−Zµ x 3

3¶µ 1

x¶ d x

=1

3x 3 ln x− 1

3Z x 2 d x = 1

3x 3 ln x− 1

9x 3 +C

Example 1.12. Evaluate Z 1

0

sin−1 x d x .

Solution 1.12. Let u= sin−1 x→ d u = 1

p1 −x2 d x and d v =d x → v= x . Inte-

gration by parts now produces

Zsin −1 x d x =xsin−1 x − Zx

p1 −x2 d x

=xsin −1 x +1

2Z 2x

p1 −x2 d x

=xsin −1 x +p 1 −x2 +C

Using this anti-derivative, you can evaluate the definite integral as follows.

Z1

0

sin−1 x d x =h x sin−1 x+p 1−x 2 i 1

0=π

2− 1

Some integrals require starting by substitution method then integrate by

parts, may repeatedly.

Example 1.13. Find Z 1

3sin 3

px d x .

12 CHAPTER 1. TECHNIQUES OF INTEGRATION

Solution 1.13. First we use the substitution x= y 3 → d x = 3y 2 d y to solve this

integral, and we obtain

Z1

3sin 3

px d x = Z y 2 sin y d y

Let u= y 2 → du = 2 yd y and d v = sin y d y → v = − cos y . Integration by parts

now produces Z y 2 sin y d y =− y 2 cos y+Z 2 y cos y d y

This first use of integration by parts has succeeded in simplifying the original

integral, but the integral on the right still doesn't fit a basic integration rule. To

evaluate that integral, you can apply integration by parts again. This time, let

u=2 y→ du =2 d y and d v = cos yd y → v= sin y . Now, integration by parts

produces

Z2ycos y d y = 2y sin y− Z2sin yd y =2 y sin y+ 2 cos y+ C

Combining these two results, you can write

Z1

3sin 3

px d x = Z y 2 sin y d y

=−y2 cos y +2y sin y + 2 cos y + C

=− 3

px 2 cos 3

px +23

pxsin 3

px +2cos 3

px +C

The following example will require a technique that deserves special at-

tention.

Example 1.14. Evaluate Z ex cos x d x .

Solution 1.14. Let u= cos x→ d u = − sin xd x and d v = ex d x → v= ex . Thus,

Zex cos x d x = ex cos x+ Zex sin x d x

Since the integral Z ex sin x d x is similar in form to the original integral Z ex cos x d x

, it seems that nothing has been accomplished. However, let us integrate this

new integral by parts. We let u= sinx→ d u = cos x d x and d v = ex d x → v =

ex . Thus, Z e x sin x d x =ex sin x −Z e x cos x d x

1.2. INTEGRATION BY PARTS 13

Combining these two results, you can write

Zex cos x d x = ex cos x+ ex sin x− Zex cos x d x

which is an equation we can solve for the unknown integral. We obtain

2Z ex cos x d x = ex cos x+ ex sin x

and hence Z ex cos x d x = 1

2ex cos x+ 1

2ex sin x+ C

Example 1.15. Find Z sec3 x d x.

Solution 1.15. The most complicated portion of the integrand that can be

easily integrated is sec2 xso you should let u= sec x→ du = sec x tan xd x and

d v = sec2 x d x → v= tan x . Integration by parts produces

Zsec3 x d x = sec2 x tan x− Zsec x tan2 x d x

=sec2 x tan x −Z sec x ¡ sec2 x −1¢ d x

=sec2 x tan x −Z sec3 x d x +Z sec x d x

2Z sec3 x d x = sec2 x tan x+Z sec x d x

2Z sec3 x d x = sec2 x tan x+ ln | sec x+ tan x|+ C

Zsec3 x d x = 1

2sec 2 x tan x+ 1

2ln | sec x+ tan x|+ C

In each of the following problems, the integration by parts is a bit more

challenging.

Example 1.16. Evaluate Z ¡ sin−1 x¢ 2 d x .

14 CHAPTER 1. TECHNIQUES OF INTEGRATION

Solution 1.16. Let θ= sin−1 x . So, x= sin θ and d x = cos θd θ . Thus,

Z¡ sin −1 x ¢ 2 d x = Zθ 2 cos θd θ (let u=θ 2 and d v = cos θd θ )

=θ2 sin θ −Z 2θ sin θd θ (let u =2θ and d v =sin θd θ )

=θ2 sin θ +2θ cos θ −2Z cos θd θ

=θ2 sin θ +2θ cos θ −2sin θ +C

=x¡ sin −1 x¢ 2 +2sin −1 x

p1 +x2 −2x +C

1

x

p1 +x 2

θ

Example 1.17. Evaluate Z x 2 e x

(x+ 2)2 d x .

Solution 1.17. let u= x 2 ex and d v = 1

(x+ 2)2 d x

Zx 2 ex

(x+ 2)2 d x = − x 2 e x

x+2 + Z 



(x+ 2)ex





x+2 d x

=− x 2 e x

x+2 + Z e x d x

=− x 2 e x

x+2 + e x + C

Exercise 1.2. Evaluate each of the following integrals.

1. Z¡ x 2 −x + 1¢ ex d x . Hint: by parts, let u= x 2 −x + 1

2. Z xp x− 5 d x . Hint: by substitution, let y= x− 5

3. Z π /8

0

xsec2 x d x . Hint: by parts, let u= x

1.3. TRIGONOMETRIC INTEGRALS 15

4. Z cos ( ln x ) d x . Hint: Start by substituting y= ln x

5. Z xe x sin x d x . Hint: by parts, let u= x

6. Z ln ³ x+p x 2 + 1´ d x . Hint: by parts, let u= ln ³ x+p x 2 + 1 ´

7. Z x

1+ sin xd x . Hint: Multiply by 1− sin x

1− sin x

8. Z ln x− 1

(ln x )2 d x .

9. Z x ( 1+ ln x )2 d x . Hint: by parts, let u= ( 1+ ln x ) 2

10. Z ( ln2x ) ( ln x ) d x . Hint: ln(ab )= ln a+ ln b

11. Z ln µ x+ 1

x−1¶ d x .Hint: ln ¡ a

b¢=ln a− ln b

12. Z p x tan−1 p x d x .

1.3 Trigonometric Integrals

In this section you will study techniques for evaluating integrals of the form

Zsin m xcosn x d x and Zsec m tan n d x

where either m or n is a positive integer. To find anti-derivatives for these

forms, try to break them into combinations of trigonometric integrals to which

you can apply the Power Rule. To break up R sinm xcosn x d x into forms to

which you can apply the Power Rule, use the following identities.

sin2 θ+ cos2 θ= 1

sin2 θ= 1− cos(2θ)

2

cos2 θ= 1+ cos(2θ)

2

16 CHAPTER 1. TECHNIQUES OF INTEGRATION

Algorithm 1.1. Guidelines for Evaluating Integrals Involving Powers of Sine

and Cosine

1. If the power of the sine is odd and positive, save one sine factor and

convert the remaining factors to cosines. Then, expand and integrate.

Zsin

Odd

z}|{

2k+ 1xcos nx d x =Z

Convert to cos

z }| {

¡sin2 x ¢k cos n x

Save for du

z }| {

sin x d x

=Z¡ 1 −cos2 x ¢ kcosn x sin x d x

2. If the power of the cosine is odd and positive, save one cosine factor and

convert the remaining factors to sines. Then, expand and integrate.

Zsin m xcos

Odd

z}|{

2k+ 1x d x =Z sin mx

Convert to sin

z }| {

¡cos2 x ¢k

Save for du

z }| {

cos x d x

=Z sinm x¡ 1 −sin2 x ¢ k cos x d x

3. If the powers of both the sine and cosine are even and non-negative,

make repeated use of the identities

sin2 x= 1− cos(2x)

2and cos 2 x= 1+cos(2x)

2

to convert the integrand to odd powers of the cosine. Then proceed as

in guideline 2.

Example 1.18. Evaluate Z sin3 xcos4 x d x .

Solution 1.18. Because you expect to use the Power Rule with u= cos x , save

1.3. TRIGONOMETRIC INTEGRALS 17

one sine factor to form du and convert the remaining sine factors to cosines.

Zsin3 xcos4 x d x = Zsin2 xcos4 x sin x d x

=Z¡ 1 −cos2 x¢ cos4 x sin x d x

=Z¡ cos4 x −cos6 x¢ sin x d x Let u =cos x

=Z¡ u6 −u4 ¢ du

=1

7u 7 − 1

5u 5 +C

=1

7cos 7 x− 1

5cos 5 x+C

In the next example the power of the cosine is 3, but the power of the sine

is − 1

2.

Example 1.19. Find Z π /2

π/6

cos3 x

psin xd x .

Solution 1.19. Because you expect to use the Power Rule with u= sin x , save

one cosine factor to form du and convert the remaining cosine factors to

sines.

Zπ/2

π/6

cos3 x

psin xd x =Z π/2

π/6

cos2 x cos x

psin xd x

=Z π/2

π/6 ¡ 1−sin 2 x¢ cos x

psin xd x Let u= sin x

=Z 1

1/2

u− 1

2¡ 1−u 2 ¢du

=Z 1

1/2 ³u − 1

2−u 3

2´du = · 2u 1

2− 2

5u 5

2¸ 1

1/2 = 32 − 19p2

20

Example 1.20. Evaluate Z cos4 x d x .

18 CHAPTER 1. TECHNIQUES OF INTEGRATION

Solution 1.20. Because m and n are both even and non-negative (m= 0) you

can replace cos4 x by h 1+cos(2x)

2i 2 .

Zcos4 x d x = Z·1+cos(2x)

2¸ 2

d x

=Z·1

4+ cos(2x)

2+ cos 2 (2x)

4¸ d x

=Z·1

4+ cos(2x)

2+ 1+cos(4x)

8¸ d x

=3

8Z d x + 1

2Z cos(2x) d x + 1

8Z cos(4x) d x

=3

8x+ 1

4sin(2x)+ 1

32 sin(4x)+ C

Theorem 1.3.1. WALLIS'S FORMULAS

1. If n is odd ( n≥ 3), then

Zπ/2

0

cosn x d x =µ 2

3¶µ 4

5¶µ 6

7¶ ··· µ n−1

n¶ .

2. If n is even ( n≥ 2), then

Zπ/2

0

cosn x d x =µ 1

2¶µ 3

4¶µ 5

6¶ ··· µ n−1

n¶³ π

2´ .

These formulas are also valid if cos nx is replaced by sin n x.

Example 1.21. Evaluate Z π /2

0¡ 8cos 4 x−3sin 5 x¢ d x.

Solution 1.21. By using Wallis's Formulas, we have

Zπ/2

0¡ 8cos 4 x−3sin 5 x¢ d x =8Z π /2

0

cos4 x d x − 3Z π /2

0

sin5 x d x

=8µ 1

2¶µ 3

4¶³ π

2´ − 3 µ 2

3¶µ 4

5¶

=15π− 16

10

1.3. TRIGONOMETRIC INTEGRALS 19

The following guidelines can help you evaluate integrals of the form R secm xtann x d x .

Algorithm 1.2. Guidelines for Evaluating Integrals Involving Powers of Se-

cant and Tangant

1. If the power of the secant is even and positive, save a secant-squared

factor and convert the remaining factors to tangents. Then expand and

integrate.

Zsec

Even

z}|{

2kxtan nx d x =Z

Convert to tan

z }| {

¡sec2 x ¢k−1 tan n x

Save for du

z }| {

sec2 x d x

=Z¡ 1 +tan2 x¢ k −1 tann xsec2 x d x

2. If the power of the tangent is odd and positive, save a secant-tangent

factor and convert the remaining factors to secants. Then expand and

integrate.

Zsec m xtan

Odd

z}|{

2k+ 1x d x =Z sec m− 1x

Convert to sec

z }| {

¡tan2 x ¢k

Save for du

z }| {

sec x tan x d x

=Z secm −1 x¡ sec2 x −1¢ k sec x tan x d x

3. If there are no secant factors and the power of the tangent is even and

positive, convert a tangent-squared factor to a secant-squared factor,

then expand and repeat if necessary.

Ztan n x d x = Ztan n−2 x

Convert to sec

z }| {

¡tan2 x ¢ d x

=Z tann −2 x¡ sec2 x −1¢ d x

4. If the integral is of the form R secm x d x where m is odd and positive,

use integration by parts, as illustrated in Example 1.15 in the preceding

section.

5. If none of the above applies, try converting to sines and cosines.

20 CHAPTER 1. TECHNIQUES OF INTEGRATION

Example 1.22. Evaluate Z tan 3 x

psec xd x .

Solution 1.22. Because you expect to use the Power Rule with u= sec x , save

a factor of sec x tan x to form d u and convert the remaining tangent factors to

secants.

Ztan3 x

psec xd x =Z (sec x ) − 1

2tan 3 x d x

=Z (sec x ) − 3

2tan 2 xsec xtan x d x

=Z (sec x ) − 3

2¡ sec 2 x−1¢ sec x tan x d x Let u =sec x

=Z u − 3

2¡u 2 −1¢ du =Z ³ u 1

2−u− 3

2´du

=2

3u 3

2+2u− 1

2+C

=2

3sec 3

2x+2sec − 1

2x+C

Example 1.23. Find Z sec4(3x) tan3(3x ) d x .

Solution 1.23. Let u= tan(3x ) then d u = 3 sec2(3x ) d x and you can write

Zsec4(3x) tan3(3x ) d x = Zsec2(3x)tan3(3x)sec2(3x ) d x

=Z¡ 1 +tan2(3x )¢ tan3(3x) sec2(3x ) d x

=1

3Z¡ 1+u 2 ¢ u 3 du = 1

3Z¡ u 3 +u 5 ¢ du

=1

12 u 4 + 1

18 u 6 + C

=1

12 tan 4 (3x)+ 1

18 tan 6 (3x)+ C

Example 1.24. Evaluate Z π /4

0

tan4 x d x .

1.3. TRIGONOMETRIC INTEGRALS 21

Solution 1.24. Because there are no secant factors, you can begin by convert-

ing a tangent-squared factor to a secant-squared factor.

Ztan4 x d x = Ztan2 xtan2 x d x = Ztan2 x ¡ sec2 x−1 ¢ d x

=Z tan2 xsec2 x d x −Z tan2 x d x

=Z tan2 xsec2 x d x −Z¡ sec2 x −1¢ d x

=1

3tan 3 x−tan x+x+ C

You can evaluate the definite integral as follows.

Zπ/4

0

tan4 x d x =· 1

3tan 3 x−tan x+ x¸ π/4

0=π

4− 2

3

For integrals involving powers of cotangents and cosecants, you can fol-

low a strategy similar to that used for powers of tangents and secants. Also,

when integrating trigonometric functions, remember that it sometimes helps

to convert the entire integrand to powers of sines and cosines.

Example 1.25. Find Z sec x

tan2 xd x .

Solution 1.25. Because the guidelines do not apply, try converting the inte-

grand to sines and cosines. In this case, you are able to integrate the resulting

powers of sine and cosine as follows.

Zsec x

tan2 xd x =Zµ 1

cos x ¶µ cos 2 x

sin2 x¶ d x

=Z cos x

sin2 xd x Let u= sin x→ d u = cos x d x

=Z 1

u2 du =− 1

u+ C

=− 1

sin x+ C =− csc x+ C

Integrals involving the products of sines and cosines of two different an-

gles occur in many applications. In such instances you can use the following

product-to-sum identities.

22 CHAPTER 1. TECHNIQUES OF INTEGRATION

sin(mx) sin(nx )= 1

2{ cos [ (m− n) x ] − cos [ (m+ n) x ] }

sin(mx) cos(nx )= 1

2{ sin [ (m− n) x ] + sin [ (m+ n) x ] }

cos(mx) cos(nx )= 1

2{ cos [ (m− n) x ] + cos [ (m+ n) x ] }

Example 1.26. Find Z sin(5x) cos(4x ) d x .

Solution 1.26. Considering the second product-to-sum identity above, you

can write

Zsin(5x) cos(4x ) d x = 1

2Z (sin x+ sin(9x)) d x

=− 1

2cos x− 1

18 cos(9x)+ C

Exercise 1.3. Evaluate the following integrals.

1. Z sin5 x d x

2. Z sin5 x cos x d x

3. Z sin x tan2 x d x

4. Z x sin2 x d x

5. Z ¡ tan4 x− sec4 x¢ d x

6. Z cos(2x)

cos xd x

7. Z π /2

0

sin12 x d x

8. Z sin(− 4x )sin(3x ) d x

1.4. TRIGONOMETRIC SUBSTITUTIONS 23

1.4 Trigonometric Substitutions

Now that you can evaluate integrals involving powers of trigonometric func-

tions, you can use trigonometric substitution to evaluate integrals involving

the radicals p a 2 −x 2 ,p a 2 +x 2 and p x 2 −a 2 . The objective with trigonomet-

ric substitution is to eliminate the radical in the integrand. You do this by

using the Pythagorean identities

cos2 θ= 1− sin2 θ , sec2 θ= 1+ tan2 θ , tan2 θ= sec2 θ− 1

Note 1.1. TRIGONOMETRIC SUBSTITUTION

1. For integrals involving p a 2 −x 2, let x= a sinθ . Then

pa 2 −x2 = acos θ where −π /2 ≤θ≤π /2

pa 2 −x2

xa

θ

2. For integrals involving p a 2 +x 2, let x= a tanθ . Then

pa 2 +x2 = asec θ where −π /2 <θ<π /2

a

x

pa 2 +x2

θ

3. For integrals involving p x 2 −a 2, let x= a secθ . Then

px 2 −a2 = ½ atan θ if x> a where 0 ≤θ<π /2

−atan θ if x <−a where π /2 <θ ≤ π

24 CHAPTER 1. TECHNIQUES OF INTEGRATION

a

px 2 −a2

x

θ

The restrictions on θensure that the function that defines the substitution

is one-to-one. In fact, these are the same intervals over which the arcsine,

arctangent, and arcsecant are defined.

Example 1.27. Find Z 1

x2 p 9− x2 d x .

Solution 1.27. First, note that none of the basic integration rules applies. To

use trigonometric substitution, you should observe that p 9−x 2 is of the form

pa 2 −x2 . So, you can use the substitution x= asin θ= 3 sin θ . Using differen-

tiation and the triangle shown below, you obtain

d x = 3 cos θ dθ,p 9− x2 = 3 cos θ , x2 = 9 sin2 θ

So, trigonometric substitution yields

Z1

x2 p 9− x2 d x =Z 3 cos θ

¡9sin2 θ ¢ (3cos θ)d θ

=1

9Z 1

sin2 θd θ= 1

9Z csc 2 θdθ=− 1

9cot θ+C

=− p 9 −x 2

9x+ C

p9 −x 2

x

3

θ

Example 1.28. Find Z 1

p4x2 +1 d x .

1.4. TRIGONOMETRIC SUBSTITUTIONS 25

Solution 1.28. Let 2x= tan θ then d x = 1

2sec 2 θd x and p 4 x 2 + 1=sec θ . Trigono-

metric substitution produces

Z1

p4x2 +1 d x = 1

2Z sec 2 θ

sec θd θ= 1

2Z sec θd θ

=1

2ln |sec θ+ tan θ |+ C

=1

2ln ¯ ¯ ¯p4x2 +1+ 2x¯ ¯ ¯+C

1

2x

p1 +4x 2

θ

Example 1.29. Evaluate Z 1

¡x 2 +1 ¢3/2 d x .

Solution 1.29. Begin by writing ¡ x 2 + 1¢ 3/2 as ³ p x 2 + 1´ 3 . Then, let x= tan θ.

Using d x = sec2 θd θ and p x 2 + 1= sec θ you can apply trigonometric substi-

tution, as follows.

Z1

¡x 2 +1 ¢3/2 d x = Z 1

³p x 2 +1 ´3 d x = Z sec2 θ

sec3 θd θ

=Z 1

sec θd θ=Z cos θd θ= sin θ+ C

=x

px 2 +1 +C

1

x

px 2 +1

θ

For definite integrals, it is often convenient to determine the integration

limits for θ that avoid converting back to x.

26 CHAPTER 1. TECHNIQUES OF INTEGRATION

Example 1.30. Evaluate Z 2

p3

px 2 −3

xd x .

Solution 1.30. Because p x 2 − 3 has the form p x 2 −a 2 , you can consider x =

p3sec θ . Then d x = p3 sec θtan θd θ and p x 2 −3= p3 tan θ . To determine

the upper and lower limits of integration, use the substitution x=p 3sec θas

follows.

when x=p 3→ sec θ= 1→θ = 0

when x= 2→ sec θ= 2

p3 →θ = π

6

So, you have

Z2

p3

px 2 −3

xd x =Z π/6

0¡p3tan θ ¢¡p3sec θtan θ ¢

p3sec θd θ

=Z π/6

0

p3tan2 θd θ

=p 3Z π/6

0¡ sec 2 θ−1¢ dθ

=p 3[ tan θ −θ] π/6

0=1−p 3π

6

Exercise 1.4. Evaluate the following integrals.

1. Z xp 1+x 2 d x

2. Z 1

p49 −x2 d x

3. Z (x+ 1)p x 2 + 2x+ 2 d x

4. Z 3/5

0p 9−25x 2 d x

5. Z 1

4+ 4x 2 +x 4 d x

6. Zr 1−x

xd x

1.5. PARTIAL FRACTIONS 27

7. Zp 1−e 2x d x

8. Z cos x p 4sin2 x+ 9 d x

9. Z 1

0

ln(x+ 1)

x2 +1 d x .Hint: sin θ+ cos θ=p 2cos ¡ θ− π

4¢

1.5 Partial Fractions

This section examines a procedure for decomposing a rational function into

simpler rational functions to which you can apply the basic integration for-

mulas. This procedure is called the method of partial fractions . Its use de-

pends on the ability to factor the denominator, and to find the partial frac-

tions.

Recall from algebra that every polynomial with real coefficients can be fac-

tored into linear and irreducible quadratic factors. For instance, the polyno-

mial x 5 +x 4 −x − 1 can be written as

x5 +x4 − x−1 =( x−1)( x+1)2 ¡ x2 +1¢

where (x− 1) is a linear factor, ( x+ 1 )2 is a repeated linear factor, and ¡ x 2 + 1 ¢

is an irreducible quadratic factor. Using this factorization, you can write the

partial fraction decomposition of the rational expression as follows

P( x)

x5 + x4 − x−1 = A

x−1 + B

x+1 + C

(x+ 1)2 + Dx +E

x2 +1

where P (x ) is a polynomial of degree less than 5, and A ,B ,C ,D,E are con-

stants.

Note 1.2. Decomposition of N ( x)

D( x) Into Partial Fractions

1. Divide if improper: If N (x )/D (x ) is an improper fraction (that is, if the

degree of the numerator is greater than or equal to the degree of the

denominator), divide the denominator into the numerator to obtain

N( x)

D( x)= (a polynomial) + N ∗ ( x)

D( x)

where the degree of N∗ (x ) is less than the degree of D (x ). Then apply

Steps 2, 3, and 4 to the proper rational expression N ∗ (x )

D( x).

28 CHAPTER 1. TECHNIQUES OF INTEGRATION

2. Factor denominator: Completely factor the denominator into factors of

the form (αx+ β )m and ¡ ax 2 + bx + c ¢ n where ax 2 + b x + c is irreducible.

3. Linear factors: For each factor of the form (αx+ β )m the partial fraction

decomposition must include the following sum of m fractions.

A1

(αx+ β )+ A 2

(αx+ β )2 ···+ A m

(αx+ β ) m

4. Quadratic factors: For each factor of the form ¡ ax 2 + bx + c ¢ n the partial

fraction decomposition must include the following sum of n fractions.

B1 x+ C1

¡ax 2 +bx +c ¢+B 2 x +C2

¡ax 2 +bx + c ¢2 +···+ Bn x+ Cn

¡ax 2 +bx + c ¢n

Example 1.31. Find Z 1

x2 −5 x+6 d x .

Solution 1.31. Because x 2 − 5x+ 6= (x− 3)(x− 2) you should include one par-

tial fraction for each factor and write

1

x2 −5 x+6 = A

x−3 + B

x−2

where A and B are to be determined. Multiplying this equation by the least

common denominator (x− 3)(x− 2) yields the basic equation

1=A (x− 2) +B (x− 3)

Because this equation is to be true for all x, you can substitute any convenient

values for xto obtain equations in A and B . The most convenient values are

the ones that make particular factors equal to 0. To solve for A , let x= 3 to

obtain A= 1. To solve for B, let x= 2 to obtain B =− 1. So,

Z1

x2 −5 x+6 d x =Z·1

x−3 − 1

x−2¸ d x

=ln |x −3 |− ln |x −2 |+ C =ln ¯ ¯ ¯ ¯

x−3

x−2¯ ¯ ¯ ¯+C

Example 1.32. Evaluate Z 5x 2 + 20x+ 6

x3 +2 x2 + xd x .

1.5. PARTIAL FRACTIONS 29

Solution 1.32. Because x 3 + 2x 2 +x =x (x+ 1)2 you should include one fraction

for each power of x and x+ 1 and write

5x 2 + 20x+ 6

x3 +2 x2 + x= A

x+ B

(x+ 1) + C

(x+ 1)2

Multiplying by the least common denominator x (x+ 1)2 yields the basic equa-

tion

5x 2 + 20x+ 6=A (x+ 1)2 + B x (x+ 1) + C x

To solve for A let x= 0. This eliminates the B and C terms and yields A= 6.

To solve for C let x =− 1. This eliminates the A and B terms and yields C= 9.

The most convenient choices for xhave been used, so to find the value of B,

you can use any other value of xalong with the calculated values of A and C.

Using x= 1, A= 6, and C= 9 produces B =−1. So, it follows that

Z5x2 +20 x+6

x3 +2 x2 + xd x =Z·6

x− 1

(x+ 1) + 9

(x+ 1)2 ¸ d x

=6ln |x |− ln |x +1 |+ 9(x+1) −1

−1 + C

=ln ¯ ¯ ¯ ¯

x6

x+1¯ ¯ ¯ ¯−9

x+1 + C

When using the method of partial fractions with linear factors, a conve-

nient choice of ximmediately yields a value for one of the coefficients. With

quadratic factors, a system of linear equations usually has to be solved, re-

gardless of the choice of x.

Example 1.33. Find Z 2x 3 − 4x− 8

¡x 2 −x ¢¡x 2 +4 ¢ d x .

Solution 1.33. Because ¡ x 2 −x ¢¡ x 2 + 4¢ =x ( x− 1)¡ x 2 + 4¢ you should include

one partial fraction for each factor and write

2x 3 − 4x− 8

¡x 2 −x ¢¡x 2 +4 ¢= A

x+ B

x−1 + C x +D

x2 +4

Multiplying by the least common denominator x ( x− 1)¡ x 2 + 4¢ yields the ba-

sic equation

2x 3 − 4x− 8=A (x− 1) ¡ x 2 + 4¢ + B x ¡ x 2 + 4¢ + (C x + D )x (x− 1)

30 CHAPTER 1. TECHNIQUES OF INTEGRATION

To solve for A , let x= 0 and obtain A= 2. To solve for B , let x= 1 and obtain

B= −2. At this point, Cand Dare yet to be determined. You can find these

remaining constants by choosing two other values for x and solving the re-

sulting system of linear equations. If x = − 1, then, using A= 2 and B =− 2 you

can obtain −C +D = 2. If x= 2, you have 2C+ D= 8. Solving these two linear

equations yields C= 2 and consequently D= 4. It follows that

Z2x3 −4 x−8

¡x 2 −x ¢¡x 2 +4 ¢ d x = Z · 2

x− 2

x−1 + 2 x

x2 +4 + 4

x2 +4¸ d x

=2ln |x |− 2 ln |x −1 |+ ln ¡ x 2 +4¢ + 2tan −1 ³ x

2´ +C

An improper rational function can be integrated by performing a long di-

vision and expressing the function as the quotient plus the remainder over the

divisor. The remainder over the divisor will be a proper rational function.

Example 1.34. Find Z x 3 +x 2 − 1

x2 +1 d x .

Solution 1.34. The integrand is an improper rational function since the nu-

merator has degree 3 and the denominator has degree 2. Thus, we first per-

form the long division.

x+1

x2 +1¢ x3 +x2 −1

−x3 −x

x2 − x−1

−x2 −1

−x −2

It follows that the integrand can be expressed as

x3 +x2 −1

x2 +1 = x+1 − x+2

x2 +1

and hence

Zx 3 +x2 −1

x2 +1 d x =Z· x +1 − x+2

x2 +1¸ d x

=Z x d x +Z 1d x −1

2Z 2x

x2 +1 d x −2Z 1

x2 +1 d x

=1

2x 2 +x− 1

2ln ¡ x 2 + 1 ¢ − 2tan −1 x+C

1.5. PARTIAL FRACTIONS 31

Some times it is not necessary to use the partial fractions technique on all

rational functions like in the previous example. Also, if the integrand is not

in reduced form, reducing it may eliminate the need for partial fractions, as

shown in the following example.

Example 1.35. Evaluate Z x 2 −x− 2

x3 −2 x−4 d x .

Solution 1.35.

Zx 2 −x−2

x3 −2 x−4 d x =Z ( x+1)



(x− 2)

 



(x− 2) ¡ x 2 + 2x+ 2¢ d x

=1

2Z 2(x+ 1)

¡x 2 +2x +2 ¢d x = 1

2ln ¡ x 2 + 2x+2¢ + C

Finally, partial fractions can be used with some quotients involving tran-

scendental functions.

Example 1.36. Find Z cos x

sin x (sin x− 1) d x .

Solution 1.36. Let u= sin x→ d u = cos x d x . So,

Zcos x

sin x (sin x− 1) d x =Z 1

u( u−1) du

=Z·1

u−1 − 1

u¸ du ←By Partial Fractions

=ln |u −1 |− ln |u |+ C =ln ¯ ¯ ¯ ¯

u−1

u¯ ¯ ¯ ¯+C

=ln ¯ ¯ ¯ ¯

sin x− 1

sin x ¯ ¯ ¯ ¯+C=ln |1 −csc x|+ C

The previous example involves a rational expression of sin x and cos x.

If you are unable to find an appropriate method to solve an integral of this

forms, try using the following special substitution to convert the trigonomet-

ric expression to a standard rational expression.

Note 1.3. Substitution for Rational Functions of Sine and Cosine For inte-

grals involving rational functions of sine and cosine, the substitution

u=sin x

1+ cos x= tan ³ x

2´

32 CHAPTER 1. TECHNIQUES OF INTEGRATION

yields

cos x= 1−u 2

1+u 2 , sin x= 2u

1+u 2 , and d x = 2

1+u 2 du

Example 1.37. Find Z 1

1− sin x+ cos xd x .

Solution 1.37. The integrand is a rational function of sin x and cos x that does

not match any appropriate we have learned before, so we make the substitu-

tion u= tan(x /2). Thus, from Note 1.3 we obtain

Z1

1− sin x+ cos xd x =Z 1

1− 2u

1+u 2 + 1−u 2

1+u 2

2

1+u 2 du

=Z 2

¡1+u2 ¢−2u + ¡1+u2 ¢du

=Z 1

1−udu =− ln | 1−u |+ C=− ln | 1− tan(x/2)|+ C

Exercise 1.5. Evaluate the following integrals.

1. Z x 3 −x+ 3

x2 + x−2 d x

2. Z 2x 3 − 4x 2 − 15x+ 5

x2 −2 x−8 d x

3. Z 2x− 1

(x+ 1)3 d x . Hint: 2x− 1= 2(x+ 1) −3.

4. Z sin x

sin x+ tan xd x .

5. Z p 1−x 2

x3 d x .

6. Z 1

px ¡ 1 + 3

px ¢ 2 d x

7. Z cos x

sin2 x+ 3 sin x+ 2 d x

8. Z e 2x

(ex +1 )3 d x

1.6. IMPROPER INTEGRALS 33

9. Z x 5

(x− 1)10 (x+ 1)10 d x

10. Z 1

0

x4 (1 − x)4

1+x 2 d x

1.6 Improper Integrals

The definition of a definite integral Z b

a

f( x) d x requires that the interval [a ,b ]

be finite. Furthermore, the Fundamental Theorem of Calculus, by which you

have been evaluating definite integrals, requires that f be continuous on [ a ,b ].

In this section you will study a procedure for evaluating integrals that do

not satisfy these requirements, usually because either one or both of the limits

of integration are infinite, or fhas a finite number of infinite discontinuities

in the interval [a ,b ]. Integrals that possess either property are improper inte-

grals.

Definition 1.6.1. A function f is said to have an infinite discontinuity at cif,

from the right or left,

lim

x→ cf(x )=∞ or lim

x→ cf(x )=−∞

Definition 1.6.2. Improper Integrals with Infinite Integration Limits

1. If f is continuous on the interval [a , ∞), then

Z∞

a

f( x) d x = lim

b→∞ Z b

a

f( x) d x

2. If f is continuous on the interval (−∞ ,b ], then

Zb

−∞

f( x) d x = lim

a→−∞ Z b

a

f( x) d x

3. If f is continuous on the interval (−∞ , ∞), then

Z∞

−∞

f( x) d x =Z c

−∞

f( x) d x +Z ∞

c

f( x) d x where c any real number.

34 CHAPTER 1. TECHNIQUES OF INTEGRATION

In the first two cases, the improper integral converges if the limit exists, other-

wise, the improper integral diverges. In the third case, the improper integral

on the left diverges if either of the improper integrals on the right diverges.

Example 1.38. Evaluate Z ∞

1

1

xd x .

Solution 1.38.

Z∞

1

1

xd x =lim

b→∞ Z b

1

1

xd x =lim

b→∞ [ln x ] b

1=lim

b→∞(ln b− 0) =∞

Example 1.39. Evaluate Z ∞

0

e−x d x .

Solution 1.39.

Z∞

0

e−x d x = lim

b→∞ Z b

0

e−x d x = lim

b→∞ £−e − x ¤ b

0=lim

b→∞ ³ −e − b −1´ = 1

Example 1.40. Evaluate Z ∞

0

1

x2 +1 d x .

Solution 1.40.

Z∞

0

1

x2 +1 d x =lim

b→∞ Z b

0

1

x2 +1 d x =lim

b→∞ £tan −1 x¤ b

0=lim

b→∞ ¡tan −1 b−0¢ =π

2

Example 1.41. Evaluate Z ∞

1

(1 −x )e−x d x .

Solution 1.41. Use integration by parts, with u= 1−x and d v = e−x d x .

Z(1 −x )e−x d x = − e−x (1 −x )− Ze −xd x

=



−e −x + xe −x + 

e−x + C= xe −x +C

Now, apply the definition of an improper integral.

Z∞

1

(1 −x )e−x d x = lim

b→∞ Z b

1

(1 −x )e−x d x

=lim

b→∞ £ xe − x ¤ b

1

=µ lim

b→∞

b

eb ¶ − 1

e=− 1

e

1.6. IMPROPER INTEGRALS 35

Example 1.42. Evaluate Z ∞

−∞

ex

1+e 2xd x .

Solution 1.42. Note that the integrand is continuous on (−∞ , ∞). To evaluate

the integral, you can break it into two parts, choosing c= 0 as a convenient

value, and using the substitution u= ex .

Z∞

−∞

ex

1+e 2xd x =Z 0

−∞

ex

1+e 2xd x +Z ∞

0

ex

1+e 2xd x

=lim

b→−∞ £tan −1 ¡ e x ¢¤ 0

b+lim

b→∞ £tan −1¡ e x ¢¤ b

0

=lim

b→−∞ h π

4− tan −1 ³ eb ´i + lim

b→∞ htan −1 ³ e b ´−π

4i b

0

=π

4− 0+π

2− π

4

=π

2

The second basic type of improper integral is one that has an infinite dis-

continuity at or between the limits of integration.

Definition 1.6.3. Improper Integrals with Infinite Discontinuities

1. If f is continuous on the interval [a ,b ), and has an infinite discontinuity

at b , then Z b

a

f( x) d x = lim

c→ b− Z c

a

f( x) d x

2. If f is continuous on the interval (a ,b ], and has an infinite discontinuity

at a , then Z b

a

f( x) d x = lim

c→ a+ Z b

c

f( x) d x

3. If f is continuous on the interval [ a ,b ], except for some c∈ (a ,b ) at

which f has an infinite discontinuity, then

Zb

a

f( x) d x =Z c

a

f( x) d x +Z b

c

f( x) d x

In the first two cases, the improper integral converges if the limit exists, other-

wise, the improper integral diverges. In the third case, the improper integral

on the left diverges if either of the improper integrals on the right diverges.

36 CHAPTER 1. TECHNIQUES OF INTEGRATION

Example 1.43. Evaluate Z 1

0

1

3

pxd x .

Solution 1.43. The integrand has an infinite discontinuity at x= 0. You can

evaluate this integral as shown below.

Z1

0

1

3

pxd x =lim

b→0 + Z 1

b

1

3

pxd x =lim

b→0 + · x 2/3

2/3 ¸ 1

b=lim

b→0 +

3

2h 1− 3

pb 2 i = 3

2

Example 1.44. Evaluate Z 2

0

1

x3 d x .

Solution 1.44. Because the integrand has an infinite discontinuity at x= 0,

you can write

Z2

0

1

x3 d x =lim

b→0 + Z 2

b

1

x3 d x =lim

b→0 + · − 1

2x 2 ¸ 2

b=lim

b→0 + · − 1

8+ 1

2b 2 ¸ =∞

Example 1.45. Evaluate Z 2

−1

1

x3 d x .

Solution 1.45. This integral is improper because the integrand has an infinite

discontinuity at the interior point x= 0. So, you can write

Z2

−1

1

x3 d x =Z 0

−1

1

x3 d x +Z 2

0

1

x3 d x

From Example 1.44 you know that the second integral diverges. So, the origi-

nal improper integral also diverges.

The integral in the next example is improper for two reasons. One limit of

integration is infinite, and the integrand has an infinite discontinuity at the

outer limit of integration.

Example 1.46. Evaluate Z ∞

0

1

px(x+ 1) d x .

1.7. STRATEGY FOR INTEGRATION 37

Solution 1.46. To evaluate this integral, split it at a convenient point (say, x =

1) and write

Z∞

0

1

px(x+ 1) d x =Z 1

0

1

px(x+ 1) d x +Z ∞

1

1

px(x+ 1) d x

=lim

b→0 + Z 1

b

1

px(x+ 1) d x +lim

c→∞ Z c

1

1

px(x+ 1) d x

=lim

b→0 + £2tan −1 p x¤ 1

b+lim

c→∞ £2tan −1 p x¤ c

1

=2³ π

4´ − 0+2³ π

2´ − 2 ³ π

4´

=π

This section concludes with a useful theorem describing the convergence

or divergence of a common type of improper integral.

Theorem 1.6.1. A Special Type of Improper Integral

Z∞

1

1

xp d x =





1

p−1 if p >1

diverges if p ≤ 1

Exercise 1.6. Evaluate the following integrals.

(1) Z π /4

0

csc x d x (2) Z 1

0

1

3x− 5d x (3) Z 1

0

xln x d x (4) Z ∞

5

1

xpx2 −25 d x

1.7 Strategy for Integration

As we have seen, integration is more challenging than differentiation. In find-

ing the derivative of a function it is obvious which differentiation formula we

should apply. But it may not be obvious which technique we should use to

integrate a given function.

Until now individual techniques have been applied in each section. For

instance, we usually used substitution, integration by parts, and partial frac-

tions. But in this section we present a collection of miscellaneous integrals in

random order and the main challenge is to recognize which technique or for-

mula to use. No hard and fast rules can be given as to which method applies

38 CHAPTER 1. TECHNIQUES OF INTEGRATION

in a given situation, but we give some advice on strategy that you may find

useful.

A prerequisite for applying a strategy is a knowledge of the basic integra-

tion formulas. In the following table we have collected the integrals from our

previous list together with several additional formulas that we have learned in

this chapter. Most of them should be memorized. It is useful to know them

all, but the ones marked with an asterisk need not be memorized since they

are easily derived. Formula 19 can be avoided by using partial fractions, and

trigonometric substitutions can be used in place of Formula 20.

Once you are armed with these basic integration formulas, if you don't im-

mediately see how to attack a given integral, you might try the following four-

step strategy.

1.7. STRATEGY FOR INTEGRATION 39

1. Simplify the Integrand if Possible. Sometimes the use of algebraic ma-

nipulation or trigonometric identities will simplify the integrand and

make the method of integration obvious. For example,

Zp x ¡ 1+p x ¢ d x = Z¡ p x+x ¢ d x

and Z ( sin x+ cos x )2 d x =Z ¡ sin2 x+ 2 cos x sin x+ cos2 x¢ d x

=Z [1 +sin(2x)] d x

2. Look for an Obvious Substitution. Try to find some function u= g (x )

in the integrand whose differential du = g0 (x ) d x also occurs, apart from

a constant factor. For instance, in the integral

Zx

x2 −1 d x

we notice that if u= x 2 − 1, then du = 2 xd x . Therefore we use the sub-

stitution u= x 2 − 1 instead of the method of partial fractions.

3. Classify the Integrand According to Its Form. If Steps 1 and 2 have not

led to the solution, then we take a look at the form of the integrand f (x ).

a) Trigonometric functions. If is a product of powers of sin x and

cos x , of tan x and sec x , or of cot x and csc x , then we use the sub-

stitutions recommended in Section 1.3.

b) Rational functions. If f is a rational function, we use the proce-

dure of Section 1.5 involving partial fractions.

c) Integration by parts. If f (x ) is a product of a power of x(or a poly-

nomial) and a transcendental function (such as a trigonometric,

exponential, or logarithmic function), then we try integration by

parts, choosing u and d v according to the advice given in Section

1.2.

d) Radicals. Particular kinds of substitutions are recommended when

certain radicals appear.

i. If p ±x 2 ±a 2 occurs, we use a trigonometric substitution ac-

cording to the table in Section 1.4.

40 CHAPTER 1. TECHNIQUES OF INTEGRATION

ii. If n

pax +boccurs, we use the rationalizing substitution u=

n

pax +b. More generally, this sometimes works for n

pg(x ).

4. Try Again. If the first three steps have not produced the answer, remem-

ber that there are basically only two methods of integration: substitu-

tion and parts.

a) Try substitution. Even if no substitution is obvious (Step 2), some

inspiration or ingenuity (or even desperation) may suggest an ap-

propriate substitution.

b) Try parts. Although integration by parts is used most of the time

on products of the form described in Step 3(c), it is sometimes ef-

fective on single functions. Looking at Section 1.2, we see that it

works on sin−1 x.

c) Manipulate the integrand. Algebraic manipulations (perhaps ra-

tionalizing the denominator or using trigonometric identities) may

be useful in transforming the integral into an easier form. These

manipulations may be more substantial than in Step 1 and may

involve some ingenuity.

d) Relate the problem to previous problems. When you have built up

some experience in integration, you may be able to use a method

on a given integral that is similar to a method you have already

used on a previous integral. Or you may even be able to express

the given integral in terms of a previous one.

e) Use several methods. Sometimes two or three methods are re-

quired to evaluate an integral. The evaluation could involve several

successive substitutions of different types, or it might combine in-

tegration by parts with one or more substitutions.

Exercise 1.7. Evaluate the following integrals.

(1) Z· tan x

cos x ¸ 3

d x (2) Z ep x d x (3) Z x 5 + 1

x3 −3 x2 +10 d x

(4) Z 1

xp ln xd x (5) Zr 1+x

1−xd x

CHAPTER

Infinite series are sums of infinitely many terms. One of our aims in this chap-

ter is to define exactly what is meant by an infinite sum. Their importance in

calculus stems from Newton's idea of representing functions as sums of in-

finite series. For instance, in finding areas he often integrated a function by

first expressing it as a series and then integrating each term of the series. We

will pursue his idea in order to integrate such functions as e−x 2 , recall that

we have previously been unable to do this. Many of the functions that arise

in mathematical physics and chemistry, such as Bessel functions, are defined

as sums of series, so it is important to be familiar with the basic concepts of

convergence of infinite sequences and series.

Physicists also use series in another way. In studying fields as diverse as

optics, special relativity, and electromagnetism, they analyze phenomena by

replacing a function with the first few terms in the series that represents it.

2.1 Sequences

In mathematics, the word sequence is used in much the same way as in or-

dinary English. To say that a collection of objects or events is in sequence

usually means that the collection is ordered so that it has an identified first

member, second member, third member, and so on.

Mathematically, a sequence is defined as a function whose domain is the

set of positive integers. Although a sequence is a function, it is common to

represent sequences by subscript notation rather than by the standard func-

41

42 CHAPTER 2. INFINITE SERIES

tion notation. For example,

a1 ,a2 ,a3 ,··· , an ,···

The numbers a 1 ,a 2,a 3 , ··· are the terms of the sequence. The number an is

the n th term of the sequence, and the entire sequence is denoted by { an }or

{an }∞

n=1.

Example 2.1. Listing the first few terms of the given sequences.

a) an = 3+ (−1)n b) ½ n 2

2n − 1¾

Solution 2.1.

a) The terms of the sequence an = 3+ ( − 1)nare

n=1 , n=2 , n=2 , n=4 , ···

↓↓↓↓

2 , 4 , 2 , 4 , ···

b) The terms of the sequence ½ n 2

2n − 1¾ are

n=1 , n=2 , n=2 , n=4 , ···

↓↓↓↓

1 , 4

3, 9

7, 16

15 ,···

There are sequences that don't have a simple defining equation like the

one in the next example.

Example 2.2. Find the terms of the recursively defined Fibonacci sequence

fn where f1 =1, f2 =1, and fn =fn−1 +fn−2 for n≥3.

Solution 2.2. The terms of the sequence fn are

1,1,

| {z }

Initial Terms

2, 3, 5, 8, 13, ···

2.1. SEQUENCES 43

Example 2.3. Find a formula for the general term an of the sequence

½3

5,− 4

25 , 5

125 ,− 6

625 , 7

3125 , ··· ¾

assuming that the pattern of the first few terms continues.

Solution 2.3. Notice that the numerators of these fractions start with 3 and

increase by 1 whenever we go to the next term. The second term has numer-

ator 4, the third term has numerator 5; in general, the nth term will have nu-

merator n+ 1. The denominators are the powers of 5, so an has denominator

5n . The signs of the terms are alternately positive and negative, so we need to

multiply by a power of (− 1)n . Here we want to start with a positive term and

so we use (− 1)n−1 or (− 1)n+1 . Therefore,

an =( −1) n−1 n+2

5n

Definition 2.1.1. Definition of the Limit of a Sequence

A sequence { an } has the limit `and we write lim

n→∞ a n =`or a n →`as n → ∞ if

we can make the terms an as close to `as we like by taking n sufficiently large.

If lim

n→∞ a n exists, we say the sequence converges (or is convergent). Otherwise,

we say the sequence diverges (or is divergent).

Theorem 2.1.1. Let ` be a real number. Let f ( x) be a function of a real vari-

able such that lim

x→∞ f(x )=`. If { a n } is a sequence such that f (n )=a n for every

positive integer n then lim

n→∞ a n =`.

Example 2.4. Find the limit of the sequence whose n th term is an = µ 1+ 1

n¶ n

.

Solution 2.4. Previously, in Calculus 1, you learned that lim

x→∞ ³1+α

x´ βx

=eαβ .

So, you can apply Theorem 2.1.1 to conclude that

lim

n→∞ a n =lim

n→∞ µ1+1

n¶ n

=e

Example 2.5. Find the limit of the sequence 0, 1

2, 2

3, 3

4,···

44 CHAPTER 2. INFINITE SERIES

Solution 2.5. This is the sequence with general term an = n− 1

n=1− 1

n. Then

by Theorem 2.1.1, lim

n→∞ ·1−1

n¸ =1.

Example 2.6. Determine whether the sequence ½ n+ lnn

n2 ¾ converges or di-

verges.

Solution 2.6. Apply Theorem 2.1.1 directly on an = n+ ln n

n2 to obtain using

L'Hospital's Rule

lim

n→∞

n+ln n

n2 =lim

n→∞

1+ 1

n

2n= 0(Converges)

Example 2.7. Determine whether the sequence ½ n 2 (4n+ 1)(5n+ 3)

6n 3 + 2¾ converges

or diverges.

Solution 2.7. By Theorem 2.1.1, we obtain

lim

n→∞

n2 (4 n+A

1)(5n+ A

3)

6n 3 +A

2= lim

n→∞

n2 (4n)(5n)

6n 3

=lim

n→∞

20 



n4

6 



n3 =lim

n→∞

20n

6=∞ (Diverges)

The following properties of limits of sequences parallel those given for lim-

its of functions of a real variable in Calculus I.

Theorem 2.1.2. Properties of Limits of Sequences

Let lim

n→∞ a n =A and lim

n→∞ b n =B. Then

(1) lim

n→∞ [ a n ±b n ]=A ±B (2) lim

n→∞ [ c a n ]=c A

(3) lim

n→∞ [ a n b n ]=A B (4) lim

n→∞ · a n

bn ¸ = A

Bif b n , B6= 0

Example 2.8. Determine whether the sequence ½ 2−3e −n

6+ 4e−n ¾ converges or di-

verges.

Solution 2.8. Observe that lim

n→∞ £2−3 e − n ¤=2 and lim

n→∞ £6+4 e − n ¤=6. Ac-

2.1. SEQUENCES 45

cording to Theorem 2.1.2, we have

lim

n→∞

2− 3e−n

6+ 4e−n = 2

6= 1

3(Converges)

Theorem 2.1.3. Sequences of the Forms rn and 1

nr .

1. Suppose r is a nonzero constant. The sequence © rn ª converges to 0 if | r | <

1and diverges if | r |> 1.

2. The sequence 1

nr converges to 0for r any positive rational number.

Example 2.9. Determine the convergence or divergence of the sequence with

the given n th term.

(1) an = e −n (2) an =µ 3

2¶n

(3) an = 4

pn 5

Solution 2.9. By Theorem 2.1.3,

1. since an = e −n =µ 1

e¶ n

and r= 1

e<1, then an =e −n converges to 0.

2. since r= 3

2> 1, then the sequence an =µ 3

2¶ n

diverges.

3. the sequence an = 4

pn 5 = 4

n5

2

converges to 0 since r= 5

2is positive ra-

tional number.

Theorem 2.1.4. Squeeze Theorem for Sequences

If

lim

n→∞ a n =`=lim

n→∞ b n

and there exists an integer N such that an ≤ cn ≤ bn for all n ≥N , then

lim

n→∞ c n =`

Example 2.10. Show that the sequence n cos n

n2 o converges.

46 CHAPTER 2. INFINITE SERIES

Solution 2.10. We have,

−1

n2 ≤ cos n

n2 ≤ 1

n2

Since both ½ − 1

n2 ¾ and ½ 1

n2 ¾ tend to 0, then by Theorem 2.1.4, n cos n

n2 o con-

verges to 0.

Example 2.11. Determine whether the sequence ½ 2 n

n!¾ converges or diverges.

Solution 2.11. Even though lim

n→∞ µ2 n

n!¶ =∞

∞we can not use L'Hospital's Rule

since we have studied no function f (x )=x !. We can use Theorem 2.1.4 as

follows.

0≤ 2 n

n!=

nfactors of 2

z }| {

2·2· 2··· 2· 2· 2

1·2· 3··· (n− 2) · (n− 1) · n

| {z }

nfactors

=

nfractions

z }| {

2

1· 2

2· 2

3··· 2

n−2 · 2

n−1 · 2

n

≤2 ·1 ·2

3· 2

3··· 2

3· 2

3

| {z }

n−2fractions

=2µ 2

3¶ n−2

=9

2µ 2

3¶ n

Since lim

n→∞ · 9

2µ 2

3¶ n ¸ = 0 by Theorem 2.1.3, then by Theorem 2.1.4 the sequence

½2 n

n!¾ converges to 0.

Theorem 2.1.5. Absolute Value Theorem

For the sequence { an } , if lim

n→∞ | a n |=0then lim

n→∞ a n =0.

Example 2.12. Determine whether the sequence ½ (−1)n

pn ¾ converges or di-

verges.

Solution 2.12. Since lim

n→∞ ¯ ¯ ¯ ¯

(−1)n

pn ¯ ¯ ¯ ¯=lim

n→∞ · 1

pn ¸ =0, then by Theorem 2.1.5,

the sequence ½ ( − 1) n

pn ¾ converges to 0.

Example 2.13. Determine whether the sequence an = ( − 1)n converges or di-

verges.

2.1. SEQUENCES 47

Solution 2.13. If we write out the terms of the sequence, we obtain

{− 1,1, −1,1, ··· }

Since the terms oscillate between 1 and − 1 infinitely often, an does not ap-

proach any number. Thus lim

n→∞ £(−1) n ¤does not exist; that is, the sequence

an =( −1) n is divergent.

Theorem 2.1.6. If f ( x) is continuous and the limit lim

n→∞ a n =`exists, then

lim

n→∞ f( a n )=f³ lim

n→∞ a n ´=f(`).

Example 2.14. Find lim

n→∞ sin ³ π

n´.

Solution 2.14. Because the sine function is continuous at 0, Theorem 2.1.6

enables us to write

lim

n→∞ sin ³ π

n´ =sin ³lim

n→∞

π

n´ =sin0 =0.

Example 2.15. Show that the sequence n (1 +n ) 1

noconverges.

Solution 2.15. By Theorem 2.1.1 and Theorem 2.1.6,

lim

n→∞(1 +n)1

n=lim

n→∞ e ln h (1+ n)1

ni

=elim

n→∞ ln h (1 +n ) 1

ni=elim

n→∞

ln(1 +n )

n

=elim

n→∞

1

1+n =e 0 = 1

Theorem 2.1.7. If { an } converges to `, then lim

n→∞ a n+1 =lim

n→∞ a n =`.

Example 2.16. Assuming that the sequence defined by the recurrence relation

a1 =2, an+1 =6 + a n

2, for n= 1, 2, 3, ···

converges, show that the limit is 6.

48 CHAPTER 2. INFINITE SERIES

Solution 2.16. Since the sequence { an } converges, then lim

n→∞ a n =`exists.

Theorem 2.1.7 does not tell us what the value of the limit is. But we can use

the given recurrence relation to write

`=lim

n→∞ a n+1 =lim

n→∞ ·6+a n

2¸ = 6+`

2=⇒ `=6

The next theorem gives some limits that arise frequently.

Theorem 2.1.8. The following sequences converge to the limits listed below:

(1) lim

n→∞

ln n

n=0(2) lim

n→∞

n

pn = 1

(3) lim

n→∞ x n =0 (|x |< 1) (4) lim

n→∞ x 1

n=1, (x >0)

(5) lim

n→∞ ³1+x

n´ n

=ex ( any x ) (6) lim

n→∞

xn

n!= 0 ( any x )

Exercise 2.1.

1. Determine whether the sequence converges or diverges.

(1) 11 −2en

3en (2) cos(nπ ) (3) 2 n

3n + 1

(4) n 2/(n+1) (5) sin(nπ ) (6) µ − 1

3¶n

(7) n 3 e −n (8) sin 2 n

4n (9) sin ³ nπ

2´

(10) ln µ 4n+ 1

3n− 1¶ (11) n sin µ 6

n¶ (12) n!

nn

(13) en −e −n

en +e−n (14) (−1)n

n2 +1(15) ( −1) n 2 n 3

n2 +1

(16) µ n+ 3

n+1¶ n

(17) µ 1− 2

n¶ n

(18) p n+ 1−p n

(19) 1+( −1)n

n(20) ( n−2)!

n!(21) µ 1+ 1

n2 ¶ n

2.1. SEQUENCES 49

(22) µ 2+ 4

n2 ¶ 1

3

(23) n

n+ n

pn (24) en +3 n

5n

(25) 3−4 n

2+7· 3n (26) tan −1 µ n 2

n+1¶ (27) tan µ 2 nπ

1+ 8n¶

(28) r n+ 1

9n+ 1(29) µ 3

n¶ 1

n

(30) · 2− 1

3n ¸· 3+ 1

2n ¸

(31) 1

nZ n

1

1

xd x (32) 1

3, −1

9, 1

27 , − 1

81 , ··· (33) 2ln(3n)− ln ¡ 1+ n 2 ¢

2. The recursively defined sequence an+1 = 1

2· an + 5

an ¸ is know to con-

verge to a given initial value a 1 > 0. Find the limit of the sequence.

hHint: lim

n→∞ a n+1 =lim

n→∞ a n =`i

3. For what positive values of bdoes the following sequence converge?

b, 0, b2 , 0, b3 , 0, b4 ,···

hHint: lim

n→∞ b n =0i .

4. Evaluate lim

n→∞

n

p2 n +3n .h Hint: Show that 3 ≤ n

p2 n +3n ≤ 3 n

p2 i .

5. Give an example of a divergent sequence { an } such that {| an | } converges.

[Hint: See Exercise 2.1.1.15 ].

6. Show, by giving an example, that there exists divergent sequences { an }

and { bn } such that { an + bn } converges. · Hint: What about n 2 , 1

n− n 2 ? ¸.

7. Determine whether the sequence defined as follows is convergent or di-

vergent.

a1 =1 an+1 =4 − an for n≥1

What happens if the first term is a 1 = 2? [ Hint: Write the first few ele-

ments of the sequence ].

50 CHAPTER 2. INFINITE SERIES

2.2 Series and Convergence

The purpose of this section is to discuss sums that contain infinitely many

terms. The most familiar examples of such sums occur in the decimal repre-

sentations of real numbers. For example, when we write 1

3in the decimal form

1

3=0.3333 ··· , we mean

1

3= 0.3 + 0.03 + 0.003 + 0.0003 +···

which suggests that the decimal representation of 1

3can be viewed as a sum

of infinitely many real numbers.

One important application of infinite sequences is in representing infinite

summations. Informally, if { an } is an infinite sequence, then

∞

X

n=1

an = a1 +a2 +a3 +···

is an infinite series (or simply a series). The numbers a 1 ,a 2,a 3 , ··· are the

terms of the series. For some series it is convenient to begin the index at n= 0

(or some other integer). As a typesetting convention, it is common to repre-

sent an infinite series as simply P an . In such cases, the starting value for the

index must be taken from the context of the statement.

To find the sum of an infinite series, consider the following sequence of

partial sums.

S1 = a1

S2 = a1 +a2

S3 = a1 +a2 +a3

.

.

.

Sn = a1 +a2 +a3 +···+ an

If this sequence of partial sums converges, the series is said to converge and

has the sum indicated in the following definition.

Definition 2.2.1. Definitions of Convergent and Divergent Series

For the infinite series ∞

X

n=1

an , the nth partial sum is given by

Sn = a1 +a2 +a3 +···+ an

2.2. SERIES AND CONVERGENCE 51

If the sequence of partial sums { Sn }converges to Sthen the series ∞

X

n=1

an con-

verges to S . If { Sn } diverges, then the series diverges.

Example 2.17. Determine whether the following series converges or diverges.

If it converges, find the sum.

∞

X

n=1

(− 1)n+1 = 1−1+ 1− 1+1−1 +···

Solution 2.17. The partial sums are

S1 =1

S2 =1 −1 =0

S3 =1 −1+1 =1

Sn =1 −1+1 −1 =0

and so forth. Thus, the sequence of partial sums is

1,0,1,0,1,0, ···

Since this is a divergent sequence, the given series diverges and consequently

has no sum.

Telescoping Sums

Atelescoping series is a series whose partial sums eventually only have a fixed

number of terms after cancellation. Such a technique is also known as the

method of differences. The next example treats a convergent telescoping se-

ries, where the partial sums are particularly easy to evaluate.

Example 2.18. Determine whether the series ∞

X

n=1

1

n(1 + n)converges or di-

verges. If it converges, find the sum.

Solution 2.18. We will begin by rewriting Sn in closed form. This can be ac-

complished by using the method of partial fractions to obtain (verify)

1

n(1 + n)= 1

n− 1

1+n

52 CHAPTER 2. INFINITE SERIES

from which we obtain the sum

Sn =∞

X

n=1· 1

n− 1

1+n¸

=· 1 −1

2¸ + · 1

2− 1

3¸ + · 1

3− 1

4¸ +···+ · 1

n−1 − 1

n¸ + · 1

n− 1

1+n¸

=1 −1

1+n

Thus, ∞

X

n=1

1

n(1 + n)= lim

n→∞ S n =lim

n→∞ ·1−1

1+n¸ = 1.

Example 2.19. Determine whether the series ∞

X

n=1

ln ³ n

n+1´ converges or di-

verges. If it converges, find the sum.

Solution 2.19. We will begin by rewriting Sn in closed form by writing

ln ³ n

n+1´ =ln(n )− ln( n+1)

from which we obtain the sum

Sn =∞

X

n=1

ln ³ n

n+1´

=[ln(1) −ln(2)] +[ln(2) −ln(3)] +···+ [ ln(n ) −ln(n +1)]

=− ln(n + 1)

Thus, ∞

X

n=1

ln ³ n

n+1´ =lim

n→∞ S n =lim

n→∞ [ −ln(1 +n )] = −∞ . Hence, the sum di-

verges.

Geometric Series

In many important series, each term is obtained by multiplying the preceding

term by some fixed constant. Thus, if the initial term of the series is a and

each term is obtained by multiplying the preceding term by r , then the series

has the form

a+ ar + a r 2 +ar 3 +··· = ∞

X

n=0

ar n

Such series are called geometric series, and the number ris called the ratio

for the series.

2.2. SERIES AND CONVERGENCE 53

Theorem 2.2.1. Convergence of a Geometric Series

A geometric series with ratio r diverges if | r | ≥ 1 . If 0 < | r | < 1 then the series

converges to the sum ∞

X

n=0

ar n = a

1−r.

Example 2.20. Determine whether the series ∞

X

n=0

5

4n converges, and if so find

its sum.

Solution 2.20. This is a geometric series with a= 5 and r= 1

4. Since |r | < 1,

the series converges and the sum is a

1−r = 5

1− 1

4= 20

3.

Example 2.21. Determine whether the series ∞

X

n=0µ 5

4¶n

converges, and if so

find its sum.

Solution 2.21. This is a geometric series with a= 1 and r= 5

4. Since |r | > 1,

the series diverges.

Example 2.22. Determine whether the series ∞

X

n=1¡ 2 2 n 5 1−n¢ converges, and if

so find its sum.

Solution 2.22. This is a geometric series in concealed form, since we can

rewrite it as ∞

X

n=1 ¡2 2 n 5 1−n¢ = ∞

X

n=1

4n · 5

5n =∞

X

n=1

5µ 4

5¶ n

with a= 4 and r= 4

5. Since |r | < 1, the series converges and the sum is

a

1−r = 4

1− 4

5=20

Example 2.23. Use a geometric series to write 0.08 as the ratio of two integers.

Solution 2.23. We can write

0.08 = 0.08 + 0.0008 + 0.000008 +···

=8

100 + 8

1002 + 8

1003 +···= ∞

X

n=1

8

100n

54 CHAPTER 2. INFINITE SERIES

So the given decimal is the sum of a geometric series with a= 8

100 and r=

1

100 . Thus,

0.08 =∞

X

n=1

8

100n =

8

100

1− 1

100 = 8

99

Example 2.24. Find all values of x for which the series ∞

X

n=0

3h −x

2i n converges,

and find the sum of the series for those values of x.

Solution 2.24. This is a geometric series with a= 3 and r =− x

2. It converges

if ¯ ¯ ¯−x

2¯ ¯ ¯<1, or equivalently, when |x |< 2. When the series converges its sum is

∞

X

n=0

3h −x

2i n

=3

1+x

2= 6

2+x

The following properties are direct consequences of the corresponding

properties of limits of sequences.

Theorem 2.2.2. Properties of Infinite Series

Let P an and P bn be convergent series, and let A, B and c be real numbers.

If P an = A and P bn = B , then the following series converge to the indicated

sums.

1. P can = c A

2. P ( an ± bn ) = A± B

Example 2.25. Find the sum of the series ∞

X

n=1· 3

4n − 2

5n−1 ¸ .

Solution 2.25. The series ∞

X

n=1

3

4n is a convergent geometric series ¡ a= 3

4,r= 1

4¢,

and the series ∞

X

n=1

2

5n−1 is also a convergent geometric series ¡ a=2, r = 1

5¢.

Thus, from Theorems 2.2.2 the given series converges and

∞

X

n=1· 3

4n − 2

5n−1 ¸ =∞

X

n=1

3

4n −∞

X

n=1

2

5n−1 =

3

4

1− 1

4− 2

1− 1

5=− 3

2

2.2. SERIES AND CONVERGENCE 55

The following theorem states that if a series converges, the limit of its term

must be 0.

Theorem 2.2.3. Limit of the n th Term of a Convergent Series

If ∞

X

n=1

an converges, then lim

n→∞ a n =0.

The contrapositive of Theorem 2.2.3 provides a useful test for divergence.

This n th-Term Test for Divergence states that if the limit of the term of a series

does not converge to 0, the series must diverge.

Theorem 2.2.4. n th-Term Test for Divergence

If lim

n→∞ a n 6= 0, then ∞

X

n=1

an diverges .

Example 2.26. Determine whether the series ∞

X

n=1

n

1+n converges or diverges.

Solution 2.26. Since lim

n→∞

n

1+n = 16= 0 then the series diverges.

Example 2.27. Determine whether the series ∞

X

n=0

2n converges or diverges.

Solution 2.27. Since lim

n→∞ 2 n =∞6= 0 then the series diverges.

Example 2.28. Determine whether the series ∞

X

n=1

1

nconverges or diverges.

Solution 2.28. Since lim

n→∞

1

n=0 then the nth-Term Test for Divergence does

not apply and you can draw no conclusions about convergence or divergence.

(In the next section, you will see that this particular series diverges.)

Example 2.29. A ball is dropped from a height of 6 feet and begins bouncing,

as shown in the figure below. The height of each bounce is three-fourths the

height of the previous bounce. Find the total vertical distance travelled by the

ball.

56 CHAPTER 2. INFINITE SERIES

Solution 2.29. When the ball hits the ground for the first time, it has travelled

a distance of D 1 = 6 feet. For subsequent bounces, let Di be the distance trav-

elled up and down. For example, D 2 and D 3are as follows.

D2 =6µ 3

4¶

|{z}

up

+6µ 3

4¶

|{z}

down

=12 µ 3

4¶

D3 =6µ 3

4¶µ 3

4¶

| {z }

up

+6µ 3

4¶µ 3

4¶

| {z }

down

=12 µ 3

4¶ 2

By continuing this process, it can be determined that the total vertical dis-

tance is

D=6 +12 µ 3

4¶ + 12 µ 3

4¶ 2

+12 µ 3

4¶ 3

+···

=6 +12 ∞

X

n=1µ 3

4¶n

=6 +12 " 3

4

1− 3

4#=6+ 12 × 3= 42 feet.

Exercise 2.2.

1. Find the sum of the series if it converges.

(1) ∞

X

n=1

1

n2 +7 n+12 (2) ∞

X

n=1

(−1)n

2n−1 (3) ∞

X

n=0

5n4−n

(4) ∞

X

n=1

2n − 1

4n (5) ∞

X

n=1

10 (6) ∞

X

n=1

ln h n

3n+ 1i

(7) ∞

X

n=1

4n+2

7n−1 (8) ∞

X

n=1

2n +4n

3n +4n (9) ∞

X

n=1

n

p1 +n 2

2.2. SERIES AND CONVERGENCE 57

(10) ∞

X

n=1

tan−1 n (11) ∞

X

n=1

cos · 1

n¸ (12) ∞

X

n=0

cos ( nπ )

5n

(13) ∞

X

n=1

ln · 1

3n ¸ (14) ∞

X

n=1 ·1−1

n¸ n

(15) ∞

X

n=1

nn

n!

2. If the n th partial sum of a series ∞

X

n=1

an is Sn = n−1

n+1, find a n and ∞

X

n=1

an .

3. Let an = 2 n

3n+ 1. Determine whether the sequence { an } and the series

Pan are convergent?

4. A sequence of terms is defined by an = (5 −n )an−1 where a 1 = 1. Find

∞

X

n=1

an .

5. Show that : ∞

X

n=1

pn +1− p n

pn 2 +n =1.

6. Write the repeating decimal number 1.314 as a quotient of integers.

7. Determine the values of xfor which the series ∞

X

n=0

2n x 2n converges.

8. The accompanying figure shows the first five of a sequence of squares.

The outermost square has an area of 4 m2 . Each of the other squares is

obtained by joining the midpoints of the sides of the squares before it.

Find the sum of the areas of all the squares.

58 CHAPTER 2. INFINITE SERIES

9. Determine whether the series 1

1.1 + 1

1.11 + 1

1.111 +··· converges?



Hint: an =1

Pn

k=0³ 1

10k ´



10. Find the sum of the series 1+ 9

25 + 1+ 27

125 + 1+ 81

625 ··· .

11. Show that for all real values of x,

sin x− 1

2sin 2 x+ 1

4sin 3 x− 1

8sin 4 x+···= 2 sin x

2+ sin x

12. Find the value of c for which the series equals the indicated sum.

∞

X

n=2

(1 +c ) −n =2

2.3 The Integral Test and p − Series

In this and the following section, you will study several convergence tests that

apply to series with positive terms.

Theorem 2.3.1. The Integral Test

If f is positive, continuous, and decreasing for x ≥ N and a n = f ( n) , then

∞

X

n= N

an and Z ∞

N

f( x) d x either both converge or both diverge.

Example 2.30. Apply the Integral Test to the series ∞

X

n=1

n

n2 +1.

Solution 2.30. The function f (x )= x

x2 +1is positive and continuous for x≥1.

To determine whether fis decreasing, find the derivative.

f0 ( x)=¡ x 2 +1¢ (1) − (x )(2x)

¡x 2 +1 ¢2 =−x2 +1

¡x 2 +1 ¢2 < 0 for x> 1

2.3. THE INTEGRAL TEST AND p − SERIES 59

It follows that fsatisfies the conditions for the Integral Test. You can integrate

to obtain

Z∞

1

x

x2 +1 d x =1

2Z ∞

1

2x

x2 +1 d x

=1

2lim

b→∞ Z b

1

2x

x2 +1 d x

=1

2lim

b→∞ £ln ¡x 2 +1¢¤ b

1= 1

2lim

b→∞ £ln ¡b 2 +1¢ − ln ( 2)¤ =∞

So, the series diverges.

Example 2.31. Apply the Integral Test to the series ∞

X

n=1

1

n2 +1.

Solution 2.31. The function f (x )= 1

x2 +1is positive and continuous for x≥1.

To determine whether fis decreasing, find the derivative.

f0 ( x)=− 2 x

¡x 2 +1 ¢2 < 0 for x> 1

It follows that fsatisfies the conditions for the Integral Test. You can integrate

to obtain

Z∞

1

1

x2 +1 d x =lim

b→∞ Z b

1

1

x2 +1 d x =lim

b→∞ £tan −1 x¤ b

1

=lim

b→∞ £tan −1 b−tan −1 1¤ =π

2− π

4= π

4

So, the series converges.

Example 2.32. Determine whether the series ∞

X

n=2

1

nln nconverges or diverges.

Solution 2.32. The function f (x )= 1

xln xis positive and continuous for x≥2.

To determine whether fis decreasing, first rewrite f as f (x )= ( x ln x ) −1 and

then find its derivative.

f0 ( x)= (− 1) ( xln x)−2 (1+ ln x) =− 1 + ln x

x2 (ln x)2 < 0 for x>2

60 CHAPTER 2. INFINITE SERIES

It follows that fsatisfies the conditions for the Integral Test. You can integrate

to obtain

Z∞

2

1

xln xd x =Z ∞

2

1/x

ln xd x = lim

b→∞ [ln ( ln x )] b

2=lim

b→∞ [ln (ln b ) −ln (ln2)] =∞

So, the series diverges.

In the remainder of this section, you will investigate a second type of series

that has a simple arithmetic test for convergence or divergence. A series of the

form ∞

X

n=1

1

np = 1

1p + 1

2p + 1

3p +···

is a p −series , where pis a positive constant. For p= 1, the series

∞

X

n=1

1

n=1+ 1

2+ 1

3+ 1

4+···

is the harmonic series. A general harmonic series is of the form X 1

an + b . The

Integral Test is convenient for establishing the convergence or divergence of

p−series.

Theorem 2.3.2. Convergence of p −Series

The p − series ∞

X

n=1

1

np = 1

1p + 1

2p + 1

3p +··· converges if p >1 and diverges if 0 <

p≤1 .

Example 2.33. Determine whether the series ∞

X

n=1

1

3

pn converges or diverges.

Solution 2.33. The series ∞

X

n=1

1

3

pn = ∞

X

n=1

1

n1

3

diverges since it is a p −series with

p=1

3<1.

Exercise 2.3.

1. Determine whether the series converges or diverges.

(1) ∞

X

n=1

1

n+3(2) ∞

X

n=1

3−n (3) ∞

X

n=1

ne − n

2

2.4. COMPARISONS OF SERIES 61

(4) ∞

X

n=1

tan−1 n

n2 +1(5) ∞

X

n=1

n+2

n+1(6) ∞

X

n=2

1

np ln n

(7) ∞

X

n=1· sin n

n¸2

(8) ∞

X

n=1

2+ cos n

n(9) ∞

X

n=1

3

3

pn 5

(10) ∞

X

n=1

2

np n(11) ∞

X

n=2

pn

ln n (12) ∞

X

n=1

2

en +e−n

2. Find the sum of the series ∞

X

n=2

ln · 1− 1

n2 ¸ . £ Hint : X =− ln2¤.

3. Find all positive values of bfor which the series ∞

X

n=1

bln n converges.

hHint : b ln n = ¡ e ln b ¢ lnn = ¡ e ln n ¢ ln b =nln b i

2.4 Comparisons of Series

For the convergence tests developed so far, the terms of the series have to be

fairly simple and the series must have special characteristics in order for the

convergence tests to be applied. A slight deviation from these special charac-

teristics can make a test non-applicable. For example, in the following pairs,

the second series cannot be tested by the same convergence test as the first

series even though it is similar to the first.

∞

X

n=0

1

2n is geometric, but ∞

X

n=0

n

2n is not.

∞

X

n=1

1

n3 is a p−series, but ∞

X

n=1

1

n3 +1is not.

n

¡n 2 +3 ¢2 is easily integrated, but n 2

¡n 2 +3 ¢2 is not.

In this section you will study two additional tests for positive-term series.

These two tests greatly expand the variety of series you are able to test for con-

vergence or divergence. They allow you to compare a series having compli-

cated terms with a simpler series whose convergence or divergence is known.

62 CHAPTER 2. INFINITE SERIES

Theorem 2.4.1. Direct Comparison Test

Let 0< an ≤ bn for all n.

1. If ∞

X

n=1

bn converges, then ∞

X

n=1

an converges.

2. If ∞

X

n=1

an diverges, then ∞

X

n=1

bn diverges.

Example 2.34. Determine the convergence or divergence of ∞

X

n=1

1

2+p n 3n .

Solution 2.34. This series resembles ∞

X

n=1

1

3n which is convergent geometric

series. Term-by-term comparison yields

an =1

2+p n 3n ≤ 1

3n =bn for n≥1

So, by the Direct Comparison Test, the series converges.

Example 2.35. Determine the convergence or divergence of ∞

X

n=1

1

2+p n .

Solution 2.35. This series resembles ∞

X

n=1

1

n1

2

which is divergent p − series. Term-

by-term comparison yields

1

2+p n ≤ 1

pn for n≥ 1

which does not meet the requirements for divergence. (Remember that if

term-by-term comparison reveals a series that is smaller than a divergent se-

ries, the Direct Comparison Test tells you nothing.) Still expecting the series

to diverge, you can compare the given series with ∞

X

n=1

1

nwhich is divergent

harmonic series. In this case, term-by-term comparison yields

an =1

n≤ 1

2+p n =bn for n ≥4

and, by the Direct Comparison Test, the given series diverges.

2.4. COMPARISONS OF SERIES 63

Example 2.36. Determine the convergence or divergence of ∞

X

n=1

cos2 n

n2 +1.

Solution 2.36. This series resembles ∞

X

n=1

1

n2 which is convergent p−series.

Term-by-term comparison yields

an =cos2 n

n2 +1 ≤ 1

n2 +1 ≤ 1

n2 = b n for n≥1

So, by the Direct Comparison Test, the series converges.

Example 2.37. Determine the convergence or divergence of ∞

X

n=1

tan−1 n

pn 6 +5n 3 + 6.

Solution 2.37. This series resembles ∞

X

n=1

π/2

n3 which is convergent p−series.

Term-by-term comparison yields

an =tan −1 n

pn 6 +5n 3 +6 ≤ π /2

pn 6 ≤π /2

n3 = b n for n≥1

So, by the Direct Comparison Test, the series converges.

Example 2.38. Does ∞

X

n=1

ln n

n3/2 converge?

Solution 2.38. Because ln n grows more slowly than nc for any positive con-

stant c , we can compare the series to a convergent p − series by choosing c> 0

such that 3

2−c>1 =⇒ 0 <c < 1

2

To get the p − series, we see that

an =ln n

n3/2 ≤ n 1/4

n3/2 = 1

n5/4 = b n for n≥1

Since 1

n5/4 is a convergent p−series, then by the Direct Comparison Test, the

series ∞

X

n=1

ln n

n3/2 converges.

64 CHAPTER 2. INFINITE SERIES

Often a given series closely resembles a series or a geometric series, yet

you cannot establish the term-by-term comparison necessary to apply the Di-

rect Comparison Test. Under these circumstances you may be able to apply a

second comparison test, called the Limit Comparison Test.

Theorem 2.4.2. Limit Comparison Test

Suppose that an >0 , b n >0 , and lim

n→∞

an

bn =`where `is finite and positive.

Then the two series P an and P bn either both converge or both diverge.

Example 2.39. Show that the following general harmonic series diverges.

∞

X

n=1

1

an + b, a>0, b >0.

Solution 2.39. By comparison with the divergent harmonic series ∞

X

n=1

1

nyou

have

lim

n→∞

1/(an + b )

1/n= lim

n→∞

n

an + b= 1

a

Because this limit is greater than 0, you can conclude from the Limit Compar-

ison Test that the given series diverges.

The Limit Comparison Test works well for comparing a messy algebraic

series with a p − series. In choosing an appropriate p − series, you must choose

one with an nth term of the same magnitude as the nth term of the given

series. In other words, when choosing a series for comparison, you can disre-

gard all but the highest powers of in both the numerator and the denominator.

Example 2.40. Determine the convergence or divergence of ∞

X

n=1

pn

n2 +1.

Solution 2.40. Disregarding all but the highest powers of in the numerator

and the denominator, you can compare the series with

∞

X

n=1

pn

n2 =∞

X

n=1

1

n3/2 convergent p−series.

Because

lim

n→∞

an

bn =lim

n→∞ µ pn

n2 +1¶µ n 3/2

1¶ = lim

n→∞

n2

n2 +1 =1

you can conclude by the Limit Comparison Test that the given series con-

verges.

2.5. ALTERNATING SERIES 65

Example 2.41. Determine the convergence or divergence of ∞

X

n=1

n2 n

1+ 4n 3 .

Solution 2.41. A reasonable comparison would be with the divergent series

∞

X

n=1

2n

n2 . Note that this series diverges by the nth-Term Test. From the limit

lim

n→∞

an

bn =lim

n→∞ µ n2 n

1+ 4n 3 ¶µ n 2

2n ¶ = lim

n→∞

n3

4n 3 + 1= 1

4

you can conclude that the given series diverges.

Exercise 2.4.

1. Determine whether the series converges or diverges.

(1) ∞

X

n=3

ln n

n5 (2) ∞

X

n=2

1+ 2n

nln n(3) ∞

X

n=1

2+ sin n

3

pn 4 + 1

(4) ∞

X

n=1

pn +1− p n

n(5) ∞

X

n=1

n2 − n+2

3n 5 +n 2 (6) ∞

X

n=2

n

(4n+ 1)3/2

(7) ∞

X

n=1

sin µ 1

n¶ (8) ∞

X

n=2

1

n!(9) ∞

X

n=1

1

43

pn − 1

2. Show that if the series P an of positive terms converges, then P ln ( 1+an )

converges.

3. The meaning of decimal representation of a number 0.d 1 d 2 d 3 ··· is that

0.d 1 d 2 d 3 ···= d 1

101 + d 2

102 + d 3

103 +···

Show that this series always converges. · Hint : d i

10i ≤ 9

10i ¸ .

2.5 Alternating Series

So far, most series you have dealt with have had positive terms. In this section

and the following section, you will study series that contain both positive and

66 CHAPTER 2. INFINITE SERIES

negative terms. The simplest such series is an alternating series , whose terms

alternate in sign.

Theorem 2.5.1. Alternating Series Test

Let an >0 . The alternating series ∞

X

n=1

(− 1)n an and ∞

X

n=1

(− 1)n+1 an converge if the

following two conditions are met.

1. lim

n→∞ a n =0

2. an+1 ≤ an for all n.

It is not essential for condition (2) in Theorem 2.5.1 to hold for all terms; an

alternating series will converge if condition (1) is true and condition (2) holds

eventually. If an alternating series violates condition (1) of the alternating se-

ries test, then the series must diverge by the nth-Term Test.

Example 2.42. Determine the convergence or divergence of ∞

X

n=1

(−1)n+1 1

n.

Solution 2.42. Note that lim

n→∞

1

n=0. So, the first condition of Theorem 2.5.1

is satisfied. Also note that the second condition of Theorem 2.5.1 is satisfied

because

an+1

an = 1/( n+1)

1/n= n

n+1 ≤1 for all n =⇒ an +1 ≤an

So, applying the Alternating Series Test, you can conclude that the series con-

verges.

Example 2.43. Determine the convergence or divergence of ∞

X

n=1

n

(− 2)n−1 .

Solution 2.43. To apply the Alternating Series Test, note that, for n≥ 1,

an+1

an = ( n+1)/2n

n/2 n−1 = n+1

2n≤ 1=⇒ an +1 ≤an .

Furthermore, by L'Hopital's Rule,

lim

n→∞

n

2n−1 = lim

n→∞

1

2n−1 ln2 = 0

Therefore, by the Alternating Series Test, the series converges.

2.5. ALTERNATING SERIES 67

Example 2.44. Determine the convergence or divergence of ∞

X

n=1

(−1)n+1 2n+ 1

3n− 1.

Solution 2.44. The series diverges by the nth-Term Test since

lim

n→∞

2n+ 1

3n− 1= 2

36= 0.

Example 2.45. Determine the convergence or divergence of ∞

X

n=1

(− 1)n+1 p n

n+1.

Solution 2.45. In order to show that the terms of the series satisfy the condi-

tion an+1 ≤an , let us consider the function f (x )=p x

x+1for which a n = f(n ).

From the derivative we see that

f0 ( x)=− x−1

2p x (x+ 1)2 < 0 for x> 1

and hence, the function f (x ) decreases for x> 1. Thus, an+1 ≤an is true for

n≥1. Moreover, lim

n→∞

pn

n+1 =lim

n→∞

1

2p n= 0. Therefore, by the Alternating

Series Test, the series converges.

Example 2.46. Determine the convergence or divergence of ∞

X

n=1

cos ( πn )

n.

Solution 2.46. Note that ∞

X

n=1

cos ( πn )

n=∞

X

n=1

(− 1)n+1

nwhich is convergent as

shown in Example 2.42.

We have convergence tests for series with positive terms and for alternat-

ing series. But what if the signs of the terms switch back and forth irregularly?

Given any series P an , we can consider the corresponding series

∞

X

n=1 | a n |=| a 1 |+| a 2 |+| a 3 |+···

whose terms are the absolute values of the terms of the original series.

Definition 2.5.1. Absolute Convergence

A series P an is called absolutely convergent if the series of absolute values

P| an | is convergent.

68 CHAPTER 2. INFINITE SERIES

Example 2.47. Determine whether the series ∞

X

n=1

(−1)n−1

n2 is absolutely con-

vergent or not.

Solution 2.47. The series is absolutely convergent since ∞

X

n=1 ¯ ¯ ¯ ¯

(−1)n−1

n2 ¯ ¯ ¯ ¯=∞

X

n=1

1

n2

is a convergent p − series.

The next theorem shows that absolute convergence implies convergence.

Theorem 2.5.2. Absolute Convergence Test

If the series P | an | converges, then the series P an also converges.

The converse of Theorem 2.5.2 is not true. For instance, the alternating

harmonic series ∞

X

n=1

(−1)n+1 1

nconverges by the Alternating Series Test. Yet

the harmonic series diverges. This type of convergence is called conditional.

Definition 2.5.2. Conditional Convergence

A series P an is conditionally convergent if P an converges but P | an | diverges.

Example 2.48. Determine whether the series ∞

X

n=1

(− 1)n

pn is convergent, diver-

gent or conditionally convergent series.

Solution 2.48. The given series can be shown to be convergent by the Alter-

nating Series Test. Moreover, because

∞

X

n=1 ¯ ¯ ¯ ¯

(−1)n

pn ¯ ¯ ¯ ¯=∞

X

n=1

1

pn

is a divergent p −series, the given series is conditionally convergent.

Example 2.49. Determine whether the series ∞

X

n=1

(− 1) n (n+1)

2

3n is convergent, di-

vergent or conditionally convergent series.

Solution 2.49. This is not an alternating series. However, because

∞

X

n=1 ¯ ¯ ¯ ¯ ¯

(−1) n (n+1)

2

3n ¯ ¯ ¯ ¯ ¯=∞

X

n=1

1

3n

is a convergent geometric series, you can apply Theorem 2.5.2 to conclude

that the given series is absolutely convergent, and therefore convergent.

2.6. THE RATIO AND ROOT TESTS 69

Example 2.50. Determine whether the series ∞

X

n=1

(− 1)n n !

2n is convergent, di-

vergent or conditionally convergent series.

Solution 2.50. By the n th-Term Test for Divergence, you can conclude that

this series diverges.

Exercise 2.5.

1. Determine whether the series is convergent, divergent or conditionally

convergent.

(1) ∞

X

n=1

(− 1)n+1 n

3n+ 1(2) ∞

X

n=1· −1

e¸ n

(3) ∞

X

n=1

(− 1)n n

ln(1 +n )

(4) ∞

X

n=1

(− 1)n+1 1+p n

n+1(5) ∞

X

n=1

(− 1)n tan −1 n

n2 (6) ∞

X

n=1

1

nsin · (2 n−1)π

2¸

2. Explain why the following series converges for every positive value of x.

e−x sin( x)+ e−2x sin(2 x)+ e−3x sin(3 x)+···

·Hint : Show that ∞

X

n=1 ¯ ¯e−nx sin(n x )¯ ¯converges.¸

2.6 The Ratio and Root Tests

The comparison test and the limit comparison test hinge on first making a

guess about convergence and then finding an appropriate series for compar-

ison, both of which can be difficult. In such cases the next tests can often be

used, since it works exclusively with the terms of the given series, it requires

neither an initial guess about convergence nor the discovery of a series for

comparison.

The Ratio Test measures the rate of growth (or decline) of a series by ex-

amining the ratio an+1

an

. For a geometric series P ar n , this rate is a constant r

and the series converges if and only if its ratio is less than 1 in absolute value.

The Ratio Test is a powerful rule extending that result.

70 CHAPTER 2. INFINITE SERIES

Theorem 2.6.1. Ratio Test

Let P an be a series with non-zero terms.

1. P an converges absolutely if lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯<1.

2. P an diverges if lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯>1or lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=∞.

3. The Ratio Test is inconclusive if lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=1.

Example 2.51. Determine the convergence or divergence of ∞

X

n=1

2n

n!.

Solution 2.51. Because an = 2 n

n!, you can write the following.

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ ·2 n+1

(n+ 1)! ÷ 2 n

n!¸ = lim

n→∞ ·2 n+1

(n+ 1)! × n !

2n ¸

=lim

n→∞ ·2· 



2n

(n+ 1) 



n!× 



n!





2n ¸ = lim

n→∞ · 2

n+1¸ =0<1

∴This series converges.

Example 2.52. Determine whether the series ∞

X

n=1

n2 2 n+1

3n converges or diverges.

Solution 2.52. This series converges because

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ ·(n+ 1)2 µ2 n+2

3n+1 ¶µ 3 n

n2 2 n+1 ¶¸

=lim

n→∞

2(n+ 1)2

3n 2 = 2

3< 1

Example 2.53. Determine whether the series ∞

X

n=1

nn

n!converges or diverges.

Solution 2.53. This series diverges because

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ ·(n+1) n+1

(n+ 1)! µ n !

nn ¶¸ =lim

n→∞ · (n+ 1) n+1

n+1µ 1

nn ¶¸

=lim

n→∞

(n+ 1)n

nn =lim

n→∞ · n+1

n¸ n

=lim

n→∞ ·1+1

n¸ n

=e>1

2.6. THE RATIO AND ROOT TESTS 71

Example 2.54. Determine the convergence or divergence of ∞

X

n=1

(−1)n p n

n+1.

Solution 2.54. Here, the Ratio Test is inconclusive because

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ "Ã p n+1

n+2!µ n+1

pn ¶# =lim

n→∞ "r n+1

nµ n+1

n+2¶# =1

To determine whether the series converges, you need to try a different test. In

this case, you can apply the Alternating Series Test to show that this series is

convergent. See Example 2.45.

The next test for convergence or divergence of series works especially well

for series involving nth powers.

Theorem 2.6.2. Root Test

Let P an be a series.

1. P an converges absolutely if lim

n→∞

n

p| an |<1.

2. P an diverges if lim

n→∞

n

p| an |>1or lim

n→∞

n

p| an | =∞.

3. The Root Test is inconclusive if lim

n→∞

n

p| an |=1.

Note 2.1. The Root Test is always inconclusive for any p − series.

Example 2.55. Determine the convergence or divergence of ∞

X

n=1

e2n

nn .

Solution 2.55. The series converges, since

lim

n→∞

n

p| an |=lim

n→∞

n

se 2n

nn =lim

n→∞

e2

n=0<1

Example 2.56. Determine the convergence or divergence of ∞

X

n=1· 4n−5

2n+ 1¸n

.

Solution 2.56. The series diverges, since

lim

n→∞

n

p| an |=lim

n→∞

n

s· 4n−5

2n+ 1¸n

=lim

n→∞

4n− 5

2n+ 1= 2>1

72 CHAPTER 2. INFINITE SERIES

Exercise 2.6.

1. Determine whether the series is convergent or divergent.

(1) ∞

X

n=1· 1

ln(1 +n ) ¸ n

(2) ∞

X

n=1

1

n!(3) ∞

X

n=1

(2n)!

n5

(4) ∞

X

n=1

n· 3

4¸ n−1

(5) ∞

X

n=1 £2 n

pn +1 ¤n (6) ∞

X

n=0

(− 1)n+1

(n+ 1)2n

(7) ∞

X

n=1

n4

4n (8) ∞

X

n=1 ·1−1

n¸ n 2

(9) ∞

X

n=1

ln n

en

2. For what positive values of αdoes the series ∞

X

n=1

αn

nα converge?

3. The terms of a series are defined recursively by the equation an+1 =

5n+ 1

4n+ 3an where a 1 = 2. Determine whether P an converges or diverges?

2.7 Strategies for Testing Series

You have now studied 10 tests for determining the convergence or divergence

of an infinite series, (see the summary in the next page). Skill in choosing and

applying the various tests will come only with practice. In some instances,

more than one test is applicable. However, your objective should be to learn

to choose the most efficient test. Below is a set of guidelines for choosing an

appropriate test.

Note 2.2. Guidelines for Testing a Series for Convergence or Divergence

1. Does the nth term approach 0? If not, the series diverges.

2. Is the series one of the special types: geometric, p − series, telescoping,

or alternating?

3. Can the Integral Test, the Root Test, or the Ratio Test be applied?

4. Can the series be compared favorably to one of the special types?

2.7. STRATEGIES FOR TESTING SERIES 73

Exercise 2.7. Determine the convergence or divergence of each series.

(1) ∞

X

n=1

n+1

3n+ 1(2) ∞

X

n=1h π

6i n (3) ∞

X

n=1

ne −n2 (4) ∞

X

n=1· n+1

2n+ 1¸n

(5) ∞

X

n=1

1

1+ 3n (6) ∞

X

n=0

n!

10n (7) ∞

X

n=1

(−1)n 3

4n+ 1

74 CHAPTER 2. INFINITE SERIES

2.8 Power Series

Now that we can test many infinite series of numbers for convergence, we can

study sums that look like infinite polynomials . We call these sums power se-

ries because they are defined as infinite series of powers of some variable, in

our case x. Like polynomials, power series can be added, subtracted, multi-

plied, differentiated, and integrated to give new power series.

We begin with the formal definition, which specifies the notation and terms

used for power series.

Definition 2.8.1. Definition of Power Series

If x is a variable, then an infinite series of the form

∞

X

n=0

an xn = a0 +a1 x+ a2 x2 + a3 x3 +···

is called a power series. More generally, an infinite series of the form

∞

X

n=0

an ( x− c) n = a0 +a1 ( x− c)+ a2 ( x− c)2 + a3 ( x− c)3 +···

is called a power series centered at c , where cis a constant.

Radius and Interval of Convergence

A power series in can be viewed as a function of xas

f( x)=∞

X

n=0

an ( x− c)n

where the domain of fis the set of all xfor which the power series converges.

Determination of the domain of a power series is the primary concern in this

section. Of course, every power series converges at its center because

f( c)=∞

X

n=0

an ( c− c) n = a0 +0 +0 +···= a0

So, c always lies in the domain of f . The following important theorem states

that the domain of a power series can take three basic forms: a single point,

an interval centered at c, or the entire real line.

2.8. POWER SERIES 75

Theorem 2.8.1. Convergence of a Power Series

For a power series centered at c, precisely one of the following is true.

1. The series converges only at c.

2. There exists a real number R >0 such that the series converges absolutely

for | x− c | < R, and diverges for | x− c | > R. The series may or may not

converge at either of the endpoints x = c± R .

3. The series converges absolutely for all x.

The number R is the radius of convergence of the power series. If the series

converges only at c, the radius of convergence is R =0 , and if the series converges

for all x, the radius of convergence is R =∞. The set of all values of x for which

the power series converges is the interval of convergence of the power series.

Note that for a power series whose radius of convergence is a finite num-

ber R , Theorem 2.8.1 says nothing about the convergence at the endpoints

of the interval of convergence. Each endpoint must be tested separately for

convergence or divergence. As a result, the interval of convergence of a power

series can take any one of the six forms shown in the figure below.

The usual procedure for finding the radius and interval of convergence of

a power series is to apply the Ratio (or Root) Test for absolute convergence.

The following examples illustrates how this works.

Example 2.57. Find the interval and radius of convergence of the power series

∞

X

n=0

n! xn .

76 CHAPTER 2. INFINITE SERIES

Solution 2.57. For x= 0, you obtain f (0) =∞

X

n=0

n!0 n =1. For any fixed value of

x, let an = n! xn . Then

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ ¯ ¯ ¯ ¯

(n+ 1)!xn+1

n! xn ¯ ¯ ¯ ¯=|x|lim

n→∞(n+1) =∞

Therefore, by the Ratio Test, the series diverges for x 6= 0, and converges only

at its center, 0. So, the radius of convergence is R= 0.

Example 2.58. Find the interval and radius of convergence of the power series

∞

X

n=0

(−1)n x 2n+1

(2n+ 1)! .

Solution 2.58. Let an = x 2n+1

(2n+ 1)! . Then

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ ¯ ¯ ¯ ¯

x2n+3

(2n+ 3)! × (2n+ 1)!

x2n+1 ¯ ¯ ¯ ¯=lim

n→∞

x2

(2n+ 3)(2n+ 2) = 0

For any fixed value of x , this limit is 0. So, by the Ratio Test, the series con-

verges for all x. Therefore, the radius of convergence is R = ∞ and the interval

of convergence is (−∞ , ∞).

Example 2.59. Find the interval and radius of convergence of the power series

∞

X

n=0

(− 1)n (x+ 1)n

2n .

Solution 2.59. Letting an = (x+ 1) n

2n produces

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ ¯ ¯ ¯ ¯

(x+ 1)n+1

2n+1 × 2 n

(x+ 1)n ¯ ¯ ¯ ¯=lim

n→∞ ¯ ¯ ¯ ¯

x+1

2¯ ¯ ¯ ¯= ¯ ¯ ¯ ¯

x+1

2¯ ¯ ¯ ¯

By the Ratio Test, the series converges if ¯ ¯ ¯ ¯

x+1

2¯ ¯ ¯ ¯<1 or |x + 1 | < 2. So, the ra-

dius of convergence is R= 2. Because the series is centered at x = − 1, it will

converge in the interval (− 3,1). Furthermore, at the endpoints you have

when x =− 3 =⇒ ∞

X

n=0

(− 1)n (− 2)n

2n =∞

X

n=0

1 Diverges.

when x= 1 =⇒ ∞

X

n=0

(− 1)n 2 n

2n =∞

X

n=0

(− 1)nDiverges.

both of which diverge. So, the interval of convergence is (− 3,1)

2.8. POWER SERIES 77

Example 2.60. Find the interval and radius of convergence of the power series

∞

X

n=1

(x− 5)n

n2 .

Solution 2.60. Letting an = (x− 5) n

n2 produces

lim

n→∞ ¯ ¯ ¯ ¯

an+1

an ¯ ¯ ¯ ¯=lim

n→∞ ¯ ¯ ¯ ¯

(x− 5)n+1

(n+ 1)2 × n 2

(x− 5)n ¯ ¯ ¯ ¯=|x−5 |lim

n→∞ h n

n+1i 2

=| x −5 |

By the Ratio Test, the series converges if | x− 5|< 1. So, the radius of conver-

gence is R= 1. Because the series is centered at x= 5, it will converge in the

interval (4,6). Furthermore, at the endpoints you have

when x= 4 =⇒ ∞

X

n=1

(− 1)n

n2 Converges.

when x= 6 =⇒ ∞

X

n=1

1

n2 Converges.

both of which converge. So, the interval of convergence is [4,6]

Example 2.61. Find the interval and radius of convergence of the power series

∞

X

n=1

nn

2n xn .

Solution 2.61. Letting an = n n

2n xn produces

lim

n→∞

n

p| an |=lim

n→∞ ¯ ¯ ¯ ¯

nn

2n xn ¯ ¯ ¯ ¯

1

n

=|x | lim

n→∞

n

2=∞ for x6= 0

So, the series only converges only it x= 0, and the radius of converges is R= 0.

Differentiation and Integration of Power Series

Power series representation of functions has played an important role in the

development of calculus. In fact, much of Newton's work with differentia-

tion and integration was done in the context of power series, especially his

work with complicated algebraic functions and transcendental functions. Eu-

ler, Lagrange, Leibniz, and the Bernoulli all used power series extensively in

calculus.

78 CHAPTER 2. INFINITE SERIES

Once you have defined a function with a power series, it is natural to won-

der how you can determine the characteristics of the function. Is it continu-

ous? Differentiable? Theorem 2.8.2 answers these questions.

Theorem 2.8.2. Properties of Functions Defined by Power Series

If the function f given by

f( x)=∞

X

n=0

an ( x− c)n

=a0 +a1 ( x −c) +a2 ( x −c)2 +a3 ( x −c)3 +a4 ( x −c)4 +···

has a radius of convergence of R >0 , then, on the interval ( c− R, c+ R) , f

is differentiable (and therefore continuous). Moreover, the derivative and an-

tiderivative of f are as follows.

1. f 0 ( x)=∞

X

n=1

nan ( x− c ) n−1 = a1 +2 a2 ( x− c)+ 3 a3 ( x− c )2 +··· .

2. Z f( x) d x = C+∞

X

n=0

an

(x− c )n+1

n+1 = C+a0 ( x −c) +a1

(x− c )2

2+··· .

The radius of convergence of the series obtained by differentiating or integrating

a power series is the same as that of the original power series. The interval of

convergence, however, may differ as a result of the behaviour at the endpoints.

Example 2.62. Consider the function given by

f( x)=∞

X

n=1

xn

n= x+ x 2

2+ x 3

3+ x 4

4+···

Find the interval of convergence for each of the following.

(1) Z f (x ) d x (2) f (x ) (3) f0 (x )

Solution 2.62. By Theorem 2.8.2, you have

f0 ( x)=∞

X

n=1

xn−1 =1 + x+ x2 + x3 +···

2.8. POWER SERIES 79

and Z f (x) d x = C+∞

X

n=1

xn+1

n( n+1) = C+ x 2

1· 2+ x 3

2· 3+ x 4

3· 4···

By the Ratio Test, you can show that each series has a radius of convergence

of R= 1. Considering the interval (−1, 1), you have the following.

(1) For Z f (x ) d x , the series ∞

X

n=1

xn+1

n( n+1) converges for x=±1, and its inter-

val of convergence is [− 1,1].

(2) For f (x ), the series ∞

X

n=1

xn

nconverges for x= −1 and diverges for x = 1.

So, its interval of convergence is [− 1,1).

(3) For f0 (x ), the series ∞

X

n=1

xn−1 diverges for x=±1, and its interval of con-

vergence is (− 1,1).

Exercise 2.8.

1. Find the radius and interval of convergence of the series.

(1) ∞

X

n=1

xn

2n− 1(2) ∞

X

n=0

xn

n!(3) ∞

X

n=0

(x− 2)n

n2 +1

(4) ∞

X

n=1

n!(2 x−1) n (5) ∞

X

n=1

n

4n (x+1) n (6) ∞

X

n=1

3n (x+ 4)n

pn

2. Find all values of x∈ [0,2π ] for which ∞

X

n=1· 2

p3 ¸ n

sinn xconverges.

·Hint : The answer is h 0, π

3´ ∪ µ 2π

3, 4π

3¶ ∪ µ 5π

3,2π¸¸

3. Give an example of a power series that converges for x∈ [2, 6).

·Hint : For example, ∞

X

n=1

(x− 4)n

n2 n ¸

4. Let f (x )=∞

X

n=1

xn

n2 . Find the interval of convergence of : f,f0 , and f00 .

80 CHAPTER 2. INFINITE SERIES

2.9 Representation of Functions by Power Series

In this section we learn how to represent certain types of functions as sums

of power series by manipulating geometric series or by differentiating or inte-

grating such a series. You might wonder why we would ever want to express

a known function as a sum of infinitely many terms. We will see later that

this strategy is useful for integrating functions that don't have elementary an-

tiderivatives, for solving differential equations, and for approximating func-

tions by polynomials.

We start with the geometric power series (a= 1, r= x )

∞

X

n=0

xn =1 + x+ x2 +x3 +···= 1

1−x for |x |< 1 (2.9.1)

Example 2.63. Find a power series for f (x )= 1

1+x 2 , centered at 0.

Solution 2.63. Replacing x by −x 2 in Equation 2.9.1, we have

1

1+x 2 = 1

1−¡ −x 2 ¢ =∞

X

n=0 ¡ −x 2 ¢ n = ∞

X

n=0

(−1 )n x 2n

Because this is a geometric series, it converges when ¯ ¯−x2 ¯ ¯<1, that is, x 2 <1,

or |x |< 1. Therefore the interval of convergence is (− 1,1).

Example 2.64. Find a power series for f (x )= 4

2+x , centered at 0.

Solution 2.64. In order to put this function in the form of the left side of Equa-

tion 2.9.1, we first factor a 2 from the denominator:

4

2+x = 4

2h 1+x

2i = 2

1−h −x

2i =∞

X

n=0

2h −x

2in

=∞

X

n=0

(−1)n x n

2n−1

This power series converges when ¯ ¯ ¯−x

2¯ ¯ ¯<1 which implies that the interval of

convergence is (− 2,2).

Example 2.65. Find a power series for f (x )= 1

x, centered at 1.

2.9. REPRESENTATION OF FUNCTIONS BY POWER SERIES 81

Solution 2.65. In order to put this function in the form of the left side of Equa-

tion 2.9.1, we first add and subtract 1 from the denominator:

1

x= 1

1− 1+x = 1

1− (1 −x )= ∞

X

n=0

[−(x − 1) ]n =∞

X

n=0

(− 1)n (x− 1)n

This power series converges when |x − 1 |< 1 which implies that the interval of

convergence is (0,2).

Example 2.66. Find a power series for f (x )= 4x 3

2+x , centered at 0.

Solution 2.66. Since this function is just x 3times the function in Example

2.64, all we have to do is to multiply that series by x 3:

4x 3

2+x =x 3 × 4

2+x =x 3 ×∞

X

n=0

(−1)n x n

2n−1 = ∞

X

n=0

(− 1)n xn+3

2n−1

As in Example 2.64, the interval of convergence is (− 2,2).

The versatility of geometric power series will be shown later in this section,

following a discussion of power series operations. These operations, used

with differentiation and integration, provide a means of developing power se-

ries for a variety of elementary functions. For simplicity, the following prop-

erties are stated for a series centered at 0.

Theorem 2.9.1. Operations with Power Series

Let f ( x)=∞

X

n=0

an xn and g ( x)=∞

X

n=0

bn xn .

1. f (mx )=∞

X

n=0

an mn xn .

2. f ¡ xm ¢ =∞

X

n=0

an xmn .

3. f ( x)± g( x)=∞

X

n=0

(an ± bn )xn .

The operations described above can change the interval of convergence

for the resulting series. For example, in the following addition, the interval of

82 CHAPTER 2. INFINITE SERIES

convergence for the sum is the intersection of the intervals of convergence of

the two original series.

∞

X

n=0

xn

| {z }

(− 1,1)

+∞

X

n=0³ x

2´n

| {z }

(−2,2)

=∞

X

n=0 µ1+1

2n ¶ xn

| {z }

(−1,1)∩(−2,2)=(−1,1)

Example 2.67. Find a power series, centered at 0, for f (x )= 3x− 1

x2 −1.

Solution 2.67. Using partial fractions, you can write as f (x )= 2

x+1 + 1

x−1.

By adding the two geometric power series

2

x+1 = 2

1− (−x )= 2 ∞

X

n=0

(−1)n xn where |x |< 1

and 1

x−1 = −1

1−x =− ∞

X

n=0

xn where | x|< 1

you obtain the power series 3x− 1

x2 −1 = ∞

X

n=0 £2(−1) n −1¤ x n , where the interval of

convergence for this power series is (− 1,1).

Example 2.68. Find a power series for f (x )= ln x , centered at 1.

Solution 2.68. Since Z 1

xd x =ln x +C, and from Example 2.65, you know that

1

x=∞

X

n=0

(− 1)n (x− 1)n , then

ln x= C+Z 1

xd x =C +Z· ∞

X

n=0

(− 1)n (x− 1)n ¸ d x

=C +∞

X

n=0

(− 1)n (x− 1) n+1

n+1

By letting x= 1, you can conclude that C= 0. Therefore,

ln x=∞

X

n=0

(− 1)n (x− 1) n+1

n+1

Example 2.69. Find a power series for g (x )= tan−1 x , centered at 0.

2.9. REPRESENTATION OF FUNCTIONS BY POWER SERIES 83

Solution 2.69. Because d

d x £ tan −1 x ¤ = 1

1+x 2 , and from Example 2.63 you

know that 1

1+x 2 =∞

X

n=0

(−1 )n x 2n , you obtain

tan−1 x= C+Z 1

1+x 2 d x =C +Z· ∞

X

n=0

(−1 )n x 2n ¸ d x

=C +∞

X

n=0

(−1 )n x 2n+1

2n+ 1

By letting x= 0, you can conclude that C= 0. Therefore,

tan−1 x=∞

X

n=0

(−1 )n x 2n+1

2n+ 1

Exercise 2.9.

1. Find a power series for the function, centered at c, and determine the

interval of convergence.

a) f (x )= 1

(1 −x )2 at c= 0. · Hint : 1

(1 −x )2 = d

d x µ 1

1−x¶¸ ..

b) f (x )= ln(1 +x ) at c= 0. · Hint : ln(1 +x )=Z 1

1+xd x ¸ ..

c) f (x )= 1

3−x at c= 1. " Hint : 1

3−x = 1/2

1−¡ x−1

2¢# ..

d) f (x )= 4 x

x2 +2 x−3at c=0. · Hint : 4 x

x2 +2 x−3 = 1

x−1 + 3

x+3¸ ..

e) f (x )= 2

(1 +x )3 at c= 0. · Hint : 2

(1 +x )3 = d 2

d x 2 µ 1

1+x¶¸ ..

f) f (x )= ln ¡ 1−x 2 ¢ at c= 0. £ Hint : ln ¡ 1−x 2 ¢ = ln ( 1−x ) + ln ( 1+x ) ¤.

.

g) f (x )= x

(1 + 4x )2 at c= 0.

h) f (x )=³ x

2−x´ 3 at c= 0.

2. Suppose that the series P an xn has radius of convergence 2, and the se-

ries P bn xn has radius of convergence 3. What is the radius of conver-

gence of the series P ( an + bn ) xn ?

84 CHAPTER 2. INFINITE SERIES

3. Suppose that the radius of convergence of the power series P an xn is R.

What is the radius of convergence of the power series P an x 2n ?

4. Evaluate ∞

X

n=1

n

2n . · Hint : Show that 1

(1 −x )2 = ∞

X

n=1

nxn−1 ¸ .

2.10 Taylor and Maclaurin Series

We saw in the previous section that functions such as f (x )= tan−1 x can be

represented as power series. These power series give us a certain tangible in-

sight into the function represented and they allow us to approximate the val-

ues of f (x ) to any desired degree of accuracy. Thus, it is desirable to develop

general methods for finding power series representations.

Theorem 2.10.1. The Form of a Convergent Power Series

If f is represented by a power series f ( x)=∞

X

n=0

an ( x− c) n for all x in an open

interval I containing c , then a n = f (n) ( c )

n! and f (x )=∞

X

n=0

f(n) ( c)

n!( x−c) n .

Definition 2.10.1. Definition of Taylor and Maclaurin Series

If a function f has derivatives of all orders at x= c , then the series f (x ) =

∞

X

n=0

f(n) ( c)

n!( x−c) n is called the Taylor series for f (x ) at c . Moreover, if c= 0,

then the series is the Maclaurin series for f.

If you know the pattern for the coefficients of the Taylor polynomials for a

function, you can extend the pattern easily to form the corresponding Taylor

series.

Example 2.70. Find the Maclaurin series for f (x )=ex .

Solution 2.70. The n th derivative f (x ) is f (n) (x )=ex for all nand thus

f(0) = f0 (0) = f00 (0) = ···= e0 =1.

Therefore, the coefficients of the Maclaurin series are an = f (n) (0)

n!= 1

n!and

the Maclaurin series is f (x )=∞

X

n=0

xn

n!.

2.10. TAYLOR AND MACLAURIN SERIES 85

Example 2.71. Find the Maclaurin series for f (x )= sin x.

Solution 2.71. For f (x )= sin x , we have

f(2n) ( x)= (− 1) n sin xand f(2n+ 1) ( x)= (− 1) n cos x

Therefore, f (2n) (0) = 0 and f (2n+ 1) (0) = (−1)n . Hence, the coefficients of the

Maclaurin series are a 2n+1 = ( − 1) n

(2n+ 1)! and the Maclaurin series is

f( x)=∞

X

n=0

(− 1)n

(2n+ 1)! x 2n+1

Since a Taylor series is a power series, we may differentiate and integrate

a Taylor series term by term within its interval of convergence. We may also

multiply two Taylor series or substitute one Taylor series into another. This

leads to shortcuts for generating new Taylor series from known ones. The fol-

lowing list provides the Maclaurin series for several elementary functions with

the corresponding intervals of convergence.

86 CHAPTER 2. INFINITE SERIES

Example 2.72. Find the Maclaurin series for f (x )=x 2 ex .

Solution 2.72. We obtain the Maclaurin series of f (x ) by multiplying the known

Maclaurin series for ex by x 2:

x2 ex = x2∞

X

n=0

xn

n!=∞

X

n=0

xn+2

n!=∞

X

n=2

xn

(n− 2)!

Example 2.73. Find the Maclaurin series for f (x )= cos p x.

Solution 2.73. Using the power series cos x=∞

X

n=0

(−1)n

(2n )! x 2n you can replace

xby p xto obtain the series cos p x=∞

X

n=0

(− 1)n

(2n )! xn . This series converges for

x≥0.

Example 2.74. Find the Maclaurin series for f (x )= sin2 x.

Solution 2.74. Write sin2 xas

sin2 x= 1

2− 1

2cos(2x)

and then, use the Maclaurin series for cos xas follows.

sin2 x= 1

2− 1

2cos(2x)= 1

2− 1

2

∞

X

n=0

(−1)n

(2n )! (2x)2n

=1

2−∞

X

n=0

(− 1)n

(2n )! 2 2n−1 x 2n

Example 2.75. Find Taylor series for f (x )= ln x at c= 1.

Solution 2.75. We begin by letting t= x− 1 that transforms f to ln(t+ 1), and

now looking for the Maclaurin series at c= 0. So,

f( x)= ln x=ln( t+1) =∞

X

n=1

(−1)n−1 t n

n=∞

X

n=1

(−1)n−1 (x− 1) n

n

Example 2.76. Evaluate Z e−x 2 d x as an infinite series.

2.10. TAYLOR AND MACLAURIN SERIES 87

Solution 2.76. the Maclaurin series for e−x 2 is ∞

X

n=0

(−1)n x 2n

n!. Now we integrate

to obtain

Ze −x2 d x =C + Z·∞

X

n=0

(− 1)n x 2n

n!¸ d x =C +∞

X

n=0

(− 1)n x 2n+1

(2n+ 1)n!

Example 2.77. If f (x )= sin ¡ x 3 ¢ , find f (15)(0).

Solution 2.77. It would be far too much to compute 15 derivatives of f . The

key idea is to remember that f (n) (0) occurs in the coefficient of xn in the

Maclaurin series of f . Since

f( x)= sin ¡ x3 ¢ =∞

X

n=0

(− 1)n x 6n+3

(2n+ 1)! =x 3 − x 9

3! + x 15

5! −···

then the coefficient of x 15 is f (15) (0)

15! = 1

5! , and hence

f(15) (0) = 15!

5! = 10897286400

Example 2.78. Find the sum of the series ∞

X

n=1

(− 1)n−1 3 n

n5 n .

Solution 2.78. Remember that ln(1 +x )=∞

X

n=1

(−1)n−1 x n

n. So,

∞

X

n=1

(− 1)n−1 3 n

n5 n =∞

X

n=1

(− 1)n−1 ¡ 3

5¢ n

n=ln µ 1+ 3

5¶ = ln µ 8

5¶

Exercise 2.10.

1. Find Maclaurin series for the given function.

(1) sin x cos x (2) 1

1+ 5x (3) cos 2 x

(4) ln · 1+x

1−x¸ (5) x 2 e −3x

88 CHAPTER 2. INFINITE SERIES

2. Find Taylor series for the given function at the indicated value of c.

(1) 1

1+x , (c= 4) (2) sin x , (c=π /2) (3) ln(1 +x ) , (c= 2)

(4) ex , (c= 1) (5) p x , (c= 1)

3. Find f (10) (0) for f (x )=x 4 sin ¡ x 2 ¢.

4. Does f (x )= cot x possess a Maclaurin series representation ?

5. Use Maclaurin series to evaluate lim

x→0

1+x −ex

1− cos x .

6. Find the sum of the series ∞

X

n=0

(−1)n π 2n

36n(2n)!

7. Find the interval of convergence of ∞

X

n=1

n3 xn and find its sum.

8. Find the sum of the series

1+ 1

2+ 1

3+ 1

4+ 1

6+ 1

8+ 1

9+ 1

12 +···

where the terms are reciprocals of the positive integers whose only prime

factors are 2's and 3's.

CHAPTER

In this chapter, you will analyze and write equations of conics using their

properties. You will also learn how to write parametric equations and polar

equations. In addition to the rectangular equations of conics, you will also

study polar equations of conics.

3.1 Introduction to Conics

Each conic section (or simply conic) can be described as the intersection of a

plane and a double-napped cone. Notice in Figure 3.1 that for the four basic

conics, the intersecting plane does not pass through the vertex of the cone.

Figure 3.1:

89

90

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

There are several ways to study conics. You could begin as the Greeks did

by defining the conics in terms of the intersections of planes and cones, or you

could define them algebraically in terms of the general second-degree equa-

tion

Ax2 + B x y + C y 2 + Dx + E y + F= 0

The discriminant B 2 − 4 AC determines the shape of the conic section.

CONIC DISCRIMINANT

Parabola B 2 − 4 AC = 0

Ellipse B 2 − 4 AC < 0

Hyperbola B 2 − 4 AC > 0

Example 3.1. Determine what type of conic corresponds to the equations

x2

4+y 2 = 1.

Solution 3.1. Note that the discriminant B 2 − 4 AC = 02 − 4µ 1

4¶ (1) = −1 < 0.

Since the discriminant is negative, the equation above is that of an ellipse.

However, a third approach, in which each of the conics is defined as a lo-

cus (collection) of points satisfying a certain geometric property, works best.

The following table summarizes the geometric definition for the four conics:

circle, ellipse, parabola, and hyperbola.

CONIC GEOMETRIC DEFINITION: THE SET OF ALL POINTS

Circle equidistant to a point

Parabola equidistant to both a line and a point

Ellipse the sum of whose distances to two fixed points is constant

Hyperbola the difference of whose distances to two fixed points is a

constant

Exercise 3.1. Identify the conic section as a parabola, ellipse, circle, or hyper-

bola.

1. x 2 + x y − y 2 + 2x =−3

2. x 2 + x y + y 2 + 2x =−3

3.2. PARABOLAS 91

3. 5x 2 + 20 y 2 =25

4. x 2 −y − 1= 0

3.2 Parabolas

Aparabola is the set of all points (x ,y ) that are equidistant from a fixed line

called the directrix and a fixed point called the focus not on the line. The

midpoint between the focus and the directrix is the vertex, and the line pass-

ing through the focus and the vertex is the axis of the parabola. Note in Figure

3.2 that a parabola is symmetric with respect to its axis.

Figure 3.2:

Note 3.1. EQUATION OF A PARABOLA WITH VERTEX AT THE POINT (h ,k )

Example 3.2. Find the focus and directrix of a parabola whose equation is

y2 =8 x.

92

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

Solution 3.2. Comparing this equation with (y− k )2 = 4p (x− h ), you can con-

clude that h= 0, k= 0, and 4p= 8→p = 2> 0. The focus is (2, 0) and the

directrix is x =−2.

Example 3.3. Find the standard form of the equation of a parabola whose

focus is at the point ¡ 0, 1

2¢ and whose directrix is y= − 1

2.

Solution 3.3. The midpoint of the segment joining the focus and the directrix

along the axis of symmetry is the vertex. So,

vertex = midpoint between µ 0, 1

2¶ and µ 0, − 1

2¶ = Ã 0+0

2,

1

2+ −1

2

2! = (0,0)

A parabola with vertex at (0,0), focus at ¡ 0, 1

2¢ , and directrix y= − 1

2corre-

sponds to the equation x 2 = 4 p y . Since the focus ¡ 0, 1

2¢=(0, p ) then p= 1

2.

Hence, the parabola is x 2 = 2y .

Example 3.4. Find the vertex, focus, and directrix of a parabola whose equa-

tion is y 2 − 6y− 2x+ 8= 0.

3.2. PARABOLAS 93

Solution 3.4. First, convert to standard form by completing the square.

y2 −6 y−2 x+8 =0

y2 −6 y=2 x−8

y2 −6 y+9 =2 x−8 +9

(y− 3)2 = 2µ x+ 1

2¶

Compare with (y− k )2 = 4p (x− h ) we have: vertex = (h ,k )= ¡ − 1

2,3¢ , 4p= 2 →

p=1

2>0 so the parabola opens to the right, focus = (p+ h ,k )= (0, 3) and

directrix x =− p +h =−1.

Example 3.5. Find the equation of a parabola whose vertex is located at the

point (− 1,1) and whose focus is located at the point ¡ − 1, 1

2¢.

Solution 3.5. Draw a Cartesian plane and label the vertex and focus. The ver-

tex and focus share the same axis of symmetry x =−1, and indicate a parabola

opening downward. The standard equation of a parabola opening downward

with vertex (−1,1) is (x+ 1)2 = 4p (y− 1). Since the focus is ¡ − 1, 1

2¢=(−1,1 +p )

then p= 1

2−1=− 1

2. Hence, the parabola is (x+ 1) 2 =−2(y− 1).

94

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

Example 3.6. A satellite dish is 24 feet in diameter at its opening and 4 feet

deep in its center. Where should the receiver be placed?

Solution 3.6. Draw a parabola with a vertex at the origin representing the cen-

ter cross section of the satellite dish. The standard equation of a parabola

opening upward with vertex at (0,0) is x 2 = 4 py where p> 0. Since the point

(12,4) lies on the parabola, substitute it into the equation to obtain p= 9.

Hence, the focus is (h ,p+ k )= (0,9), i.e., the receiver should be placed 9 feet

from the vertex of the dish.

Exercise 3.2.

1. Find an equation for the parabola whose vertex at (2,4) and its directrix

equation is y= 9.

2. Find an equation for the parabola whose vertex at (2,4) and focus at

(0,4).

3. Find the focus, vertex, and directrix of the parabola x 2 − 6x− 4y+ 10 =0.

4. A bridge with a parabolic shape has an opening 80 feet wide at the base

(where the bridge meets the water), and the height in the center of the

bridge is 20 feet. A sailboat whose mast reaches 17 feet above the water

is travelling under the bridge 10 feet from the center of the bridge. Will

it clear the bridge without scraping its mast? Justify your answer.

3.3. ELLIPSES 95

5. If one parabolic segment of a suspension bridge is 300 feet and if the

cables at the vertex are suspended 10 feet above the bridge, whereas the

height of the cables 150 feet from the vertex reaches 60 feet, find the

equation of the parabolic path of the suspension cables.

6. Find two parabolas with focus (1,2) and vertices (1,1) and (1, − 1) that

intersect each other.

3.3 Ellipses

An ellipse is the set of all points in a plane the sum of whose distances from

two fixed points is constant. These two fixed points are called foci (plural of

focus). A line segment through the foci called the major axis intersects the

ellipse at the vertices. The midpoint of the line segment joining the vertices

is called the center. The line segment that intersects the center and joins two

points on the ellipse and is perpendicular to the major axis is called the minor

axis.

96

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

Note 3.2. Equation of an Ellipse with Center at the Point (h ,k )

Example 3.7. Describe the ellipse given by x 2

25 + y 2

9= 1.

Solution 3.7. The center of this ellipse is the origin (0,0). Since 25 > 9 the

major axis is horizontal. Also, a 2 = 25 →a = 5, b 2 = 9, and c=p a 2 −b 2 =

p16 = 4. Hence,

vertices are :(h± a ,k )= (0 ± 5,0) = (−5,0) and (5,0)

foci are :(h± c ,k )= (0 ± 4,0) = (−4,0) and (4,0)

Example 3.8. Find the center, vertices, and foci of the ellipse given by

9y 2 + 18y+ 16 x 2 −64x− 71 =0

3.3. ELLIPSES 97

Solution 3.8. By completing the square, you can write the original equation

in standard form.

9y 2 + 18y+ 16 x 2 −64x− 71 =0

9¡ y 2 + 2y¢ + 16 ¡ x 2 − 4x¢ = 71

9¡ y 2 + 2y+ 1¢ + 16 ¡ x 2 − 4x+ 4¢ = 71 + 9+ 64

9¡ y+ 1¢ 2 + 16 ( x− 2)2 = 144

(x−2 )2

9+ ¡ y+1¢ 2

16 = 1

The center of this ellipse is (2, − 1). Since 16 > 9 the major axis is vertical. Also,

a2 =16 → a=4, b2 =9, and c=p a2 − b2 =p 7. Hence,

vertices are :(h ,k± a )= (2, − 1± 4) = (2, − 5) and (2,3)

foci are :(h ,k± c )= (2, − 1±p 7) = (2, − 1−p 7) and (2, − 1+p 7)

Example 3.9. Find the standard form of the equation of an ellipse with foci at

(− 3,0) and (3,0) and vertices (−4,0) and (4,0).

Solution 3.9. The major axis lies along the x-axis, since it contains the foci and

vertices. The distance between the foci is 3 − ( − 3) = 6= 2c→c= 3. Also, the

distance between the vertices is 4 − ( − 4) = 2a= 8→a = 4. Since c 2 =a 2 −b 2 →

b2 =16 −9 =7. The center is the midpoint between the foci as equal as the

midpoint of the vertices. So, the center = µ − 4+ 4

2, 0+0

2¶ = (0,0). Hence, the

ellipse is

x2

16 + y 2

7= 1

98

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

Exercise 3.3.

1. Find the standard form of the equation of an ellipse with foci (−2,5) and

(6,5), vertices (−3,5) and (7,5).

2. Find the standard form of the equation of an ellipse with vertical major

axis of length 8, minor axis of length 4, and centered at (3,2).

3. Find the center, vertices, and foci of the ellipse given by

y2 +6 y+5 x2 +20 x=21

4. Find the equation of an ellipse centered at (1,− 2) that passes through

the points (1, −4) and (2, −2).

5. As illustrated in the accompanying figure, the tank of an oil truck is 18

ft long and has elliptical cross sections that are 6 ft wide and 4 ft high.

Show that the volume V of oil in the tank (in cubic feet) when it is filled

to a depth of h feet is

V=27 ·4sin −1 µ h−2

2¶ + (h−2)p 4 h− h 2 + 2π¸

3.4 Hyperbolas

The definition of a hyperbola is similar to the definition of an ellipse. A hy-

perbola is the set of all points in a plane the difference of whose distances

from two fixed points is a positive constant. These two fixed points are called

3.4. HYPERBOLAS 99

foci. The hyperbola has two separate curves called branches. The two points

where the hyperbola intersects the line joining the foci are called vertices . The

line segment joining the vertices is called the transverse axis of the hyperbola.

The midpoint of the transverse axis is called the center.

Note 3.3. Equation of a Hyperbola with Center at the Point (h ,k )

100

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

An important aid in sketching the graph of a hyperbola is the determina-

tion of its asymptotes. Each hyperbola has two asymptotes that intersect at

the center of the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions 2aby 2bwith its center at (h ,k ).

To determine whether the transverse axis is parallel to the x −axis or the

y−axis from the standard equation of a hyperbola, determine the location of

the minus sign in the equation. The transverse axis is parallel to the x −axis

when the minus sign precedes the y 2 −term, and it is parallel to the y −axis

when the minus sign precedes the x 2 −term. Note that a 2always precedes the

minus sign while b 2follows it.

Example 3.10. Find the center, vertices, foci and asymptotes of the hyperbola

given by

9x 2 − 18x− 16 y 2 +32y− 151 =0

Solution 3.10. By completing the square, you can write the original equation

in standard form.

9x 2 − 18x− 16 y 2 +32y− 151 =0

9¡ x 2 − 2x¢ − 16 ¡ y 2 − 2y¢ = 151

9¡ x 2 − 2x+ 1¢ − 16 ¡ y 2 − 2y+ 1¢ = 151 + 9− 16

9( x− 1)2 − 16 ¡ y− 1¢ 2 = 144

(x−1 )2

16 − ¡ y−1 ¢ 2

9= 1

The center of this hyperbola is (1,1). Since the minus sign precedes the y 2 − term,

then the transverse axis is parallel to the x − axis. Also, a 2 = 16 →a = 4, b 2 =9,

and c=p a 2 +b 2 =p 25 = 5. Hence,

vertices are :(h± a ,k )= (1 ± 4,1) = (−3,1) and (5,1)

foci are :(h± c ,k )= (1 ± 5,1) = (−4,1) and (6,1)

asymptotes are :y= 3

4x+ 1

4and y=− 3

4x+ 7

4

3.4. HYPERBOLAS 101

Example 3.11. Find the standard form of an equation of the hyperbola with

center (0,0), y −axis is the transverse axis, and asymptotes y= 2x and y =− 2x .

Solution 3.11. From the asymptotes we have a

b=2→a =2b . Hence, the

equation of the hyperbola has the form

y2

a2 − x 2

b2 =1→ y 2

4b 2 − x 2

b2 =1→ y 2

4−x 2 =b 2

Exercise 3.4.

1. Find the standard form of the equation of a hyperbola with vertices

(− 2,5) and (6,5), and foci (−3,5) and (7,5).

2. Analyze the hyperbola x 2 − 6x− 4y 2 − 16y− 8= 0

3. Find the standard form of the equation of a hyperbola whose graph is

given below.

102

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

3.5 Plane Curves and Parametric Equations

Until now, you have been representing a graph by a single equation involving

variables. In this section you will study situations in which variables are used

to represent a curve in the plane.

The equation y = −x 2 /72 +x , for example, does not tell the whole story. Al-

though it does tell you where the object has been, it doesn't tell you when the

object was at a given point (x ,y ). To determine this time, you can introduce a

third variable t called a parameter . By writing both x and y as functions of t

you obtain parametric equations.

Definition 3.5.1. Definition of a Plane Curve

If f and g are continuous functions of on an interval I, then the equations

x= f( t) and y= g( t)

are called parametric equations and tis called the parameter. The set of

points (x ,y ) obtained as tvaries over the interval Iis called the graph of

the parametric equations. Taken together, the parametric equations and the

graph are called a plane curve, denoted by C.

Example 3.12. Sketch the curve described by the parametric equations

x= t2 −4 and y= t

2;−2 ≤t ≤3.

Solution 3.12. For values of ton the given interval, the parametric equations

yield the points (x ,y ) shown in the table. By plotting these points in order

t−2−1 0 1 2 3

x0− 3−4− 3 0 5

y−1 − 1

20 1

21 3

2

of increasing tand using the continuity of f and g , you obtain the curve C

shown in the figure. Note that the arrows on the curve indicate its orientation

as t increases from − 2 to 3.

3.5. PLANE CURVES AND PARAMETRIC EQUATIONS 103

It often happens that two different sets of parametric equations have the

same graph. For example, the set of parametric equations

x=4 t2 −4 and y= t;− 1≤ t≤3

2.

But when comparing the values of tin these sets, you can see that the second

graph is traced out more rapidly (considering tas time) than the first graph.

So, in applications, different parametric representations can be used to repre-

sent various speeds at which objects travel along a given path.

Definition 3.5.2. Eliminating the Parameter

Finding a rectangular equation that represents the graph of a set of parametric

equations is called eliminating the parameter.

The range of x and y implied by the parametric equations may be altered

by the change to rectangular form. In such instances the domain of the rect-

angular equation must be adjusted so that its graph matches the graph of the

parametric equations.

Example 3.13. Obtain the standard form of the rectangular equation for the

curve represented by the equations

x=1

pt +1and y= t

t+1; t>−1.

104

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

Solution 3.13. Begin by solving one of the parametric equations for t . For

instance, you can solve the first equation for as follows.

x=1

pt +1 →x2 =1

t+1 → t=1

x2 −1= 1−x 2

x2

Now, substituting into the parametric equation for y produces

y= t

t+1 = h 1−x2

x2 i

h1−x2

x2 i+1= 1−x 2

1−x 2 +x 2 = 1−x 2

∴The rectangular equation, y= 1−x 2 , is defined for all values of x> 0 since

t>−1.

Example 3.14. Obtain the standard form of the rectangular equation for the

curve represented by the equations

x=3cos θ and y=4sin θ ; 0 ≤θ≤2π .

Solution 3.14. Begin by solving for sin θand cos θin the given equations.

cos θ= x

3and sin θ=y

4

Next, make use of the identity sin2 θ+ cos2 θ= 1 to form an equation involving

only x and y.

sin2 θ+ cos2 θ= 1→h y

4i 2

+h x

3i 2

=1 →x 2

9+ y 2

16 = 1

From this rectangular equation you can see that the graph is an ellipse cen-

tered at (0,0) with vertices at (0,4) and (0, −4) and minor axis of length 2b= 6.

Definition 3.5.3. Definition of a Smooth Curve

A curve C represented by x= f (t ) and y= g (t ) on an interval I is called

smooth if f0 and g0 are continuous on Iand not simultaneously 0, except

possibly at the endpoints of I. The curve is called piecewise smooth if it is

smooth on each subinterval of some partition of I.

3.6. POLAR COORDINATES 105

Exercise 3.5. Obtain the standard form of the rectangular equation for the

curve represented by the given parametric equations.

1. x= t 3 ,y= t 2

2

2. x= e−t ,y= e 2t − 1

3. x= tan θ ,y= sec2 θ

4. x= 4+ 2cos θ ,y =− 1+ sin θ

3.6 Polar Coordinates

So far, you have been representing graphs as collections of points (x ,y ) on the

rectangular coordinate system. The corresponding equations for these graphs

have been in either rectangular or parametric form. In this section you will

study a coordinate system called the polar coordinate system.

To form the polar coordinate system in the plane, fix a point O , called the

pole (or origin), and construct from O an initial ray called the polar axis, as

shown in the figure. Then each point Pin the plane can be assigned polar

coordinates (r ,θ ) as follows.

r=directed distance from Oto P.

θ=directed angle, counter-clockwise from polar axis to segment OP .

106

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

If P is the pole, then r= 0, but there is no clearly defined polar angle. We

will agree that an arbitrary angle can be used in this case; that is, (0, θ) are

polar coordinates of the pole for all choices of θ.

In the figure below, the points (6, π /4), (5, 2π/3), (3,5π /4), and (4, 11π/6)

are plotted in polar coordinate systems.

The polar coordinates of a point are not unique. For example, the polar

coordinates (1,7π /4), (1, −π /4), and (1,15π /4) all represent the same point,

see the next figure. In general, if a point Phas polar coordinates (r ,θ ), then

(r ,θ± 2nπ ) are also polar coordinates of Pfor any nonnegative integer n . Thus,

every point has infinitely many pairs of polar coordinates.

As defined above, the radial coordinate rof a point P is nonnegative, since

it represents the distance from Pto the pole. However, it will be convenient

to allow for negative values of r as well. The minus sign serving to indicate

that the point is on the extension of the angle's terminal side rather than on

the terminal side itself, see the figure below. In general, the terminal side of

the angle θ+ π is the extension of the terminal side of θ, so we define negative

radial coordinates by agreeing that (−r ,θ ) and (r ,θ+ π ) are polar coordinates

of the same point.

3.6. POLAR COORDINATES 107

Frequently, it will be useful to superimpose a rectangular x y − coordinate

system on top of a polar coordinate system, making the positive x −axis co-

incide with the polar axis. If this is done, then every point Pwill have both

rectangular coordinates (x ,y ) and polar coordinates (r ,θ ). As shown by the

figure, these coordinates are related by the equations suggested in the next

theorem.

Theorem 3.6.1. Coordinate Conversion

The polar coordinates ( r,θ ) of a point are related to the rectangular coordinates

of the point ( x, y) as follows.

1. x = r cos θ ; y = r sinθ .

2. tan θ= y

x; r 2 = x 2 + y 2 .

Example 3.15. Find the rectangular coordinates of the point Pwhose polar

coordinates are (r ,θ )= (6,2π/3).

Solution 3.15. Substituting the polar coordinates r= 6 and θ= 2π /3 in Theo-

rem 3.6.1(1) yields

x= rcos θ= 6cos 2 π

3= 6×µ − 1

2¶ =−3

y= rsin θ= 6sin 2 π

3= 6×à p 3

2! = 3 p 3

Thus, the rectangular coordinates of P are (x ,y )= (−3, 3p3).

Example 3.16. Find polar coordinates of the point P whose rectangular coor-

dinates are (− 2, − 2 p3).

108

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

Solution 3.16. We will find the polar coordinates (r ,θ ) of P that satisfy the

conditions r> 0 and 0 ≤θ< 2π . From Theorem 3.6.1(2),

r2 = x2 + y2 =( −2)2 + (−2 p 3)2 =4 +12 =16 → r=4

tan θ= y

x=−2 p 3

−2 = p 3

From this and the fact that (− 2, − 2 p 3) lies in the third quadrant, it follows that

θ=4 π/3. Thus, (r , θ)= (4, 4 π/3 + 2n π) and (− 4, π/3 + 2n π) are polar coordi-

nates of P where n is an integer.

We will now consider the problem of graphing equations in r and θ , where

θis assumed to be measured in radians.

Example 3.17. Identify the curve r= 2 by transforming the given polar equa-

tion to rectangular coordinates.

3.6. POLAR COORDINATES 109

Solution 3.17. Using Theorem 3.6.1(2) we have r= 2→r 2 = 4→x 2 +y 2 = 4,

which is the circle with center (0,0) and radius 2.

Example 3.18. Identify the curve r= 2cos θ by transforming the given polar

equation to rectangular coordinates.

Solution 3.18. Using Theorem 3.6.1(1) we have

r=2cos θ → r2 =2 rcos θ→ x2 + y2 =2 x→( x−1)2 + y2 =1

which is the circle with center (1,0) and radius 1.

Example 3.19. Express the equation x 2 +y 2 + 6y= 0 in polar coordinates.

Solution 3.19. Using Theorem 3.6.1 we have

¡x 2 +y2 ¢+6y= 0→r 2 +6r sin θ →r2 =−6r sin θ →r =−6 sin θ

Exercise 3.6.

1. Find the corresponding rectangular coordinates for the polar point.

(A) (8, π/2) (B) (− 2,5π/3)

2. Find two sets of polar coordinates for the Cartesian point for 0 ≤θ< 2π .

(A) (0,6) (B) (4, −2)

3. Convert the rectangular equation to polar form.

(A) x 2 −y 2 = 9 (B) x y = 4

4. Convert the polar equation to rectangular form.

(A) r= 3sec θ

(B) r= 6

3cos θ+ 2sin θ

(C) θ= 5π /6

110

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

3.7 Arc Length

In this section, definite integrals are used to find the arc lengths of curves.

To avoid some complications that would otherwise occur, we will impose the

requirement that f0 be continuous on [ a ,b ], in which case we will say that

y= f( x) is a smooth curve on [ a, b] or that fis a smooth function on [ a, b].

Definition 3.7.1. Arc Length

Let the function given by y= f (x ) represent a smooth curve on the interval

[a ,b ]. The arc length of f between a and bis

s=Z b

aq1+£ f 0 (x )¤ 2 d x .

Example 3.20. Find the arc length of the graph of f (x )=x 3

6+ 1

2x on the inter-

val · 1

2,2 ¸ .

Solution 3.20. Using f0 (x )= 3x 2

6− 1

2x 2 = 1

2µ x 2 − 1

x2 ¶ yields an arc length of

s=Z b

aq1+£ f 0 (x )¤ 2 d x =Z 2

1/2 s 1+· 1

2µ x 2 − 1

x2 ¶¸2

d x

=Z 2

1/2 s 1

4· x 4 + 2+ 1

x4 ¸ d x

=Z 2

1/2

1

2· x 2 + 1

x2 ¸ d x =33

16 .

Example 3.21. Find the arc length of the graph of f (x )= ln ( cos x ) from x= 0

to x=π /4.

Solution 3.21. Using f0 (x ) =− sin x

cos x =− tan x yields an arc length of

s=Z b

aq1+£ f 0 (x ) ¤ 2 d x =Z π/4

0q 1+[tan x ] 2 d x

=Z π/4

0p sec 2 x d x =Z π /4

0

sec2 x d x

=[ln | sec x +tan x |]π/4

0=ln ³ 1+p 2´ .

3.7. ARC LENGTH 111

You have seen how parametric equations can be used to describe the path

of a particle moving in the plane. You will now develop a formula for deter-

mining the distance travelled by the particle along its path.

Definition 3.7.2. Arc Length in Parametric Form

If a smooth curve Cis given by x= f (t ) and y= g (t ) such that C does not

intersect itself on the interval a≤ t≤ b (except possibly at the endpoints),

then the arc length of Cover the interval is given by

s=Z b

asµ d x

d t ¶ 2

+µ d y

d t ¶ 2

d t =Z b

aq£ f 0 (t)¤ 2 +£ g 0 (t)¤ 2 d t

Example 3.22. Find the arc length of the curve given by the parametric equa-

tions x= 5 cos t− cos(5t ) and x= 5 sin t− sin(5t ) from t= 0 to t= 2π .

Solution 3.22. Using d x

d t = −5 sin t+ 5 sin(5t ) and y= 5 cos t− 5cos(5t ) yields

an arc length of

s=Z b

asµ d x

d t ¶ 2

+µ d y

d t ¶ 2

d t

=Z 2π

0q [−5 sin t + 5sin(5t)] 2 + [5 cos t− 5cos(5t ) ] 2 d t

=Z 2π

0p 2−2 sin t sin(5t ) −2 cos t cos(5t ) d t

=Z 2π

0p 2−2cos(4t ) d t =Z 2π

0q 4sin 2 (2t) d t

=Z 2π

0| 2sin(2t)| d t = 40

The formula for the length of a polar arc can be obtained from the arc

length formula for a curve described by parametric equations.

Definition 3.7.3. Arc Length of a Polar Curve

Let f be a function whose derivative is continuous on an interval α≤ θ≤ β .

The length of the graph of r= f (θ ) from θ= α to θ= β is

s=Z β

αq£f(θ ) ¤ 2 +£ f 0 (θ)¤ 2 dθ

112

CHAPTER 3. CONICS, PARAMETRIC EQUATIONS, AND POLAR

COORDINATES

Example 3.23. Find the length of the arc from θ= 0 to θ= 2π for the polar

curve r= f (θ )= 2− 2cos θ.

Solution 3.23. Using f0 (θ )= 2sin θ yields an arc length of

s=Z β

αq£f(θ ) ¤ 2 +£ f 0 (θ)¤ 2 dθ

=Z 2π

0q [2− 2cos θ ] 2 + [2sin θ ] 2 dθ

=2 p 2Z 2π

0

p1 − cos θd θ=2 p2 Z 2π

0s 2sin 2 µθ

2¶ dθ

=4Z 2π

0

θsin µ θ

2¶ dθ=16.

Exercise 3.7.

1. Find the arc length of the graph of the function f (x )= 3

2x 2

3over the

interval [1,8].

2. Find the arc length of the graph of the function f (x )= ln · ex + 1

ex −1¸ over

the interval [ln2,ln3].

3. Find the arc length of the curve given by the parametric equations x =

e−t cos tand y= e−t sin ton h 0, π

2i .

4. Find the arc length of the polar curve given by r= f (θ )= 4sin θ on [ 0, π ].

APPENDIX

ˆ

John Bernoulli discovered a rule for calculating limits of fractions whose nu-

merators and denominators both approach zero or +∞. The rule is known

today as l'H ˆ

ospital's Rule, after Guillaume de l'H ˆ

ospital. He was a French no-

bleman who wrote the first introductory differential calculus text, where the

rule first appeared in print.

A.1 Indeterminate Form 0/ 0, ∞/∞

If the functions f (x ) and g (x ) are both zero or both ±∞ at x= a then

lim

x→ a

f( x)

g( x)

cannot be found by substituting x= a . The substitution produces 0/0 or

∞/∞, a meaningless expressions (indeterminate forms), that we cannot eval-

uate. Sometimes, but not always, limits that lead to indeterminate forms may

be found by cancelation, rearrangement of terms, or other algebraic manip-

ulations. L'H ˆ

ospital's Rule enables us to draw on our success with derivatives

to evaluate limits that otherwise lead to indeterminate forms.

113

114 APPENDIX A. INDETERMINATE FORMS AND L'H ˆ

oSPITAL'S RULE

Theorem A.1.1. Suppose f and g are differentiable and g 0 ( x) 6= 0 on an open

interval I that contains a (except possibly at a). Suppose that

lim

x→ a

f( x)

g( x)= 0

0or ±∞

±∞

(In other words, we have an indeterminate form of type 0/0 or ∞/∞ .) Then

lim

x→ a

f( x)

g( x)= lim

x→ a

f0 ( x)

g0 ( x)

if the limit on the right side exists or ±∞.

L'H ˆ

ospital's Rule says that the limit of a quotient of functions is equal to the

limit of the quotient of their derivatives, provided that the given conditions are

satisfied. It is especially important to verify the conditions regarding the limits

of and before using L'H ˆ

ospital's Rule.

L'H ˆ

ospital's Rule is also valid for one-sided limits and for limits at infinity

or negative infinity.

Example A.1. Find

lim

x→1

ln x

x−1

Solution A.1. Since limx→1 ln x= 0 and lim x→1 (x− 1) = 0, then we can apply

L'H ˆ

ospital's Rule:

lim

x→1

ln x

x−1 =lim

x→1

d

d x (ln x)

d

d x (x−1) =lim

x→1

1/x

1= lim

x→1

1

x=1

Example A.2. Calculate

lim

x→∞

ex

x2

Solution A.2. We have limx→∞ ex = ∞ and limx→∞ x 2 = ∞, so L'H ˆ

ospital's

Rule gives:

lim

x→∞

ex

x2 =lim

x→∞

d

d x ( e x )

d

d x ¡ x 2 ¢=lim

x→∞

ex

2x

Since ex →∞ and 2 x →∞ as x → ∞ , the limit on the right side is also indeter-

minate, but a second application of L'H ˆ

ospital's Rule gives

lim

x→∞

ex

x2 =lim

x→∞

ex

2x= lim

x→∞

ex

2=∞

A.2. INDETERMINATE PRODUCTS 0 ·±∞ 115

Example A.3. Calculate

lim

x→∞

ln x

3

px

Solution A.3. Since limx→∞ ln x = ∞ and limx→∞ 3

px =∞ as x → ∞, L'H ˆ

ospital's

Rule applies:

lim

x→∞

ln x

3

px =lim

x→∞

1/x

1

3x −2/3

Notice that the limit on the right side is now indeterminate of type 0

0. But in-

stead of applying L'H ˆ

ospital's Rule a second time as we did in the previous

example, we simplify the expression and see that a second application is un-

necessary:

lim

x→∞

ln x

3

px =lim

x→∞

1/x

1

3x − 2/3 =lim

x→∞

3

3

px =0

Example A.4. Find

lim

x→0

tan x− x

x3

Solution A.4. Noting that both tan x− x→ 0 and x 3 →0 as x→ 0, we use

L'H ˆ

ospital's Rule:

lim

x→0

tan x− x

x3 =lim

x→0

sec2 x− 1

3x 2

Since the limit on the right side is still indeterminate of type 0

0, we apply L'H ˆ

ospital's

Rule again:

lim

x→0

tan x− x

x3 =lim

x→0

sec2 x− 1

3x 2

=lim

x→0

2sec2 x tan x

6x

=2

6× lim

x→0sec 2 x×lim

x→0

tan x

x

=1

3× 1×1 = 1

1

A.2 Indeterminate Products 0 ·±∞

If limx→a f (x )= 0 and limx→a g (x ) = ±∞ , then it is not clear what the value

of limx→a £ f (x )g (x )¤ , if any, will be. There is a struggle between f and g . If f

116 APPENDIX A. INDETERMINATE FORMS AND L'H ˆ

oSPITAL'S RULE

wins, the answer will be 0; if gwins, the answer will be ±∞. Or there may be a

compromise where the answer is a finite nonzero number. This kind of limit

is called an indeterminate form of type 0 ·∞ . We can deal with it by writing

the product as a quotient:

f g = f

1/g or f g = g

1/ f

This converts the given limit into an indeterminate form of type 0

0or ∞/ ∞ so

that we can use L'H ˆ

ospital's Rule.

Example A.5. Evaluate

lim

x→0 + xln x

Solution A.5. The given limit is indeterminate because, as x→ 0+ , the first

factor x approaches 0 while the second factor ln x approaches −∞ . Writing x

as 1

1/x we have 1/x→ ∞ as x→0 + , so L'H ˆ

ospital's Rule gives:

lim

x→0 + xln x= lim

x→0 +

ln x

1/x= lim

x→0 +

1/x

−1/x 2 = lim

x→0 + (−x) = 0

A.3 Indeterminate Differences ∞−∞

If limx→a f (x ) = ∞ and limx→a g (x ) =∞, then the limit

lim

x→ a£ f(x )−g (x )¤

is called an indeterminate form of type ∞−∞ . Again there is a contest be-

tween f and g . Will the answer be ∞ (f wins) or will it be ∞ (g wins) or

will they compromise on a finite number? To find out, we try to convert the

difference into a quotient (for instance, by using a common denominator, or

rationalization, or factoring out a common factor) so that we have an indeter-

minate form of type 0

0or ∞/∞.

Example A.6. Evaluate

lim

x→(π/2) − (sec x− tan x )

A.4. INDETERMINATE POWERS 00 , ∞ 0 ,1∞ 117

Solution A.6. First notice that sec x → ∞ and tan x → ∞ as x→ (π /2)− , so the

limit is indeterminate. Here we use a common denominator:

lim

x→(π/2) − (sec x− tan x ) = lim

x→(π/2) − µ 1

cos x− sin x

cos x ¶

=lim

x→(π/2) −

1− sin x

cos x

=lim

x→(π/2) − −cos x

−sin x =0

A.4 Indeterminate Powers 00 , ∞ 0 , 1∞

These several indeterminate forms arise from the limit

lim

x→ a£ f(x )¤ g( x)

Each of these three cases can be treated by writing the function as an expo-

nential: £ f (x)¤ g(x ) = eg(x )ln f (x )

and then

lim

x→ a£ f(x )¤ g(x )=e lim x→a g( x) ln f( x)

where the indeterminate product g (x )ln f (x ) is of type 0 ·∞.

Example A.7. Calculate

lim

x→0 + (1 + sin4x ) cot x

Solution A.7. First notice that as x→ 0+ , we have 1 + sin4x→ 1 and cot x →∞,

so the given limit is indeterminate. Let

(1 + sin4x)cot x =e cot x ln(1+sin 4x)

Then

lim

x→0 + (1 + sin4x ) cot x =e lim x→0 + cot xln(1+sin 4 x)

Since

lim

x→0 + cot x ln(1 + sin4x )= lim

x→0 +

ln(1 + sin4x)

tan x

=lim

x→0 +

4cos4x

1+ sin4x

sec2 x= 4

118 APPENDIX A. INDETERMINATE FORMS AND L'H ˆ

oSPITAL'S RULE

then

lim

x→0 + (1 + sin4x)cot x =e 4

Example A.8. Find

lim

x→0 + x x

Solution A.8. Notice that this limit is indeterminate since 0x =0 for any x >

0 but x 0 =1 for any x 6= 0. We could proceed by writing the function as an

exponential

xx = exln x

and then

lim

x→0 + x x =e lim x→0 + xln x=e 0= 1

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