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C H A P T E R 5Integration

Section 5.1 Antiderivatives and Indefinite Integration . . . . . . . . . 395

Section 5.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407

Section 5.3 Reimann Sums and Definite Integrals . . . . . . . . . . . 420

Section 5.4 The Fundamental Theorem of Calculus . . . . . . . . . . 429

Section 5.5 Integration by Substitution . . . . . . . . . . . . . . . . . 441

Section 5.6 Numerical Integration . . . . . . . . . . . . . . . . . . . 460

Section 5.7 The Natural Logarithmic Function: Integration . . . . . . 470

Section 5.8 Inverse Trigonometric Functions: Integration . . . . . . . 480

Section 5.9 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 489

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510

C H A P T E R 5Integration

Section 5.1 Antiderivatives and Indefinite Integration

395

1.ddx�

3x3 � C� �

ddx

�3x�3 � C� � �9x�4 ��9x 4

3.ddx�

13

x3 � 4x � C� � x2 � 4 � �x � 2��x � 2�

7.

Check:ddx�

25

x5�2 � C� � x3�2

y �25

x5�2 � C

dydx

� x3�2

2.ddx�x4 �

1x

� C� � 4x3 �1x2

4.

� x1�2 � x�3�2 �x2 � 1

x3�2

ddx�

2�x2 � 3�3�x

� C� �ddx�

23

x3�2 � 2x�1�2 � C�

5.

Check:ddt

t3 � C � 3t2

y � t3 � C

dydt

� 3t2 6.

Check:dd�

�� � C � �

r � �� � C

drd�

� �

8.

Check:ddx�

�1x2 � C� � 2x�3

y �2x�2

�2� C �

�1x2 � C

dydx

� 2x�3

Given Rewrite Integrate Simplify

9.34

x 4�3 � Cx 4�3

4�3� C�x1�3 dx�3�x dx

11. �2

�x� C

x�1�2

�1�2� C�x�3�2 dx� 1

x�x dx

13. �1

4x2 � C12�

x�2

�2� � C12�x�3 dx� 1

2x3 dx

10. �1x

� Cx�1

�1� C�x�2 dx� 1

x2 dx

12.14

x 4 �32

x2 � Cx 4

4� 3�x2

2 � � C��x3 � 3x� dx�x�x2 � 3� dx

14.�19x

� C19�

x�1

�1� � C19�x�2 dx� 1

�3x2� dx

396 Chapter 5 Integration

15.

Check:ddx�

x2

2� 3x � C� � x � 3

��x � 3� dx �x2

2� 3x � C

17.

Check:ddx�

14

x 4 � 5x � C� � x3 � 5

��x3 � 5� dx �14

x 4 � 5x � C

19.

Check:ddx�

25

x5�2 � x2 � x � C� � x3�2 � 2x � 1

��x3�2 � 2x � 1� dx �25

x5�2 � x2 � x � C

16.

Check:ddx�5x �

x2

2� C� � 5 � x

��5 � x� dx � 5x �x2

2� C

18.

Check:ddx

x 4 � 2x 3 � x � C � 4x 3 � 6x 2 � 1

��4x3 � 6x2 � 1� dx � x 4 � 2x 3 � x � C

20.

Check:ddx�

47

x7�4 � x � C� � x3�4 � 1 � 4�x3 � 1

��4�x3 � 1� dx � ��x3�4 � 1� dx �47

x7�4 � x � C

21.

Check:ddx��

12x2 � C� �

1x3

� 1x3 dx � �x�3 dx �

x�2

�2� C � �

12x2 � C 22.

Check:ddx��

13x3 � C� �

1x4

� 1x4 dx � �x�4 dx �

x�3

�3� C � �

13x3 � C

25.

Check:

� �x � 1��3x � 2�

ddx�x3 �

x2

2� 2x � C� � 3x2 � x � 2

� x3 �x2

2� 2x � C

��x � 1��3x � 2� dx � ��3x2 � x � 2� dx

23.

Check: �x2 � x � 1

�x

ddx�

25

x5�2 �23

x3�2 � 2x1�2 � C� � x3�2 � x1�2 � x�1�2

�2

15x1�2�3x2 � 5x � 15� � C�

25

x5�2 �23

x3�2 � 2x1�2 � C�x2 � x � 1

�x dx � ��x3�2 � x1�2 � x�1�2� dx

24.

Check: �x2 � 2x � 3

x4 ddx��

1x

�1x2 �

1x3 � C� � x�2 � 2x�3 � 3x�4

��1x

�1x2 �

1x3 � C�

x�1

�1�

2x�2

�2�

3x�3

�3� C �x2 � 2x � 3

x4 dx � ��x�2 � 2x�3 � 3x�4� dx

26.

Check:

� �2t2 � 1�2

ddt�

45

t5 �43

t3 � t � C� � 4t4 � 4t2 � 1

�45

t5 �43

t3 � t � C

��2t2 � 1�2 dt � ��4t4 � 4t2 � 1� dt

27.

Check:ddy�

27

y7�2 � C� � y5�2 � y2�y

�y2�y dy � �y5�2 dy �27

y7�2 � C 28.

Check:ddt�

13

t3 �34

t4 � C� � t2 � 3t3 � �1 � 3t�t2

��1 � 3t�t2 dt � ��t2 � 3t3� dt �13

t3 �34

t4 � C

29.

Check:ddx

�x � C� � 1

�dx � �1 dx � x � C 30.

Check:ddt

�3t � C� � 3

�3 dt � 3t � C

Section 5.1 Antiderivatives and Indefinite Integration 397

31.

Check:ddx

��2 cos x � 3 sin x � C� � 2 sin x � 3 cos x

��2 sin x � 3 cos x� dx � �2 cos x � 3 sin x � C 32.

Check:ddt�

13

t3 � cos t � C� � t2 � sin t

��t2 � sin t� dt �13

t3 � cos t � C

33.

Check:ddt

�t � csc t � C� � 1 � csc t cot t

��1 � csc t cot t� dt � t � csc t � C 34.

Check:d

d��13

� 3 � tan � � C� � � 2 � sec2 �

��� 2 � sec2 �� d� �13

� 3 � tan � � C

35.

Check:ddx

��2 cos x � 5e x � C� � 2 sin x � 5e x

� �2 sin x � 5e x� dx � �2 cos x � 5e x � C

37.

Check:dd�

�tan � � cos � � C� � sec2 � � sin �

��sec2 � � sin �� d� � tan � � cos � � C

39.

Check:ddy

�tan y � C� � sec2 y � tan2 y � 1

��tan2 y � 1� dy � �sec2 y dy � tan y � C

41.

Check:ddx�x2 �

4x

ln 4� C� � 2x � 4x

� �2x � 4x� dx � x2 �4x

ln 4� C

36.

Check:ddx

�x3 � 2e x � C� � 3x2 � 2e x

� �3x2 � 2e x� dx � x3 � 2e x � C

38.

Check:

� sec y�tan y � sec y�

ddy

�sec y � tan y � C� � sec y tan y � sec2 y

� sec y � tan y � C

�sec y�tan y � sec y� dy � ��sec y tan y � sec2 y� dy

40.

�cos x

1 � cos2 x

Check: ddx

�csc x � C � csc x cot x �1

sin x�

cos xsin x

� �csc x cot x dx � �csc x � C

� cos x1 � cos2 x

dx � �cos xsin2 x

dx � �� 1sin x��

cos xsin x� dx

42.

Check:ddx

�sin x �3x

ln 3� C� � cos x � 3x

� �cos x � 3x� dx � sin x �3x

ln 3� C

43.

Check:ddx �

x2

2� 5 ln�x� � C� � x �

5x

� �x �5x� dx �

x2

2� 5 ln�x� � C 44.

Check:ddx

�4 ln�x� � tan x � C� �4x

� sec2 x

� �4x

� sec2 x� dx � 4 ln�x� � tan x � C

45.

2

2

3

3

2 2

x

3C

2C

0C

y

f �x� � cos x 46.

x4 6 8−2

−2

−4

2

4

6

C = 0

C = 3

C = 2−

y

f �x� � �x 47.

x

6

4

2

−2

−4

6

C = 3

C = 0

C = 2−8

y

f �x� � ln x

398 Chapter 5 Integration

48.

y

6

4

2

−4

x2−2−4

C = 3

C = 0

C = 2−

4

f �x� �12

e x 49.

Answers will vary.

5

4

3

32123

y

x

2x)x)

22x) )xf

ff ′

f �x� � 2x � C

f� �x� � 2 50.

x

y

4

6

8

−2−4 2 4

f x( ) 2= +x2

2

f x( ) = x2

2

f ′

f �x� �x2

2� C

f��x� � x

51.

Answers will vary.

3

4

3

2

231

2

x

y3

3

3x

xx)f

xx3

x)f )3

)

f

f �x� � x �x3

3� C

f��x� � 1 � x2 52.

x−2−4

−2

−4

4

f x( ) = −

f x( ) 1= − +

1

1

x

x

f ′

y

f �x� � �1x

� C

f��x� �1x2 53.

y � x2 � x � 1

1 � �1�2 � �1� � C ⇒ C � 1

y � ��2x � 1� dx � x2 � x � C

dydx

� 2x � 1, �1, 1�

54.

y � x2 � 2x � 1

2 � �3�2 � 2�3� � C ⇒ C � �1

y � �2�x � 1� dx � x2 � 2x � C

dydx

� 2�x � 1� � 2x � 2, �3, 2� 55.

y � sin x � 4

4 � sin 0 � C ⇒ C � 4

y � �cos x dx � sin x � C

dydx

� cos x, �0, 4� 56.

y � 3 ln x

3 � 3 ln e � C ⇒ C � 0

y � �3x dx � 3 ln x � C

dydx

�3x, x > 0, �e, 3�

57. (a) Answers will vary.

x

y

−3 5

−3

5

(b)

y �x2

4� x � 2

2 � C

2 �42

4� 4 � C

y �x2

4� x � C

−4 8

−2

6dydx

�12

x � 1, �4, 2�

Section 5.1 Antiderivatives and Indefinite Integration 399

58. (a)

x− 4

−5

5

4

y (b)

y �x3

3� x �

73

C �73

3 � �13

� 1 � C

3 ���1�3

3� ��1� � C

y �x3

3� x � C

−4 4

−5

( 1, 3)−

5dydx

� x2 � 1, ��1, 3�

59. (a)

3

−2

6

−4

y

x

(b)

y � sin x � 4

4 � sin�0� � C ⇒ C � 4

y � �cos x dx � sin x � C

7

6

−1

−6

dydx

� cos x, �0, 4�

60. (a)

1 7x

y

(1, 3)

(b)

y �1x

� 2

3 �11

� C ⇒ C � 2

y � ��1x 2 dx � ��x�2 dx �

�x�1

�1� C �

1x

� C

−1

5

8−1

dydx

��1x2 , x > 0, �1, 3�

61. (a) 9

3

−9

−3

(b)

y � x2 � 6

�2 � ��2�2 � C � 4 � C ⇒ C � �6

y � �2x dx � x2 � C

dydx

� 2x, ��2, �2�

62. (a) 20

60

0

(b)

y �43

x3�2 �43

12 �43

�4�3�2 � C �43

�8� � C �323

� C ⇒ C �43

y � �2x1�2 dx �43

x 3� 2�C

dydx

� 2�x, �4, 12� (c) 20

60

0

(c) 12

15

−8

−15

400 Chapter 5 Integration

63.

f �x� � 2x2 � 6

f�0� � 6 � 2�0�2 � C ⇒ C � 6

f �x� � �4x dx � 2x2 � C

f��x� � 4x, f �0� � 6

65.

h�t� � 2t4 � 5t � 11

h�1� � �4 � 2 � 5 � C ⇒ C � �11

h�t� � ��8t3 � 5� dt � 2t 4 � 5t � C

h��t� � 8t3 � 5, h�1� � �4

67.

f �x� � x2 � x � 4

f �2� � 6 � C2 � 10 ⇒ C2 � 4

f �x� � ��2x � 1� dx � x2 � x � C2

f��x� � 2x � 1

f��2� � 4 � C1 � 5 ⇒ C1 � 1

f��x� � �2 dx � 2x � C1

f �2� � 10

f��2� � 5

f � �x� � 2

69.

f �x� � �4x1�2 � 3x � �4�x � 3x

f �0� � 0 � 0 � C2 � 0 ⇒ C2 � 0

f �x� � ���2x�1�2 � 3� dx � �4x1�2 � 3x � C2

f��x� � �2

�x� 3

f��4� � �22

� C1 � 2 ⇒ C1 � 3

f��x� � �x�3�2 dx � �2x�1�2 � C1 � �2

�x� C1

f �0� � 0

f��4� � 2

f � �x� � x�3�2

64.

g�x� � 2x3 � 1

g�0� � �1 � 2�0�3 � C ⇒ C � �1

g�x� � �6x2 dx � 2x3 � C

g��x� � 6x2, g�0� � �1

66.

f �s� � 3s 2 � 2s 4 � 23

f �2� � 3 � 3�2�2 � 2�2�4 � C � 12 � 32 � C ⇒ C � 23

f �s� � ��6s � 8s 3� ds � 3s 2 � 2s 4 � C

f��s� � 6s � 8s 3, f �2� � 3

68.

f �x� �1

12x4 � 6x � 3

f �0� � 0 � 0 � C2 � 3 ⇒ C2 � 3

f �x� � ��13

x3 � 6� dx �1

12x4 � 6x � C2

f��x� �13

x3 � 6

f��0� � 0 � C1 � 6 ⇒ C1 � 6

f��x� � �x2 dx �13

x3 � C1

f �0� � 3

f��0� � 6

f ��x� � x2

70.

f �x� � �sin x � 2x � 6

f �0� � 0 � 0 � C2 � 6 ⇒ C2 � 6

f �x� � ���cos x � 2� dx � �sin x � 2x � C2

f��x� � �cos x � 2

f��0� � �1 � C1 � 1 ⇒ C1 � 2

f��x� � �sin x dx � �cos x � C1

f �0� � 6

f��0� � 1

f ��x� � sin x

Section 5.1 Antiderivatives and Indefinite Integration 401

71.

f �x� � e x � x � 4

f �0� � 5 � e0 � 0 � C2 ⇒ C2 � 4

f �x� � � �e x � 1� dx � e x � x � C2

f��0� � 2 � e0 � C1 ⇒ C1 � 1

f��x� � � e x dx � e x � C1

f � �x� � e x 72.

f �x� � �2 ln�x� � 6x � 3

f �1� � 3 � 6 � C2 ⇒ C2 � �3

f �x� � � ��2x

� 6� dx � �2 ln�x� � 6x � C2

f��1� � 4 � �2 � C1 ⇒ C1 � 6

f��x� � � 2x2 dx � � 2x�2 dx �

�2x

� C1

f � �x� �2x2

73. (a)

(b) h�6� � 0.75�6�2 � 5�6� � 12 � 69 cm

h�t� � 0.75t2 � 5t � 12

h�0� � 0 � 0 � C � 12 ⇒ C � 12

h�t� � ��1.5t � 5� dt � 0.75t2 � 5t � C 74.

P�7� � 100�7�3�2 � 500 2352 bacteria

P�t� �23

�150�t3�2 � 500 � 100t3�2 � 500

P�1� �23

k � 500 � 600 ⇒ k � 150

P�0� � 0 � C � 500 ⇒ C � 500

P�t� � �kt1�2 dt �23

kt 3�2 � C

dPdt

� k�t, 0 ≤ t ≤ 10

75. Graph of is given.

(a)

(b) No. The slopes of the tangent lines are greater than 2on Therefore, f must increase more than 4 unitson

(c) No, because f is decreasing on

(d) f is an maximum at because andthe first derivative test.

(e) f is concave upward when is increasing on and f is concave downward on Pointsof inflection at x � 1, 5.

�1, 5�.�5, �.��, 1�f�

f��3.5� 0x � 3.5

4, 5.f �5� < f �4�

0, 4.0, 2.

f��4� �1.0

f�f �0� � �4. (f) is a minimum at

(g)

x

2

2

4

4

6

6 8−2

−6

y

x � 3.f �

76. Since is negative on is decreasing onSince is positive on is increasing

on has a relative minimum at Since ispositive on f is increasing on

x1 2 3−2−3

1

2

3

−2

−3f

f ′

f ′′

y

��, �.��, �,f��0, 0�.f��0, �.

f��0, �,f ���, 0�.f���, 0�,f � 77.

The ball reaches its maximum height when

s�158 � � �16�15

8 �2� 60�15

8 � � 6 � 62.25 feet

t �158 seconds

32t � 60

v�t� � �32t � 60 � 0

s�t� � �16t2 � 60t � 6, Position function

s�0� � 6 � C2

s�t� � ���32t � 60� dt � �16t 2 � 60t � C2

v�0� � 60 � C1

v�t� � ��32 dt � �32t � C1

a�t� � �32 ft�sec2

402 Chapter 5 Integration

78.

f �t� � �16t2 � v0t � s0

f �0� � 0 � 0 � C2 � s0 ⇒ C2 � s0

f �t� � s�t� � ���32t � v0� dt � �16t2 � v0t � C2

f��t� � �32t � v0

f��0� � 0 � C1 � v0 ⇒ C1 � v0

f��t� � v�t� � ��32 dt � �32t � C1

f �0� � s0

f��0� � v0

f� �t� � a�t� � �32 ft�sec2 79. From Exercise 78, we have:

when time to reachmaximum height.

v0 187.617 ft�sec

v02 � 35,200

�v0

2

64�

v02

32� 550

s� v0

32� � �16� v0

32�2

� v0� v0

32� � 550

t �v0

32�s��t� � �32t � v0 � 0

s�t� � �16t2 � v0t

80.

(a)

Choosing the positive value,

(b)

� �8�65 �64.498 ft�sec

v�1 � �654 � � �32�1 � �65

4 � � 8

v�t� � s��t� � �32t � 8

t �1 � �65

4 2.266 seconds.

t �1 ± �65

4

�8�2t2 � t � 8� � 0

s�t� � �16t2 � 8t � 64 � 0

s0 � 64 ft

v0 � 8 ft�sec

82. From Exercise 81, (Using the canyon floor as position 0.)

t2 �16004.9

⇒ t �326.53 18.1 sec

4.9t2 � 1600

f �t� � 0 � �4.9t2 � 1600

f �t� � �4.9t2 � 1600.

81.

f �0� � s0 � C2 ⇒ f �t� � �4.9t2 � v0t � s0

f �t� � ���9.8t � v0� dt � �4.9t2 � v0t � C2

v�0� � v0 � C1 ⇒ v�t� � �9.8t � v0

v�t� � ��9.8 dt � �9.8t � C1

a�t� � �9.8

83. From Exercise 81,

(Maximum height when )

f � 109.8� 7.1 m

t �109.8

9.8t � 10

v � 0.v �t� � �9.8t � 10 � 0

f �t� � �4.9t2 � 10t � 2.

Section 5.1 Antiderivatives and Indefinite Integration 403

84. From Exercise 81, If

then

for this t-value. Hence, and we solve

v02 � 3880.8 ⇒ v0 62.3 m�sec.

4.9 v02 � �9.8�2 198

�4.9 v02 � 9.8 v0

2 � �9.8�2 198

�4.9 v0

2

�9.8�2 �v0

2

9.8� 198

�4.9� v0

9.8�2

� v0� v0

9.8� � 2 � 200

t � v0�9.8

v�t� � �9.8t � v0 � 0

f �t� � 200 � �4.9t2 � v0t � 2,

f �t� � �4.9t2 � v0t � 2.

85.

since the stone was dropped,

Thus, the height of the cliff is 320 meters.

v�20� � �32 m�sec

v�t� � �1.6t

s0 � 320

s�20� � 0 ⇒ �0.8�20�2 � s0 � 0

s�t� � ���1.6t� dt � �0.8t2 � s0

v0 � 0. v�t� � ��1.6 dt � �1.6t � v0 � �1.6t,

a � �1.6

86.

When

v2 � v02 � 2GM�1

y�

1R�

v2 �2GM

y� v0

2 �2GM

R

12

v2 �GM

y�

12

v02 �

GMR

C �12

v02 �

GMR

12

v02 �

GMR

� C

y � R, v � v0.

12

v2 �GM

y� C

�v dv � �GM� 1y2 dy 87.

(a)

(b) when or

(c) when

v�2� � 3�1���1� � �3

t � 2.a�t� � 6�t � 2� � 0

3 < t < 5.0 < t < 1v�t� > 0

a�t� � v��t� � 6t � 12 � 6�t � 2�

� 3�t2 � 4t � 3� � 3�t � 1��t � 3�

v�t� � x��t� � 3t2 � 12t � 9

0 ≤ t ≤ 5x�t� � t3 � 6t2 � 9t � 2,

404 Chapter 5 Integration

88.

(a)

(b) when and

(c) when

v�73� � �3�7

3� � 5��73

� 3� � 2��23� � �

43

t �73

.a�t� � 6t � 14 � 0

3 < t < 5.0 < t < 53

v�t� > 0

a�t� � v��t� � 6t � 14

v�t� � x��t� � 3t2 � 14t � 15 � �3t � 5��t � 3� � t3 � 7t2 � 15t � 9

0 ≤ t ≤ 5x�t� � �t � 1��t � 3�2, 89.

a�t� � v��t� � �12

t�3�2 ��1

2t 3�2, acceleration

x�t� � 2t1�2 � 2, position function

x�1� � 4 � 2�1� � C ⇒ C � 2

x�t� � �v�t� dt � 2t1�2 � C

t > 0v�t� �1

�t� t�1�2,

90. (a)

(b) for k � 0, 1, 2, . . .t � k�,v�t� � 0 � sin t

f �t� � �cos t � 4

f �0� � 3 � �cos�0� � C2 � �1 � C2 ⇒ C2 � 4

f �t� � �v�t� dt � �sin t dt � �cos t � C2

v�t� � �a�t� dt � �cos t dt � sin t � C1 � sin t �since v0 � 0�

a�t� � cos t

91. (a)

(b)

s�13� �275234

�13�2

2�

25036

�13� 189.58 m

s�t� � at 2

2�

25036

t �s�0� � 0�

a �550468

�275234

1.175 m�sec2

55036

� 13a

v�13� �80036

� 13a �25036

v�0� �25036

⇒ v�t� � at �25036

v�t� � at � C

a�t� � a �constant acceleration�

v�13� � 80 km�hr � 80 �10003600

�80036

m�sec

v�0� � 25 km�hr � 25 �10003600

�25036

m�sec

Section 5.1 Antiderivatives and Indefinite Integration 405

92.

s�t� � �8.25t 2 � 66t

v�t� � �16.5t � 66

a�t� � �16.5

� 132 when a �332

� 16.5.

s�66a � � �

a2�

66a �

2

� 66�66a �

�at � 66 � 0 when t �66a

.

v�t� � 0 after car moves 132 ft.

s�t� � �a2

t2 � 66t �Let s�0� � 0.�

v�t� � �at � 66

a�t� � �a

15 mph � 22 ft�sec

30 mph � 44 ft�sec

v�0� � 45 mph � 66 ft�sec (a)

(b)

(c)

It takes 1.333 seconds to reduce the speed from 45 mphto 30 mph, 1.333 seconds to reduce the speed from 30 mph to 15 mph, and 1.333 seconds to reduce the speedfrom 15 mph to 0 mph. Each time, less distance is neededto reach the next speed reduction.

0

73.33 117.33

132

feet feet

45 m

ph =

66

ft/se

c

30 m

ph =

44

ft/se

c15

mph

= 2

2 ft/

sec

0 m

ph

s� 4416.5� 117.33 ft

t �44

16.5 2.667

�16.5t � 66 � 22

s� 2216.5� 73.33 ft

t �22

16.5 1.333

�16.5t � 66 � 44

93. Truck:

Automobile:

At the point where the automobile overtakes the truck:

when sec.t � 10 0 � 3t�t � 10�

0 � 3t2 � 30t

30t � 3t2

s�t� � 3t2 �Let s�0� � 0.�

v�t� � 6t �Let v�0� � 0.�

a�t� � 6

s�t� � 30t �Let s�0� � 0.�

v�t� � 30 (a)

(b) v�10� � 6�10� � 60 ft�sec 41 mph

s�10� � 3�10�2 � 300 ft

94.

(a) (b)

(c)

The second car was going faster than the first until the end.

s2�30� 1970.3 feet

s1�30� 953.5 feet

In both cases, the constant of integration is 0 because s1�0� � s2�0� � 0.

s2�t� � �v2�t� dt � �0.1208t3

3�

6.7991t2

2� 0.0707t

s1�t� � �v1�t� dt �0.1068

3 t 3 �

0.04162

t2 � 0.3679t

v2�t� � �0.1208t2 � 6.7991t � 0.0707

v1�t� � 0.1068t2 � 0.0416t � 0.3679

�1 mi�hr��5280 ft�mi��3600 sec�hr� �

2215

ft�sec

t 0 5 10 15 20 25 30

0 3.67 10.27 23.47 42.53 66 95.33

0 30.8 55.73 74.8 88 93.87 95.33v2�ft�sec�

v1�ft�sec�

406 Chapter 5 Integration

95. True 96. True 97. True 98. True

99. False. For example, because x3

3� C �x2

2� C1��x2

2� C2�.�x � x dx �x dx � �x dx

100. False. has an infinite number of antiderivatives, eachdiffering by a constant.

f 101.

Answer: f �x� �x3

3 � 4x �

163

f �2� � 0 ⇒ 83

� 8 � C1 � 0 ⇒ C1 �163

f �x� �x3

3 � 4x � C1

f��2� � 0 ⇒ 4 � C � 0 ⇒ C � �4

f��x� � x2 � C

f ��x� � 2x

102.

continuous at

continuous at

f �x� � ��x � 12x � 51

,,

,

0 ≤ x < 22 ≤ x < 33 ≤ x ≤ 4

x � 3 ⇒ 6 � 5 � C3 � 1f

x � 2 ⇒ �2 � 1 � 4 � C2 ⇒ C2 � �5f

f �0� � 1 ⇒ C1 � 1

f �x� � ��x � C1

2x � C2

C3

,,

,

0 ≤ x < 22 < x < 33 < x ≤ 4

1 2

1

2

3 4x

y

f��x� � ��120

,,,

0 ≤ x < 22 < x < 33 < x ≤ 4

103.

is continuous: Values must agree at

The left and right hand derivatives at do not agree. Hence is not differentiable at x � 2.f

x � 2

f �x� � �x � 2,3x2

� 2,2

0 ≤ x < 2 2 ≤ x ≤ 5

4 � 6 � C2 ⇒ C2 � �2

x � 2:f

f �1� � 3 ⇒ 1 � C1 � 3 ⇒ C1 � 2

f �x� � �x � C1,3x2

� C22 ,

0 ≤ x < 2 2 ≤ x ≤ 5

f��x� � �1,3x,

0 ≤ x < 2 2 ≤ x ≤ 5

104.

Thus, for some constant k. Since,and

Therefore,

[Note that and satisfy theseproperties.]

c�x� � cos xs�x� � sin x

s�x�2 � c�x�2 � 1.

k � 1.c�0� � 1,s�0� � 0s�x�2 � c�x�2 � k

� 0

� 2s�x�c�x� � 2c�x�s�x�

ddx

s�x�2 � c�x�2 � 2s�x�s��x� � 2c�x�c��x�

Section 5.2 Area 407

105.ddx

�ln�Cx�� �ddx

�ln�C� � ln�x�� � 0 �1x

�1x

106.ddx

�ln�x� � C� �1x

� 0 �1x

107.

Note: and satisfy these conditions.

Differentiate with respect to

Differentiate with respect to

Letting

Hence,

Adding,

Integrating,

Clearly for if then which contradicts that are nonconstant.

Now,

Thus, and we have f �x�2 � g�x�2 � 1.C � 1

� C 2.

� � f �x�2 � g�x�2�� f � y�2 � g� y�2�

� f �x�2f � y�2 � g �x�2g� y�2 � f �x�2g� y�2 � g�x�2f � y�2

C � f �x � y�2 � g�x � y�2 � � f �x� f �y� � g�x�g�y��2 � � f �x�g� y� � g�x�f �y��2

f, gf �x�2 � �g�x�2 ⇒ f �x� � g�x� � 0,C � 0,C � 0,

f �x�2 � g�x�2 � C.

2 f �x�f��x� � 2g �x�g��x� � 0.

2g�x�g��x� � 2g�x�f �x�g��0�.

2 f �x�f��x� � �2 f �x�g�x�g��0�

g� �x� � f �x�g��0� � g�x�f��0� � f �x�g��0�.

f� �x� � f �x�f��0� � g�x�g��0� � �g�x�g��0�y � 0,

y.��g� �x � y� � f �x�g�� y� � g�x�f�� y�

y.��f� �x � y� � f �x�f�� y� � g�x�g�� y�

�g�x� � sin xf �x� � cos x�

f� �0� � 0

g�x � y� � f �x�g� y� � g�x�f � y�

f �x � y� � f �x�f � y� � g�x�g� y�

Section 5.2 Area

1. �5

i�1�2i � 1� � 2�

5

i�1i � �

5

i�11 � 2�1 � 2 � 3 � 4 � 5� � 5 � 35

2. �6

k�3k�k � 2� � 3�1� � 4�2� � 5�3� � 6�4� � 50 3. �

4

k�0

1k2 � 1

� 1 �12

�15

�1

10�

117

�15885

4. �5

j�3 1j

�13

�14

�15

�4760

5. �4

k�1c � c � c � c � c � 4c

6. �4

i�1��i � 1�2 � �i � 1�3� � �0 � 8� � �1 � 27� � �4 � 64� � �9 � 125� � 238

7. �9

i�1 13i

8. �15

i�1

51 � i

9. �8

j�1�5� j

8 � 3 10. �4

j�1�1 � � j

42

11.2n

�n

i�1��2i

n 3

� �2in 12.

2n

�n

i�1�1 � �2i

n� 1

2

13.3n

�n

i�1�2�1 �

3in

2

14.1n

�n�1

i�0�1 � � i

n2

15. � 2�20�21�2 � 420�

20

i�12i � 2�

20

i�1i 16. � 2�15�16�

2 � 45 � 195�15

i�1�2i � 3� � 2�

15

i�1i � 3�15�

408 Chapter 5 Integration

17.

� �19�20��39�6 � 2470

�20

i�1�i � 1�2 � �

19

i�0i2 18.

� �10�11��21�6 � 10 � 375

�10

i�1�i2 � 1� � �

10

i�1i2 � �

10

i�11

19.

� 12,040

� 14,400 � 2480 � 120

�152�16�2

4� 2

15�16��31�6

�15�16�

2

�15

i�1 i�i � 1�2 � �

15

i�1i 3 � 2�

15

i�1i 2 � �

15

i�1i 20.

�102�11�2

4�

10�11�2

� 3080

�10

i�1i�i2 � 1� � �

10

i�1i3 � �

10

i�1i

21. sum seq (TI-82)

��20��21��41�

6� 60 � 2930

�20

i�1�i2 � 3� �

20�20 � 1��2�20� � 1�6

� 3�20�

2 � 3, x, 1, 20, 1� � 2930�x 22. sum seq (TI-82)

��15�2�16�2

4� 15�16� � 14,160

�15

i�1�i3 � 2i� �

�15� 2�15 � 1� 2

4� 2

15�15 � 1�2

3 � 2x, x, 1, 15, 1� � 14,160�x

23.

s � �1 � 3 � 4 �92��1� �

252 � 12.5

S � �3 � 4 �92 � 5��1� �

332 � 16.5 24.

s � �4 � 4 � 2 � 0��1� � 10

S � �5 � 5 � 4 � 2��1� � 16

25.

s � �2 � 2 � 3��1� � 7

S � �3 � 3 � 5��1� � 11 26.

s � �2 �43 � 1 �

45 �

23� �

295 � 5.8

S � �4 � 2 �43 � 1 �

45� �

13715 � 9.13

27.

s�4� � 0�14 ��1

4 �14 ��1

2 �14 ��3

4�14 �

1 � �2 � �38

� 0.518

S�4� ��14 �

14 ��1

2 �14 ��3

4 �14 � �1�1

4 �1 � �2 � �3 � 2

8� 0.768

28.

s�4� � 4�e�0.5 � e�1 � e�1.5 � e�2� 12 � 2.666

S�4� � 4�e�0 � e�0.5 � e�1 � e�1.5� 12 � 4.395

29.

s�5� �1

6 5�15 �

17 5�

15 �

18 5�

15 �

19 5�

15 �

12�

15 �

16

�17

�18

�19

�1

10� 0.646

S�5� � 1�15 �

16 5�

15 �

17 5�

15 �

18 5�

15 �

19 5�

15 �

15

�16

�17

�18

�19

� 0.746

30.

s�5� ��1 � �15

2

�15 ��1 � �2

52

�15 ��1 � �3

52

�15 ��1 � �4

52

�15 � 0 � 0.659

�15�1 �

�245

��21

5�

�165

��95 � 0.859

S�5� � 1�15 ��1 � �1

52

�15 ��1 � �2

52

�15 ��1 � �3

52

�15 ��1 � �4

52

�15

>>

Section 5.2 Area 409

31. limn→�

��81n4n2�n � 1�2

4 �814

limn→�

�n4 � 2n3 � n2

n4 �814

�1� �814

32. limn→�

��64n3n�n � 1��2n � 1�

6 �646

limn→�

�2n3 � 3n2 � nn3 �

646

�2� �643

33. limn→�

��18n2n�n � 1�

2 �182

limn→�

�n2 � nn2 �

182

�1� � 9 34. limn→�

�� 1n2n�n � 1�

2 �12

limn→�

�n2 � nn2 �

12

�1� �12

35.

S�10,000� � 1.0002

S�1000� � 1.002

S�100� � 1.02

S�10� �1210

� 1.2

�n

i�1

2i � 1n2 �

1n2 �

n

i�1�2i � 1� �

1n2�2

n�n � 1�2

� n �n � 2

n� S�n�

36.

S�10,000� � 2.0003

S�1000� � 2.003

S�100� � 2.03

S�10� �2310

� 2.3

�n

j�1

4j � 1n2 �

1n2 �

n

j�1�4j � 1� �

1n2�4n�n � 1�

2� n �

2n � 3n

� S�n�

37.

S�10,000� � 1.99999998

S�1000� � 1.999998

S�100� � 1.9998

S�10� � 1.98

�6n2�2n2 � 3n � 1 � 3n � 3

6 �1n2�2n2 � 2� � S�n�

�n

k�1

6k�k � 1�n3 �

6n3 �

n

k�1�k2 � k� �

6n3�n�n � 1��2n � 1�

6�

n�n � 1�2

38.

S�10,000� � 1.000066657

S�1000� � 1.00066567

S�100� � 1.006566

S�10� � 1.056

�1

3n3�3n3 � 2n2 � 3n � 2� � S�n�

�1

3n3�3n3 � 6n2 � 3n � 4n2 � 6n � 2�

�4n3�n3 � 2n2 � n

4�

2n2 � 3n � 16

�n

i�1

4i2�i � 1�n4 �

4n4 �

n

i�1�i3 � i2� �

4n4�n2�n � 1�2

4�

n�n � 1��2n � 1�6

410 Chapter 5 Integration

39. � 8 limn→�

�1 �1n � 8� lim

n→� �8�n2 � n

n2 � limn→�

16n2�n�n � 1�

2 limn→�

�n

i�1�16i

n2 � limn→�

16n2 �

n

i�1i

40. � limn→�

42�1 �

1n � 2� lim

n→� 4n2�n�n � 1�

2 limn→�

�n

i�1�2i

n �2n � lim

n→� 4n2 �

n

i�1i

41.

� limn→�

� 16�

2 � �3 n� � �1 n2�1 �

13

� limn→�

16�

2n3 � 3n2 � nn3

� limn→�

1n3��n � 1��n��2n � 1�

6 limn→�

�n

i�1 1n3�i � 1�2 � lim

n→� 1n3 �

n�1

i�1i 2

42.

� 2�1 � 2 �43 �

263

� 2 limn→�

�1 � 2 �2n

�43

�2n

�2

3n2

� limn→�

2n3�n3 � �4n��n�n � 1�

2 �4�n��n � 1��2n � 1�

6

� limn→�

2n3��

n

i�1n2 � 4n �

n

i�1i � 4 �

n

i�1i 2

limn→�

�n

i�1�1 �

2in

2

�2n � lim

n→� 2n3 �

n

i�1�n � 2i�2

43. � 2 limn→�

�1 �n2 � n

2n2 � 2�1 �12 � 3� 2 lim

n→� 1n�n �

1n�

n�n � 1�2 lim

n→� �

n

i�1�1 �

in�

2n � 2 lim

n→� 1n��

n

i�11 �

1n

�n

i�1i

44.

� 2 limn→�

�10 �13n

�4n2 � 20

� 2 limn→�

�1 � 3 �3n

� 4 �6n

�2n2 � 2 �

4n

�2n2

� 2 limn→�

1n4 �n4 � 6n2�n�n � 1�

2 � 12n�n�n � 1��2n � 1�6 � 8�n2�n � 1�2

4

� 2 limn→�

1n4 �

n

i�1�n3 � 6n2i � 12ni 2 � 8i 3�

limn→�

�n

i�1�1 �

2in

3

�2n � 2 lim

n→� 1n4 �

n

i�1�n � 2i�3

45. (a)

(c) Since is increasing, on

—CONTINUED—

� �n

i�1 f �2i � 2

n �2n � �

n

i�1��i � 1��2

n�2n

s�n� � �n

i�1 f �xi�1� �x

�xi�1, xi�.f �mi� � f �xi�1�y � x

x31

3

2

1

y (b)

Endpoints:

(d) on

S�n� � �n

i�1 f �xi� �x � �

n

i�1 f �2i

n 2n

� �n

i�1�i�2

n�2n

�xi�1, xi�f �Mi� � f �xi�

0 < 1�2n < 2�2

n < . . . < �n � 1��2n < n�2

n � 2

�x �2 � 0

n�

2n

Section 5.2 Area 411

45. —CONTINUED—

(e) (f)

� limn→�

2�n � 1�

n� 2

� limn→�

� 4n2n�n � 1�

2

limn→�

�n

i�1�i�2

n�2n � lim

n→� 4n2 �

n

i�1i

� limn→�

�2�n � 1�n

�4n � 2

� limn→�

4n2�n�n � 1�

2� n

limn→�

�n

i�1��i � 1��2

n�2n � lim

n→� 4n2 �

n

i�1�i � 1�x 5 10 50 100

1.6 1.8 1.96 1.98

2.4 2.2 2.04 2.02S�n�

s�n�

46. (a)

(c) Since is increasing, on

(d) on

(e)

S�n� � �n

i�1 f �xi � �x � �

n

i�1 f �1 � i�2

n�2n � �

n

i�1�1 � i�2

n�2n

[xi�1, xi�f �Mi� � f �xi�

� � n

i�1f �1 � �i � 1��2

n�2n � �

n

i�1�1 � �i � 1��2

n�2n

s�n� � �n

i�1 f �xi�1� �x

�xi�1, xi�.f �mi � � f �xi�1�y � x

x2 4

1

2

3

4

y

x 5 10 50 100

3.6 3.8 3.96 3.98

4.4 4.2 4.04 4.02S�n�

s�n�

(b)

Endpoints:

1 < 1 � 1�2n < 1 � 2�2

n < . . . < 1 � �n � 1��2n < 1 � n�2

n

1 < 1 �2n

< 1 �4n

< . . . < 1 �2nn

� 3

�x �3 � 1

n�

2n

(f)

� limn→�

�2 �2�n � 1�

n � limn→�

�4 �2n � 4

limn→�

�n

i�1�1 � i�2

n�2n � lim

n→� 2n�n � �2

nn�n � 1�

2

� limn→�

�2 �2n � 2

n�

4n � lim

n→� �4 �

2n � 4

limn→�

�n

i�1�1 � �i � 1��2

n�2n � lim

n→� �2

n�n �2n�

n�n � 1�2

� n

412 Chapter 5 Integration

47.

Area � limn→�

s�n� � 2

� 3 �2n2 �

n

i�1i � 3 �

2�n � 1�n2n2 � 2 �

1n

s�n� � �n

i�1 f � i

n�1n � �

n

i�1��2� i

n � 3�1n

321x

2

1

y�Note: �x �1 � 0

n�

1n�0, 1�y � �2x � 3,

48.

Area � limn→�

S�n� � 6 �272

�392

� 6 �27n2��n � 1�n

2 � 6 �272 �1 �

1n

� �n

i�1�3�2 �

3in � 4�3

n � 18 � 3�3n

2

�n

i�1i � 12

S�n� � � n

i�1f �2 �

3in �

3n

x4 6 8 10 12

2

4

6

8

10

12

y�Note: �x �5 � 2

n�

3n�2, 5�y � 3x � 4,

49.

Area � limn→�

S�n� �73

� � 1n3 �

n

i�1i 2 � 2 �

n�n � 1��2n � 1�6n3 � 2 �

16�2 �

3n

�1n2 � 2

S�n� � � n

i�1f � i

n�1n � �

n

i�1�� i

n2

� 2�1n

x2 3

3

1

1

y

�Note: �x �1n�0, 1�y � x2 � 2,

50.

Area � limn →�

S�n� �92

�2� � 3 � 12

�27n3

n�n � 1��2n � 1�6

�3n

�n� �92

2n2 � 3n � 1

n2 � 3

�27n 3 �

n

i�1i 2 �

3n

�n

i�11

S�n� � �n

i�1 f �3i

n �3n � �

n

i�1��3i

n 2

� 1�3n

y

x1 2

2

4

6

8

10

12

3

�0, 3� �Note: �x �3 � 0

n�

3ny � x2 � 1,

Section 5.2 Area 413

51.

Area � limn →�

s�n� � 30 �83

� 4 �703

� 2313

� 30 �8

6n2 �n � 1��2n � 1� �4n

�n � 1�

�2n

�15n �4

n2 n�n � 1��2n � 1�

6�

4n

n�n � 1�

2

�2n

�n

i�1 �15 �

4i2

n2 �4in

s�n� � �n

i�1 f �1 �

2in �

2n � �

n

i�1 �16 � �1 �

2in

2

�2n

x

y

1−1

2468

101214

18

2 3

�1, 3� �Note: �x �2ny � 16 � x2,

52. Find area of region over the interval

Area �43

12

Area � limn→�

s�n� � 1 �13

�23

� 1 �1n3 �

n

i�1i 2 � 1 �

n�n � 1��2n � 1�6n3 � 1 �

16�2 �

3n

�1n2

s�n� � �n

i�1 f � i

n�1n � �

n

i�1�1 � � i

n2

�1n

x

2

3

−2 2

−1

y

�Note: �x �1n�0, 1�.��1, 1�;y � 1 � x2,

53.

Area � limn →�

s�n� � 189 �814

� 27 �272

�5134

� 128.25

� 189 �814n2�n � 1�2 �

816n2�n � 1��2n � 1� �

272

n � 1

n

�3n�63n �

27n3

n2�n � 1�2

4�

27n2

n�n � 1��2n � 1�6

�9n

n�n � 1�

2

�3n

�n

i�1 �63 �

27i3

n3 �27i2

n2 �9in

s�n� � �n

i�1 f�1 �

3in �

3n � �

n

i�1�64 � �1 �

3in

3

�3n

x5 6

30

y

40

50

60

70

20

10

1 2−1 3

�1, 4� �Note: �x �4 � 1

n�

3ny � 64 � x3,

54.

Since y both increases and decreases on is neither an upper nor lower sum.

Area � limn→�

T�n� � 1 �14

�34

� 1 �1n

�14

�2

4n�

14n2

�2n2 �

n

i�1i �

1n4 �

n

i�1i3 �

n�n � 1�n2 �

1n4�n2�n � 1�2

4

T�n� � �n

i�1 f � i

n�1n � �

n

i�1�2� i

n � � in

3

�1n

T�n��0, 1�,

x0.5 1.0 2.0

0.5

1.0

1.5

2.0

y

�Note: �x �1 � 0

n�

1n�0, 1�y � 2x � x3,

414 Chapter 5 Integration

55.

Again, is neither an upper nor a lower sum.

Area � limn→�

T�n� � 4 � 10 �323

� 4 �23

� 4 � 10�1 �1n �

163 �2 �

3n

�1n2 � 4�1 �

2n

�1n2

�4n

�n� �20n2 �

n�n � 1�2

�32n3 �

n�n � 1��2n � 1�6

�16n4 �

n2�n � 1�2

4

� �n

i�1�2 �

10in

�16i 2

n2 �8i3

n3 �2n �

4n�

n

i�11 �

20n2 �

n

i�1i �

32n3 �

n

i�1i 2 �

16n4 �

n

i�1i 3

� �n

i�1��1 �

4in

�4i2

n2 � ��1 �6in

�12i 2

n2 �8i3

n3 �2n

T�n� � �n

i�1 f ��1 �

2in �

2n � �

n

i�1���1 �

2in

2

� ��1 �2in

3

�2n

T�n�

x1

2

1

1

y

�Note: �x �1 � ��1�

n�

2n��1, 1�y � x2 � x3,

57.

Area � limn→�

S�n� � limn→�

�6 �6n � 6

�6�n � 1�

n� 6 �

6n

�12n2 �

n

i�1i � �12

n2 �n�n � 1�

2

� �n

i�1 f �2i

n �2n � �

n

i�13�2i

n �2nS�n� � �

n

i�1 f �mi� �y

64x

4

2

2

2

y

�Note: �y �2 � 0

n�

2n0 ≤ y ≤ 2f �y� � 3y,

56.

Area � limn→�

s�n� � 2 �52

�43

�14

�7

12

� 2 �52

�5

2n�

43

�2n

�2

3n2 �14

�1

2n�

14n2

� �n

i�1��2 �

5in

�4i 2

n2 �i 3

n3�1n � 2 �

5n2 �

n

i�1i �

4n3 �

n

i�1i 2 �

1n4 �

n

i�1i 3

s�n� � �n

i�1 f ��1 �

in�

1n � �

n

i�1���1 �

in

2

� ��1 �in

3

�1n

x1 2−1−2

−1

1

2

3

y

�Note: �x �0 � ��1�

n�

1n��1, 0�y � x2 � x3,

58.

Area � limn →�

S�n� � 2 � 1 � 3

�2n�n �

1n

n�n � 1�

2 � 2 �n � 1

n

� �n

i�1 12�2 �

2in �

2n �

2n

�n

i�1�1 �

in

S�n� � �n

i�1 g�2 �

2in �

2n

y

x1

1

2

3

4

5

2 3 4 5

g�y� �12

y, 2 ≤ y ≤ 4 �Note: �y �4 � 2

n�

2n

Section 5.2 Area 415

59.

Area � limn→�

S�n� � limn→�

�9 �272n

�9

2n2 � 9

�9n2�2n2 � 3n � 1

2 � 9 �272n

�9

2n2 �27n3 �

n�n � 1��2n � 1�6

� �n

i�1 �3i

n 2

�3n �

27n3 �

n

i�1i 2S�n� � �

n

i�1 f �3i

n �3n

x2 4 6 8 10

2

4

6

−2

−4

y�Note: �y �3 � 0

n�

3n0 ≤ y ≤ 3f �y� � y2,

60.

Area � limn →�

S�n� � 3 � 1 �13

�113

� 3 �n � 1

n�

�n � 1��2n � 1�6n2

�1n

�3n �2n

n�n � 1�

2�

1n2

n�n � 1��2n � 1�

6

�1n

�n

i�1 �3 �

2in

�i2

n2

�1n

�n

i�1 �4 �

4in

� 1 �2in

�i2

n2

�1n

�n

i�1 �4�1 �

in � �1 �

in

2

S�n� � �n

i�1 f�1 �

in�

1n

1

1

2

3

5

2 3 4 5

y

x

f �y� � 4y � y2, 1 ≤ y ≤ 2 �Note: �y �2 � 1

n�

1n

61.

Area � limn →�

S�n� � 6 � 10 �83

� 4 �443

�2n

�n

i�1�3 �

10in

�4i2

n2 �8i3

n3 �2n�3n �

10n

n�n � 1�

2�

4n2

n�n � 1��2n � 1�6

�8n3

n2�n � 1�2

4

�2n

�n

i�14�1 �

4in

�4i2

n2 � �1 �6in

�12i2

n2 �8i3

n3

� �n

i�1�4�1 �

2in

2

� �1 �2in

3

2n

S�n� � �n

i�1 g�1 �

2in �

2n

x

6

y

8

10

2

−2

−4

−2−4

g�y� � 4y2 � y3, 1 ≤ y ≤ 3 �Note: �y �3 � 1

n�

2n

416 Chapter 5 Integration

62.

Area � limn →�

S�n� � 2 �14

� 1 �32

�194

� 2 ��n � 1�2

n24�

12

�n � 1��2n � 1�

n2 �3�n � 1�

2n

�1n�2n �

1n3

n2�n � 1�2

4�

3n2

n�n � 1��2n � 1�6

�3n

n�n � 1�

2

�1n

�n

i�1�2 �

i3

n3 �3i2

n2 �3in

� �n

i�1��1 �

in

3

� 11n

S�n� � �n

i�1h�1 �

in�

1n

2

1

2

3

4

5

4 6 8 10

y

x

h�y� � y3 � 1, 1 ≤ y ≤ 2 �Note: �y �1n

63.

Let

�698

�12��

116

� 3 � � 916

� 3 � �2516

� 3 � �4916

� 3

Area � � n

i�1f �ci� �x � �

4

i�1�ci

2 � 3��12

c4 �74

c3 �54

,c2 �34

,c1 �14

,�x �12

,

ci �xi � xi�1

2.

n � 40 ≤ x ≤ 2,f �x� � x2 � 3, 64.

Let

� 53

� ��14

� 2 � �94

� 6 � �254

� 10 � �494

� 14

Area � �n

i�1 f �ci� �x � �

4

i�1�ci

2 � 4ci��1�

c4 �72

c3 �52

,c2 �32

,c1 �12

,�x � 1,

ci �xi � xi�1

2.

n � 40 ≤ x ≤ 4,f �x� � x2 � 4x,

65.

Let

16�tan

32� tan

3

32� tan

5

32� tan

7

32 � 0.345

Area � � n

i�1f �ci� �x � �

4

i�1�tan ci��

16

c4 �7

32c3 �

5

32,c2 �

3

32,c1 �

32,�x �

16,

ci �xi � xi�1

2.

n � 40 ≤ x ≤

4,f �x� � tan x, 66.

Let

8�sin

16� sin

3

16� sin

5

16� sin

7

16 � 1.006

Area � �n

i�1 f �ci� �x � �

4

i�1�sin ci��

8

c4 �7

16c3 �

5

16,c2 �

3

16,c1 �

16,�x �

8,

ci �xi � xi�1

2.

n � 40 ≤ x ≤

2,f �x� � sin x,

67.

�Exact value is 16 3.�

�0, 4�f �x� � �x,

n 4 8 12 16 20

5.3838 5.3523 5.3439 5.3403 5.3384Approximate area

Section 5.2 Area 417

68. �2, 6�f �x� �8

x2 � 1,

n 4 8 12 16 20

2.3397 2.3755 2.3824 2.3848 2.3860Approximate area

70. �0, 2�f �x� � cos�x,

n 4 8 12 16 20

1.1041 1.1053 1.1055 1.1056 1.1056Approximate area

69. �1, 3�f �x� � tan�x8 ,

n 4 8 12 16 20

2.2223 2.2387 2.2418 2.2430 2.2435Approximate area

71. �1, 5�f �x� � ln x,

n 4 8 12 16 20

4.0786 4.0554 4.0509 4.0493 4.0485Approximate area

72. �0, 2�f �x� � xe x,

n 4 8 12 16 20

8.1711 8.3341 8.3646 8.3753 8.3802Approximate area

74.

(a) square unitsA � 3

x1 2 3 4

1

2

3

4

y73.

(b) square unitsA � 6

x1 2 3 4

1

2

3

4

y

75. We can use the line bounded by and The sum of the areas of these inscribed rectangles is thelower sum.

x

y

a b

x � b.x � ay � x The sum of the areas of these circumscribed rectangles is theupper sum.

We can see that the rectangles do not contain all of the area inthe first graph and the rectangles in the second graph covermore than the area of the region. The exact value of the arealies between these two sums.

x

y

a b

418 Chapter 5 Integration

76. See the definition of area, page 297.

80. True. (Theorem 5.3)79. True. (Theorem 5.2 (2))

78.

Let area bounded by the x-axis, and Let area of the rectangle bounded by and Thus,In this program, the computer is generating pairs of random points in the rectangle whose area is represented by It is keeping track of how many of these points, lie in the regionwhose area is represented by Since the points are randomly generated, we assume that

The larger is the better the approximation to A1.N2

A1

A2�

N1

N2 ⇒ A1 �

N1

N2A2.

A1.N1,A2.

N2

A2 � � 2��1� � 1.570796.x � 2.x � 0,y � 0,y � 1,A2 �x � 2.x � 0f �x� � sin x,A1 �

x

0.25

0.5

0.75

1.0

π4

π2

π2( ), 1f x x( ) sin( )=

y�0,

2f �x� � sin x,

81. Suppose there are n rows and columns in the figure. The stars on the left total as do the stars on the right. There are stars in total, hence

1 � 2 � . . . � n �12�n��n � 1�.

2�1 � 2 � . . . � n� � n�n � 1�

n�n � 1�1 � 2 � . . . � n,n � 1

77. (a)

Lower sum:

(c)

Midpoint Rule:

(e)

(f) increases because the lower sum approaches the exact value as n increases. decreases because the upper sumapproaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than theexact value, whereas the upper sum is always larger.

S�n�s�n�

M�4� � 223 � 44

5 � 557 � 62

9 �6112315 � 19.403

x1

2

2 3

4

4

6

8

y

s�4� � 0 � 4 � 513 � 6 � 151

3 �463 � 15.333

x1

2

2 3

4

4

6

8

y (b)

Upper sum:

(d) In each case, The lower sum uses left end-points, The upper sum uses right endpoints,

The Midpoint Rule uses midpoints, �i �12��4 n�.�i��4 n�.

�i � 1��4 n�.�x � 4 n.

S�4� � 4 � 513 � 6 � 62

5 � 211115 �

32615 � 21.733

x1

2

2 3

4

4

6

8

y

n 4 8 20 100 200

15.333 17.368 18.459 18.995 19.06

21.733 20.568 19.739 19.251 19.188

19.403 19.201 19.137 19.125 19.125M�n�

S�n�

s�n�

Section 5.2 Area 419

83. (a)

(c) Using the integration capability of a graphing utility,you obtain

A � 76,897.5 ft2.

y � ��4.09 10�5�x3 � 0.016x2 � 2.67x � 452.9 (b)

00 350

500

82. (a)

(b)

(c)

Let As

limn→�

An � limx→0

r2�sin xx � r2�1� � r 2

x → 0.n →�,x � 2 n.

An � n�12

r2 sin 2

n �r2n2

sin 2

n� r2�sin�2 n�

2 n

A �12

bh �12

r�r sin �� �12

r2 sin �

h � r sin �

sin � �hr θ

r

r

h

� �2

n

84. For n odd,

1 row, 1 block

2 rows, 4 blocks

3 rows, 9 blocks

�n � 12

2 blocks

n � 12

rows,n,

n � 5,

n � 3,

n � 1,

For n even,

1 row, 2 blocks

2 rows, 6 blocks

3 rows, 12 blocks

n2 � 2n4

blocksn2

rows,n,

n � 6,

n � 4,

n � 2,

85. (a)

The formula is true for

Assume that the formula is true for

Then we have

which shows that the formula is true for n � k � 1.

� �k � 1��k � 2�

� k�k � 1� � 2�k � 1�

�k�1

i�12i � �

k

i�12i � 2�k � 1�

�k

i�12i � k�k � 1�.

n � k:

n � 1: 2 � 1�1 � 1� � 2.

�n

i�12i � n�n � 1� (b)

The formula is true for because

Assume that the formula is true for

Then we have

which shows that the formula is true for n � k � 1.

��k � 1�2

4�k � 2�2

��k � 1�2

4�k2 � 4�k � 1��

�k2�k � 1�2

4� �k � 1�3

�k�1

i�1i3 � �

k

i�1i3 � �k � 1�3

�k

i�1 i3 �

k2�k � 1�2

4.

n � k:

13 �12�1 � 1�2

4�

44

� 1.

n � 1

�n

i�1i3 �

n2�n � 1�2

4

420 Chapter 5 Integration

86. Assume that the dartboard has corners at

A point in the square is closer to the center than the top edge if

By symmetry, a point in the square is closer to the center than the rightedge if

In the first quadrant, the parabolas and intersect at There are 8 equal regions that make up the total region, as indicated in the figure.

Area of shaded region

Probability �8S

Area square� 2�2�2

3�

56� �

4�23

�53

S � ��2�1

0 �1

2�1 � x2� � x� dx �

2�23

�56

−1 1

−1

1

x

y

2 − 1, 2 − 1( (

(1, 1)

��2 � 1, �2 � 1�.x �12�1 � y2�y �

12�1 � x2�

x ≤12

�1 � y2�.

�x, y�

y ≤12

�1 � x2�.

x2 � y2 ≤ 1 � 2y � y2

�x2 � y2 ≤ 1 � y

�x, y�

(0, 0)

(x, 1)

(x, y)

−1 1

−1

1

x

y�±1, ±1�.

Section 5.3 Reimann Sums and Definite Integrals

1.

� 3�3�23

� 0� � 2�3 � 3.464

� limn →�

3�3��n � 1��2n � 1�3n2 �

n � 12n2 �

� limn →�

3�3

n3 �2n�n � 1��2n � 1�

6�

n�n � 1�2 �

� limn →�

3�3

n3 �n

i�1�2i2 � i�

limn →�

�n

i�1f �ci��xi � lim

n →� �

n

i�1�3i2

n2 3n2�2i � 1�

�xi �3i2

n2 �3�i � 1�2

n2 �3n2�2i � 1�

3n2

3(2)2

n23(n − 1)2

n2. . .

. . .

3

1

2

3

y

x

y = x

f �x� � �x, y � 0, x � 0, x � 3, ci �3i2

n2

Section 5.3 Reimann Sums and Definite Integrals 421

2.

� 2 limn→�

1n4�3n4

4�

n3

2�

n2

4 � � 2 limn→�

�34

�1

2n�

14n2� �

32

� 2 limn→�

1n4�3n4 � 6n3 � 3n2

4�

2n3 � 3n2 � n2

�n2 � n

2 �

� 2 limn→�

1n4�3n2�n � 1�2

4 � 3n�n � 1��2n � 1�6 �

n�n � 1�2 �

� 2 limn→�

1n4 �

n

i�1�3i 3 � 3i 2 � i�

limn→�

�n

i�1 f �ci� �xi � 2 lim

n→� �

n

i�1

3�i 3

n3 �3i 2 � 3i � 1n3 �

�xi �i 3

n3 ��i � 1�3

n3 �3i 2 � 3i � 1

n3

ci �i 3

n3x � 1,x � 0,y � 0,f �x� � 2 3�x,

3. on

�10

4 6 dx � lim

n→� 36 � 36

� �n

i�166

n � �n

i�1

36n

� 36� n

i�1f �ci� �xi � �

n

i�1 f 4 �

6in

6n

Note: �x �10 � 4

n�

6n

, ��� → 0 as n →��4, 10 .y � 6

4. on

�3

�2x dx � lim

n→� 5

2�

252n �

52

� �10 � 25n2n�n � 1�

2� �10 �

252 1 �

1n �

52

�252n

�n

i�1 f �ci � �xi � �

n

i�1 f �2 �

5in

5n � �

n

i�1�2 �

5in

5n � �10 �

25n2 �

n

i�1i

Note: �x �3 � ��2�

n�

5n

, ��� → 0 as n →���2, 3 .y � x

5. on

�1

�1x3 dx � lim

n→� 2n

� 0

� �2 � 61 �1n � 42 �

3n

�1n2 � 41 �

2n

�1n2 �

2n

� �2 �12n2 �

n

i�1i �

24n3 �

n

i�1i 2 �

16n4 �

n

i�1i 3

�n

i�1 f �ci� �xi � �

n

i�1 f �1 �

2in

2n � �

n

i�1�1 �

2in

3

2n � �

n

i�1��1 �

6in

�12i 2

n2 �8i 3

n3 �2n

Note: �x �1 � ��1�

n�

2n

, ��� → 0 as n →���1, 1 .y � x3

422 Chapter 5 Integration

6. on

� 6 � 12 � 8 � 26

�3

13x2 dx � lim

n →� �6 �

12�n � 1�n

�4�n � 1��2n � 1�

n2 �

� 6 � 12 n � 1

n� 4

�n � 1��2n � 1�n2

�6n�n �

4n

n�n � 1�

2�

4n2

n�n � 1��2n � 1�6 �

�6n�

n

i�11 �

4in

�4i2

n2

� �n

i�131 �

2in

2

2n

�n

i�1 f �ci� �xi � �

n

i�1 f 1 �

2in

2n

�1, 3 . Note: �x �3 � 1

n�

2n

, ��� → 0 as n →�y � 3x2

7. on

�2

1�x2 � 1� dx � lim

n→� 10

3�

32n

�1

6n2 �103

� 2 �2n2 �

n

i�1i �

1n3 �

n

i�1i2 � 2 � 1 �

1n �

162 �

3n

�1n2 �

103

�3

2n�

16n2

�n

i�1 f �ci� �xi � �

n

i�1 f 1 �

in

1n � �

n

i�1�1 �

in

2

� 1�1n � �

n

i�1�1 �

2in

�i2

n2 � 1�1n

Note: �x �2 � 1

n�

1n

, ��� → 0 as n →��1, 2 .y � x2 � 1

8. on

� 15 � 27 � 27 � 15

�2

�1 �3x2 � 2� dx � lim

n →� �15 � 27

�n � 1�n

�272

�n � 1��2n � 1�

n2 �

� 15 �27�n � 1�

n�

272

�n � 1��2n � 1�

n2

�3n�3n �

18n

n�n � 1�

2�

27n2

n�n � 1��2n � 1�6

� 2n�

�3n

�n

i�1�31 �

6in

�9i2

n2 � 2�

�3n�

n

i�1�3�1 �

3in

2

� 2

�n

i�1 f �ci� �xi � �

n

i�1 f �1 �

3in

3n

��1, 2 . Note: �x �2 � ��1�

n�

3n

; ��� → 0 as n →�y � 3x2 � 2

9.

on the interval ��1, 5 .

lim���→0

�n

i�1�3ci � 10� �xi � �5

�1�3x � 10� dx 10.

on the interval �0, 4 .

lim���→0

�n

i�16ci�4 � ci�2 �xi � �4

06x�4 � x�2 dx

Section 5.3 Reimann Sums and Definite Integrals 423

11.

on the interval �0, 3 .

lim���→0

�n

i�1

�ci2 � 4 �xi � �3

0

�x2 � 4 dx 12.

on the interval �1, 3 .

lim���→0

�n

i�1 3

ci2 �xi � �3

1

3x2 dx

13. �5

1 1 �

3x dx 14. ��

0 2�x sin x dx 15. �5

03 dx 16. �2

0x2 dx

17. �4

1 2x dx 18. �2

02e�x dx 19. ��

0sin x dx 20. ���4

0tan x dx

21. �2

0y3 dy 22. �2

0�y � 2�2 dy

23. Rectangle

x5

5

3

2

42 31

1

Rectangle

y

A � �3

04 dx � 12

A � bh � 3�4�

24. Rectangle

x

1

2

3

5

−a a

Rectangle

y

A � �a

�a

4 dx � 8a

A � bh � 2�4��a�

26. Triangle

x1 2 3 4

−1

1

2

3

Triangle

y

A � �4

0

x2

dx � 4

A �12

bh �12

�4��2�

25. Triangle

x

Triangle

4

2

42

y

A � �4

0x dx � 8

A �12

bh �12

�4��4�

27. Trapezoid

x

9

6

321

3

Trapezoid

y

A � �2

0�2x � 5� dx � 14

A �b1 � b2

2h � 5 � 9

2 2

29. Triangle

A � �1

�1�1 � �x�� dx � 1

A �12

bh �12

�2��1�1 Triangle

11x

y

28. Triangle

2

2

4

6

8

10

4 6 8 10

y

x

A � �8

0�8 � x�dx � 32

A �12

bh �12

�8��8� � 32

30. Triangle

A � �a

�a

�a � �x�� dx � a2

A �12

bh �12

�2a�a

x

Triangle

−a a

a

y

424 Chapter 5 Integration

32. Semicircle

A � �r

�r

�r2 � x2 dx

A �12

�r2

x−r

−r

r

r Semicircle

y31. Semicircle

A � �3

�3

�9 � x2 dx �9�

2

A �12

�r 2 �12

��3�2

x

4 Semicircle

4224

y

33. �2

4x dx � ��4

2x dx � �6

35. �4

24x dx � 4�4

2x dx � 4�6� � 24

In Exercises 33–40, �4

2x3 dx � 60, �4

2x dx � 6, �4

2dx � 2

37. �4

2�x � 8� dx � �4

2x dx � 8�4

2dx � 6 � 8�2� � �10

39.

�12

�60� � 3�6� � 2�2� � 16

�4

21

2x3 � 3x � 2 dx �

12�

4

2x3 dx � 3�4

2x dx � 2�4

2dx

34. �2

2x3 dx � 0

36. �4

215 dx � 15�4

2dx � 15�2� � 30

38. �4

2�x3 � 4� dx � �4

2x3 dx � 4�4

2dx � 60 � 4�2� � 68

40.

� 6�2� � 2�6� � 60 � �60

�4

2�6 � 2x � x3� dx � 6�4

2dx � 2�4

2x dx � �4

2x3 dx

42. (a)

(b)

(c)

(d) �6

3�5f �x� dx � �5�6

3f �x� dx � �5��1� � 5

�3

3f �x� dx � 0

�3

6f �x� dx � ��6

3f �x� dx � ���1� � 1

�6

0f �x� dx � �3

0f �x� dx � �6

3f �x� dx � 4 � ��1� � 3

44. (a)

(b)

(c)

(d) �1

03f �x� dx � 3�1

0f �x� dx � 3�5� � 15

�1

�13f �x� dx � 3�1

�1f �x� dx � 3�0� � 0

�1

0f �x� dx � �0

�1f �x� dx � 5 � ��5� � 10

�0

�1f �x� dx � �1

�1f �x� dx � �1

0f �x� dx � 0 � 5 � �5

41. (a)

(b)

(c)

(d) �5

03f �x� dx � 3�5

0f �x� dx � 3�10� � 30

�5

5f �x� dx � 0

�0

5f �x� dx � ��5

0f �x� dx � �10

�7

0f �x� dx � �5

0f �x� dx � �7

5f �x� dx � 10 � 3 � 13

43. (a)

(b)

(c)

(d) �6

23f �x� dx � 3�6

2f �x� dx � 3�10� � 30

�6

22g�x� dx � 2�6

2g�x� dx � 2��2� � �4

� �2 � 10 � �12

�6

2�g �x� � f �x� dx � �6

2g�x� dx � �6

2f �x� dx

� 10 � ��2� � 8

�6

2� f �x� � g�x� dx � �6

2f �x� dx � �6

2g�x� dx

45. Lower estimate:

Upper estimate: �32 � 24 � 12 � 4 � 20 �2� � 88

�24 � 12 � 4 � 20 � 36 �2� � �48 46. (a) left endpoint estimate

(b) right endpoint estimate

(c) midpoint estimate

If f is increasing, then (a) is below the actual value and (b) is above.

�0 � 18 � 50 �2� � 136,

�8 � 30 � 80 �2� � 236,

��6 � 8 � 30 (2) � 64,

Section 5.3 Reimann Sums and Definite Integrals 425

49. (a)

(b)

(c)

(d) � f odd��5

�5f �x� dx � 0

� f even��5

�5f �x� dx � 2�5

0f �x� dx � 2�4� � 8

�Let u � x � 2.��3

�2f �x � 2� dx � �5

0f �x� dx � 4

�5

0� f �x� � 2 dx � �5

0f �x� dx � �5

02 dx � 4 � 10 � 14

47. (a) Quarter circle below x-axis:

(b) Triangle:

(c) Triangle Semicircle below x-axis:

(d) Sum of parts (b) and (c):

(e) Sum of absolute values of (b) and (c):

(f) Answer to (d) plus �3 � 2�� � 20 � 23 � 2�2�10� � 20:

4 � �1 � 2�� � 5 � 2�

4 � �1 � 2�� � 3 � 2�

�12 �2��1� �

12 ��2�2 � ��1 � 2���

12 bh �

12 �4��2� � 4

�14 �r 2 � �

14 ��2�2 � ��

50.

4 8

4

8(8, 8)

x

y

�8

0 f �x� dx � 4�4� � 4�4� �

12

�4��4� � 40

f �x� � �4,x,

x < 4 x ≥ 4

48. (a)

(b)

(c)

(d)

(e)

(f) �10

4f (x) dx � 2 � 4 � �2

�11

0f (x) dx � �

12

� 2 � 2 � 2 � 4 �12

� 2

�11

5f (x) dx �

12

�12

�4��2� �12

� �3

�7

0f (x) dx � �

12

�12

�2��2� � 2 �12

�2��2� � 12

� 5

�4

33f (x) dx � 3�2� � 6

x

y

(4, 2)

(11, 1)

(3, 2)

(0, �1)

(8, �2)

−2 2 4 8 12

−1

−2

1

2

�1

0�f �x� dx � ��1

0f �x� dx �

12

51. The left endpoint approximation will be greater than the actual area: >

52. The right endpoint approximationwill be less than the actual area: <

53.

is not integrable on the intervalbecause f has a discontinuity

at x � 4.�3, 5

f �x� �1

x � 4

54. is integrable on but is not continuous on There is discontinuity at To see that

is integrable, sketch a graph of the region bounded by and the x-axis forYou see that the integral equals 0.�1 ≤ x ≤ 1.

f �x� � �x��x

�1

�1

�x�x

dx

x � 0.

x1 2−2

−2

1

2

y��1, 1 .��1, 1 ,f �x� � �x��x

426 Chapter 5 Integration

56.

(b) square unitsA � 43

x1 1314 42

1

2

3

4

y55.

(a) square unitsA � 5

x1 2 3 4

1

2

3

4

y

57.

�2

02e�x 2

dx �12

�2��2� � 2

1

3

1 2 3

y

x

58.

Area �13

1

1 2

−1

y

x

�2

1ln x dx

59. �3

0x�3 � x dx

4 8 12 16 20

3.6830 3.9956 4.0707 4.1016 4.1177

4.3082 4.2076 4.1838 4.1740 4.1690

3.6830 3.9956 4.0707 4.1016 4.1177R�n�

M�n�

L�n�

n

60. �3

0

5x2 � 1

dx

4 8 12 16 20

7.9224 7.0855 6.8062 6.6662 6.5822

6.2485 6.2470 7.2460 6.2457 6.2455

4.5474 5.3980 5.6812 5.8225 5.9072R�n�

M�n�

L�n�

n

61. �3

1 1x dx

4 8 12 16 20

1.2833 1.1865 1.1562 1.1414 1.1327

1.0898 1.0963 1.0976 1.0980 1.0982

0.9500 1.0199 1.0451 1.0581 1.0660R�n�

M�n�

L�n�

n

Section 5.3 Reimann Sums and Definite Integrals 427

64. �3

0x sin x dx

4 8 12 16 20

2.8186 2.9985 3.0434 3.0631 3.0740

3.1784 3.1277 3.1185 3.1152 3.1138

3.1361 3.1573 3.1493 3.1425 3.1375R�n�

M�n�

L�n�

n

62. �4

0 e x dx

4 8 12 16 20

31.1929 41.3106 45.1605 47.1772 48.4169

51.4284 53.0439 53.3508 53.4588 53.5089

84.7910 68.1097 63.0265 60.5768 59.1365R�n�

M�n�

L�n�

n

63. ���2

0sin2 x dx

4 8 12 16 20

0.5890 0.6872 0.7199 0.7363 0.7461

0.7854 0.7854 0.7854 0.7854 0.7854

0.9817 0.8836 0.8508 0.8345 0.8247R�n�

M�n�

L�n�

n

65. True 66. False

�1

0x�x dx � �1

0x dx�1

0

�x dx67. True

68. True 69. False

�2

0��x� dx � �2

70. True. The limits of integration arethe same.

71.

� �4��1� � �10��2� � �40��4� � �88��1� � 272

�4

i�1 f �ci� �x � f �1� �x1 � f �2� �x2 � f �5� �x3 � f �8� �x4

c4 � 8c3 � 5,c2 � 2,c1 � 1,

�x4 � 1�x3 � 4,�x2 � 2,�x1 � 1,

x4 � 8x3 � 7,x2 � 3,x1 � 1,x0 � 0,

�0, 8 f �x� � x2 � 3x,

428 Chapter 5 Integration

72.

� 12

4 � �32 �

12 � �32 2�

3 � ��1���� � �0.708

�4

i�1 f �ci� �xi � f �

6 �x1 � f �

3 �x2 � f 2�

3 �x3 � f 3�

2 �x4

c4 �3�

2c3 �

2�

3,c2 �

3,c1 �

6,

�x4 � ��x3 �2�

3,�x2 �

12,�x1 �

4,

x4 � 2�x3 � �,x2 ��

3,x1 �

4,x0 � 0,

�0, 2� f �x� � sin x,

73.

��b � a��a � b�

2�

b2 � a2

2

� �b � a��a �b � a

2 �

� a�b � a� ��b � a�2

2

� limn→�

�a�b � a� ��b � a�2

n n � 1

2 �

� limn→�

�b � an

�an� � b � an 2

n�n � 1�

2 �

� limn→�

�b � an �

n

i�1a � b � a

n 2 �

n

i�1i�

� limn→�

�n

i�1�a � ib � a

n �b � an

�b

a

x dx � lim���→0

�n

i�1 f �ci� �x

�x �b � a

n, ci � a � i��x� � a � ib � a

n

74.

�13

�b3 � a3�

� a2�b � a� � a�b � a�2 �13

�b � a�3

� limn→�

�a2�b � a� �a�b � a�2�n � 1�

n�

�b � a�3

6 �n � 1��2n � 1�

n2 �

� limn→�

b � an �na2 �

2a�b � a�n

n�n � 1�

2� b � a

n 2 n�n � 1��2n � 1�

6 �

� limn→�

�b � an �

n

i�1a2 �

2ai�b � a�n

� i2b � an 2�

� limn→�

�n

i�1�a � ib � a

n �2b � an

�b

a

x 2 dx � lim���→0

�n

i�1 f �ci� �x

�x �b � a

n, ci � a � i��x� � a � ib � a

n

Section 5.4 The Fundamental Theorem of Calculus 429

75.

is not integrable on the interval As or in each subinterval since there

are an infinite number of both rational and irrational numbers in any interval, no matter how small.

f �ci� � 0f �ci� � 1��� → 0,�0, 1�.

f �x� � �1,0,

x is rationalx is irrational

76.

The limit

does not exist. This does notcontradict Theorem 5.4 becausef is not continuous on �0, 1�.

lim���→0 �

n

i�1 f �ci� �xi

1 2

1

2

x

f(x) = 1x

y

f �x� � �0,1x

,

x � 0 0 < x ≤ 1

77. The function f is nonnegative between and

Hence,

is a maximum for and b � 1.a � �1

�b

a

�1 � x2� dx

−2 1 2

−1

−2

2

x

y

f(x) = 1 − x2

x � 1.x � �1 78. To find use a geometric approach.

Thus,

�2

0x dx � 1�2 � 1� � 1.

x1 2 3−1

−1

1

2

3

y

�20 x dx,

79. Let and The appropriate Riemann Sum is

� limn→�

2n2 � 3n � 1

6n2 � limn→�

�13

�1

2n�

16n2 �

13

limn→�

1n3�12 � 22 � 32 � . . . � n2� � lim

n→� 1n3 �

n�2n � 1��n � 1�6

�n

i�1f �ci � �xi � �

n

i�1� i

n 2 1

n�

1n3 �

n

i�1i 2.

�xi � 1�n.f �x� � x2, 0 ≤ x ≤ 1,

Section 5.4 The Fundamental Theorem of Calculus

1.

is positive.��

0

4x2 � 1

dx

−2

−5 5

5f �x� �

4x2 � 1

3.

�2

�2x�x2 � 1 dx � 0

−5

−5 5

5f �x� � x�x2 � 1

5. �1

02x dx � �x2�

1

0� 1 � 0 � 1

2.

��

0cos x dx � 0 0

−2

2f �x� � cos x

4.

is negative.�2

�2x�2 � x dx −2 2

−5

5f �x� � x�2 � x

6. �7

23 dv � �3v�

7

2� 3�7� � 3�2� � 15

430 Chapter 5 Integration

17. �1

�1�3�t � 2� dt � �3

4t 4�3 � 2t�

1

�1� �3

4� 2 � �3

4� 2 � �4

18. �8

1�2

x dx � �2�8

1x�1�2 dx � ��2�2�x1�2�

8

1� �2�2x�

8

1� 8 � 2�2

19. �1

0

x � �x3

dx �13�

1

0�x � x1�2� dx �

13�

x2

2�

23

x3�2�1

0�

13�

12

�23 � �

118

20. �2

0�2 � t��t dt � �2

0�2t1�2 � t3�2� dt � �4

3t3�2 �

25

t5�2�2

0� �t�t

15�20 � 6t��

2

0�

2�215

�20 � 12� �16�2

15

21. �0

�1�t1�3 � t2�3� dt � �3

4t 4�3 �

35

t5�3�0

�1� 0 � �3

4�

35 � �

2720

22.

�12�

35

x5�3 �38

x8�3��1

�8� �x5�3

80�24 � 15x��

�1

�8� �

180

�39� �3280

�144� �456980

��1

�8

x � x2

2 3�x dx �

12�

�1

�8�x2�3 � x5�3� dx

7. �0

�1�x � 2� dx � �x2

2� 2x�

0

�1� 0 � �1

2� 2 � �

52

9. �1

�1�t2 � 2� dt � �t3

3� 2t�

1

�1� �1

3� 2 � ��

13

� 2 � �103

8. �5

2��3v � 4� dv � ��

32

v2 � 4v�5

2� ��

752

� 20 � ��6 � 8� � �392

10. �3

1

�3x2 � 5x � 4� dx � �x3 �5x2

2� 4x�

3

1� �27 �

45

2� 12 � �1 �

5

2� 4 � 38

11. �1

0�2t � 1�2 dt � �1

0�4t2 � 4t � 1� dt � �4

3t3 � 2t2 � t�

1

0�

43

� 2 � 1 �13

13. �2

1� 3

x2 � 1 dx � ��3x

� x�2

1� ��

32

� 2 � ��3 � 1� �12

15. �4

1

u � 2

�u du � �4

1�u1�2 � 2u�1�2� du � �2

3u3�2 � 4u1�2�

4

1� �2

3��4�3

� 4�4� � �23

� 4� �23

12. �1

�1�t3 � 9t� dt � �1

4t4 �

92

t2�1

�1� �1

4�

92 � �1

4�

92 � 0

14. ��1

�2�u �

1u2 du � �u2

2�

1u�

�1

�2� �1

2� 1 � �2 �

12 � �2

16. �3

�3v1�3 dv � �3

4v 4�3�

3

�3�

34

��3�3 �4 � �3��3 �4� � 0

Section 5.4 The Fundamental Theorem of Calculus 431

23. split up the integral at the zero

� �3x � x2�3�2

0� �x2 � 3x�

3

3�2� �9

2�

94 � 0 � �9 � 9� � �9

4�

92 � 2�9

2�

94 �

92

x �32 ��3

0�2x � 3� dx � �3�2

0�3 � 2x� dx � �3

3�2�2x � 3� dx

24.

� �4 � 0� � ��35 �252 � �28 � 8�� �

132

� �x2

2� x�

4

2� �7x �

x2

2 �5

4

� �4

2�x � 1� dx � �5

4�7 � x� dx

�5

2�3 � �x � 4�� dx � �4

2�3 � �x � 4�� dx � �5

4�3 � �x � 4�� dx

25.

� �8 �83 � �9 � 12� � �8

3� 8 �

233

� �4x �x3

3 �2

0� �x3

3� 4x�

3

2

�3

0�x2 � 4� dx � �2

0�4 � x2� dx � �3

2�x2 � 4� dx

26.

�43

� 0 �43

�43

� 0 � 4

� �13

� 2 � 3 � �9 � 18 � 9� � �13

� 2 � 3 � �643

� 32 � 12 � �9 � 18 � 9�

� �x3

3� 2x2 � 3x�

1

0� �x3

3� 2x2 � 3x�

3

1� �x3

3� 2x2 � 3x�

4

3

�4

0�x2 � 4x � 3� dx � �1

0�x2 � 4x � 3� dx � �3

1�x2 � 4x � 3� dx � �4

3�x2 � 4x � 3� dx split up the integral at

the zeros x � 1, 3��

27. ��

0�1 � sin x� dx � �x � cos x�

0� �� � 1� � �0 � 1� � 2 � �

29. ���6

���6sec2 x dx � �tan x�

��6

���6�

�33

� ���33 �

2�33

31. �e

1 �2x �

1x dx � �x2 � ln x�

e

1� �e2 � 1� � �1 � 0� � e2 � 2

28. ���4

0

1 � sin2 �cos2 �

d� � ���4

0d� � ���

��4

0�

4

30. ���2

��4�2 � csc2 x� dx � �2x � cot x�

��2

��4� �� � 0� � ��

2� 1 �

2� 1 �

� � 22

32. �5

1 x � 1

x dx � �5

1 �1 �

1x dx � �x � ln x�

5

1� �5 � ln 5� � �1 � ln 1� � 4 � ln 5

33. ���3

���34 sec � tan � d� � �4 sec ��

��3

���3� 4�2� � 4�2� � 0 34.

� �� 2

4� 1 � �� 2

4� 1 � 2

���2

���2�2t � cos t� dt � �t 2 � sin t�

��2

���2

432 Chapter 5 Integration

47. Since on

A � �2

0�x3 � x� dx � �x4

4�

x2

2 �2

0� 4 � 2 � 6.

�0, 2�,y ≥ 0 48. Since on

A � �3

0�3x � x2� dx � �3

2x2 �

x3

3 �3

0�

92

.

�0, 3�,y ≥ 0

49. �e

1 4x dx � 4 ln x�

e

1� 4 ln e � 4 ln 1 � 4 50. �2

0 e x dx � e x�

2

0� e2 � e0 � e2 � 1

51.

c � 0.4380 or c � 1.7908

c � �1 ± �6 � 4�23 �2

�c � 1 � ±�6 � 4�23

��c � 1�2�

6 � 4�23

c � 2�c � 1 �3 � 4�2

3� 1

c � 2�c �3 � 4�2

3

f �c��2 � 0� �6 � 8�2

3

�2

0�x � 2�x � dx � �x2

2�

4x3�2

3 �2

0� 2 �

8�23

52.

c � 3�92

� 1.6510

c3 �92

9c3 � 2

f �c��3 � 1� � 4

�3

1

9x3 dx � ��

92x2�

3

1� �

12

�92

� 4

432 Chapter 5 Integration

35. �3

ln 2� 12�2

0 �2x � 6� dx � � 2x

ln 2� 6x�

2

0� � 4

ln 2� 12 � � 1

ln 2� 0

37. �1

�1 �e� � sin �� d� � �e� � cos ��

1

�1� �e � cos 1� � �e�1 � cos��1�� � e �

1e

36. �92

�124ln 5

� �72.546 �3

0 �t � 5t� dt � �t2

2�

5t

ln 5�3

0� �9

2�

125ln 5 � �0 �

1ln 5

38. �2e

e

�cos x �1x dx � �sin x � ln x�

2e

e� �sin�2e� � ln�2e�� � �sin�e� � ln�e�� � sin�2e� � sin�e� � ln 2

39. A � �1

0�x � x2� dx � �x2

2�

x3

3 �1

0�

16

40. A � �1

�1�1 � x4� dx � �x �

15

x5�1

�1�

85

41. A � �3

0�3 � x��x dx � �3

0�3x1�2 � x3�2� dx � �2x3�2 �

25

x5�2�3

0� �x�x

5�10 � 2x��

3

0�

12�35

42. A � �2

1

1x2 dx � ��

1x�

2

1� �

12

� 1 �12

44. A � ��

0�x � sin x� dx � �x2

2� cos x�

0�

� 2

2� 2 �

� 2 � 42

43. A � ���2

0cos x dx � �sin x�

��2

0� 1

45. Since on

A � �2

0�3x2 � 1� dx � �x3 � x�

2

0� 8 � 2 � 10.

�0, 2�,y ≥ 0 46. Since on

Area � �8

0�1 � x1�3� dx � �x �

34

x 4�3�8

0� 8 �

34

�16� � 20.

�0, 8�,y ≥ 0

Section 5.4 The Fundamental Theorem of Calculus 433

53.

� ±arccos ��

2� ±0.4817

c � ±arcsec� 2

��

sec c � ±2

��

sec2 c �4�

2 sec2 c �8�

f �c���

4� ��

4 � � 4

���4

���42 sec2 x dx � �2 tan x�

��4

���4� 2�1� � 2��1� � 4 54.

c � ±0.5971

cos c �3�32�

f �c���

3� ��

3 � � �3

���3

���3cos x dx � �sin x�

��3

���3� �3

55.

c �3

ln 4� 2.1640

3c

� ln 4

15 �3c

� 15 � ln 4

�5 �1c �3� � 15 � ln 4

f �c��4 � 1� � 15 � ln 4

� �20 � ln 4� � �5 � 0� � 15 � ln 4

�4

1

�5 �1x dx � �5x � ln x�

4

1

�1, 4�f �x� � 5 �1x, 56.

c � log2� 73 ln 2 � 1.7512

2c �7

3 ln 2

3�2c� �7

ln 2

�10 � 2c��3� � 30 �7

ln 2

f �c��3 � 0� � 30 �7

ln 2

� �30 �8

ln 2 � �0 �1

ln 2 � 30 �7

ln 2

�3

0 �10 � 2x� dx � �10x �

2x

ln 2�3

0

�0, 3�f �x� � 10 � 2x,

57.

Average value

when or x � ±2�3

3� ±1.155.x2 � 4 �

83

4 � x2 �83

�83

5

−1

−3 3

2 33 3

8,−( ( 2 33 3

8,( (12 � ��2��

2

�2�4 � x2� dx �

14�4x �

13

x3�2

�2�

14��8 �

83 � ��8 �

83 � �

83

58.

� 2�3 �13 �

163

1

3 � 1�3

1

4�x2 � 1�x2 dx � 2�3

1�1 � x�2� dx � 2�x �

1x�

3

1

434 Chapter 5 Integration

60.

0

1

40

x �3

ln 4� 2.1640

2x �6

ln 4

12x

�16

ln 4

�16

ln x�4

1�

16

ln 4 � 0.2310

Average value �1

4 � 1 �4

1 12x

dx

�1, 4�f �x� �12x

,

61.

x � 0.690, 2.451

sin x �2�

−1

2

20.690,

2.451,

(((

π

2

�23�

2

Average value �2�

1� � 0�

0sin x dx � ��

1�

cos x��

0�

2�

62.

x � 0.881

cos x �2�

−0.5

0 2.71

0.881, 2π ((

1.5

Average value �2�

1���2� � 0

���2

0cos x dx � � 2

� sin x�

��2

0�

2�

63. The distance traveled is The area under the curve from is approximately�18 squares��30� � 540 ft.

0 ≤ t ≤ 8�8

0 v�t� dt. 64. The distance traveled is The area under the curve from is approximately�29 squares��5� � 145 ft.

0 ≤ t ≤ 5�5

0 v�t� dt.

65. If is continuous on and then �b

a

f �x� dx � F�b� � F�a�.F�x� � f �x� on �a, b�,�a, b�f

66. (a)

x1 2 3 4 5 6 7

1

2

3

4

A

A A A

1

2 3 4

y

� 8

�12

�3 � 1� �12

�1 � 2� �12

�2 � 1� � �3��1�

� A1 � A2 � A3 � A4

�7

1f �x� dx � Sum of the areas (b)

(c)

Average value

x1 2 3 4 5 6 7

1

2

3

4

5

6

7

y

�206

�103

A � 8 � �6��2� � 20

Average value ��7

1 f �x� dx

7 � 1�

86

�43

434 Chapter 5 Integration

59.

−1

5

1

−1

(0.16, 2.35)

x � ln�e � e�1

2 � 0.1614

e x �12

�e � e�1�

2e x � e � e�1

� e x�1

�1� e � e�1 � 2.3504

Average value �1

1 � ��1� �1

�1 2e x dx � �1

�1 e x dx

��1, 1�f �x� � 2e x,

Section 5.4 The Fundamental Theorem of Calculus 435

67. �2

0f �x� dx � ��area of region A� � �1.5 68.

� 3.5 � ��1.5� � 5.0

�6

2f �x� dx � �area of region B� � �6

0f �x� dx � �2

0f �x� dx

69. �6

0� f �x�� dx � ��2

0f �x� dx ��6

2f �x� dx � 1.5 � 5.0 � 6.5

71.

� 12 � 3.5 � 15.5

�6

0�2 � f �x�� dx � �6

02 dx � �6

0f �x� dx

73. (a)

F �x� � 500 sec2 x

F�0� � k � 500

F�x� � k sec2 x (b)

� 826.99 newtons � 827 newtons

�1500

���3 � 0�

1��3 � 0�

��3

0500 sec2 x dx �

1500� �tan x�

��3

0

70. �2

0�2f �x� dx � �2�2

0f �x� dx � �2��1.5� � 3.0

72. Average value �16�

6

0f �x� dx �

16

�3.5� � 0.5833

74.1

R � 0�R

0k�R2

� r2� dr �kR�R2r �

r3

3�R

0�

2kR2

3

75.1

5 � 0�5

0�0.1729t � 0.1522t2 � 0.0374t3� dt �

15�0.08645t2 � 0.05073t3 � 0.00935t4�

5

0� 0.5318 liter

76. (a)

The area above the -axis equals the area below the-axis. Thus, the average value is zero.x

x

24

−1

0

1 (b)

The average value of appears to be g.S

240

0

10

77. (a)

(b) (c) meters�60

0v�t� dt � ��8.61 10�4t4

4�

0.0782t3

3�

0.208t2

2� 0.0952t�

60

0� 2476

70

−10

−10

90

v � �8.61 10�4t3 � 0.0782t2 � 0.208t � 0.0952

78. (a) histogram

—CONTINUED—

t2

21

4

43

6

8

10

12

14

16

18

65 8 97

N (b)

� 4980 customers

�6 � 7 � 9 � 12 � 15 � 14 � 11 � 7 � 2�60 � �83�60

436 Chapter 5 Integration

78. —CONTINUED—

(c) Using a graphing utility, you obtain

(d)

(e)

The estimated number of customers is

(f) Between 3 P.M. and 7 P.M., the number of customers is approximately

Hence, per minute.3017�240 � 12.6

��7

3N�t� dt �60� � �50.28��60� � 3017.

�85.162��60� � 5110.

�9

0N�t� dt � 85.162

−2

−2 10

16

N�t� � �0.084175t3 � 0.63492t2 � 0.79052 � 4.10317.

79.

F�8� �642

� 5�8� � �8

F�5� �252

� 5�5� � �252

F�2� �42

� 5�2� � �8

F�x� � �x

0�t � 5� dt � �t2

2� 5t�

x

0�

x2

2� 5x

80.

F�8� �84

4� 64 � 16 � 4 � 1068

F�5� �6254

� 25 � 10 � 4 � 167.25

�Note: F�2� � �2

2�t3 � 2t � 2� dt � 0�F�2� � 4 � 4 � 4 � 4 � 0

�x4

4� x2 � 2x � 4

� �x4

4� x2 � 2x � �4 � 4 � 4�

F�x� � �x

2�t3 � 2t � 2� dt � �t4

4� t2 � 2t�

x

2

81.

F�8� � 10�78 �

354

F�5� � 10�45 � 8

F�2� � 10�12 � 5

� �10x

� 10 � 10�1 �1x

F�x� � �x

1

10v2 dv � �x

110v�2 dv �

�10v �

x

182.

F�8� �1

64�

14

� �1564

F�5� �1

25�

14

� �21100

� �0.21

F�2� �14

�14

� 0

F�x� � �x

2

�2t3

dt � ��x

22t�3 dt �

1t2�

x

2�

1x2 �

14

436 Chapter 5 Integration

Section 5.4 The Fundamental Theorem of Calculus 437

83.

F�8� � sin 8 � sin 1 � 0.1479

F�5� � sin 5 � sin 1 � �1.8004

F�2� � sin 2 � sin 1 � 0.0678

F�x� � �x

1cos � d� � sin ��

x

1� sin x � sin 1

84.

F�8� � 1 � cos 8 � 1.1455

F�5� � 1 � cos 5 � 0.7163

F�2� � 1 � cos 2 � 1.4161

F�x� � �x

0sin � d� � �cos ��

x

0� �cos x � cos 0 � 1 � cos x

85.

(a)

(b) g increasing on and decreasing on

(c) g is a maximum of 9 at

(d)

2 4 6 8

2

4

6

8

10

x

y

x � 4.

�4, 8��0, 4�

g�8� � �8

0f �t� dt � 8 � 3 � 5

g�6� � �6

0f �t� dt � 9 � ��1� � 8

g�4� � �4

0f �t� dt � 7 � 2 � 9

g�2� � �2

0f �t� dt � 4 � 2 � 1 � 7

g�0� � �0

0f �t� dt � 0

g�x� � �x

0f �t� dt 86.

(a)

(b) g decreasing on and increasing on

(c) g is a minimum of at

(d)

x

y

−2 2 4 8 10−2

−4

−6

−8

2

4

x � 4.�8

�4, 8��0, 4�

g�8� � �8

0f �t� dt � �2 � 6 � 4

g�6� � �6

0f �t� dt � �8 � 2 � 4 � �2

g�4� � �4

0f �t� dt � �

12�4��4� � �8

g�2� � �2

0f �t� dt � �

12�2��4� � �4

g�0� � �0

0f �t� dt � 0

g�x� � �x

0f �t�dt

87. (a)

(b)ddx�

12

x2 � 2x� � x � 2

�x

0�t � 2� dt � �t2

2� 2t�

x

0�

12

x2 � 2x

88. (a)

(b)ddx�

14

x4 �12

x2� � x3 � x � x�x2 � 1�

�x

0t�t2 � 1� dt � �x

0�t3 � t� dt � �1

4t4 �

12

t2�x

0�

14

x4 �12

x2 �x2

4�x2 � 2�

89. (a)

(b)ddx�

34

x 4�3 � 12� � x1�3 � 3�x

�x

8

3�t dt � �34

t 4�3�x

8�

34

�x4�3 � 16� �34

x4�3 � 12 90. (a)

(b)ddx�

23

x3�2 �163 � � x1�2 � �x

�x

4

�t dt � �23

t3�2�x

4�

23

x3�2 �163

�23

�x3�2 � 8�

438 Chapter 5 Integration

91. (a)

(b)ddx

�tan x � 1� � sec2 x

�x

��4sec2 t dt � �tan t�

x

��4� tan x � 1 92. (a)

(b)ddx

�sec x � 2� � sec x tan x

�x

��3sec t tan t dt � �sec t�

x

��3� sec x � 2

93. (a)

(b)ddx

�e x � e�1� � e x

F �x� � �x

�1 et dt � et�

x

�1� e x � e�1 94. (a)

(b)ddx

�ln x� �1x

F �x� � �x

1 1t dt � ln t�

x

1� ln x

95.

F�x� � x2 � 2x

F�x� � �x

�2�t 2 � 2t� dt 96.

F�x� �x2

x2 � 1

F�x� � �x

1

t2

t2 � 1 dt 97.

F�x� � �x4 � 1

F�x� � �x

�1

�t4 � 1 dt

98.

F�x� � 4�x

F�x� � �x

1

4�t dt 99.

F�x� � x cos x

F�x� � �x

0t cos t dt 100.

F�x� � sec3 x

F�x� � �x

0sec3 t dt

101.

Alternate Solution:

F�x� � ��4x � 1� � 4�x � 2� � 1 � 8

� ��x

0�4t � 1� dt � �x�2

0�4t � 1� dt

� �0

x

�4t � 1� dt � �x�2

0�4t � 1� dt

F�x� � �x�2

x

�4t � 1� dt

F�x� � 8

� 8x � 10

� �2�x � 2�2 � �x � 2�� � �2x2 � x�

� �2t2 � t�x�2

x

F�x� � �x�2

x

�4t � 1� dt 102.

Alternate Solution:

F�x� � ���x�3��1� � �x3� � 0

� ���x

0t3 dt � �x

0t3 dt

� �0

�x

t3 dt � �x

0t3 dt

F�x� � �x

�x

t3 dt

F�x� � 0

F�x� � �x

�x

t3 dt � �t4

4�x

�x� 0

103.

Alternate Solution:

F�x� � �sin x ddx

�sin x� � �sin x�cos x�

F�x� � �sin x

0

�t dt

F�x� � �sin x�1�2 cos x � cos x�sin x

F�x� � �sin x

0

�t dt � �23

t3�2�sin x

0�

23

�sin x�3�2

438 Chapter 5 Integration

Section 5.4 The Fundamental Theorem of Calculus 439

105.

F�x� � sin�x3�2 � 3x2 � 3x2 sin x 6

F�x� � �x3

0sin t 2 dt 106.

F�x� � sin�x2�2�2x� � 2x sin x 4

F�x� � �x2

0sin � 2 d�

107.

has a relative maximum at x � 2.g

g�0� � 0, g�1� �12

, g�2� � 1, g�3� �12

, g�4� � 0x

y

1

1 2 3 4

2

−2

−1

f g

g�x� � �x

0f �t� dt

108. (a)

Horizontal asymptote:

(b)

The graph of does not have a horizontal asymptote.A�x�

limx→�

A�x� � limx→�

�4x �4x

� 8 � � � 0 � 8 � �

�4x2 � 8x � 4

x�

4�x � 1�2

x

� �4t �4t�

x

1� 4x �

4x

� 8

A�x� � �x

1�4 �

4t2 dt

y � 4

limt→�

g�t� � 4

g�t� � 4 �4t2

109.

� 28 units

� 4 � 4 � 20

� 3�1

0�t 2 � 4t � 3� dt � 3�3

1�t 2 � 4t � 3� dt � 3�5

3�t 2 � 4t � 3� dt

� �5

03��t � 3��t � 1�� dtTotal distance � �5

0�x�t�� dt

� 3�t � 3��t � 1�

� 3�t 2 � 4t � 3�

x�t� � 3t 2 � 12t � 9

x�t� � t 3 � 6t 2 � 9t � 2

104.

Alternate Solution: F�x� � �x2��3�2x� � 2x�5

F�x� � �x2

2t�3 dt � � t�2

�2�x2

2� ��

12t2�

x2

2�

�12x4 �

18

⇒ F�x� � 2x�5

440 Chapter 5 Integration

112. P �2��

��2

0sin � d� � ��

2�

cos ����2

0� �

2�

�0 � 1� �2�

� 63.7%

113. True 114. True

115. The function is not continuous on

Each of these integrals is infinite. has a nonremovable discontinuity at x � 0.

f �x� � x�2

�1

�1x�2 dx ��0

�1x�2 dx ��1

0x�2 dx

��1, 1�.f �x� � x�2 116. Let be an antiderivative of Then,

. � f �v�x��v�x� � f �u�x��u�x�

� F�v�x��v�x� � F�u�x��u�x�

ddx ��

v�x�

u�x�f �t� dt� �

ddx

�F�v�x�� � F�u�x��

�v�x�

u�x�f �t� dt � �F�t��

v�x�

u�x�� F�v�x�� � F�u�x��

f �t�.F�t�

117.

By the Second Fundamental Theorem of Calculus, we have

Since must be constant.f �x�f�x� � 0,

� �1

1 � x2 �1

x2 � 1� 0.

f�x� �1

�1�x�2 � 1��1x2 �

1x2 � 1

f �x� � �1�x

0

1t2 � 1

dt � �x

0

1t2 � 1

dt

118.

(a)

(c)

(d) G��0� � 0 � f �0� � �0

0f �t� dt � 0

G��x� � x � f �x� � �x

0f �t� dt

G�0� � �0

0�s�s

0f �t� dt� ds � 0

G�x� � �x

0�s�s

0f �t� dt� ds

(b) Let

G�0� � 0�0

0f �t� dt � 0

G�x� � F�x� � x�x

0f �t� dt

G�x� � �x

0F�s� ds

F�s� � s�s

0f �t� dt.

440 Chapter 5 Integration

110.

Using a graphing utility,

Total distance � �5

0�x�t�� dt � 27.37 units.

x�t� � 3t2 � 14t � 15

x�t� � �t � 1��t � 3�2 � t 3 � 7t 2 � 15t � 9 111.

� 2�2 � 1� � 2 units

� 2t1�2�4

1

� �4

1

1

�t dt

� �4

1�v�t�� dt

Total distance � �4

1�x�t�� dt

Section 5.5 Integration by Substitution

Section 5.5 Integration by Substitution 441

1. 10x dx

2.

3. 2x dx

4. 2x 2 dx

5. tan x

6. sin x cos x dx�cos xsin3 x

dx

sec2 x dx�tan2 x sec2 x dx

�sec 2x tan 2x dx

x2 � 1� x

�x2 � 1 dx

3x2 dxx3 � 1�x2�x3 � 1 dx

5x2 � 1��5x2 � 1�2�10x� dx

du � g��x� dxu � g�x��f �g�x��g��x� dx

7.

Check:ddx�

�1 � 2x�5

5� C� � 2�1 � 2x�4

��1 � 2x�4�2� dx ��1 � 2x�5

5� C 8.

Check:

� �x2 � 9�3�2x�

ddx�

�x2 � 9�4

4� C� �

4�x2 � 9�3

4�2x�

��x2 � 9�3�2x� dx ��x2 � 9�4

4� C

9.

Check:ddx�

23

�9 � x2�3�2 � C� �23

�32

�9 � x2�1�2��2x� � �9 � x2��2x�

��9 � x2�1�2��2x� dx ��9 � x2�3�2

3�2� C �

23

�9 � x2�3�2 � C

10.

Check:ddx�

34

�1 � 2x2�4�3 � C� �34

�43

�1 � 2x2�1�3��4x� � �1 � 2x2�1�3��4x�

��1 � 2x2�1�3��4x� dx �3

4�1 � 2x2�4�3 � C

11.

Check:ddx�

�x4 � 3�3

12� C� �

3�x4 � 3�2

12�4x3� � �x4 � 3�2�x3�

�x3�x4 � 3�2 dx �1

4��x4 � 3�2�4x3� dx �1

4 �x4 � 3�3

3� C �

�x4 � 3�3

12� C

12.

Check:ddx�

�x3 � 5�5

15� C� �

5�x3 � 5�4�3x2�15

� �x3 � 5�4x2

�x2�x3 � 5�4 dx �13��x3 � 5�4�3x2� dx �

13

�x3 � 5�5

5� C �

�x3 � 5�5

15� C

13.

Check:ddx�

�x3 � 1�5

15� C� �

5�x3 � 1�4�3x2�15

� x2�x3 � 1�4

�x2�x3 � 1�4 dx �13��x3 � 1�4�3x2� dx �

13�

�x3 � 1�5

5 � � C ��x3 � 1�5

15� C

442 Chapter 5 Integration

14.

Check:ddx�

�4x2 � 3�3

24� C� �

3�4x2 � 3�2�8x�24

� x�4x2 � 3�2

�x�4x2 � 3�2 dx �18��4x2 � 3�2�8x� dx �

18�

�4x2 � 3�3

3 � � C ��4x2 � 3�3

24� C

15.

Check:ddt�

�t2 � 2�3�2

3� C� �

3�2�t2 � 2�1�2�2t�3

� �t2 � 2�1�2t

�t�t2 � 2 dt �12��t2 � 2�1�2�2t� dt �

12

�t2 � 2�3�2

3�2� C �

�t2 � 2�3�2

3� C

16.

Check:ddt�

16

�t4 � 5�3�2 � C� �16

�32

�t4 � 5�1�2�4t3� � �t4 � 5�1�2�t3�

�t 3�t4 � 5 dt �14��t 4 � 5�1�2�4t3� dt �

14

�t 4 � 5 �3�2

3�2� C �

16

�t 4 � 5�3�2 � C

17.

Check:ddx��

158

�1 � x2�4�3 � C� � �158

�43

�1 � x2�1�3��2x� � 5x�1 � x2�1�3 � 5x 3�1 � x2

�5x�1 � x2�1�3 dx � �52��1 � x2�1�3��2x� dx � �

52

��1 � x2�4�3

4�3� C � �

158

�1 � x2�4�3 � C

18.

Check:ddu�

2�u3 � 5�3�2

9� C� �

29

� 32

�u3 � 5�1�2�3u2� � �u3 � 5�1�2�u2�

�u2�u3 � 5 du �13��u3 � 5�1�2�3u2� du �

13

�u3 � 5�3�2

3�2� C �

2�u3 � 5�3�2

9� C

19.

Check:ddx�

14�1 � x2�2 � C� �

14

��2��1 � x2��3��2x� �x

�1 � x2�3

� x�1 � x2�3 dx � �

12��1 � x2��3��2x� dx � �

12

�1 � x2��2

�2� C �

14�1 � x2�2 � C

20.

Check:ddx�

�14�1 � x 4� � C� �

14

�1 � x 4��2�4x3� �x3

�1 � x 4�2

� x3

�1 � x4�2 dx �14��1 � x4��2�4x3� dx � �

14

�1 � x4��1 � C ��1

4�1 � x4� � C

21.

Check:ddx��

13�1 � x3� � C� � �

13

��1��1 � x3��2�3x2� �x2

�1 � x3�2

� x2

�1 � x3�2 dx �13��1 � x3��2�3x2� dx �

13�

�1 � x3��1

�1 � � C � �1

3�1 � x3� � C

22.

Check:ddx�

13�9 � x3� � C� �

13

��1��9 � x3��2��3x2� �x2

�9 � x3�2

� x2

�9 � x3�2 dx � �13��9 � x3��2��3x2� dx � �

13�

�9 � x3��1

�1 � � C �1

3�9 � x3� � C

Section 5.5 Integration by Substitution 443

23.

Check:d

dx���1 � x2�1�2 � C � �

1

2�1 � x2��1�2��2x� �

x�1 � x2

� x�1 � x2

dx � �1

2��1 � x2��1�2��2x� dx � �

1

2 �1 � x2�1�2

1�2� C � ��1 � x2 � C

24.

Check:d

dx��1 � x4

2� C� �

1

2�

1

2�1 � x4��1�2�4x3� �

x3

�1 � x4

� x3

�1 � x4 dx �

1

4��1 � x4��1�2�4x3� dx �

1

4 �1 � x4�1�2

1�2� C �

�1 � x4

2� C

25.

Check:ddt��

�1 � �1�t�4

4� C� � �

14

�4�1 �1t �

3

�1t 2� �

1t 21 �

1t �

3

�1 �1t �

3

1t 2� dt � ��1 �

1t �

3

�1t 2� dt � �

�1 � �1�t�4

4� C

26.

Check:ddx�

13

x3 �19

x�1 � C� � x2 �19

x�2 � x2 �1

�3x�2

��x2 �1

�3x�2� dx � �x2 �19

x�2� dx �x3

3�

19

x�1

�1� � C �x3

3�

19x

� C �3x4 � 1

9x� C

27.

Check:ddx

��2x � C �12

�2x��1�2�2� �1

�2x

� 1

�2x dx �

12��2x��1�22 dx �

12�

�2x�1�2

1�2 � � C � �2x � C

28.

Check:ddx

��x � C �1

2�x

� 1

2�x dx �

12�x�1�2 dx �

12

x1�2

1�2� � C � �x � C

29.

Check:ddx�

25

x5�2 � 2x3�2 � 14x1�2 � C� �x2 � 3x � 7

�x

�x2 � 3x � 7�x

dx � ��x3�2 � 3x1�2 � 7x�1�2� dx �25

x5�2 � 2x3�2 � 14x1�2 � C �25�x�x2 � 5x � 35� � C

30.

Check:ddt�

23

t3�2 �45

t5�2 � C� � t1�2 � 2t3�2 �t � 2t2

�t

�t � 2t2

�t dt � ��t1�2 � 2t3�2� dt �

23

t3�2 �45

t5�2 � C �2

15t3�2�5 � 6t� � C

31.

Check:ddt�

14

t 4 � t2 � C� � t3 � 2t � t2t �2t �

�t 2t �2t � dt � ��t 3 � 2t� dt �

14

t 4 � t 2 � C

444 Chapter 5 Integration

32.

Check:ddt�

112

t4 �14t

� C� �13

t3 �1

4t2

�t3

3�

14t2� dt � �1

3t3 �

14

t�2� dt �13

t4

4� �14

t�1

�1� � C �1

12t4 �

14t

� C

33.

Check:ddy�

25

y3�2�15 � y� � C� �ddy�6y3�2 �

25

y5�2 � C� � 9y1�2 � y3�2 � �9 � y��y

��9 � y��y dy � ��9y1�2 � y3�2� dy � 923

y3�2� �25

y5�2 � C �25

y3�2�15 � y� � C

34.

Check:ddy�

4�y2

7�14 � y3�2� � C� �

ddy�2�4y2 �

27

y7�2� � C� � 16�y � 2�y5�2 � �2�y��8 � y3�2�

�2�y�8 � y3�2� dy � 2���8y � y5�2� dy � 2�4y2 �27

y7�2� � C �4�y2

7�14 � y3�2� � C

35.

� 2x2 � 4�16 � x2 � C

� 4x2

2 � � 2��16 � x2�1�2

1�2 � � C

� 4�x dx � 2��16 � x2��1�2��2x� dx

y � ��4x �4x

�16 � x2� dx 36.

�203�1 � x3 � C

�103 ��1 � x3�1�2

1�2 � � C

�103 ��1 � x3��1�2�3x2� dx

y � � 10x2

�1 � x3 dx

37.

� �1

2�x2 � 2x � 3� � C

�12�

�x2 � 2x � 3��1

�1 � � C

�12��x2 � 2x � 3��2�2x � 2� dx

y � � x � 1�x2 � 2x � 3�2 dx 38.

� �x2 � 8x � 1 � C

�12�

�x2 � 8x � 1�1�2

1�2 � � C

�12��x2 � 8x � 1��1�2�2x � 8� dx

y � � x � 4�x2 � 8x � 1

dx

39. (a)

x

y

−2 2

−1

3

(b)

−2 2

−1

2

y � �13

�4 � x2�3�2�2

2 � �13

�4 � 22�3�2 � C ⇒ C � 2�2, 2�:

� �12

�23

�4 � x2�3�2 � C � �13

�4 � x2�3�2 � C

y � �x�4 � x2 dx � �12��4 � x2�1�2��2x dx�

dydx

� x�4 � x2, �2, 2�

Section 5.5 Integration by Substitution 445

(b)

y �19

�x3 � 1�3

0 � C

�13

�x3 � 1�3

3� C �

19

�x3 � 1�3 � C

y � �x2�x3 � 1�2 dx �13��x3 � 1�2�3x2 dx�, �u � x3 � 1�

dydx

� x2�x3 � 1�2, �1, 0�40. (a)

x

y

2

−2

−2 2

41. (a)

(b)

5

6

−3

−6

y �12

sin�x2� � 1

1 �12

sin�0� � C ⇒ C � 1�0, 1�:

�12

sin�x2� � C

y � �x cos x2 dx �12�cos�x2�2x dx

dydx

� x cos x2, �0, 1�

−4 4

−4

4

x

y 42. (a)

(b)

y � �sec�2x�

�1 � �sec�0� � C ⇒ C � 0

� �sec�2x� � C

�u � 2x� y � ��2 sec�2x� tan�2x� dx,

dydx

� �2 sec�2x� tan�2x�, �0, �1�

x

y

3

−3

3−3

43. (a)

(b)

−4 8

−2

6y � �4e�x�2 � 5

�0, 1�: 1 � �4e0 � C � �4 � C ⇒ C � 5

� �4e�x�2 � C

y � �2e�x�2 dx � �4�e�x�2�12

dx�

dydx

� 2e�x�2, �0, 1�

x−2

−2

5

5

y

(0, 1)

44. (a)

(b)

y � �2.5e�0.2x2� 1

0, �32�: �

32

� �2.5e0 � C � �2.5 � C ⇒ C � 1

� �1

0.4e�0.2x2

� C � �2.5e�0.2x2� C

y � �xe�0.2x2 dx �

1�0.4

�e�0.2x2 ��0.4x�dx

dydx

� xe�0.2x2, 0, �

32�

x

y

−4

−4

4

4

446 Chapter 5 Integration

45. (a)

(b)

−6

−4

6

4

y � 3ex�3 �52

0, 12�:

12

� 3e0�3 � C ⇒ C � �52

y � �ex�3 dx � 3ex�3 � C

dydx

� ex�3, 0, 12�

−5

−5

5

5

y

x

46. (a)

(b)

y � esin x � 1

��, 2�: 2 � esin � � C � 1 � C ⇒ C � 1

y � �esin x cos x dx � esin x � C

dydx

� esin x cos x, ��, 2�

x

2

6

4

−2

y

10

47. �� sin �x dx � �cos �x � C

49. �sin 2x dx �12��sin 2x��2� dx � �

12

cos 2x � C

51. � 1� 2 cos

1�

d� � ��cos 1��

1� 2� d� � �sin

1�

� C

53. Let

� e5x�5� dx � e5x � C

du � 5 dx.u � 5x, 55.

��13

e�x3� C

� x2e�x3 dx � �

13

� e�x3��3x2� dx

48. �4x3 sin x4 dx � �sin x4�4x3� dx � �cos x4 � C

50. �cos 6x dx �16��cos 6x��6� dx �

16

sin 6x � C

52. �x sin x2 dx �12��sin x2��2x� dx � �

12

cos x2 � C

54. �e�x3��3x2� dx � e�x3�C

56. �12

ex2�2x � C � �x � 1�ex2�2x dx �12

� ex2�2x �2x � 2� dx

57. OR

OR

�sin 2x cos 2x dx �12

�2 sin 2x cos 2x dx �12

�sin 4x dx � �18

cos 4x � C2

�sin 2x cos 2x dx � �12��cos 2x���2 sin 2x� dx � �

12

�cos 2x�2

2� C1 � �

14

cos2 2x � C1

�sin 2x cos 2x dx �12��sin 2x��2 cos 2x� dx �

12

�sin 2x�2

2� C �

14

sin2 2x � C

58. �sec�2 � x� tan�2 � x� dx � ���sec�2 � x� tan�2 � x���1� dx � �sec�2 � x� � C

59. �tan4 x sec2 x dx �tan5 x

5� C �

15

tan5 x � C 60. ��tan x sec2 x dx ��tan x�3�2

3�2� C �

23

�tan x�3�2 � C

Section 5.5 Integration by Substitution 447

61.

� ��cot x��2

�2� C �

12 cot2 x

� C �12

tan2 x � C �12

�sec2 x � 1� � C �12

sec2 x � C1

�csc2 xcot3 x

dx � ���cot x��3��csc2 x� dx

63. �cot2 x dx � ��csc2 x � 1� dx � �cot x � x � C

65. � e x�e x � 1�2 dx ��e x � 1�3

3� C

67. Let

� �23

�1 � ex�3�2 � C

�ex�1 � ex dx � ���1 � ex�1�2��ex� dx

u � 1 � ex, du � �ex dx.

69.

� �52

e�2x � e�x � C

�5 � ex

e2x dx � �5e�2x dx � �e�x dx

62. � sin xcos3 x

dx � ���cos x��3��sin x� dx � ��cos x��2

�2� C �

12 cos2 x

� C �12

sec2 x � C

64. �csc2x2� dx � 2�csc2x

2�12� dx � �2 cotx

2� � C

66.

���1 � 3e x�2

6

� e x�1 � 3e x� dx ��13

� �1 � 3e x���3e x� dx

68. Let

��2

ex � e�x � C

�2ex � 2e�x

�ex � e�x�2 dx � 2��ex � e�x��2�ex � e�x� dx

u � ex � e�x, du � �ex � e�x� dx.

70.

� ex � 2x � e�x � C

� e2x � 2ex � 1

ex dx � � �ex � 2 � e�x� dx

71.

�1�

esin �x � C

�e sin �x cos �x dx �1��e sin �x�� cos �x� dx

73.

� �tan�e�x� � C

�e�x sec2�e�x� dx � ��sec2�e�x���e�x� dx

75.

�2

ln 33x�2 � C

� 3x�2 dx � 2� 3x�212� dx � 2

3x�2

ln 3� C

77.

��1

2 ln 5�5�x2 � � C

� �12�

5�x2

ln 5� C

�x 5�x2 dx � �

12�5�x2��2x� dx

72.

�12

etan 2x � C

� etan 2x sec2 2x dx �12�etan 2x�2 sec2 2x� dx

74.

� x2 � x � C

�ln�e2x�1� dx � � �2x � 1� dx

76. �4�x dx � �4�x

ln 4� C

78.

� �1

2 ln 7�7�3�x�2 � C

��3 � x�7�3�x�2 dx � �

12��2�3 � x�7�3�x�2

dx

448 Chapter 5 Integration

79.

f �x� � �13

�4 � x2�3�2 � 2

f �2� � 2 � �13

�0� � C ⇒ C � 2

� �12

�4 � x2�3�223� � C � �

13

�4 � x2�3�2 � C

f �x� � � x�4 � x2 dx � �12� �4 � x2�1�2 ��2x� dx

81.

Since Thus,

f �x� � 2 sin x2

� 3.

C � 3.f �0� � 3 � 2 sin 0 � C,

f �x� � �cos x2

dx � 2 sin x2

� C

83.

f �x� � �8e�x�4 � 9

f �0� � 1 � �8 � C ⇒ C � 9

� �8e�x�4 � C

f �x� � � 2e�x�4 dx � �8� e�x�4�14� dx

80.

y �3

ln 0.40.4x�3 �

12

�3

ln 0.4

f �0� �3

ln 0.4� C �

12 ⇒ C �

12

�3

ln 0.4

�3

ln 0.4 0.4x�3 � C

y � �0.4x�3 dx � 3�0.4x�313�dx

dydx

� 0.4x�3, 0, 12�

82.

Since Thus

f �x� � sec �x � 1.

C � �1.f �1�3� � 1 � sec���3� � C,

f �x� � �� sec �x tan �x dx � sec �x � C

84.

f �x� � �53

e�0.2x3�

196

0, 32� :

32

��53

� C ⇒ C �196

��53

e�0.2x3� C

�1

�0.6 �e�0.2x3��0.6x2� dx

f �x� � �x2e�0.2x3 dx

85.

f �x� �12

�ex � e�x�

f �0� � 1 � C2 � 1 ⇒ C2 � 0

f �x� � �12

�ex � e�x� dx �12

�ex � e�x� � C2

f��0� � C1 � 0

f��x� � � 12

�ex � e�x� dx �12

�ex � e�x� � C1 86.

f �x� � x � sin x �14

e2x

f �0� �14

� C2 �14

⇒ C2 � 0

� �sin x �14

e2x � x � C2

f �x� � � �cos x �12

e2x � 1� dx

f��x� � �cos x �12

e2x � 1

f��0� � �1 �12

� C1 �12

⇒ C1 � 1

f��x� � ��sin x � e2x� dx � �cos x �12

e2x � C1

Section 5.5 Integration by Substitution 449

87.

�2

15�x � 2�3�2�3x � 4� � C

�2

15�x � 2�3�2�3�x � 2� � 10 � C

�2u3�2

15�3u � 10� � C

�25

u5�2 �43

u3�2 � C

� ��u3�2 � 2u1�2� du

�x�x � 2 dx � ��u � 2��u du

dx � dux � u � 2,u � x � 2, 88.

�15

�2x � 3�3�2�x � 1� � C

�1

20�2x � 3�3�2�4x � 4� � C

�1

20�2x � 3�3�2�2�2x � 3� � 10 � C

�u3�2

20�2u � 10 � C

�14�

25

u5�2 � 2u3�2� � C

�14

��u3�2 � 3u1�2� du

�x�2x � 3 dx � �u � 32

�u 12

du

dx �12

dux �u � 3

2,u � 2x � 3,

89.

� �2

105�1 � x�3�2�15x2 � 12x � 8� � C

� �2

105�1 � x�3�2�35 � 42�1 � x� � 15�1 � x�2 � C

� �2u3�2

105�35 � 42u � 15u2� � C

� �23

u3�2 �45

u5�2 �27

u7�2� � C

� ���u1�2 � 2u3�2 � u5�2� du

�x2�1 � x dx � ���1 � u�2�u du

dx � �dux � 1 � u,u � 1 � x,

90.

� �25

�2 � x�3�2�x � 3� � C

� �25

�2 � x�3�2�5 � �2 � x� � C

� �2u3�2

5�5 � u� � C

� �2u3�2 �25

u5�2� � C

� ���3u1�2 � u3�2� du

��x � 1��2 � x dx � ���3 � u��u du

dx � �dux � 2 � u,u � 2 � x,

450 Chapter 5 Integration

91.

�1

15�2x � 1�3x2 � 2x � 13� � C

�1

60�2x � 1�12x2 � 8x � 52� � C

��2x � 1

60�3�2x � 1�2 � 10�2x � 1� � 45 � C

�u1�2

60�3u2 � 10u � 45� � C

�18

25

u5�2 �43

u3�2 � 6u1�2� � C

�18��u3�2 � 2u1�2 � 3u�1�2� du

�18�u�1�2��u2 � 2u � 1� � 4 du

� x2 � 1

�2x � 1 dx � ���1�2��u � 1�2 � 1

�u 12

du

dx �12

dux �12

�u � 1�,u � 2x � 1,

92. Let

�23�x � 4�2x � 13� � C

�23�x � 4�2�x � 4� � 21 � C

�23

u1�2�2u � 21� � C

�43

u3�2 � 14u1�2 � C

� ��2u1�2 � 7u�1�2� du

� 2x � 1�x � 4

dx � �2�u � 4� � 1�u

du

u � x � 4, x � u � 4, du � dx. 93.

where C1 � �1 � C.

� ��x � 2�x � 1� � C1

� �x � 2�x � 1 � 1 � C

� ��x � 1� � 2�x � 1 � C

� �u � 2�u � C

� ��u � 2u1�2� � C

� ���1 � u�1�2� du

� ����u � 1���u � 1��u��u � 1� du

� �x

�x � 1� � �x � 1 dx � ���u � 1�

u � �u du

dx � dux � u � 1,u � x � 1,

94.

�37

�t � 4�4�3�t � 3� � C

�37

�t � 4�4�3��t � 4� � 7 � C

�3u4�3

7�u � 7� � C

�37

u7�3 � 3u4�3 � C

� ��u4�3 � 4u1�3� du

�t 3�t � 4 dt � ��u � 4�u1�3 du

dt � dut � u � 4,u � t � 4, 95. Let

� �18

�x2 � 1�4�1

�1� 0

�1

�1x�x2 � 1�3 dx �

12�

1

�1�x2 � 1�3�2x� dx

du � 2x dx.u � x2 � 1,

Section 5.5 Integration by Substitution 451

96. Let

�19

��64 � 8�3 � ��8 � 8�3 � 41,472

�4

�2x2�x3 � 8�2 dx �

13�

4

�2�x3 � 8�2�3x2� dx � �1

3 �x3 � 8�3

3 �4

�2

u � x3 � 8, du � 3x2 dx.

97. Let

�49

�27 � 2�2 � 12 �89�2�

49��x3 � 1�3�2�

2

1� �2

3 �x3 � 1�3�2

3�2 �2

1 �2

12x2�x3 � 1 dx � 2 �

13�

2

1�x3 � 1�1�2�3x2� dx

u � x3 � 1, du � 3x2 dx.

98. Let

�2

0x�4 � x2 dx � �

12�

2

0�4 � x2�1�2��2x� dx � ��

13

�4 � x2�3�2�2

0� 0 �

83

�83

du � �2x dx.u � 4 � x2,

99. Let

�4

0

1

�2x � 1 dx �

12�

4

0�2x � 1��1�2�2� dx � ��2x � 1�

4

0� �9 � �1 � 2

du � 2 dx.u � 2x � 1,

100. Let

�2

0

x

�1 � 2x2 dx �

14�

2

0�1 � 2x2��1�2�4x� dx � �1

2�1 � 2x2�

2

0�

32

�12

� 1

du � 4x dx.u � 1 � 2x2,

101. �1

0 e�2x dx � �

12�

1

0 e�2x��2� dx � �

12

e�2x�1

0� �

12

e�2 �12

102. �2

1 e1�x dx � ��2

1 e1�x��1� dx � �e1�x�

2

1� �e�1 � 1

103. � �13

�e � e3� �e3

�e2 � 1�� ��13

e3�x�3

1 �3

1 e3�x

x2 dx � �13

�3

1 e3�x�

3x2� dx

104. Let

��2

0 xe�x2�2 dx � ���2

0 e�x2�2��x� dx � ��e�x2�2�

�2

0 � 1 � e�1 �e � 1

e

u ��x2

2, du � �x dx.

105. Let

�9

1

1

�x�1 � �x �2 dx � 2�9

1�1 � �x ��2 1

2�x� dx � ��2

1 � �x�9

1� �

12

� 1 �12

du �1

2�x dx.u � 1 � �x,

106. Let

�2

0x 3�4 � x2 dx �

12�

2

0�4 � x2�1�3�2x� dx � �3

8�4 � x2�4�3�

2

0�

38

�84�3 � 44�3� � 6 �32

3�4 � 3.619

du � 2x dx.u � 4 � x2,

452 Chapter 5 Integration

107.

When When

� �0

1�u3�2 � u1�2� du � �2

5u5�2 �

23

u3�2�0

1� ��2

5�

23� �

415�2

1�x � 1��2 � x dx � �0

1���2 � u� � 1�u du

u � 0.x � 2,u � 1.x � 1,

dx � �dux � 2 � u,u � 2 � x,

108. Let

When When

�163

�14�

23

�27� � 2�3�� � 23

� 2��

�14�

23

u3�2 � 2u1�2�9

1 �5

1

x�2x � 1

dx � �9

1

1�2�u � 1��u

1

2 du �

1

4�9

1�u1�2 � u�1�2� du

x � 5, u � 9.x � 1, u � 1.

u � 2x � 1, du � 2 dx, x �12

�u � 1�.

109. ���2

0cos2

3x� dx � �3

2 sin2

3x��

��2

0�

32�32 � �

3�34

111. �7

2 ln 2�

7ln 4

�1

ln 2 �4 �

12��2

�12x dx � � 2x

ln 2�2

�1

110. ���2

��3�x � cos x� dx � �x2

2� sin x�

��2

��3� �2

8� 1� � �2

18�

�32 � �

5�2

72�

2 � �32

112.

� �2x�0

�2� 4 � �0

�22 dx

�0

�2�33

� 52� dx � �0

�2�27 � 25� dx

113.

y � �2x3 � 1�3 � 3

4 � 13 � C ⇒ C � 3

y � 3�2x3 � 1�3

3� C � �2x3 � 1�3 � C

y � 3��2x3 � 1�2�6x2� dx, �u � 2x3 � 1�

dydx

� 18x2�2x3 � 1�2, �0, 4� 114.

y �8

�3x � 5�2 � 1

3 �8

�3��1� � 5�2 � C �84

� C ⇒ C � 1

�8

�3x � 5�2 � C

��16�3x � 5��2

�2� C

� ��48� 13��3x � 5��3 3 dx

y � �48��3x � 5��3 dx

dydx

��48

�3x � 5�3, ��1, 3�

115.

y � �2x2 � 1 � 3

4 � �49 � C � 7 � C ⇒ C � �3

y �12

�2x2 � 1�1� 2

1�2� C � �2x2 � 1 � C

y �12��2x2 � 1��1� 2�4x dx�, �u � 2x2 � 1�

dydx

�2x

�2x2 � 1, �5, 4�

Section 5.5 Integration by Substitution 453

116.

y � 2x2 �2

�3x3 � 1� 4

2 � 0 �21

� C ⇒ C � 4

� 2x2 �2

�3x3 � 1� C

� 2x 2 ��3x3 � 1��1� 2

��1�2� � C

y � ��4x � �3x3 � 1��3� 2 9x2� dx

dydx

� 4x �9x2

�3x3 � 1�3�2, �0, 2� 117.

When When

�120928

� 3847

� 12� � 37

�34�

� �37

u7�3 �34

u4�3�8

1

� �8

1�u4�3 � u1�3� du

Area � �7

0x 3�x � 1 dx � �8

1�u � 1� 3�u du

u � 8.x � 7,u � 1.x � 0,

dx � dux � u � 1,u � x � 1,

119. A � ��

0�2 sin x � sin 2x� dx � ��2 cos x �

12

cos 2x��

0� 4

118.

When When

� �8

0�u7�3 � 4u4�3 � 4u1�3� du � � 3

10u10�3 �

127

u7�3 � 3u4�3�8

0�

475235

Area � �6

�2x2 3�x � 2 dx � �8

0�u � 2�2 3�u du

u � 8.x � 6,u � 0.x � �2,

dx � dux � u � 2,u � x � 2,

120. A � ��

0�sin x � cos 2x� dx � ��cos x �

12

sin 2x��

0� 2

121. Area � �2��3

��2sec2x

2� dx � 2�2��3

��2sec2x

2�12� dx � �2 tanx

2��2��3

��2� 2��3 � 1�

122. Let

Area � ���4

��12csc 2x cot 2x dx �

12�

��4

��12csc 2x cot 2x�2� dx � ��

12

csc 2x���4

��12�

12

du � 2 dx.u � 2x,

123.

0 60

150

�5

0 ex dx � �ex�

5

0� e5 � 1 � 147.413

125.

−4.5

−3

4.5

3

� �2e�3�2 � 2 � 1.554

��6

0xe�x2�4 dx � ��2e�x2�4��6

0

124.

a b

�b

a

e�x dx � ��e�x�b

a� e�a � e�b

126.

−20

4

4

� �12

e�4 � 4 �12

� 4.491

�2

0 �e�2x � 2� dx � ��

12

e�2x � 2x�2

0

454 Chapter 5 Integration

127.

−1

−1 5

3

�4

0

x

�2x � 1 dx � 3.333 �

103

128.

00

2

20

�2

0x3�x � 2 dx � 7.581 129.

00 8

15

�7

3x�x � 3 dx � 28.8 �

1445

130.

1 50

50

�5

1x2�x � 1 dx � 67.505 131.

−1

−1 4

5

�3

0� � cos

6� d� � 7.377 132.

0

−1

2

�2

���2

0sin 2x dx � 1.0

133.

−1 2

−1

2

� �e�1 � 1 � 0.632

��2

0 xe�x2�2 dx � ��e��x2�2��

�2

0134.

−20

4

4

� �12

e�4 � 4 �12

� 4.491

�2

0 �e�2x � 2� dx � ��

12

e�2x � 2x�2

0

135.

They differ by a constant: C2 � C1 �16

��2x � 1�2 dx � ��4x2 � 4x � 1� dx �43

x3 � 2x2 � x � C2

��2x � 1�2 dx �12��2x � 1�22 dx �

16

�2x � 1�3 � C1 �43

x3 � 2x2 � x �16

� C1

136.

They differ by a constant: C2 � C1 �12

�cos2 x

2� C2 � �

�1 � sin2 x�2

� C2 �sin2 x

2�

12

� C2

�sin x cos x dx � ���cos x�1��sin x dx� � �cos2 x

2� C2

�sin x cos x dx � ��sin x�1�cos x dx� �sin2 x

2� C1

137. is even.

� 2�325

�83� �

27215

�2

�2x2�x2 � 1� dx � 2�2

0�x4 � x2� dx � 2�x5

5�

x3

3 �2

0

f �x� � x2�x2 � 1� 138. is even.

� 2�sin3 x3 �

��2

0�

23

���2

���2sin2 x cos x dx � 2���2

0sin2 x�cos x� dx

f �x� � sin2 x cos x

Section 5.5 Integration by Substitution 455

139. is odd.

�2

�2x�x2 � 1�3 dx � 0

f �x� � x�x2 � 1�3 140. is odd.

���4

���4sin x cos x dx � 0

f �x� � sin x cos x

141. the function is an even function.

(a)

(c)�2

0��x2� dx � ��2

0x2 dx � �

83

�0

�2x2 dx � �2

0x2 dx �

83

x2�2

0x2 dx � �x3

3 �2

0�

83

;

(b)

(d) �0

�23x2 dx � 3�2

0x2 dx � 8

�2

�2x2 dx � 2�2

0x2 dx �

163

142. (a) since sin x is symmetric to the origin.

(b) since cos x is symmetric to the y-axis.

(c)

(d) since and hence is symmetric to the origin.sin x cos xsin��x� cos��x� � �sin x cos x,���2

���2 sin x cos x dx � 0

���2

���2 cos x dx � 2���2

0cos x dx � �2 sin x�

��2

0� 2

���4

���4 cos x dx � 2���4

0cos x dx � �2 sin x�

��4

0� �2

���4

���4 sin x dx � 0

143. � 0 � 2�4

0�6x2 � 3� dx � 2�2x3 � 3x�

4

0� 232�4

�4�x3 � 6x2 � 2x � 3� dx � �4

�4�x3 � 2x� dx � �4

�4�6x2 � 3� dx

144. ��

��

�sin 3x � cos 3x� dx � ��

��

sin 3x dx � ��

��

cos 3x dx � 0 � 2��

0cos 3x dx � �2

3 sin 3x�

0� 0

145. If then and �x�5 � x2�3 dx � �12��5 � x2�3��2x� dx � �

12�u3 du.du � �2x dxu � 5 � x2,

147.

Thus,

When $250,000.Q�50� �t � 50,

Q�t� � 2�100 � t�3.

Q�0� � �k3

�100�3 � 2,000,000 ⇒ k � �6

Q�t� � �k3

�100 � t�3

Q�100� � C � 0

Q�t� � �k�100 � t�2 dt � �k3

�100 � t�3 � C

dQdt

� k�100 � t�2

146. is odd. Hence, �2

�2x�x2 � 1�2 dx � 0.f �x� � x�x2 � 1�2

148.

Solving this system yields and. Thus,

When V�4� � $340,000.t � 4,

V�t� �200,000

t � 1� 300,000.

C � 300,000k � �200,000

V�1� � �12

k � C � 400,000

V�0� � �k � C � 500,000

V�t� � � k�t � 1�2 dt � �

kt � 1

� C

dVdt

�k

�t � 1�2

456 Chapter 5 Integration

149.

(a)

Relative minimum: or June

Relative maximum: or January

(b) inches

(c) inches13�

12

9R�t� dt �

13

�13� � 4.33

�12

0R�t� dt � 37.47

�0.4, 5.5��6.4, 0.7�

0 120

8

R � 3.121 � 2.399 sin�0.524t � 1.377�

151. (a)

Maximum flow: at .is a relative maximum.

(b) thousand gallonsVolume � �24

0R�t� dt � 1272

��18.861, 61.178�t � 9.36R � 61.713

0 240

70

150.

(a) thousand units

(b) thousand units

(c) thousand units112�74.50t �

262.5�

cos �t6 �

12

0�

112894 �

262.5�

�262.5

� � � 74.5

13�74.50t �

262.5�

cos �t6 �

6

3�

13447 �

262.5�

� 223.5� � 102.352

13�74.50t �

262.5�

cos �t6 �

3

0�

13223.5 �

262.5� � � 102.352

1b � a�

b

a�74.50 � 43.75 sin

�t6 � dt �

1b � a�74.50t �

262.5�

cos �t6 �

b

a

152.

(a) amps

(b)

amps

(c) amps1

�1�30� � 0��1

30� cos�60�t� �

1120�

sin�120� t��1�30

0� 30� 1

30�� � �1

30��� � 0

�2�

�5 � 2�2� � 1.382

1�1�240� � 0��

130�

cos�60� t� �1

120� sin�120� t��

1�240

0� 240��

1

30�2��

1120�� � �

130���

1�1�60� � 0��

130�

cos�60�t� �1

120� sin�120�t��

1�60

0� 60� 1

30�� 0� � �

130��� �

4�

� 1.273

1b � a�

b

a

�2 sin�60� t� � cos�120� t� dt �1

b � a��1

30� cos�60�t� �

1120 �

sin�120�t��b

a

Section 5.5 Integration by Substitution 457

153.

When When

(a)

(b)

b � 0.586 � 58.6%

�1 � b�3�2�3b � 2� � 1

P0, b � ���1 � x�3�2

2�3x � 2��

b

0

P0.5, 0.75 � ���1 � x�3�2

2�3x � 2��

0.75

0.5� 0.353 � 35.3%

� ���1 � x�3�2

2�3x � 2��

b

a�

154 �2u3�2

15�3u � 5��

1�b

1�a �

154 �

1�b

1�a

�u3�2 � u1�2� du �154 �2

5u5�2 �

23

u3�2�1�b

1�a

Pa, b � �b

a

154

x�1 � x dx �154 �

1�b

1�a

��1 � u��u du

x � b, u � 1 � b.x � a, u � 1 � a.

dx � �dux � 1 � u,u � 1 � x,

154.

When When

(a)

(b) P0.5, 1 � �u5�2

16�105u3 � 385u2 � 495u � 231��

0

0.5� 0.736 � 73.6%

P0, 0.25 � �u5�2

16�105u3 � 385u2 � 495u � 231��

0.75

1� 0.025 � 2.5%

�115532 �2u5�2

1155�105u3 � 385u2 � 495u � 231��

1�b

1�a� �u5�2

16�105u3 � 385u2 � 495u � 231��

1�b

1�a

�115532 �1�b

1�a

�u9�2 � 3u7�2 � 3u5�2 � u3�2� du �115532 � 2

11u11�2 �

23

u9�2 �67

u7�2 �25

u5�2�1�b

1�a

Pa, b � �b

a

115532

x3�1 � x�3�2 dx �115532 �1�b

1�a

��1 � u�3u3�2 du

x � b, u � 1 � b.x � a, u � 1 � a.

dx � �dux � 1 � u,u � 1 � x,

156.

ex � 1 ≥ x ⇒ ex ≥ 1 � x for x ≥ 0

�et�x

0 ≥ �t�

x

0

�x

0 et dt ≥ �x

0 1 dt155.

Graphing utility: 0.4772 � 47.72%

0.0665�60

48 e�0.0139�t�48�2

dt

157. (a)

(b) is nonnegative because the graph of is positive atthe beginning, and generally has more positivesections than negative ones.

(c) The points on that correspond to the extrema of are points of inflection of

(d) No, some zeros of like do not correspond to an extrema of The graph of continues toincrease after because remains above the

axis.x-fx � ��2gg.

x � ��2,f,

g.fg

fg

0 9.4

−4

f

g

4 (e) The graph of h is that of g shifted two units downward.

� ���2

0f �x� dx � �t

��2f �x� dx � 2 � h�t�

g�t� � �t

0f �x� dx

0 9.4

−4

4

458 Chapter 5 Integration

158. Let

Let and use righthand endpoints.

� �1�

��1 � 1� �2�

� �1�

cos �x�1

0

� �1

0 sin �x dx

limn→

n

i�1

sin�i��n�n

� lim�x�→0

n

i�1f �ci� x

ci �in

, i � 1, 2, . . . , n

x �1n

f �x� � sin �x, 0 ≤ x ≤ 1. 159. (a) Let

(b) Let

� �1

0xb�1 � x�a dx

� �1

0ub�1 � u�a du

�1

0xa�1 � x�b dx � �0

1�1 � u�aub��du�

x � 0 ⇒ u � 1, x � 1 ⇒ u � 0

u � 1 � x, du � �dx, x � 1 � u.

� �1

0x5�1 � x�2 dx

� �1

0u5�1 � u�2 du

�1

0x2�1 � x�5 dx � �0

1�1 � u�2u5��du�

x � 0 ⇒ u � 1, x � 1 ⇒ u � 0

u � 1 � x, du � �dx, x � 1 � u.

160. (a) and

Let

(b) Let as in part (a).

� ���2

0cosn u du � ���2

0cosn x dx

���2

0sinn x dx � ���2

0cosn�

2� x� dx � �0

��2cosn u��du�

u ��

2� x

� ���2

0cos2 u du � ���2

0cos2 x dx���2

0sin2 x dx � ���2

0cos2�

2� x� dx � �0

��2cos2 u��du�

u ��

2� x, du � �dx, x �

2� u.

cos x � sin�

2� x�sin x � cos�

2� x�

162. False

�x�x2 � 1� dx �12��x2 � 1��2x� dx �

14

�x2 � 1�2 � C

161. False

��2x � 1�2 dx �12��2x � 1�22 dx �

16

�2x � 1�3 � C

163. True

Odd Even

�10

�10�ax3 � bx2 � cx � d� dx � �10

�10�ax3 � cx� dx � �10

�10�bx2 � d� dx � 0 � 2�10

0�bx2 � d� dx

Section 5.5 Integration by Substitution 459

165. True

4�sin x cos x dx � 2�sin 2x dx � �cos 2x � C

164. True

�b

a

sin x dx � ��cos x�b

a� �cos b � cos a � �cos�b � 2�� � cos a � �b�2�

a

sin x dx

166. False

�sin2 2x cos 2x dx �12��sin 2x�2�2 cos 2x� dx �

12

�sin 2x�3

3� C �

16

sin3 2x � C

167. Let

� �cb

ca

f �x� dx� �cb

ca

f �u� duc�b

a

f �cx� dx � c�cb

ca

f �u� duc

u � cx, du � c dx.

168. (a)

Thus,

(b) Let

(part (a))

� 2�

� 2��� cos���

� 2�sin u � u cos u��

0

� 2��

0u sin u du

��2

0sin�x dx � ��

0sin u�2u du�

u � �x, u2 � x, 2 u du � dx.

�u sin u du � sin u � u cos u � C.

� u sin ud

du�sin u � u cos u � C � cos u � cos u � u sin u

169. Because f is odd, Then

Let in the first integral. When When

� ��a

0f �u� du � �a

0f �x� dx � 0

�a

�a

f �x� dx � ��a

0f ��u���du� � �a

0f �x� dx

x � �a, u � a.x � 0, u � 0.x � �u, dx � �du

� ���a

0f �x� dx � �a

0f �x� dx. �a

�a

f �x� dx � �0

�a

f �x� dx � �a

0f �x� dx

f ��x� � �f �x�.

170. Let then When When Thus,

�b

a

f �x � h� dx � �b�h

a�h

f �u� du � �b�h

a�h

f �x� dx.

u � b � h.x � b,u � a � h.x � a,du � dx.u � x � h,

460 Chapter 5 Integration

171. Let

(Given)

By the Mean Value Theorem for Integrals, there exists in such that

Thus, the equation has at least one real zero.

0 � f �c�.

�1

0f �x� dx � f �c��1 � 0�

�0, 1�c

� a0 � a1

2�

a2

3� . . . �

an

n � 1� 0

�1

0f �x� dx � �a0 x � a1

x2

2� a2

x3

3� . . . � an

xn�1

n � 1�1

0

f �x� � a0 � a1x � a2 x2 � . . . � anxn. 172.

Adding,

Since Hence, there are no such functions.

�� � x�2 ≥ 0, f � 0.

�1

0f �x��� � x�2 dx � 0.

�1

0��2 f �x� � 2�x f �x� � x2f �x�� dx � 0

�1

0f �x�x2 dx � �2

�2��1

0f �x�x dx � �2���� � �2�2

�2�1

0f �x� dx � �2�1� � �2

Section 5.6 Numerical Integration

1. Exact:

Trapezoidal:

Simpson’s: �2

0x2 dx �

16�0 � 41

22

� 2�1�2 � 432

2

� �2�2� �83

� 2.6667

�2

0x2 dx �

14�0 � 21

22

� 2�1�2 � 232

2

� �2�2� �114

� 2.7500

�2

0x2 dx � �1

3x3�

2

0�

83

� 2.6667

2. Exact:

Trapezoidal:

Simpson’s: �1

0x2

2� 1 dx �

112�1 � 4�1�4�2

2� 1 � 2�1�2�2

2� 1 � 4�3�4�2

2� 1 � 12

2� 1� �

76

� 1.1667

�1

0x2

2� 1 dx �

18�1 � 2�1�4�2

2� 1 � 2�1�2�2

2� 1 � 2�3�4�2

2� 1 � 12

2� 1� �

7564

� 1.1719

�1

0x2

2� 1 dx � �x3

6� x�

1

0�

76

� 1.1667

3. Exact:

Trapezoidal:

Simpson’s: �2

0x3 dx �

16�0 � 41

23

� 2�1�3 � 432

3

� �2�3� �246

� 4.0000

�2

0x3 dx �

14�0 � 21

23

� 2�1�3 � 232

3

� �2�3� �174

� 4.2500

�2

0x3 dx � �x4

4 �2

0� 4.000

4. Exact:

Trapezoidal:

Simpson’s: �2

1 2x2 dx �

112�2 � 432

25 � 289 � 432

49 �12� � 1.0008

�2

1 2x2 dx �

18�2 � 232

25 � 289 � 232

49 �12� � 1.0180

�2

1 2x2 dx � ��2

x �2

1� �1 � 2 � 1

Section 5.6 Numerical Integration 461

5. Exact:

Trapezoidal:

Simpson’s: �2

0x3 dx �

112�0 � 41

43

� 224

3

� 434

3

� 2�1�3 � 454

3

� 264

3

� 474

3

� 8� � 4.0000

�2

0x3 dx �

18�0 � 21

43

� 224

3

� 234

3

� 2�1�3 � 254

3

� 264

3

� 274

3

� 8� � 4.0625

�2

0x3 dx � �1

4x4�

2

0� 4.0000

7. Exact:

Trapezoidal:

Simpson’s: �9

4

�x dx �5

24�2 � 4�378

� �21 � 4�478

� �26 � 4�578

� �31 � 4�678

� 3� � 12.6667

� 12.6640

�9

4

�x dx �5

16�2 � 2�378

� 2�214

� 2�478

� 2�264

� 2�578

� 2�314

� 2�678

� 3��9

4

�x dx � �23

x3�2�9

4� 18 �

163

�383

� 12.6667

9. Exact:

Trapezoidal:

Simpson’s:

�1

1214

�6481

�8

25�

64121

�19 � 0.1667

�2

1

1�x � 1�2 dx �

112�

14

� 4 1��5�4� � 1�2 � 2 1

��3�2� � 1�2 � 4 1��7�4� � 1�2 �

19�

�18

14

�3281

�8

25�

32121

�19 � 0.1676

�2

1

1�x � 1�2 dx �

18�

14

� 2 1��5�4� � 1�2 � 2 1

��3�2� � 1�2 � 2 1��7�4� � 1�2 �

19�

�2

1

1�x � 1�2 dx � ��

1x � 1�

2

1� �

13

�12

�16

� 0.1667

6. Exact:

Trapezoidal:

Simpson’s: �8

0 3�x dx �

13

�0 � 4 � 2 3�2 � 4 3�3 � 2 3�4 � 4 3�5 � 2 3�6 � 4 3�7 � 2� � 11.8632

�8

0 3�x dx �

12

�0 � 2 � 2 3�2 � 2 3�3 � 2 3�4 � 2 3�5 � 2 3�6 � 2 3�7 � 2� � 11.7296

�8

0

3 �x dx � �34

x4�3�8

0� 12.0000

8. Exact:

Trapezoidal:

Simpson’s: �3

1�4 � x2� dx �

16�3 � 44 �

94 � 0 � 44 �

254 � 5� � �0.6667

�3

1�4 � x2� dx �

14 3 � 2�4 � 3

22

� � 2�0� � 2�4 � 52

2

� � 5� � �0.7500

�3

1�4 � x2� dx � �4x �

x3

3 �3

1� 3 �

113

� �23

� �0.6667

10. Exact:

Trapezoidal:

Simpson’s: �2

0x�x2 � 1 dx �

16�0 � 41

2��1�2�2 � 1 � 2�1��12 � 1 � 432��3�2�2 � 1 � 2�22 � 1� � 3.392

�2

0x�x2 � 1 dx �

14�0 � 21

2��1�2�2 � 1 � 2�1��12 � 1 � 232��3�2�2 � 1 � 2�22 � 1� � 3.457

�2

0x�x2 � 1 dx �

13��x2 � 1�3�2�

2

0�

13

�53�2 � 1� � 3.393

462 Chapter 5 Integration

11. Trapezoidal:

Simpson’s:

Graphing utility: 3.241

�2

0

�1 � x3 dx �16

�1 � 4�1 � �1�8� � 2�2 � 4�1 � �27�8� � 3� � 3.240

�2

0

�1 � x3 dx �14

�1 � 2�1 � �1�8� � 2�2 � 2�1 � �27�8� � 3� � 3.283

12. Trapezoidal:

Simpson’s:

Graphing utility: 1.402

�2

0

1

�1 � x3 dx �

16�1 � 4 1

�1 � �1�2�3 � 2 1

�1 � 13 � 4 1

�1 � �3�2�3 �13� � 1.405

�2

0

1

�1 � x3 dx �

14�1 � 2 1

�1 � �1�2�3 � 2 1

�1 � 13 � 2 1

�1 � �3�2�3 �13� � 1.397

13.

Trapezoidal:

Simpson’s:

Graphing utility: 0.393

�1

0

�x�1 � x� dx �1

12�0 � 4�141 �

14 � 2�1

21 �12 � 4�3

41 �34 � � 0.372

�1

0

�x�1 � x� dx �18�0 � 2�1

41 �14 � 2�1

21 �12 � 2�3

41 �34 � � 0.342

�1

0

�x�1 � x dx � �1

0

�x�1 � x� dx

14. Trapezoidal:

Simpson’s:

Graphing utility: 1.458

��

��2

�x sin x dx ��

24 ���

2� 4�5�

8 sin5�

8 � 2�3�

4 sin3�

4 � 4�7�

8 sin7�

8 � 0� � 1.458

��

��2

�x sin x dx ��

16 ���

2�1� � 2�5�

8 sin5�

8 � 2�3�

4 sin3�

4 � 2�7�

8 sin7�

8 � 0� � 1.430

15. Trapezoidal:

Simpson’s:

Graphing utility: 0.977

� 0.978

����2

0cos�x2� dx �

���212 �cos 0 � 4 cos���2

4 2

� 2 cos���22

2

� 4 cos3���24

2

� cos��

22

� 0.957

����2

0cos�x2� dx �

���28 �cos 0 � 2 cos���2

4 2

� 2 cos���22

2

� 2 cos3���24

2

� cos��

22

16. Trapezoidal:

Simpson’s:

Graphing utility: 0.256

� 0.257

����4

0tan�x2� dx �

���412 �tan 0 � 4 tan���4

4 2

� 2 tan���42 2

� 4 tan3���44 2

� tan��

42

� 0.271

����4

0tan�x2� dx �

���48 �tan 0 � 2 tan���4

4 2

� 2 tan���42 2

� 2 tan3���44 2

� tan��

42

Section 5.6 Numerical Integration 463

17. Trapezoidal:

Simpson’s:

Graphing utility: 0.089

�1.1

1sin x2 dx �

1120

�sin�1� � 4 sin�1.025�2 � 2 sin�1.05�2 � 4 sin�1.075�2 � sin�1.1�2� � 0.089

�1.1

1sin x2 dx �

180

�sin�1� � 2 sin�1.025�2 � 2 sin�1.05�2 � 2 sin�1.075�2 � sin�1.1�2� � 0.089

18. Trapezoidal:

Simpson’s:

Graphing utility: 1.910

���2

0

�1 � cos2 x dx ��

24��2 � 4�1 � cos2���8� � 2�1 � cos2���4� � 4�1 � cos2�3��8� � 1� � 1.910

���2

0

�1 � cos2 x dx ��

16��2 � 2�1 � cos2���8� � 2�1 � cos2���4� � 2�1 � cos2�3��8� � 1� � 1.910

19.

Trapezoidal: 1.684

Simpson’s: 1.649

Graphing utility: 1.648

�2

0 x ln�x � 1� dx

20. Trapezoidal:

Simpson’s:

Graphing utility: 1.296

�3

1 ln x dx �

16

�0 � 4 ln�1.5� � 2 ln 2 � 4 ln�2.5� � ln 3� �7.7719

6� 1.295

�3

1 ln x dx �

14

�0 � 2 ln�1.5� � 2 ln 2 � 2 ln�2.5� � ln 3� �5.1284

4� 1.282

21. Trapezoidal:

Simpson’s:

Graphing utility: 0.186

���4

0x tan x dx �

48�0 � 4 �

16 tan �

16 � 22�

16 tan2�

16 � 43�

16 tan3�

16 ��

4� � 0.186

���4

0x tan x dx �

32�0 � 2 �

16 tan �

16 � 22�

16 tan2�

16 � 23�

16 tan3�

16 ��

4� � 0.194

22. Trapezoidal:

Simpson’s:

Graphing utility: 1.852

��

0 sin x

x dx �

12�1 �4 sin���4�

��4�

2 sin���2���2

�4 sin�3��4�

3��4� 0� � 1.852

��

0 sin x

x dx �

8�1 �2 sin���4�

��4�

2 sin���2���2

�2 sin�3��4�

3��4� 0� � 1.836

23. Trapezoidal:

Simpson’s:

Graphing utility: 92.744

�4

0 �xex dx �

13

�0 � 4e1 � 2�2e2 � 4�3e3 � 2e 4� � 93.375

�4

0 �xe x dx �

12

�0 � 2e1 � 2�2e2 � 2�3e3 � 2e4� � 102.555

24. Trapezoidal:

Simpson’s:

Graphing utility: 0.594

�2

0 2xe�x dx �

16

�0 � 2e�1�2 � 2e�1 � 6e�3�2 � 2e�2� �3.5583

6� 0.5930

�2

0 xe�x dx �

14

�0 � e�1�2 � 2e�1 � 3e�3�2 � 2e�2� �2.2824

4� 0.5706

464 Chapter 5 Integration

25.

The Trapezoidal Rule overestimates the area if the graphof the integrand is concave up.

xa b

y f x= ( )

y 26. Trapezoidal: Linear polynomials

Simpson’s: Quadratic polynomials

27.

(a) Trapezoidal: since is maximum in when

(b) Simpson’s: since f �4��x� � 0.Error ≤ �2 � 0�5

180�44� �0� � 0

x � 2.�0, 2�� f� �x��Error ≤ �2 � 0�3

12�42� �12� � 0.5

f �4��x� � 0

f���x� � 6

f � �x� � 6x

f��x� � 3x2

f �x� � x3

28.

The error is 0 for both rules.

f � �x� � 0

f��x� � 2

f �x� � 2x � 3

29.

(a) Trapezoidal: because is at most 1 on

(b) Simpson’s: because is at most 1 on �0, ��.� f �4��x��Error ≤ �� � 0�5

180�44� �1� ��5

46,080� 0.006641

�0, ��.� f � �x��Error ≤ �� � 0�3

12�42� �1� ��3

192� 0.1615

f �4��x� � cos x

f�� �x� � sin x

f � �x� � �cos x

f��x� � �sin x

f �x� � cos x

30.

(a) Trapezoidal: since on

(b) Simpson’s: since on �0, 1�.� f �4� �x�� ≤ � 4Error ≤ �1 � 0�5

180�44� � 4 �� 4

46,080� 0.0021

�0, 1�.� f � �x�� ≤ � 2Error ≤ �1 � 0�3

12�42� � 2 ��2

192� 0.0514

f �4��x� � � 4 sin��x�

f�� �x� � �� 3 cos��x�

f � �x� � �� 2 sin��x�

f��x� � � cos��x�

f �x� � sin��x�

Section 5.6 Numerical Integration 465

31.

(a) Maximum of is

Trapezoidal:

(b) Maximum of is

Simpson’s:

n ≥ 6.2. Let n � 8 (even).

n4 ≥32�15��2180�256� 105 �

�296

105

Error ≤ 25

180n415�2256 ≤ 0.00001

15�2256

� 0.0829.

� f �4� �x�� � � �1516�x � 2�7�2�

n ≥ 76.8. Let n � 77.

n2 ≥8�2

12�16�105 ��224

105

Error ≤ �2 � 0�3

12n2 �216 ≤ 0.00001

�216

� 0.0884.� f � �x�� � � �14�x � 2�3�2�

f �4��x� ��1516

�x � 2��7�2

f�� �x� �38

�x � 2��5�2

f � �x� � �14

�x � 2��3�2

f��x� �12

�x � 2��1�2

f �x� � �x � 2�1�2, 0 ≤ x ≤ 2 32.

(a) Maximum of is on

Trapezoidal:

(b) Maximum of is on

Simpson’s:

n ≥ 18.5. Let n � 20 (even).

n4 ≥76

� 105

Error ≤ 25

180n410516 ≤ 0.00001

�1, 3�.10516� f �4� �x�� � � 105

16x9�2� n ≥ 223.6. Let n � 224.

n2 ≥12

� 105

Error ≤ �3 � 1�3

12n2 34 ≤ 0.00001

�1, 3�.34� f � �x�� � � 3

4x5�2�f �4��x� �

10516

x�9�2

f���x� ��15

8x�7�2

f��x� �34

x�5�2

f��x� ��12

x�3�2

1 ≤ x ≤ 3 f �x� � x�1�2,

33.

(a) Maximum of is

Trapezoidal:

(b) Maximum of is

Simpson’s:

n ≥ 15.3. Let n � 16.

n4 ≥�4

180� 105

Error ≤ 1

180n4 � 4 ≤ 0.00001

� 4.� f �4� �x�� � �� 4cos��x��

n ≥ 286.8. Let n � 287.

n2 ≥�2

12� 105

Error ≤ �1 � 0�3

12n2 �2 ≤ 0.00001

� 2.� f � �x�� � ��� 2 cos��x��f �4��x� � � 4 cos��x�

f�� �x� � � 3 sin��x�

f � �x� � �� 2 cos��x�

f��x� � �� sin��x�

f �x� � cos��x�, 0 ≤ x ≤ 1 34.

All derivatives are bounded by 1.

(a) Trapezoidal:

(b) Simpson’s:

n ≥ 8.5. Let n � 10 (even).

n4 ≥�5

5760� 105

Error ≤ ���2�5

180n4 �1� ≤ 0.00001

n ≥ 179.7. Let n � 180.

n2 ≥�3

96� 105

Error ≤ ���2�3

12n2 �1� ≤ 0.00001

f �4��x� � sin x

f�� �x� � �cos x

f � �x� � �sin x

f��x� � cos x

f �x� � sin x, 0 ≤ x ≤�

2

466 Chapter 5 Integration

35.

(a) in

is maximum when and

Trapezoidal: Error ≤ < 0.00001, > 16,666.67, n > 129.10; let

(b) in

is maximum when and

Simpson’s: Error ≤ < 0.00001, > 16,666.67, n > 11.36; let n � 12.n432180n415

16� f �4��0�� �

1516

.x � 0� f �4��x��

�0, 2�f �4��x� ��15

16�1 � x�7�2

n � 130.n2812n21

4� f � �0�� �

14

.x � 0� f � �x��

�0, 2�f� �x� � �1

4�1 � x�3�2

f �x� � �1 � x

36.

(a) in

is maximum when and

Trapezoidal: Error ≤ < 0.00001, > 14,814.81, n > 121.72; let

(b) in

is maximum when and

Simpson’s: Error ≤ < 0.00001, > 12,290.81, n > 10.53; let (In Simpson’s Rule n must be even.)n � 12.n432180n456

81� f �4��0�� �

5681

.x � 0� f �4��x��

�0, 2�f �4��x� � �56

81�x � 1�10�3

n � 122.n2812n42

9�f � �0�� �

29

.x � 0� f� �x��

�0, 2�f � �x� � �2

9�x � 1�4�3

f �x� � �x � 1�2�3

38.

(a) in

is maximum when and

Trapezoidal: Error ≤ < 0.00001, > 19,044.17, n > 138.00; let

(b) in

is maximum when and

Simpson’s: Error ≤ < 0.00001, > 15,793.61, n > 11.21; let n � 12.n4�1 � 0�5

180n4 �28.4285�

� f �4��0.852�� � 28.4285.x � 0.852� f �4��x���0, 1�f �4��x� � �16x4 � 12� sin�x2� � 48x2 cos�x2�

n � 139.n2�1 � 0�3

12n2 �2.2853�

�f ��1�� � 2.2853.x � 1�f ��x���0, 1�f � �x� � 2��2x2 sin�x2� � cos�x2��

f �x� � sin�x2�

37.

(a) in

is maximum when and

Trapezoidal: Error ≤ < 0.00001, > 412,754.17, n > 642.46; let

(b) in

is maximum when and

Simpson’s: Error ≤ < 0.00001, > 5,102,485.22, n > 47.53; let n � 48.n4�1 � 0�5

180n4 �9184.4734�

� f �4��1�� � 9184.4734.x � 1� f �4��x���0, 1�f �4��x� � 8 sec2�x2��12x2 � �3 � 32x4� tan�x2� � 36x2 tan2�x2� � 48x4 tan3�x2��

n � 643.n2�1 � 0�3

12n2 �49.5305�

� f ��1�� � 49.5305.x � 1� f ��x���0, 1�f ��x� � 2 sec2�x2��1 � 4x2 tan�x2��

f �x� � tan�x2�

Section 5.6 Numerical Integration 467

39. (a)

(b)

�773

� 25.67

�4

0f �x� dx �

412

�3 � 4�7� � 2�9� � 4�7� � 0�

�12

�49� �492

� 24.5

�4

0f �x� dx �

48

�3 � 2�7� � 2�9� � 2�7� � 0�

b � a � 4 � 0 � 4, n � 4 40.

(a)

(b)

�13

�134� �1343

� 4�10� � 2�9� � 4�6� � 0�

�8

0f �x� dx �

824

�0 � 4�1.5� � 2�3� � 4�5.5� � 2�9�

�12

�88� � 44

� 2�10� � 2�9� � 2�6� � 0�

�8

0f �x� dx �

816

�0 � 2�1.5� � 2�3� � 2�5.5� � 2�9�

n � 8, b � a � 8 � 0 � 8

41. The program will vary depending upon the computer or programmable calculator that you use.

42. on �0, 4�f �x� � �2 � 3x2

n

4 12.7771 15.3965 18.4340 15.6055 15.4845

8 14.0868 15.4480 16.9152 15.5010 15.4662

10 14.3569 15.4544 16.6197 15.4883 15.4658

12 14.5386 15.4578 16.4242 15.4814 15.4657

16 14.7674 15.4613 16.1816 15.4745 15.4657

20 14.9056 15.4628 16.0370 15.4713 15.4657

S�n�T�n�R�n�M�n�L�n�

43. on �0, 1�f �x� � �1 � x2

n

4 0.8739 0.7960 0.6239 0.7489 0.7709

8 0.8350 0.7892 0.7100 0.7725 0.7803

10 0.8261 0.7881 0.7261 0.7761 0.7818

12 0.8200 0.7875 0.7367 0.7783 0.7826

16 0.8121 0.7867 0.7496 0.7808 0.7836

20 0.8071 0.7864 0.7571 0.7821 0.7841

S�n�T�n�R�n�M�n�L�n�

44. on �0, 4�f �x� � sin�x

n

4 2.8163 3.5456 3.7256 3.2709 3.3996

8 3.1809 3.5053 3.6356 3.4083 3.4541

10 3.2478 3.4990 3.6115 3.4296 3.4624

12 3.2909 3.4952 3.5940 3.4425 3.4674

16 3.3431 3.4910 3.5704 3.4568 3.4730

20 3.3734 3.4888 3.5552 3.4643 3.4759

S�n�T�n�R�n�M�n�L�n�

48.

Simpson’s Rule:

� 4001512�125 � 15

123

� . . . � 0� � 10,233.58 ft � lb

�5

0100x�125 � x3 dx �

53�12��0 � 400 5

12�125 � 512

3

� 2001012�125 � 10

123

n � 12

W � �5

0100x�125 � x3 dx

468 Chapter 5 Integration

45. on �1, 2�f �x� �sin x

x

n

4 0.7070 0.6597 0.6103 0.6586 0.6593

8 0.6833 0.6594 0.6350 0.6592 0.6593

10 0.6786 0.6594 0.6399 0.6592 0.6593

12 0.6754 0.6594 0.6431 0.6593 0.6593

16 0.6714 0.6594 0.6472 0.6593 0.6593

20 0.6690 0.6593 0.6496 0.6593 0.6593

S�n�T�n�R�n�M�n�L�n�

46. on �0, 2�f �x� � 6e�x2�2

n

4 8.4410 7.1945 5.8470 7.1440 7.1770

8 7.8178 7.1820 6.5208 7.1693 7.1777

10 7.6911 7.1804 6.6535 7.1723 7.1777

12 7.6063 7.1796 6.7416 7.1740 7.1777

16 7.4999 7.1788 6.8514 7.1756 7.1777

20 7.4358 7.1784 6.9170 7.1764 7.1777

S�n�T�n�R�n�M�n�L�n�

47. on �0, 3�f �x� � �x ln�x � 1�

n

4 2.5311 3.3953 4.3320 3.4316 3.4140

8 2.9632 3.4026 3.8637 3.4135 3.4074

10 3.0508 3.4037 3.7711 3.4109 3.4068

12 3.1094 3.4044 3.7097 3.4095 3.4065

16 3.1829 3.4050 3.6331 3.4050 3.4062

20 3.2273 3.4054 3.5874 3.4073 3.4061

S�n�T�n�R�n�M�n�L�n�

Section 5.6 Numerical Integration 469

49. (a) Trapezoidal:

Simpson’s:

�2

0f �x� dx �

23�8��4.32 � 4�4.36� � 2�4.58� � 4�5.79� � 2�6.14� � 4�7.25� � 2�7.64� � 4�8.08� � 8.14� � 12.592

�2

0f �x� dx �

22�8��4.32 � 2�4.36� � 2�4.58� � 2�5.79� � 2�6.14� � 2�7.25� � 2�7.64� � 2�8.08� � 8.14� � 12.518

(b) Using a graphing utility,

Integrating, �2

0y dx � 12.53.

y � �1.3727x3 � 4.0092x2 � 0.6202x � 4.2844.

50. Simpson’s Rule,

�1

36�113.098� � 3.1416

� �1

2� 0

3�6� �6 � 4�6.0209� � 2�6.0851� � 4�6.1968� � 2�6.3640� � 4�6.6002� � 6.9282�

n � 6�1�2

0

6

�1 � x2 dx,

52. Area �10002�10��125 � 2�125� � 2�120� � 2�112� � 2�90� � 2�90� � 2�95� � 2�88� � 2�75� � 2�35�� � 89,250 sq m

51. Simpson’s Rule:

� 3.14159

� � 4�1

0

11 � x2 dx �

43�6��1 �

41 � �1�6�2 �

21 � �2�6�2 �

41 � �3�6�2 �

21 � �46�2 �

41 � �5�6�2 �

12�

n � 6

53.

� 7435 sq m

Area �120

2�12��75 � 2�81� � 2�84� � 2�76� � 2�67� � 2�68� � 2�69� � 2�72� � 2�68� � 2�56� � 2�42� � 2�23� � 0�

55.

By trial and error, we obtain t � 2.477.

n � 10�t

0sin�x dx � 2,

56. The quadratic polynomial

passes through the three points.

p�x� ��x � x2��x � x3�

�x1 � x2��x1 � x3� y1 �

�x � x1��x � x3��x2 � x1��x2 � x3�

y2 ��x � x1��x � x2�

�x3 � x1��x3 � x2� y3

54. Let Then

Simpson’s: Error ≤

Therefore, Simpson’s Rule is exact when approximatingthe integral of a cubic polynomial.

Example:

This is the exact value of the integral.

�1

0x3 dx �

16�0 � 41

23

� 1� �14

�b � a�5

180n4 �0� � 0

f �4��x� � 0.f �x� � Ax3 � Bx2 � Cx � D.

17.

�x3

3� 2x �

12

ln�x2 � 2� � C

�x4 � x � 4x2 � 2

dx � ��x2 � 2 �x

x2 � 2� dx 18.

� �3x �x2

2�

12

ln�x2 � 3� � C

�x3 � 3x2 � 4x � 9x2 � 3

dx � ���3 � x �x

x2 � 3� dx

470 Chapter 5 Integration

Section 5.7 The Natural Logarithmic Function: Integration

1. �5

x dx � 5�1

x dx � 5 ln�x� � C 2. �10

x dx � 10�1

x dx � 10 ln�x� � C 3.

� 1x � 1

dx � ln�x � 1� � C

du � dxu � x � 1,

4.

� 1x � 5

dx � ln�x � 5� � C

du � dxu � x � 5, 5.

� �12

ln�3 � 2x� � C

� 13 � 2x

dx � �12� 1

3 � 2x��2� dx

du � �2 dxu � 3 � 2x, 6.

�13

ln�3x � 2� � C

� 13x � 2

dx �13� 1

3x � 2�3� dx

7.

� ln�x2 � 1 � C

�12

ln�x2 � 1� � C

� xx2 � 1

dx �12� 1

x2 � 1�2x� dx

du � 2x dxu � x2 � 1, 8.

� �13

ln�3 � x3� � C

� x2

3 � x3 dx � �13� 1

3 � x3��3x2� dx

du � �3x2 dxu � 3 � x3, 9.

�x2

2� 4 ln�x� � C

�x2 � 4x

dx � ��x �4x� dx

10.

� ��9 � x2 � C

� x

�9 � x2 dx � �

12��9 � x2��1�2��2x� dx

du � �2x dxu � 9 � x2, 11.

�13

ln�x3 � 3x2 � 9x� � C

� x2 � 2x � 3x3 � 3x2 � 9x

dx �13�3�x2 � 2x � 3�

x3 � 3x2 � 9x dx

du � 3�x2 � 2x � 3� dxu � x3 � 3x2 � 9x,

12.

�13

ln�x3 � 3x2 � 4� � C

� x�x � 2�x3 � 3x2 � 4

dx �13� 3x2 � 6x

x3 � 3x2 � 4 dx, �u � x3 � 3x2 � 4�

13.

�x2

2� 4x � 6 ln�x � 1� � C

�x2 � 3x � 2x � 1

dx � ��x � 4 �6

x � 1� dx 14.

� x2 � 11x � 19 ln�x � 2� � C

�2x2 � 7x � 3x � 2

dx � ��2x � 11 �19

x � 2� dx

15.

�x3

3� 5 ln�x � 3� � C

�x3 � 3x2 � 5x � 3

dx � ��x2 �5

x � 3� dx 16.

�x3

3�

5x2

2� 19x � 115 ln�x � 5� � C

�x3 � 6x � 20x � 5

dx � ��x2 � 5x � 19 �115

x � 5� dx

Section 5.7 The Natural Logarithmic Function: Integration 471

19.

��ln x�2

x dx �

13

�ln x�3 � C

du �1x dxu � ln x, 20.

�13

ln�ln�x�� � C

� 1x ln�x3� dx �

13� 1

ln x�

1x dx

21.

� 2�x � 1 � C

� 2�x � 1�1�2 � C

� 1

�x � 1 dx � ��x � 1��1�2 dx

du � dxu � x � 1, 22.

� 3 ln�1 � x1�3� � C

� 1x2�3�1 � x1�3� dx � 3� 1

1 � x1�3� 13x2�3� dx

du �1

3x2�3 dxu � 1 � x1�3,

23.

� 2 ln�x � 1� �2

�x � 1� � C

� 2� 1x � 1

dx � 2� 1�x � 1�2 dx

� �2�x � 1��x � 1�2 dx � 2� 1

�x � 1�2 dx

� 2x�x � 1�2 dx � �2x � 2 � 2

�x � 1�2 dx 24.

� ln�x � 1� �1

2�x � 1�2 � C

� � 1x � 1

dx � � 1�x � 1�3 dx

� ��x � 1�2

�x � 1�3 dx � � 1�x � 1�3 dx

�x�x � 2��x � 1�3 dx � �x2 � 2x � 1 � 1

�x � 1�3 dx

25.

where C � C1 � 1.

� �2x � ln�1 � �2x � � C

� �1 � �2x � � ln�1 � �2x� � C1

� u � ln�u� � C1

� 1

1 � �2x dx � ��u � 1�

u du � ��1 �

1u� du

du �1

�2x dx ⇒ �u � 1� du � dxu � 1 � �2x, 26.

�23�3x �

23

ln�1 � �3x� � C1

�23

1 � �3x � ln�1 � �3x� � C

�23

u � ln�u� � C

�23��1 �

1u� du

� 1

1 � �3x dx � �1

u 2

3�u � 1� du

u � 1 � �3x, du �3

2�3x dx ⇒ dx �

2

3�u � 1� du

27.

where C � C1 � 27.

� x � 6�x � 18 ln��x � 3� � C

� ��x � 3�2� 12��x � 3� � 18 ln��x � 3� � C1

� u2 � 12u � 18 ln�u� � C1

� 2�u2

2� 6u � 9 ln�u�� � C1

� 2�u2 � 6u � 9u

du � 2��u � 6 �9u� du

� �x

�x � 3 dx � 2��u � 3�2

u du

du �1

2�x dx ⇒ 2�u � 3� du � dxu � �x � 3,

37.

� ln�cos�e�x�� � C

� ���ln�cos�e�x��� � C

�e�x tan�e�x� dx � �� tan�e�x���e�x� dx 38.

� tan t � sec t � C

�sec t�sec t � tan t� dt � �sec2 t dt � �sec t tan t dt

472 Chapter 5 Integration

28.

� 3 ln�x1�3 � 1� �3x2�3

2� 3x1�3 � x � C1

� 3��x1�3 � 1�3

3�

3�x1�3 � 1�2

2� 3�x1�3 � 1� � ln�x1�3 � 1�� � C

� 3�u3

3�

3u2

2� 3u � ln�u�� � C

� 3��u2 � 3u � 3 �1u� du

� 3�u � 1u

�u2 � 2u � 1� du

� 3�x3�x � 1

dx � �u � 1u

3�u � 1�2 du

u � x1�3 � 1, du �1

3x2�3 dx ⇒ dx � 3�u � 1�2 du

29.

�u � sin �, du � cos � d��

�cos �sin �

d� � ln�sin �� � C 30.

� �15

ln�cos 5�� � C

�tan 5� d� �15�5 sin 5�

cos 5� d�

31.

� �12

ln�csc 2x � cot 2x� � C

�csc 2x dx �12��csc 2x��2� dx 32.

� 2 ln�sec x2

� tan x2� � C

�sec x2

dx � 2�sec x2�

12� dx

33. � cos t1 � sin t

dt � ln�1 � sin t� � C 34.

�csc2 tcot t

dt � �ln�cot t� � C

u � cot t, du � �csc2 t dt

35. �sec x tan xsec x � 1

dx � ln�sec x � 1� � C 36.

� ln�sec t�sec t � tan t�� � C

� ln�sec t � tan tcos t � � C

��sec t � tan t� dt � ln�sec t � tan t� � ln�cos t� � C

Section 5.7 The Natural Logarithmic Function: Integration 473

39.

y � �3 ln�x � 2�0 � �3 ln�1 � 2� � C ⇒ C � 0�1, 0�:

� �3 ln�x � 2� � C

� �3� 1x � 2

dx−10 10

−10

(1, 0)

10y � � 3

2 � x dx 40.

−9 9

−4

(0, 4)

8y � ln�x2 � 9� � 4 � ln 9

4 � ln�0 � 9� � C ⇒ C � 4 � ln 9�0, 4�:

� ln�x2 � 9� � C

y � � 2xx2 � 9

dx

41.

s � �12

ln�cos 2�� � 2

2 � �12

ln�cos�0�� � C ⇒ C � 2�0, 2�:

� �12

ln�cos 2�� � C

�12�tan�2���2 d�� −3 3

−3

(0, 2) 4s � �tan�2�� d� 42.

−8 8

− 2

( , 4)π

10r � ln�tan t � 1� � 4

4 � ln�0 � 1� � C ⇒ C � 4��, 4�:

� ln�tan t � 1� � C

r � � sec2 ttan t � 1

dt

43.

f �x� � �2 ln x � 3x � 2

f �1� � 1 � �2�0� � 3 � C1 ⇒ C1 � �2

f �x� � �2 ln x � 3x � C1

f ��x� ��2x

� 3

f ��1� � 1 � �2 � C ⇒ C � 3

f ��x� ��2x

� C

f � �x� �2x2 � 2x�2, x > 0 44.

f �x� � 4 ln�x � 1� � x2 � 7

f �2� � 3 � 4�0� � 4 � C1 ⇒ C1 � 7

f �x� � 4 ln�x � 1� � x2 � C1

f ��x� �4

x � 1� 2x

f ��2� � 0 � 4 � 4 � C ⇒ C � 0

f ��x� �4

�x � 1� � 2x � C

f � �x� ��4

�x � 1�2 � 2 � �4�x � 1��2 � 2, x > 1

45.

(a)

x

y

3

−3

4−2

(0, 1)

�0, 1�dydx

�1

x � 2,

(b)

Hence, y � ln�x � 2� � 1 � ln 2 � ln�x � 22 � � 1.

y�0� � 1 ⇒ 1 � ln 2 � C ⇒ C � 1 � ln 2 −3 6

−3

3

y � � 1x � 2

dx � ln�x � 2� � C

474 Chapter 5 Integration

46.

(a)

(b)

Hence, y ��ln x�2

2� 2.

y�1� � �2 ⇒ �2 ��ln 1�2

2� C ⇒ C � �2

y � �ln xx

dx ��ln x�2

2� C

x

−1

−2

1

2

y

4

�1, �2�dydx

�ln x

x, 47. (a)

(b)

8

−1

−1

8

y � x � ln x � 3

4 � 1 � 0 � C ⇒ C � 3

y � x � ln x � C

dydx

� 1 �1x, �1, 4�

8−1

−2

−3

1

2

3

4

5

x

y(1, 4)

48. (a)

−4

4

x

y

2π−

(b)

y � ln�sec x � tan x� � 1

1 � ln�1 � 0� � C ⇒ C � 1

y � ln�sec x � tan x� � C

5

−3

−�2

�2

dydx

� sec x, �0, 1�

49.

�53

ln 13 4.275

�4

0

53x � 1

dx � �53

ln�3x � 1��4

050.

� ln 3 � ln 1 � ln 3

�1

�1

1x � 2

dx � �ln�x � 2��1

�151.

�73

�e

1 �1 � ln x�2

x dx � �1

3�1 � ln x�3�

e

1

du �1x dxu � 1 � ln x,

52.

� �ln�ln�x���e2

e� ln 2 �e2

e

1x ln x

dx � �e2

e� 1

ln x�1x dx

du �1x dxu � ln x, 53.

� �12

x2 � x � ln�x � 1��2

0� �ln 3

�2

0 x2 � 2x � 1

dx � �2

0�x � 1 �

1x � 1� dx

54.

� �x � 2 ln�x � 1��1

0� 1 � 2 ln 2

�1

0 x � 1x � 1

dx � �1

01 dx � �1

0

�2x � 1

dx 55.

� ln�2 � sin 21 � sin 1� 1.929

�2

1 1 � cos �� � sin �

d� � �ln�� � sin ���2

1

56.

� ��cot 2� � csc 2� � ��0.2

0.1 0.0024 � �0.2

0.1�2 csc2 2� � 2 csc 2� cot 2� � 1� d�

�0.2

0.1�csc 2� � cot 2��2 d� � �0.2

0.1�csc2 2� � 2 csc 2� cot 2� � cot2 2�� d�

Section 5.7 The Natural Logarithmic Function: Integration 475

57.

where C � C1 � 2. � 2�x � ln�1 � �x � � C

� 1

1 � �x dx � 2�1 � �x � � 2 ln�1 � �x � � C1

58.

� 4�x � x � 4 ln�1 � �x � � C where C � C1 � 5.

�1 � �x

1 � �x dx � ��1 � �x �2 � 6�1 � �x � � 4 ln�1 � �x � � C1

59. � �xx � 1

dx � ln��x � 1�x � 1� � 2�x � C 60. � x2

x � 1 dx � ln�x � 1� �

x2

2� x � C

61.

� ln��2 � 1� ��22

0.174

���2

��4�csc x � sin x� dx � ��ln�csc x � cot x� � cos x�

��2

��4

62.

� ln��2 � 1

�2 � 1� � 2�2 �1.066

���4

���4

sin2 x � cos2 xcos x

dx � �ln�sec x � tan x� � 2 sin x���4

���4

Note: In Exercises 63–66, you can use the Second Fundamental Theorem of Calculus to integrate the function.

63.

F��x� �1x

F �x� � �x

1 1t dt 64.

F� �x� � tan x

F �x� � �x

0tan t dt

65.

(by Second Fundamental Theorem of Calculus)

Alternate Solution:

F� �x� �13x

�3� �1x

F�x� � �3x

1 1t dt � �ln�t��

3x

1� ln�3x�

F� �x� �13x

�3� �1x

F�x� � �3x

1 1t dt 66.

F��x� �2xx2 �

2x

F �x� � �x2

1 1t dt

67.

Matches (d).A 1.25;

x1−1

2

12

12

y 68.

Matches (a).A 3;

x1 2 3 4

1

2

−1

−2

y 69. A � �3

1 4x dx � 4 ln�x��

3

1� 4 ln 3

476 Chapter 5 Integration

70.

� 2 ln�2 ln 2ln 2 � � 2 ln 2

� 2ln�ln 4� � ln�ln 2�

� 2 ln�ln x��4

2

A � �4

2

2x ln x

dx � 2�4

2

1ln x

1x dx 71.

� ln�2 �ln 2

2 0.3466

� �ln �22

� 0

A � ���4

0tan x dx � �ln�cos x��

��4

0

72.

� ln�3 � 2�2 �

� ln�2 � �2

2 � �2� � �ln�1 �

�22 � � ln�1 �

�22 �

� �ln�1 � cos x��3��4

��4

A � �3��4

��4

sin x1 � cos x

dx 73.

0 60

10

�152

� 8 ln 2 13.045 square units

� �x2

2� 4 ln x�

4

1� �8 � 4 ln 4� �

12

A � �4

1 x2 � 4

x dx � �4

1�x �

4x� dx

74.

00

9

6

� 3 � 4 ln 4 8.5452

� 4 � 4 ln 4 � 1

� �x � 4 ln x�4

1

A � �4

1 x � 4

x dx � �4

1�1 �

4x� dx 75.

0 40

10

�12�

ln�2 � �3 � 5.03041

�12�

ln�sec �

3� tan

3 � �12�

ln�1 � 0�

� �12�

ln�sec �x6

� tan �x6 ��2

0

�2

02 sec

�x6

dx �12��2

0sec��x

6 ��

6 dx

76.

0 5

−2

8

� �16 �103

ln cos�1.2�� � �1 �103

ln cos�0.3�� 11.7686

�4

1�2x � tan�0.3x�� dx � �x2 �

103

ln�cos�0.3x���4

1

Section 5.7 The Natural Logarithmic Function: Integration 477

77.

Trapezoid:

Simpson:

Calculator:

Exact: 12 ln 5

�5

1 12x

dx 19.3133

43�4� f �1� � 4f �2� � 2f �3� � 4f �4� � f �5� �

13

12 � 24 � 8 � 12 � 2.4 19.4667

42�4� f �1� � 2f �2� � 2f �3� � 2f �4� � f �5� �

12

12 � 12 � 8 � 6 � 2.4 � 20.2

f �x� �12x

, b � a � 5 � 1 � 4, n � 4

78.

Trapezoid:

Simpson:

Calculator:

Exact: 4 ln 5

�4

0

8xx2 � 4

dx 6.438

43�4� f �0� � 4f �1� � 2f �2� � 4f �3� � f �4� 6.4615

42�4� f �0� � 2f �1� � 2f �2� � 2f �3� � f �4� �

12

0 � 3.2 � 4 � 3.6923 � 1.6 6.2462

f �x� �8x

x2 � 4, b � a � 4 � 0 � 4, n � 4

79.

Trapezoid:

Simpson:

Calculator: �6

2 ln x dx 5.3643

43�4� f �2� � 4f �3� � 2f �4� � 4f �5� � f �6� 5.3632

42�4� f �2� � 2f �3� � 2f �4� � 2f �5� � f �6� �

12

0.6931 � 2.1972 � 2.7726 � 3.2189 � 1.7918 5.3368

f �x� � ln x, b � a � 6 � 2 � 4, n � 4

80.

Trapezoid:

Simpson:

Calculator: ���3

���3sec x dx 2.6339

2��33�4� � f ��

3� � 4f ���

6� � 2f �0� � 4f ��

6� � f ��

3�� 2.6595

2��32�4� � f ��

3� � 2f ���

6� � 2f �0� � 2f ��

6� � f ��

3�� �

122 � 2.3094 � 2 � 2.3094 � 2 2.780

f �x� � sec x, b � a ��

3� ��

3� �2�

3, n � 4

81. Power Rule 82. Substitution: andPower Rule

�u � x2 � 4� 83. Substitution: and Log Rule

�u � x2 � 4�

84. Substitution: and Log Rule

�u � tan x� 85.

� ln�sec x� � C

�ln�cos x� � C � ln� 1cos x� � C 86.

� �ln�csc x� � C

ln�sin x� � C � ln� 1csc x� � C

87.

� ln� 1sec x � tan x� � C � �ln�sec x � tan x� � C

� ln�sec2 x � tan2 xsec x � tan x � � Cln�sec x � tan x� � C � ln��sec x � tan x��sec x � tan x�

�sec x � tan x� � � C

478 Chapter 5 Integration

88.

� �ln� 1csc x � cot x� � C � ln�csc x � cot x� � C

� �ln�csc2 x � cot2 xcsc x � cot x � � C�ln�csc x � cot x� � C � �ln��csc x � cot x��csc x � cot x�

�csc x � cot x� � � C

89.

� �4�14

�12� � 1

� ��41x�

4

2

� 4�4

2x�2 dx

Average value �1

4 � 2 �4

2 8x2 dx 90.

� 2�ln 2 �14� � ln 4 �

12

1.8863

� 2�ln 4 �14

� ln 2 �12�

� 2�ln x �1x�

4

2

� 2�4

2�1

x�

1x2� dx

Average value �1

4 � 2 �4

2 4�x � 1�

x2 dx

91.

�1

2e � 2 0.291

�1

e � 1�12�

�1

e � 1��ln x�2

2 �e

1

Average value �1

e � 1�e

1

ln xx

dx 92.

�3�

ln�2 � �3�

�3�

ln�2 � �3� � ln�1 � 0�

� �12�

6�� ln�sec

�x6

� tan �x6 ��2

0

Average value �1

2 � 0�2

0sec

�x6

dx

93.

P�3� � 100012�ln 1.75� � 1 7715

� 100012 ln�1 � 0.25t� � 1

P�t� � 12,000 ln�1 � 0.25t� � 1000

C � 1000

P�0� � 12,000 ln�1 � 0.25�0�� � C � 1000

� 12,000 ln�1 � 0.25t� � C

P�t� � � 30001 � 0.25t

dt � �3000��4�� 0.251 � 0.25t

dt 94.

�10

ln 2�ln�43�� 4.1504 units of time

�10

ln 2�ln�T � 100��300

250�

10ln 2

ln 200 � ln 150

t �10

ln 2 �300

250

1T � 100

dT

95.

$168.27

1

50 � 40�50

40

90,000400 � 3x

dx � �3000 ln�400 � 3x��50

4096.

Solving this system yields and Thus,

S�t� �100 ln t

ln 2� 100 � 100� ln t

ln 2� 1�.

C � 100.k � 100�ln 2

S�4� � k ln 4 � C � 300

S�2� � k ln 2 � C � 200

S�t� � � kt

dt � k ln�t� � C � k ln t � C since t > 1.

dSdt

�kt

Section 5.7 The Natural Logarithmic Function: Integration 479

97. (a)

(c) In part (a):

Using a graphing utility the graphs intersect at The slopes are 3.295 and

respectively.�0.304 � ��1��3.295,�2.214, 1.344�.

y� �2xy

4x � 2yy� � 0

2x2 � y2 � 8

−10 10

−10

10

y2 � ��2x2 � 8

y1 � �2x2 � 8

y2 � 2x2 � 8

2x2 � y2 � 8 (b)

Let and graph

−10

−10 10

10

�y1 �2�x

, y2 � �2�x�y2 �

4x.k � 4

y2 � e���1�x� dx � e�ln x�C � eln�1�x��eC � �1x

k

98.

limk→0�

fk�x� � ln x

k � 0.1: f0.1�x� �10�x � 1

0.1� 10�10�x � 1�

k � 0.5: f0.5�x� ��x � 1

0.5� 2��x � 1�

x2 4 6 8 10

2

4

6

8

10f1

f0.5

f0.1

yk � 1: f1�x� � x � 1

99. False

�ln x�1�2

12

�ln x� � ln�x1�2�

100. False

ddx

ln x �1x

101. True

� ln�Cx�, C 0

� ln�x� � ln�C�

�1x

dx � ln�x� � C1

102. False; the integrandhas a nonremovablediscontinuity at x � 0.

103.

—CONTINUED—

5 10

0.5

1

y

x

f �x� �x

1 � x2(a) intersects

�12

ln 2 �14

� �12

ln�x2 � 1� �x2

4 �1

0

A � �1

0 �� x

1 � x2� �12

x� dx

x � 1

1 � x2 � 2

12

x �x

1 � x2

f �x� �x

1 � x2:y �12

x (b)

Hence, for the graphs of and enclose a finite region.y � mx

f0 < m < 1,

f ��0� � 1

f ��x� ��1 � x2� � x�2x�

�1 � x2�2 �1 � x2

�1 � x2�2

In part (b):

y� ��2yx2 �

�2yy2x 2 �

�2y4x

��y2x

2yy� ��4x2

y2 �4x

� 4x�1

480 Chapter 5 Integration

103. —CONTINUED—

(c)

�12

[m � ln�m� � 1�

�12

ln� 1m� �

12

�1 � m�

�12

ln�1 �1 � m

m � �12

m�1 � mm �

� �12

ln�1 � x2� �mx2

2 ���1�m�m

0

A � ��1�m�m

0 � x

1 � x2 � mx� dx, 0 < m < 1

x ��1 � mm

, intersection point

x2 �1 � m

m

1 � m � mx2

1 − mm

0.5

x

y

f (x) =

y = mx

xx2 + 1

x

1 � x2 � mx

104.

Alternate Solution:

� ln 2

� ln 2 � ln x � ln x

F�x� � ln t�2x

x� ln�2x� � ln x

F� �x� �12x

�2� �1x

� 0 ⇒ F is constant on �0, ��.

F�x� � 2x

x

1t dt, x > 0

Section 5.8 Inverse Trigonometric Functions: Integration

1. 5

�9 � x2 dx � 5 arcsin�x

3� � C 2. 3

�1 � 4x2 dx �

3

2 2

�1 � 4x2 dx �

3

2 arcsin�2x� � C

3. 716 � x2 dx �

74

arctan�x4� � C 4. 4

1 � 9x2 dx �43 3

1 � 9x2 dx �43

arctan�3x� � C

5. 1

x�4x2 � 1 dx � 2

2x��2x�2 � 1 dx � arcsec�2x� � C 6. 1

4 � �x � 1�2 dx �12

arctan�x � 12 � � C

7. (Use long division.) x3

x2 � 1 dx � �x �

xx2 � 1� dx � x dx �

12 2x

x2 � 1 dx �

12

x2 �12

ln�x2 � 1� � C

8. x4 � 1x2 � 1

dx � �x2 � 1� dx �13

x3 � x � C 9. 1

�1 � �x � 1�2 dx � arcsin�x � 1� � C

105. implies that

The second formula follows by the Chain Rule.

1x dx � ln�x� � C.

ddx

ln�x� �1x

Section 5.8 Inverse Trigonometric Functions: Integration 481

10. Let

tt 4 � 16

dt �12 1

�4�2 � �t 2�2�2t� dt �18

arctan t2

4� C

u � t 2, du � 2t dt. 11. Let

t

�1 � t4 dt �

12 1

�1 � �t2�2�2t� dt �

12

arcsin�t 2� � C

u � t 2, du � 2t dt.

12. Let

�14

arcsec x2

2� C

1

x�x4 � 4 dx �

12 1

x2��x2�2 � 22�2x� dx

u � x2, du � 2x dx. 13. Let

e2x

4 � e4x dx �

12 2e2x

4 � �e2x�2 dx �

14

arctan e2x

2� C

u � e2x, du � 2e2x dx.

14.

�1�3

arctan�x � 2�3 � � C

1

3 � �x � 2�2 dx � 1

��3 �2 � �x � 2�2 dx 15.

� 2 arcsin �x � C

1

u�1 � u2�2u du� � 2 du

�1 � u2� 2 arcsin u � C

1�x �1 � x

dx, u � �x, x � u2, dx � 2u du

16.

� 3 arctan�x � C

32 2u du

u�1 � u2� � 3 du1 � u2 � 3 arctan u � C

3

2�x�1 � x� dx, u � �x, du �

1

2�x dx, dx � 2u du 17.

�12

ln�x2 � 1� � 3 arctan x � C

x � 3x2 � 1

dx �12 2x

x2 � 1 dx � 3 1

x2 � 1 dx

18. � �4�1 � x2 � 3 arcsin x � C 4x � 3�1 � x2

dx � ��2� �2x�1 � x2

dx � 3 1�1 � x2

dx

19.

� ��6x � x2 � 8 arcsin�x3

� 1� � C � ��9 � �x � 3�2 � 8 arcsin�x � 33 � � C

x � 5�9 � �x � 3�2

dx � �x � 3��9 � �x � 3�2

dx � 8�9 � �x � 3�2

dx

20.

�12

ln�x2 � 2x � 5� �32

arctan�x � 12 � � C

x � 2�x � 1�2 � 4

dx �12 2x � 2

�x � 1�2 � 4 dx � 3

�x � 1�2 � 4 dx

21. Let .

� �13

arcsin�3x��16

0�

18

16

0

1

�1 � 9x2 dx �

1316

0

1

�1 � �3x�2�3� dx

du � 3 dxu � 3x, 22. 1

0

1

�4 � x2 dx � �arcsin

x2�

1

0�

6

23. Let

� �12

arctan�2x���32

0�

6

�32

0

11 � 4x2 dx �

12

�32

0

21 � �2x�2 dx

du � 2 dx.u � 2x, 24. 3

�3

19 � x2 dx � �1

3 arctan

x3�

3

�3�

36

482 Chapter 5 Integration

25. Let

1�2

0

arcsin x

�1 � x2 dx � �1

2 arcsin2 x�

1�2

0�

�2

32� 0.308

u � arcsin x, du �1

�1 � x2 dx. 26. Let

� ��12

arccos2 x�1�2

0�

3� 2

32� 0.925

1�2

0

arccos x

�1 � x2 dx � �1�2

0 �arccos x

�1 � x2 dx

u � arccos x, du � �1

�1 � x2 dx.

27. Let

� �0.134

� ���1 � x2�0

�12�

�3 � 22

0

�12

x

�1 � x2 dx � �

120

�12 �1 � x2��12��2x� dx

u � 1 � x2, du � �2x dx. 28. Let

� �12

ln�1 � x2��0

��3� �ln 2

0

��3

x1 � x2 dx �

120

��3

11 � x2�2x� dx

u � 1 � x2, du � 2x dx.

29. Let

� ��arctan�cos x���

�2�

4

�2

sin x1 � cos2 x

dx � ��

�2

�sin x1 � cos2 x

dx

u � cos x, du � �sin x dx. 30. �2

0

cos x1 � sin2 x

dx � arctan�sin x���2

0�

4

31.

� �arctan�x � 1��2

0�

2

2

0

dxx2 � 2x � 2

� 2

0

11 � �x � 1�2 dx 32.

�13

arctan�43�

� �13

arctan�x � 23 ��

2

�2

2

�2

dxx2 � 4x � 13

� 2

�2

dx�x � 2�2 � 9

33.

� ln�x2 � 6x � 13� � 3 arctan�x � 32 � � C

2xx2 � 6x � 13

dx � 2x � 6x2 � 6x � 13

dx � 6 1x2 � 6x � 13

dx � 2x � 6x2 � 6x � 13

dx � 6 14 � �x � 3�2 dx

34. 2x � 5x2 � 2x � 2

dx � 2x � 2x2 � 2x � 2

dx � 7 11 � �x � 1�2 dx � ln�x2 � 2x � 2� � 7 arctan�x � 1� � C

35.

� arcsin�x � 22 � � C

1

��x2 � 4x dx � 1

�4 � �x � 2�2 dx 36.

� 2 arcsin�x � 22 � � C

� 2�4 � �x � 2�2

dx

2��x2 � 4x

dx � 2�4 � �x2 � 4x � 4�

dx

37. Let

� ���x2 � 4x � C

x � 2

��x2 � 4x dx � �

12��x2 � 4x��12��2x � 4� dx

u � �x2 � 4x, du � ��2x � 4� dx. 38. Let

� �x2 � 2x � C

x � 1

�x2 � 2x dx �

12�x2 � 2x��12�2x � 2� dx

u � x2 � 2x, du � �2x � 2� dx.

Section 5.8 Inverse Trigonometric Functions: Integration 483

39.

� ��2�4x � x2 � arcsin�x � 22 ��

3

2� 4 � 2�3 �

6� 1.059

3

2

2x � 3

�4x � x2 dx � 3

2

2x � 4

�4x � x2 dx � 3

2

1

�4x � x2 dx � �3

2�4x � x2��12�4 � 2x� dx � 3

2

1

�4 � �x � 2�2 dx

40.

� arcsec�x � 1� � C

1

�x � 1��x2 � 2x dx � 1

�x � 1���x � 1�2 � 1 dx 41. Let

�12

arctan�x2 � 1� � C

xx 4 � 2x2 � 2

dx �12 2x

�x2 � 1�2 � 1 dx

u � x2 � 1, du � 2x dx.

42. Let

x

�9 � 8x2 � x 4 dx �

12 2x

�25 � �x2 � 4�2 dx �

12

arcsin� x2 � 45 � � C

u � x2 � 4, du � 2x dx.

43. Let Then and

� 2u � 2�3 arctan u

�3� C � 2�et � 3 � 2�3 arctan �et � 3

3� C

�et � 3 dt � 2u2

u2 � 3 du � 2 du � 6

1u2 � 3

du

2u duu2 � 3

� dt.u2 � 3 � et, 2u du � et dt,u � �et � 3.

44. Let

� 2u �6

�3 arctan

u

�3� C � 2�x � 2 � 2�3 arctan �x � 2

3� C

�x � 2x � 1

dx � 2u2

u2 � 3 du � 2u2 � 6 � 6

u2 � 3 du � 2du � 6 1

u2 � 3 du

2u du � dx.u2 � 2 � x,u � �x � 2,

45.

Let

� 2��

3�

4� ��

6

� 2 arctan�u���3

1

�3

1

2u duu�1 � u2� � �3

1

21 � u2 du

u � �x, u2 � x, 2u du � dx, 1 � x � 1 � u2.

3

1

dx�x �1 � x�

46.

Let

��

4�

6�

12

� arcsin��22 � � arcsin�1

2�

� arcsin�u2��

�2

1

�2

1

2u du

2�4 � u2 u� �2

1

du�4 � u2

�3 � x � �4 � u2.2u du � dx,u2 � x � 1,u � �x � 1,

1

0

dx

2�3 � x�x � 1

47. (a)

(b)

(c) cannot be evaluated using the basic

integration rules.

1

x�1 � x2 dx

x

�1 � x2 dx � ��1 � x2 � C, u � 1 � x2

1

�1 � x2 dx � arcsin x � C, u � x 48. (a) cannot be evaluated using the basic

integration rules.

(b)

(c) 1x2 e1x dx � �e1x � C, u �

1x

xex2 dx �

12

ex2� C, u � x2

ex2 dx

484 Chapter 5 Integration

49. (a)

(b) Let Then and

(c) Let Then and

Note: In (b) and (c), substitution was necessary before the basic integration rules could be used.

�23

u�u2 � 3� � C �23�x � 1�x � 2� � C x

�x � 1 dx � u2 � 1

u�2u� du � 2�u2 � 1� du � 2�u3

3� u� � C

dx � 2u du.x � u2 � 1u � �x � 1.

�2

15 u3�3u2 � 5� � C �

215

�x � 1�32 3�x � 1� � 5� � C �2

15�x � 1�32�3x � 2� � C

x�x � 1 dx � �u2 � 1��u��2u� du � 2�u4 � u2� du � 2�u5

5�

u3

3 � � C

dx � 2u du.x � u2 � 1u � �x � 1.

�x � 1 dx �23

�x � 1�32 � C, u � x � 1

50. (a) cannot be evaluated using the basic

integration rules.

(b)

(c)

�14

ln�1 � x4� � C, u � 1 � x4

x3

1 � x4 dx �14 4x3

1 � x 4 dx

�12

arctan�x2� � C, u � x2

x1 � x4 dx �

12 2x

1 � �x2�2 dx

11 � x4 dx 51.

Matches (c).

x1−1

2

32

12

12

yArea � �1��1� � 1

52. No. This integral does not correspond to any of the basic differentiation rules.

53.

y � arcsin�x2� � �

y�0� � � � C

y � 1

�4 � x2 dx � arcsin�x

2� � C

y� �1

�4 � x2, �0, �) 54.

y �12

arctan�x2� �

7�

8

��

8� C ⇒ C �

7�

8

� �12

arctan�22� � C

y � 1

4 � x2 dx �12

arctan x2

� C

y� �1

4 � x2, �2, ��

55. (a)

x

y

5

5

−5

−5(0, 0)

(b)

y � 3 arctan x

�0, 0�: 0 � 3 arctan�0� � C ⇒ C � 0

y � 3 dx

1 � x2 � 3 arctan x � C 8−8

32�−

32�

dydx

�3

1 � x2, �0, 0�

Section 5.8 Inverse Trigonometric Functions: Integration 485

56. (a)

4

−4

4

x

y (b)

y �23

arctan�x3� � 2

2 � C

y � 2

9 � x2 dx �23

arctan�x3� � C

5

4

−1

−4

y� �2

9 � x2, �0, 2�

57. (a)

−1 1 4

−2

−3

−4

1

2

3

4

x

y (b)

y �12

arcsec x2

� 1, x ≥ 2

1 �12

arcsec�1� � C � C

y � 1

x�x2 � 4 dx �

12

arcsec �x�2

� C

4

4

−4

−4

y� �1

x�x2 � 4, �2, 1�

58. (a)

x

y

−4 4

−4

4

(b)

y � 2 arcsin�x5�

� � 2 arcsin�1� � C ⇒ C � 0

y � 2

�25 � x2 dx � 2 arcsin�x

5� � C

5

−5

−5 5

y� �2

�25 � x2, �5, ��

59.

−6 12

−8

4

dydx

�10

x�x2 � 1, y�3� � 0 60.

4

6

−4

−6

dydx

�1

12 � x2, y�4� � 2

61.

−3

−1

3

3

dy

dx�

2y�16 � x2

, y�0� � 2 62.

7

60

−1

dydx

��y

1 � x2, y�0� � 4

486 Chapter 5 Integration

63.

��

8

�12

arctan�1� �12

arctan�0�

�12

arctan�x � 12 ��

3

1

A � 3

1

1x2 � 2x � 5

dx � 3

1

1�x � 1�2 � 4

dx 64.

��

4

� arctan�1� � arctan�0�

� arctan�x � 22 ��

0

�2

Area � 0

�2

2x2 � 4x � 8

dx � 0

�2

2�x � 2�2 � 4

dx

65.

��

6

� arcsin�12� � arcsin�0�

� arcsin�x2��

1

0

x

1

112

y

Area � 1

0

1�4 � x2

dx 66.

��

3�

4�

12

� arcsec�2� � arcsec� 2�2�

� arcsec x�2

2�2

Area � 2

2�2

1

x�x2 � 1 dx

67.

�3�

4�

3�

4�

3�

2

� 3 arctan�1� � 3 arctan��1�

� 3 arctan�sin x���2

��2

Area � �2

��2

3 cos x1 � sin2 x

dx � 3�2

��2

11 � sin2 x

�cos x dx� 68.

��

3�

4�

12

� arctan��3 � � arctan�1�

� arctan�ex��ln

�3

0

A � ln��3 �

0

ex

1 � e2x dx, �u � ex�

69. (a)

Thus,

(b)

�12

ln 3 �12

ln 2 ���3

9�

4� 0.3835

� �ln �3 �12

ln�4� �arctan �3

�3 � � ��12

ln 2 � arctan�1��

� �ln x �12

ln�1 � x2� �arctan x

x ��3

1

A � �3

1 arctan x

x2 dx

arctan x

x2 dx � ln x �12

ln�1 � x2� �arctan x

x� C.

�1 � x2 � x2

x�1 � x2� �1

x�1 � x2� �arctan x

x2 �arctan x

x2

ddx�ln x �

12

ln�1 � x2� �arctan x

x� C� �

1x

�x

1 � x2 � � x 1�1 � x2�� � arctan x x2 �

70. (a)

(b) �� 2

4� 2 � 0.4674� ���

2�2

� 2� � �0�� �x�arcsin x�2 � 2x � 2�1 � x2 arcsin x�1

0 A � 1

0�arcsin x�2 dx

� �arcsin x�2 � 2x�arcsin x� 1�1 � x2

� 2 �2x

�1 � x2 arcsin x � 2�1 � x2 1

�1 � x2� �arcsin x�2

ddx

x�arcsin x�2 � 2x � 2�1 � x2 arcsin x � C�

Section 5.8 Inverse Trigonometric Functions: Integration 487

71. (a)

Shaded area is given by

(b) 1

0 arcsin x dx � 0.5708

1

0arcsin x dx.

1 2

1

2

2

y

x

π

(c) Divide the rectangle into two regions.

Hence, 1

0arcsin x dx �

2� 1, ��0.5708�.

� 1

0arcsin x dx � 1

2� 1

0arcsin x dx � ��cos y��

�2

0

Area rectangle � 1

0arcsin x dx � �2

0 sin y dy

Area rectangle � �base��height� � 1��

2� ��

2

72. (a)

(b) Let

(c) 3.1415927

41

0

11 � x2 dx � 4� 1

18��1 �4

1 � �136� �2

1 � �19� �4

1 � �14� �2

1 � �49� �4

1 � �2536� �12� � 3.1415918

n � 6.

1

0

41 � x2 dx � �4 arctan x�

1

0� 4 arctan 1 � 4 arctan 0 � 4��

4� � 4�0� � �

73.

(a) represents the average value of over the interval Maximum at since the graph is greatest on

(b)

when x � �1.F��x� �1

1 � �x � 2�2 �1

1 � x2 ��1 � x2� � �x2 � 4x � 5�

�x2 � 1��x2 � 4x � 5� ��4�x � 1�

�x2 � 1��x2 � 4x � 5� � 0

F�x� � �arctan t�x�2

x� arctan�x � 2� � arctan x

�1, 1�.x � �1, x, x � 2�.f �x�F�x�

F�x� �12x�2

x

2t 2 � 1

dt

74.

(a)

(b)

� 2 arcsin� u

�6� � C � 2 arcsin��x

�6� � C

1

�6u2 � u4�2u du� � 2

�6 � u2 du

u � �x, u2 � x, 2u du � dx

1

�6x � x2 dx � dx

�9 � �x � 3�2� arcsin�x � 3

3 � � C

6x � x2 � 9 � �x2 � 6x � 9� � 9 � �x � 3�2

1

�6x � x2 dx

(c)

The antiderivatives differ by a constant,

Domain: 0, 6�

�2.

7

−2

−1

y1

y2

4

75. False, dx

3x�9x2 � 16�

112

arcsec �3x�4

� C 76. False, dx25 � x2 dx �

15

arctan x5

� C

77. True

ddx��arccos

x2

� C� �12

�1 � �x2�2�

1�4 � x2

78. False. Use substitution: u � 9 � e2x, du � �2e2x dx

488 Chapter 5 Integration

79.

Thus, du

�a2 � u2� arcsin�u

a� � C.

ddx�arcsin�u

a� � C� �1

�1 � �u2a2��u�

a � �u�

�a2 � u280.

Thus, dua2 � u2 � u�

a2 � u2 dx �1a

arctan ua

� C.

�1a2� u�

�a2 � u2�a2� �u�

a2 � u2

ddx�

1a

arctan ua

� C� �1a�

u�a1 � �ua�2�

81. Assume

The case is handled in a similar manner. Thus,

du

u�u2 � a2� u�

u�u2 � a2 dx �

1a

arcsec �u�a

� C.

u < 0

ddx�

1a

arcsec ua

� C� �1a�

u�a

�ua���ua�2 � 1� �1a�

u�

u��u2 � a2�a2� �u�

u�u2 � a2.

u > 0.

82. (a)

(b) Trapezoidal Rule:

(c) Because

you can use the Trapezoidal Rule to approximate and hence, For example, using you obtain

� � 4�0.785397� � 3.141588.

n � 200,�.�4,

1

0

11 � x2 dx � arctan x�

1

0�

4,

A � 0.7847

n � 8, b � a � 1 � 0 � 1

A � 1

0

11 � x2 dx

83. (a)

(c)

When andwe have

—CONTINUED—

v�t� ��32k

tan�arctan�500� k32� � �32k t�.

C � arctan�500�k32 �,v � 500,t � 0,

v ��32k

tan�C � �32k t�

� k32

v � tan�C � �32k t�

arctan�� k32

v� � ��32k t � C

1

�32k arctan�� k

32v� � �t � C1

132 � kv2 dv � � dt

0 200

550

v�t� � �32t � 500 (b)

When the object reaches its maximum height,

(d) When

v�t� � 0 when t0 � 6.86 sec.

00

7

500

v(t� ��32,000 tan arctan�500�0.00003125� ��0.032 t�k � 0.001:

� 3906.25 ft �Maximum height�

s�15.625� � �16�15.625�2 � 500�15.625�

t � 15.625

�32t � �500

v�t� � �32t � 500 � 0

v�t� � 0.

s�t� � �16t2 � 500t

s�0� � �16�0� � 500�0� � C � 0 ⇒ C � 0

� �16t2 � 500t � C

s�t� � v�t� dt � ��32t � 500� dt

Section 5.9 Hyperbolic Functions 489

83. —CONTINUED—

(e)

Simpson’s Rule: feet

(f) Air resistance lowers the maximum height.

h � 1088n � 10;

h � �6.86

0

�32,000 tan�arctan�500�0.00003125 � � �0.032 t� dt

84.

Since and is increasing for

for Thus,

Since and g is increasing for for Thus, Therefore,x

1 � x2 < arctan x < x.

x > arctan x.x > 0.x � arctan x > 0x > 0,g�0� � 0

g� �x� � 1 �1

1 � x2 �x2

1 � x2 > 0 for x > 0.

Let g�x� � x � arctan x

arctan x > x

1 � x2.x > 0.arctan x �x

1 � x2 > 0

x > 0,ff �0� � 0

f � �x� �1

1 � x2 �1 � x2

�1 � x2�2 �2x2

�1 � x2� > 0 for x > 0.

x2 4 6 8 10

1

3

5

4

2

y

y

y3

2

1

yLet f �x� � arctan x �x

1 � x2

Section 5.9 Hyperbolic Functions

1. (a)

(b) tanh��2� �sinh��2�cosh��2� �

e�2 � e2

e�2 � e2 � �0.964

sinh 3 �e3 � e�3

2� 10.018 2. (a)

(b) sech 1 �2

e � e�1 � 0.648

cosh 0 �e0 � e0

2� 1

3. (a)

(b)

�5 � �1�5�5 � �1�5� �

1312

coth�ln 5� �cosh�ln 5�sinh�ln 5� �

eln 5 � e�ln 5

eln 5 � e�ln 5

csch�ln 2� �2

eln 2 � e�ln 2 �2

2 � �1�2� �43

4. (a)

(b) tanh�1 0 � 0

sinh�1 0 � 0

5. (a)

(b) sech�1 23

� ln1 � �1 � �4�9�2�3 � 0.962

cosh�1 2 � ln�2 � �3 � � 1.317 6. (a)

(b) coth�1 3 �12

ln42 � 0.347

csch�1 2 � ln1 � �52 � 0.481

7. �e2x � 2 � e�2x � 4

�ex � e�x�2 �e2x � 2 � e�2x

e2x � 2 � e�2x � 1tanh2 x � sech2 x � ex � e�x

ex � e�x2

� 2ex � e�x

2

8.1 � cosh 2x

2�

1 � �e2x � e�2x��22

�e2x � 2 � e�2x

4� ex � e�x

2 2

� cosh2 x

490 Chapter 5 Integration

9.

�14

�2�ex�y � e��x�y��� �e�x�y� � e��x�y�

2� sinh�x � y�

�14

�ex�y � e�x�y � ex�y � e��x�y� � ex�y � e�x�y � ex�y � e��x�y��

sinh x cosh y � cosh x sinh y � ex � e�x

2 ey � e�y

2 � ex � e�x

2 ey � e�y

2

10. 2 sinh x cosh x � 2ex � e�x

2 ex � e�x

2 �e2x � e�2x

2� sinh 2x

11.

�12

�e3x � e�x � ex � ex � e�3x � e�x� �e3x � e�3x

2� sinh�3x�

� ex � e�x

2 �3 � e2x � 2 � e�2x� �12

�ex � e�x��e2x � e�2x � 1�

3 sinh x � 4 sinh3 x � sinh x�3 � 4 sinh2 x� � ex � e�x

2 �3 � 4ex � e�x

2 2

12.

� cosh x � cosh y

� 2�ex � ey � e�y � e�x

4 � �ex � e�x

2�

ey � e�y

2

2 cosh x � y

2 cosh

x � y2

� 2�e�x�y��2 � e��x�y��2

2 ��e�x�y��2 � e��x�y��2

2 �

13.

coth x �1

3��13�

�133

sech x �1

�13�2�

2�1313

csch x �1

3�2�

23

tanh x �3�2

�13�2�

3�1313

cosh2 x � 32

2

� 1 ⇒ cosh2 x �134

⇒ cosh x ��13

2

sinh x �32

14.

csch x �1

�3�3� �3

sinh x � tanh x cosh x � 122�3

3 ��33

coth x �1

1�2� 2

cosh x �1

�3�2�

2�33

12

2

� sech2 x � 1 ⇒ sech2 x �34

⇒ sech x ��32

tanh x �12

Putting these in order:

tanh x �12

coth x � 2

cosh x �2�3

3 sech x �

�32

sinh x ��33

csch x � �3

Section 5.9 Hyperbolic Functions 491

21.

h� �x� �12

cosh�2x� �12

�cosh�2x� � 1

2� sinh2 x

h�x� �14

sinh 2x �x2

22.

h� �t� � 1 � csch2 t � coth2 t

h�t� � t � coth t

23.

f � �t� �1

1 � sinh2 t�cosh t� �

cosh tcosh2 t

� sech t

f �t� � arctan�sinh t� 24.

� �6 sech2 3x tanh 3x

g� �x� � �2 sech�3x� sech�3x� tanh�3x��3�

g�x� � sech2 3x

15.

y� � �sech�x � 1� tanh�x � 1�

y � sech�x � 1� 16.

y� � �3 csch2�3x�

y � coth 3x 17.

f � �x� �1

sinh x�cosh x� � coth x

f �x� � ln�sinh x�

18.

� tanh x

g� �x� �1

cosh x�sinh x�

g�x� � ln�cosh x� 19.

�1

sinh x� csch x

�1

2 sinh�x�2� cosh�x�2�

y� �1�2

tanh�x�2� sech2x2

y � lntanh x2 20.

� x sinh x

y� � x sinh x � cosh x � cosh x

y � x cosh x � sinh x

25.

Tangent line:

y � �2x � 2

y � 0 � �2�x � 1�

y� �1� � �2

y� � cosh�1 � x2���2x�

y � sinh�1 � x2�, �1, 0� 26.

At

Tangent line:

Note: cosh�1� � 1.5431

y � cosh�1�x � cosh�1� � 1

y � 1 � cosh�1��x � 1�

�1, 1�, y� � cosh�1�.

y�

y�

cosh xx

� sinh x ln x

ln y � cosh x ln x

y � x cosh x, �1, 1�

27.

At

Tangent line:

y � �2x � 1

y � 1 � �2�x � 0�

�0, 1�, y� � 2�1���1� � �2.

y� � 2�cosh x � sinh x��sinh x � cosh x�

y � �cosh x � sinh x�2, �0, 1� 28.

Tangent line:

y � x � 1

y � 1 � 1�x � 0�

y� �0� � e0�1� � 1

y� � esinh x cosh x

y � esinh x, �0, 1�

29.

Relative maxima:

Relative minimum: �0, �1�

�±�, cosh ��

� 2 sin x cosh x � 0 when x � 0, ±�.

f � �x� � sin x cosh x � cos x sinh x � cos x sinh x � sin x cosh x

�−2

− 2

(0, 1)−

( , cosh )− ππ ( , cosh )π π12

�2

f �x� � sin x sinh x � cos x cosh x, �4 ≤ x ≤ 4

492 Chapter 5 Integration

30.

By the First Derivative Test,is a relative minimum.

�0, �cosh��1�� � �0, �1.543�

f � �x� � 0 for x � 0.

f � �x� � x cosh�x � 1� � sinh�x � 1� � sinh�x � 1� � x cosh�x � 1�

(0, −1.543)

−6

−2

6

6 f �x� � x sinh�x � 1� � cosh�x � 1�

31.

Using a graphing utility,

By the First Derivative Test, is a relative maximum and is a relative minimum.

−1

(1.20, 0.66)

( 1.20, 0.66)− −

1

��−

��1.1997, �0.6627��1.1997, 0.6627�

x � ±1.1997.

x tanh x � 1

� sech x�1 � x tanh x� � 0

g� �x� � sech x � x sech x tanh x

g�x� � x sech x 32.

Using a graphing utility,

From the First Derivative Test, is a relative maximum and is a relativeminimum.

−3 3

−2

(0.88, 0.53)

( 0.88, 0.53)− −

2

��0.8814, �0.5328��0.8814, 0.5328�

x � 0.8814.

sech2 x �12

h� �x� � 2 sech2 x � 1 � 0

h�x� � 2 tanh x � x

35.

f

P1

P2

3

−2

−3

2

P2�x� � f �0� � f ��0��x � 0� �12 f � �0��x � 0�2 � x

P1�x� � f �0� � f ��0��x � 0� � x

f � �x� � �2 sech2 x tanh x, f � �0� � 0

f ��x� � sech2 x, f ��0� � 1

f �x� � tanh x, f �0� � 0 36.

−2 20

f

P1

P2

3

P2�x� � 1 �12x2

P1�x� � f �0� � f � �0��x � 0� � 1

f � �x� � cosh x f � �0� � cosh�0� � 1

f � �x� � sinh x f � �0� � sinh�0� � 0

f �x� � cosh x f �0� � cosh�0� � 1

37. (a)

2010

10

20

30

−10

y

x

y � 10 � 15 cosh x

15, �15 ≤ x ≤ 15

33.

Therefore, y��� � y� � 0.

y��� � a cosh x

y� � a sinh x

y� � a cosh x

y � a sinh x 34.

Therefore, y� � y � 0.

y� � a cosh x

y� � a sinh x

y � a cosh x

(b) At

At

(c) At x � 15, y� � sinh�1� � 1.175.y� � sinh x

15.

x � 0, y � 10 � 15 cosh�0� � 25.

x � ±15, y � 10 � 15 cosh�1� � 33.146.

Section 5.9 Hyperbolic Functions 493

38. (a)

−25

−10

80

25

y � 18 � 25 cosh x

25, �25 ≤ x ≤ 25 (b) At

At

(c) At x � 25, y� � sinh�1� � 1.175.y� � sinh x

25.

x � 0, y � 18 � 25 � 43.

x � ±25, y � 18 � 25 cosh�1� � 56.577.

39. Let

� �12

cosh�1 � 2x� � C

�sinh�1 � 2x� dx � �12�sinh�1 � 2x���2� dx

u � 1 � 2x, du � �2 dx. 40. Let

�cosh �x

�x dx � 2�cosh �x 1

2�x dx � 2 sinh�x � C

u � �x, du �1

2�x dx.

41. Let

�cosh2�x � 1� sinh�x � 1� dx �13

cosh3�x � 1� � C

u � cosh�x � 1�, du � sinh�x � 1� dx. 42. Let

� �sech x � C

� sinh1 � sinh2 x

dx � � sinh xcosh2 x

dx ��1

cosh x� C

u � cosh x, du � sinh x dx.

45. Let

�x csch2 x2

2 dx � �csch2

x2

2 x dx � �coth x2

2� C

u �x2

2, du � x dx. 46. Let

� �13

sech3 x � C

�sech3 x tanh x dx � ��sech2 x��sech x tanh x� dx

u � sech x, du � �sech x tanh x dx.

47. Let

� csch 1x

� C

�csch�1�x� coth�1�x�x2 dx � ��csch

1x coth

1x�

1x2 dx

u �1x, du � �

1x2 dx. 48. Let

� arcsinex � e�x

6 � C

� cosh x

�9 � sinh2 x dx � arcsinsinh x

3 � C

u � sinh x, du � cosh x dx.

49. Let

� xx4 � 1

dx �12� 2x

�x2�2 � 1 dx �

12

arctan�x2� � C

u � x2, du � 2x dx. 50.

� 2 ln�1 � �1 � 4x2

2x � C

� 2

x�1 � 4x2 dx � 2� 1

�2x��1 � �2x�2�2� dx

43. Let

�cosh xsinh x

dx � ln sinh x � C

u � sinh x, du � cosh x dx. 44. Let

�12

tanh�2x � 1� � C

�sech2�2x � 1� dx �12�sech2�2x � 1��2� dx

u � 2x � 1, du � 2 dx.

494 Chapter 5 Integration

51.

Note: cosh�ln 2� �eln 2 � e�ln 2

2�

2 � �1�2�2

�54

� ln54 � 0 � ln5

4 � ln�cosh�ln 2� � ln�cosh�0��

� ln�cosh x��ln 2

0

�ln 2

0tanh x dx � �ln 2

0 sinh xcosh x

dx, �u � cosh x� 52.

�12

�12

sinh�1� cosh�1�

�12�1 �

12

sinh�2��

�12�x �

12

sinh�2x��1

0

�1

0 cosh2 x dx � �1

0 1 � cosh�2x�

2 dx

53.

� � 110

ln 5 � x5 � x �4

0�

110

ln 9 �15

ln 3

�4

0

125 � x2 dx �

110� 1

5 � x dx �

110� 1

5 � x dx 54. �4

0

1

�25 � x2 dx � �arcsin

x5�

4

0� arcsin

45

55. Let

� �arcsin�2x���2�4

0�

4

��2�4

0

2�1 � 4x2

dx � ��2�4

0

1�1 � �2x�2

�2� dx

u � 2x, du � 2 dx. 56.

�38

� ln 2

� �ln 2 �12

14� � �0 �

12�

� �x �12

e�2x�ln 2

0

�ln 2

02e�x cosh x dx � �ln 2

0�1 � e�2x� dx

2e�x cosh x � 2e�x�ex � e�x

2 � � 1 � e�2x

57.

y� �3

�9x2 � 1

y � cosh�1�3x� 58.

y� �1

1 � �x�2�212 �

24 � x2

y � tanh�1 x2 59.

y� �1

�tan2 x � 1�sec2 x� � sec x

y � sinh�1�tan x�

60.

since for 0 < x < ��4.sin 2x ≥ 0

y� ��1

cos 2x�1 � cos2 2x��2 sin 2x� �

2 sin 2xcos 2x sin 2x �

2cos 2x

� 2 sec 2x,

y � sech�1�cos 2x�, 0 < x < �

4

61.

y� �1

1 � sin2 2x�2 cos 2x� � 2 sec 2x

y � tanh�1�sin 2x� 62.

y� � 2 csch�1 x �1

x �1 � x2 ��2 csch�1 x

x �1 � x2

y � �csch�1 x�2

63.

� 2 sinh�1�2x�

y� � 2x 2

�1 � 4x2 � 2 sinh�1�2x� �4x

�1 � 4x2

y � 2x sinh�1�2x� � �1 � 4x2 64.

y� � x 11 � x2 � tanh�1 x �

�x1 � x2 � tanh�1 x

y � x tanh�1 x � ln�1 � x2 � x tanh�1 x �12

ln�1 � x2�

Section 5.9 Hyperbolic Functions 495

65. Answers will vary. 66. See the definitions and graphs inthe textbook.

67. limx→�

sinh x � �

68. limx→�

tanh x � 1 69. limx→�

sech x � 0 70. limx→��

csch x � 0

71. limx→0

sinh x

x� lim

x→0 ex � e�x

2x� 1 72. does not exist.

�coth x →� for x → 0�, coth x → �� for x → 0��

limx→0

coth x

73.

� �ln1 � �1 � e2x

ex � C

� 1

�1 � e2x dx � � ex

ex�1 � �ex�2 dx � �csch�1�ex� � C 74.

� �1

12 ln 3 � x2

3 � x2 � C

� �12

16 ln 3 � x2

3 � x2 � C

� x9 � x4 dx � �

12� �2x

9 � �x2�2 dx

77. � �14x � x2 dx � � 1

�x � 2�2 � 4 dx �

14

ln �x � 2� � 2�x � 2� � 2 �

14

ln x � 4x � C

78. � �12

ln2 � ��x � 2�2 � 4

x � 2 � C � dx

�x � 2��x2 � 4x � 8� � dx

�x � 2���x � 2�2 � 4

79.

�1

2�6 ln �2 �x � 1� � �3

�2 �x � 1� � �3 � C �1�2

�1

2�3 ln �3 � �2 �x � 1�

�3 � �2 �x � 1� � C

�1�2

� �2

��3 �2 � ��2�x � 1��2 dx

� 11 � 4x � 2x2 dx � � 1

3 � 2�x � 1�2 dx

80.

�1�2

� 1

�x � 1���x � 1�2 � ��3 �2dx � �

1

�6 ln�3 � ��x � 1�2 � 3

x � 1 � C

� 1

�x � 1��2x2 � 4x � 8 dx � � 1

�x � 1��2�x � 1�2 � 6 dx

75. Let

� 1

�x�1 � x dx � 2� 1

�1 � ��x �2 1

2�x dx � 2 sinh�1�x � C � 2 ln��x � �1 � x � � C

u � �x, du �1

2�x dx.

76. Let

� �x

�1 � x3 dx �

23� 1

�1 � �x3�2�232�x dx �

23 sinh�1�x3�2� � C �

23

ln�x3�2 � �1 � x3 � � C

u � x3�2, du �32�x dx.

81. Let

�14� 4

�81 � �4x � 1�2 dx �

14

arcsin4x � 19 � C y � � 1

�80 � 8x � 16x2 dx

u � 4x � 1, du � 4 dx.

496 Chapter 5 Integration

82. Let

� �1

�3 ln �3 � ��4x2 � 8x � 1

2�x � 1� � C � � 2

2�x � 1����3 �2 � �2�x � 1��2 dx

y � � 1

�x � 1���4x2 � 8x � 1 dx

u � 2�x � 1�, du � 2 dx.

83.

��x2

2� 4x �

103

ln 5 � xx � 1 � C

� �x2

2� 4x �

103

ln 1 � x5 � x � C

� �x2

2� 4x �

206

ln 3 � �x � 2�3 � �x � 2� � C

� ���x � 4� dx � 20� 132 � �x � 2�2 dx

y � � x3 � 21x5 � 4x � x2 dx � ��x � 4 �

205 � 4x � x2 dx 84.

� ln 4x � x2 �34

ln x � 4x � C

� ln 4x � x2 �34

ln �x � 2� � 2�x � 2� � 2 � C

y � � 1 � 2x4x � x2 dx � � 4 � 2x

4x � x2 dx � 3� 1�x � 2�2 � 4

dx

85.

� 8 arctan�e2� � 2� � 5.207

� �8 arctan�ex�2��4

0

� 4�4

0

ex�2

�ex�2�2 � 1 dx

� 2�4

0

2ex�2 � e�x�2 dx

A � 2�4

0sech

x2

dx 86.

� ln�e4 � e�4

2� 1.654

� �12

ln�e2x � e�2x��2

0�

12

ln�e4 � e�4� �12

ln 2

�12�2

0

1e2x � e�2x�2��e2x � e�2x� dx

� �2

0 e2x � e�2x

e2x � e�2x dxA � �2

0tanh 2x dx

87.

�52

ln�4 � �17 � � 5.237

� �52

ln�x2 � �x4 � 1 ��2

0

�52�2

0

2x

��x2�2 � 1 dx

A � �2

0

5x

�x4 � 1 dx 88.

� 6 ln5 � �21

3 � �5 � 3.626

� 6 ln�5 � �21 � � 6 ln�3 � �5 �

� �6 ln�x � �x2 � 4 ��5

3

A � �5

3

6

�x2 � 4 dx

89. (a)

(b) ��3

0

dx�x2 � 1

� sinh�1 x��3

0� sinh�1��3 � � 1.317

� ln��3 � 2� � 1.317

��3

0

dx�x2 � 1

� ln�x � �x2 � 1 ���3

090. (a)

(b)

�ln 3

2� �

ln 32 � ln 3

�1�2

�1�2

dx1 � x2 � �tanh�1 x�

1�2

�1�2

� ln 32

� ln 12

� ln 3

�12�ln

32

� ln 12

� ln 12

� ln 32�

� �12

ln 1 � x �12

ln 1 � x �1�2

�1�2

�1�2

�1�2

dx1 � x2 � �1�2

�1�2 � 1�2

1 � x�

1�21 � x� dx

Section 5.9 Hyperbolic Functions 497

91.

3kt16

� � 1�x � 6�2 � 4

dx �1

2�2� ln �x � 6� � 2�x � 6� � 2 � C �

14

ln x � 8x � 4 � C

�3k16

dt � � 1x2 � 12x � 32

dx

When

C � �14

ln�2�

t � 0

x � 0: When

k �2

15 ln7

6

30k16

�14

ln �7�3 �

14

ln�2� �14

ln76

t � 10

x � 1: When

x �10462

�5231

� 1.677 kg

62x � 104

4936

�x � 8

2x � 8

ln76

2

� ln x � 8

2x � 8

316

215 ln7

6�20� �14

ln x � 8

2x � 8

t � 20:

92. (a)

(b)

(c)

Let then

Since

� ��32

�k tanh��32k t�

��32�k ���e�32k t � e��32k t�

e�32k t � e��32k t � v �

�32�e�2�32k t � 1��k�e�2�32k t � 1� �

e�32k t

e�32k t

v��k � �k e�2�32k t� � �32�e�2�32k t � 1�

�32 � �k v � e�2�32k t��32 � �k v�

�32 � �k v

�32 � �k v� e�2�32k t

ln �32 � �k v

�32 � �k v � �2�32k t

v�0� � 0, C � 0.

1

�k�

1

2�32 ln �32 � �k v

�32 � �k v � �t � C

du � �k dv.u � �k v,

� dv32 � kv2 � ��dt

� dvkv2 � 32

� �dt

dvdt

� �32 � kv2

s�t� � �16t 2 � 400

s�0� � �16�0�2 � C � 400 ⇒ C � 400

s�t� � �v�t� dt � ���32t� dt � �16t2 � C

v�t� � �32t (d)

The velocity is bounded by

(e) Since tanh ln cosh (which can be verified by differentiation), then

When

When

when seconds

when seconds

When air resistance is not neglected, it takesapproximately 3.3 more seconds to reach the ground.

(f) As increases, the time required for the object to reachthe ground increases.

k

t � 8.3s2�t� � 0

t � 5s1�t� � 0

s1�t� � �16t2 � 400

s2�t� � 400 � 100 ln�cosh�0.32 t�k � 0.01:

� 400 ⇒ 400 � �1�k� ln�cosh��32k t��.s�0� � C

t � 0,

� �1k ln�cosh��32k t�� � C.

� ��32

�k

1

�32k ln�cosh��32k t�� � C

s�t� � ���32

�k tanh��32k t� dt

�ct��ct� dt � �1�c��

��32��k.

limt→�

���32

�k tanh��32k t� � �

�32

�k

498 Chapter 5 Integration

93.

dydx

��1

�x�a��1 � �x2�a2��

x

�a2 � x2�

�a2

x�a2 � x2�

x

�a2 � x2�

x2 � a2

x�a2 � x2�

��a2 � x2

x

y � a sech�1 xa

� �a2 � x2, a > 0

94. Equation of tangent line through

When

Hence, Q is the point

Distance from P to Q: d � �x02 � ���a2 � x0

2 �2 � a

�0, a sech�1�x0�a��.

y � a sech�1 x0

a� �a2 � x0

2 � �a2 � x02 � a sech�1

x0

a.

x � 0,

y � a sech�1 x0

a� �a2 � x0

2 � ��a2 � x0

2

x0�x � x0�

( , 0)aP

Q

L

x

a

yP � �x0, y0�:

95. Let

u �12

ln1 � x1 � x, �1 < x < 1

2u � ln1 � x1 � x

e2u �1 � x1 � x

e2u�1 � x� � 1 � x

e2u � 1 � xe2u � x

eu � e�u � xeu � xe�u

sinh ucosh u

�eu � e�u

eu � e�u � x

tanh u � x.u � tanh�1 x, �1 < x < 1, 96. Let Then,

and

Thus, Therefore,

arctan�sinh x� � arcsin�tanh x�.

y � arctan�sinh x�.

tan y �ex � e�x

2� sinh x.

sin y � tanh x �ex � e�x

ex � e�x

y

2

e e−e e+

x

x

−x

−x

y � arcsin�tanh x�.

97.

�2x sinh�xb�

�2x �

exb � e�xb

2 �

�exb

x�

e�xb

x

�b

�b

ext dt �ext

x �b

�b98.

y� �ex � e�x

2� sinh x

y � cosh x �ex � e�x

2

99.

��1

�sech y��1 � sech2 y�

�1

x�1 � x2

y� ��1

�sech y��tanh y�

��sech y��tanh y�y� � 1

sech y � x

y � sech�1 x 100.

y� �1

sinh y�

1

�cosh2 y � 1�

1

�x2 � 1

�sinh y�� y� � � 1

cosh y � x

y � cosh�1 x

Review Exercises for Chapter 5 499

Review Exercises for Chapter 5

1.

x

f

f

y 2.

x

f ′

f

y

103.

Let be a point on the catenary.

The slope at is The equation of line is

When The length of is

the ordinate of the point P.y1

�c2 sinh2�x1

c � � c2 � c � cosh x1

c� y1,

Lc �x

sinh�x1�c� ⇒ x � c sinh�x1

c �.y � 0,

y � c ��1

sinh�x1�c� �x � 0�.

Lsinh�x1�c�.P

y� � sinh xc

P�x1, y1�

x

y

P(x1, y1)

(0, c) L

y � c cosh xc

104. There is no such common normal. To see this, assume there is a common normal.

Normal line at is

Similarly, is normal at Also,

The slope between the points is Therefore,

for all Hence, But,

a contradiction.�a � c

cosh a � sinh c< 0,

c < a.x ⇒ sinh c < cosh c � sinh a < cosh a.sinh x < cosh x

cosh c > 0 ⇒ a > 0

�a � c

cosh a � sinh c� cosh c � sinh a.

sinh c � cosh ac � a

.

�1sinh a

��1

cosh c ⇒ cosh c � sinh a.

�c, sinh c�.y � sinh c ��1

cosh c�x � c�

y � cosh a ��1

sinh a�x � a�.�a, cosh a�

y � cosh x ⇒ y� � sinh x.

x

y

y = sinh x

y = cosh x

(a, cosh a)

(c, sinh c)

101.

y� �1

cosh y�

1

�sinh2 y � 1�

1

�x2 � 1

�cosh y�y� � 1

sinh y � x

y � sinh�1 x 102.

� � �2ex � e�x��ex � e�x

ex � e�x� � �sech x tanh x

y� � �2�ex � e�x��2�ex � e�x�

y � sech x �2

ex � e�x

15.

v�30� � 8�30� � 240 ft�sec

a �2�3600�

�30�2 � 8 ft�sec2.

s�30� �a2

�30�2 � 3600 or

s�t� �a2

t2

s�0� � 0 � C2 � 0 when C2 � 0.

s�t� � �at dt �a2

t2 � C2

v�t� � at

v�0� � 0 � C1 � 0 when C1 � 0.

v�t� � �a dt � at � C1

a�t� � a 16.

Solving the system

we obtain and We now solveand get Thus,

Stopping distance from 30 mph to rest is

475.2 � 264 � 211.2 ft.

s�725 � � �

55�122 �72

5 �2

� 66�725 � � 475.2 ft.

t � 72�5.��55�12�t � 66 � 0a � 55�12.t � 24�5

s(t� � �a2

t2 � 66t � 264

v�t� � �at � 66 � 44

s�t� � �a2

t2 � 66t since s�0� � 0.

v�t� � �at � 66 since v�0� � 66 ft�sec.

a�t� � �a

30 mph � 44 ft�sec

45 mph � 66 ft�sec

500 Chapter 5 Integration

3. ��2x2 � x � 1� dx �23

x3 �12

x2 � x � C 4.

� 23�3x

dx �23��3x��1�3�3� dx � �3x�2�3 � C

du � 3 dx

u � 3x

5. �x3 � 1x2 dx � ��x �

1x2� dx �

12

x2 �1x

� C

7. ��4x � 3 sin x� dx � 2x2 � 3 cos x � C

9. ��5 � ex� dx � 5x � ex � C

11. �5x dx � 5 lnx � C

13.

When

y � 2 � x2

C � 2

y � �1 � C � 1

x � �1:

f �x� � ��2x dx � �x2 � C

��1, 1�f��x� � �2x,

6.

�12

x2 � 2x �1x

� C

�x3 � 2x2 � 1x2 dx � ��x � 2 � x�2� dx

8. ��5 cos x � 2 sec2 x� dx � 5 sin x � 2 tan x � C

10. ��t � et� dt �t2

2� et � C

12. �10x

dx � 10 lnx � C

14.

Since the slope of the tangent line at is 3,

f �x� � 2ex � x � 1

f �0� � 1 � 2 � C2 ⇒ C2 � �1

f �x� � ��2ex � 1� dx � 2ex � x � C2

f��x� � 2ex � 1

⇒ C1 � 1.f��0� � 2 � C1 � 3�0, 1�

f��x� � �2ex dx � 2ex � C1

f ��x� � 2ex, �0, 1�

Review Exercises for Chapter 5 501

17.

(a) when t sec.

(b) ft

(c) when sec.

(d) fts�32� � �16�9

4� � 96�32� � 108

t �32

v�t� � �32t � 96 �962

s�3� � �144 � 288 � 144

� 3v�t� � �32t � 96 � 0

s�t� � �16t2 � 96t

v�t� � �32t � 96

a�t� � �32 18.

(a) when sec.

(b)

(c) when sec.

(d) s�2.04� � 61.2 m

t �209.8

� 2.04v�t� � �9.8t � 40 � 20

s�4.08� � 81.63 m

t �409.8

� 4.08v�t� � �9.8t � 40 � 0

�s�0� � 0�s�t� � �4.9t2 � 40t,

v�t� � �9.8t � v0 � �9.8t � 40

a�t� � �9.8 m�sec2

19. (a)

(b)

(c) 10

i�1�4i � 2�

n

i�1i3

10

i�1�2i � 1�

20.

(a)

(b)

(c)

(d) 5

i�2�xi � xi�1� � ��1 � 2� � �5 � ��1�� � �3 � 5� � �7 � 3� � 5

5

i�1�2xi � xi

2� � �2�2� � �2�2� � �2��1� � ��1�2� � �2�5� � �5�2� � �2�3� � �3�2� � �2�7� � �7�2� � �56

5

i�1 1xi

�12

� 1 �15

�13

�17

�37

210

15

5

i�1xi �

15

�2 � 1 � 5 � 3 � 7� �165

x5 � 7x4 � 3,x3 � 5,x2 � �1,x1 � 2,

22.

7 < Area of region < 14

s�3� � 1�20 � 21 � 22� � 7

S�3� � 1�21 � 22 � 23� � 14

y � 2x, �x � 1, n � 3

21.

9.0385 < Area of region < 13.0385

s�n� � s�4� �12

10�1�2�2 � 1

�10

1 � 1�

10�3�2�2 � 1

�10

22 � 1� � 9.0385

S�n� � S�4� �12

101

�10

�1�2�2 � 1�

10�1�2 � 1

�10

�3�2�2 � 1� � 13.0385

y �10

x2 � 1, �x �

1

2, n � 4

23.

� 24 � 8 � 16

� limn→�

24 � 8 n � 1

n �

� limn→�

4n 6n �

4n

n�n � 1�

2 �

� limn→�

n

i�1 �6 �

4in �

4n

Area � limn→�

n

i�1 f �ci� �x

x86

y

4

6

8

2

−2

2−2 4

y � 6 � x, �x �4n

, right endpoints

502 Chapter 5 Integration

26.

� 4 � 6 � 4 � 1 � 15

� limn→�

4n n �

3n

n�n � 1�

2�

3n2

n�n � 1��2n � 1�6

�1n3

n2�n � 1�2

4 �

� limn→�

4n

n

i�1 1 �

3in

�3i2

n2 �i3

n3�

� limn→�

12n

n

i�1 8 �

24in

�24i2

n2 �8i3

n3 �

� limn→�

n

i�1 14�2 �

2in �

3

�2n�

Area � limn→�

n

i�1 f �ci� �x

1

5

10

15

20

2 3 4

y

x

y �14

x3, �x �2n

24. right endpoints

�83

� 6 �263

� limn→�

43

�n � 1��2n � 1�

n2 � 6�

� limn→�

2n

4n2

n�n � 1��2n � 1�6

� 3n�

� limn→�

2n

n

i�1 4i2

n2 � 3�

� limn→�

n

i�1 �2i

n �2

� 3��2n�

Area � limn→�

n

i�1 f �ci� �x

1

2

4

6

8

10

12

2

y

x

y � x2 � 3, �x �2n

,

25.

� 3 � 18 � 9 � 12

� limn→�

3 � 18 n � 1

n�

92

�n � 1��2n � 1�

n2 �

� limn→�

3n n �

12n

n�n � 1�

2�

9n2

n�n � 1��2n � 1�6 �

� limn→�

3n

n

i�1 1 �

12in

�9i2

n2 �

� limn→�

n

i�1 5 � ��2 �

3in �

2

��3n�

Area � limn→�

n

i�1 f �ci� �x

x4

1

3

−4

y

4

−2

2

6

1 2−1−3 3

y � 5 � x2, �x �3n

Review Exercises for Chapter 5 503

27.

� 18 �92

� 9� �272

� limn→�

3n 6n �

3n

n�n � 1�

2�

9n2

n�n � 1��2n � 1�6 �

� limn→�

3n

n

i�1 6 �

3in

�9i2

n2 �

� limn→�

3n

n

i�1 10 �

15in

� 4 � 12in

�9i2

n2 �

Area � limn→�

n

i�1 5�2 �

3in � � �2 �

3in �

2

��3n�

1

1 2 3 4 5 6

2

3

4

6

y

x

x � 5y � y2, 2 ≤ y ≤ 5, �y �3n

28. (a)

(b)

(c) Area � limn→�

mb2�n � 1�

2n� lim

n→� mb2�n � 1�

2n�

12

mb2 �12

�b��mb� �12

�base��height�

s�n� � n�1

i�0 f �bi

n ��bn� �

n�1

i�0m�bi

n ��bn� � m�b

n�2

n�1

i�0i �

mb2

n2 ��n � 1�n2 � �

mb2�n � 1�2n

S�n� � n

i�1 f �bi

n ��bn� �

n

i�1�mbi

n ��bn� � m�b

n�2

n

i�1i �

mb2

n2 �n�n � 1�2 � �

mb2�n � 1�2n

s � m�0��b4� � m�b

4��b4� � m�2b

4 ��b4� � m�3b

4 ��b4� �

mb2

16�1 � 2 � 3� �

3mb2

8

xx b=

y mx=

yS � m�b

4��b4� � m�2b

4 ��b4� � m�3b

4 ��b4� � m�4b

4 ��b4� �

mb2

16�1 � 2 � 3 � 4� �

5mb2

8

29. lim��� →�

n

i�1�2ci � 3� �xi � �6

4�2x � 3� dx 30. lim

��� →�

n

i�13ci�9 � ci

2� �xi � �3

13x�9 � x2� dx

31.

(triangle)

�5

0�5 � x � 5� dx � �5

0�5 � �5 � x�� dx � �5

0x dx �

252

x9

y

6

9

12

3

−3

3−3 6

Triangle

32.

(semicircle)�4

�4

�16 � x2 dx �12

�4�2 � 8

1−1

1

2

3

5

6

−2

−2−3−4 2 3 4

y

x

33. (a)

(b)

(c)

(d) �6

25f �x� dx � 5�6

2f �x� dx � 5�10� � 50

�6

2�2f �x� � 3g�x�� dx � 2�6

2f �x� dx � 3�6

2g�x� dx � 2�10� � 3�3� � 11

�6

2� f �x� � g�x�� dx � �6

2f �x� dx � �6

2g�x� dx � 10 � 3 � 7

�6

2� f �x� � g�x�� dx � �6

2f �x� dx � �6

2g�x� dx � 10 � 3 � 13

504 Chapter 5 Integration

42. �2

1� 1

x2 �1x3� dx � �2

1�x�2 � x�3� dx � �

1x

�1

2x2�2

1� ��

12

�18� � ��1 �

12� �

18

43.

� 1 ��22

��2 � 2

2

�3�4

0sin d � �cos �

3�4

0� ���

�22 � � 1

45. �2

0�x � ex� dx � x2

2� ex�

2

0� 2 � e2 � 1 � 1 � e2

47.

−1−2 1 2 3 4 5 6−1

1

2

3

4

5

6

x

y

�3

1�2x � 1� dx � x2

� x�3

1� 6

44. ��4

��4sec2 t dt � tan t�

�4

��4� 1 � ��1� � 2

46. �6

1 3x dx � 3 lnx�

6

1� 3 ln 6

48.

x1 2 3 4 5−1−2

1

2

4

5

6

7

y

�2

0�x � 4� dx � x2

2� 4x�

2

0� 10

34. (a)

(b)

(c)

(d) �6

3�10 f �x� dx � �10�6

3f �x� dx � �10��1� � 10

�4

4f �x� dx � 0

�3

6f �x� dx � ��6

3f �x� dx � ���1� � 1

�6

0f �x� dx � �3

0f �x� dx � �6

3f �x� dx � 4 � ��1� � 3

35. (c)�734

,�8

1� 3�x � 1� dx � 3

4x4�3 � x�

8

1� 3

4�16� � 8� � 3

4� 1�

36. (d)�3

1 12x3 dx � 12x�2

�2 �3

1� �6

x2 �3

1�

�69

� 6 �163

, 37. �4

0�2 � x� dx � 2x �

x2

2 �4

0� 8 �

162

� 16

38. �1

�1�t2 � 2� dt � t3

3� 2t�

1

�1�

143

39. �1

�1�4t3 � 2t� dt � t4 � t2�

1

�1� 0

40.

�5215

� �325

�163

� 10� � ��325

�163

� 10�

�2

�2�x4 � 2x2 � 5� dx � x5

5�

2x3

3� 5x�

2

�2

41. �9

4x�x dx � �9

4x3�2 dx � 2

5x5�2�

9

4�

25

���9 �5� ��4 �5� �

25

�243 � 32� �4225

Review Exercises for Chapter 5 505

49.

1 2 4 5 6 7 8

1

2

3

4

5

6

7

8

x

y

� �643

� 36� � �9 � 27� �103

�4

3�x2 � 9� dx � x3

3� 9x�

4

350.

x1 3−2

2

3

−1

−2

y

�103

�76

�92

� ��83

� 2 � 4� � �13

�12

� 2�

�2

�1��x2 � x � 2� dx � �

x3

3�

x2

2� 2x�

2

�1

51.

x

1

1

1

y

�12

�14

�14�1

0�x � x3� dx � x2

2�

x4

4 �1

052.

x1

1

y

�23

�25

�4

15

� 23

x3�2 �25

x5�2�1

0

�1

0

�x�1 � x� dx � �x1�2 � x3�2� dx

53.

2−2−2

2

4

6

8

10

4 6 8 10x

y

� 8�3 � 1� � 16� 4x1�2

�1�2��9

1 Area � �9

1

4�x

dx 54.

2

3

4

5

π6

π3

y

x

� tan x��3

0� �3 Area � ��3

0sec2 x dx

55.

−1−1

1

2

3

4

5

1 2 3 4 5

y

x

Area � �3

1 2x dx � 2 ln x�

3

1� 2 ln 3 � 2 ln 1 � ln 9

y �2x

56.

4

6

8

10

1 2

y

x

� 2 � e2 � 1 � 1 � e2 � 8.3891

� x � ex�2

0 Area � �2

0�1 � ex� dx

506 Chapter 5 Integration

67. ,

�x�1 � 3x2�4 dx � �16��1 � 3x2�4��6x dx� � �

130

�1 � 3x2�5 � C �1

30�3x2 � 1�5 � C

du � �6x dxu � 1 � 3x2

68. ,

� x � 3�x2 � 6x � 5�2 dx �

12� 2x � 6

�x2 � 6x � 5�2 dx ��12

�x2 � 6x � 5��1 � C ��1

2�x2 � 6x � 5� � C

du � �2x � 6� dxu � x2 � 6x � 5

69. �sin3 x cos x dx �14

sin4 x � C 70. �x sin 3x2 dx �16��sin 3x2��6x� dx � �

16

cos 3x2 � C

71. � sin

�1 � cos d � ��1 � cos ��1�2 sin d � 2�1 � cos �1�2 � C � 2�1 � cos � C

72. � cos x

�sin x dx � ��sin x��1�2 cos x dx � 2�sin x�1�2 � C � 2�sin x � C

66. ,

�x2�x3 � 3 dx �13��x3 � 3�1�23x2 dx �

29

�x3 � 3�3�2 � C

du � 3x2 dxu � x3 � 3

57.

x �254

�x �52

1

2 4 6 8 10

2

2255,4

x

y 2

5�

1�x

�25

�3 � 2� �25

, Average value

1

9 � 4 �9

4

1�x

dx � 1

5 2�x�

9

458.

x � 3�2

x1 2

2

4

6

8

( 2 , 2)3

yx3 � 2

12 � 0

�2

0x3 dx � x4

8 �2

0� 2

59. F��x� � x2�1 � x3 60. F��x� �1x2 61. F��x� � x2 � 3x � 2 62. F��x� � csc2 x

63.

�x7

7�

35

x5 � x3 � x � C

��x2 � 1�3 dx � ��x6 � 3x 4 � 3x2 � 1� dx 64.

�x3

3� 2x �

1x

� C

��x �1x�

2

dx � ��x2 � 2 � x�2� dx

65. ,

� x2

�x3 � 3 dx � ��x3 � 3��1�2x2 dx �

13��x3 � 3��1�23x2 dx �

23

�x3 � 3�1�2 � C

du � 3x2 dxu � x3 � 3

Review Exercises for Chapter 5 507

73. �tann x sec2 x dx �tann�1 xn � 1

� C, n � �1 74. �sec 2x tan 2x dx �12��sec 2x tan 2x��2� dx �

12

sec 2x � C

75. ��1 � sec x�2 sec x tan x dx �1��1 � sec x�2� sec x tan x� dx �

13

�1 � sec x�3 � C

77. �xe�3x2 dx � �

16�e�3x2��6x� dx � �

16

e�3x 2� C

79.

�1

2 ln 55�x�1�2

� C

��x � 1�5�x�1�2 dx �

12�5�x�1�2

2�x � 1� dx

76. �cot4 � csc2 � d� � ���cot ��4��csc2 �� d� � �15

cot 5� � C

78. �e1 x

x2 dx � ��e1 x��

1x2� dx � �e1 x � C

80. �1t2

2�1 t dt � �2�1 t�t�2� dt �1

ln 2 2�1 t � C

81. �2

�1x�x2 � 4� dx �

12�

2

�1�x2 � 4��2x� dx �

12

�x2 � 4�2

2 �2

�1�

14

�0 � 9� � �94

82. �1

0x2�x3 � 1�3 dx �

13

�1

0�x3 � 1�3�3x2� dx �

112 �x3 � 1�4�

1

0�

112

�16 � 1� �54

83. �3

0

1

�1 � x dx � �3

0�1 � x��1�2 dx � 2�1 � x�1�2�

3

0� 4 � 2 � 2

85.

When When

� 2�0

1�u3�2 � 2u1�2� du � 2 2

5u5�2 �

43

u3�2�0

1�

28

15

2�1

0�y � 1��1 � y dy � 2�0

1���1 � u� � 1��u du

u � 0.y � 1,u � 1.y � 0,

dy � �duy � 1 � u,u � 1 � y,

86.

When When

� 2�1

0�u5�2 � 2u3�2 � u1�2� du � 2 2

7u7�2 �

45

u5�2 �23

u3�2�1

0�

32

105

2�0

�1 x2�x � 1 dx � 2�1

0�u � 1�2�u du

u � 1.x � 0,u � 0.x � �1,

dx � dux � u � 1,u � x � 1,

84. �6

3

x

3�x2 � 8 dx �

16

�6

3�x2 � 8��1�2�2x� dx � 1

3�x2 � 8�1�2�

6

3�

13

�2�7 � 1�

87. �

0cos�x

2� dx � 2�

0cos�x

2� 12

dx � 2 sin�x2��

0� 2 88. since sin 2x is an odd function.��4

��4 sin 2x dx � 0

508 Chapter 5 Integration

93. Trapezoidal Rule

Simpson’s Rule 0.685

Graphing utility: 0.704

�n � 4�:

��2

0

�x cos x dx � 0.637�n � 4�:

95. Trapezoidal Rule

Simpson’s Rule

Graphing utility: 1.494

�n � 4� � 1.494

�1

�1e�x2

dx � 1.463�n � 4�:

97.

� 17x � 2

dx �17� 1

7x � 2�7� dx �

17

ln7x � 2 � C

u � 7x � 2, du � 7 dx

94. Trapezoidal Rule

Simpson’s Rule 3.820

Graphing utility: 3.820

�n � 4�:

0

�1 � sin2 x dx � 3.820�n � 4�:

96. (a)

(b)

�13

4 � 4�2� � 2�1� � 4�12� �

14� � 5.417

S�4� �4 � 03�4� � f �0� � 4 f �1� � 2 f �2� � 4 f �3� � f �4��

R < I < T < L

98.

� xx2 � 1

dx �12� 2x

x2 � 1 dx �

12

lnx2 � 1 � C

u � x2 � 1, du � 2x dx

99.

� �ln1 � cos x � C

� sin x1 � cos x

dx � �� �sin x1 � cos x

dx 100.

�ln�xx

dx �12��ln x��1

x� dx �14

�ln x�2� C

u � ln x, du �1x dx

89.

(a) 2000 corresponds to

�15,000

M 1.20t � 0.02t2�11

10�

24,300M

C �15,000

M �11

10�1.20 � 0.04t� dt

t � 10.

C �15,000

M �t�1

t

p ds

p � 1.20 � 0.04t

(b) 2005 corresponds to

C �15,000

M 1.20t � 0.02t2�16

15�

27,300M

t � 15.

90. liters

Increase is liters.7

�5.1

�1.9

� 0.6048

�2

01.75 sin

t2

dt � �2 1.75 cos

t2 �

2

0� �

2

�1.75���1 � 1� �7

� 2.2282

91. Trapezoidal Rule

Simpson’s Rule

Graphing utility: 0.254

�2

1

11 � x3 dx �

112

11 � 13 �

41 � �1.25�3 �

21 � �1.5�3 �

41 � �1.75�3 �

11 � 23� � 0.254�n � 4�:

�2

1

11 � x3 dx �

18

11 � 13 �

21 � �1.25�3 �

21 � �1.5�3 �

21 � �1.75�3 �

11 � 23� � 0.257�n � 4�:

92. Trapezoidal Rule

Simpson’s Rule

Graphing utility: 0.166

�1

0

x3�2

3 � x2 dx �1

12 0 �4�1�4�3�2

3 � �1�4�2 �2�1�2�3�2

3 � �1�2�2 �4�3�4�3�2

3 � �3�4�2 �12� � 0.166�n � 4�:

�1

0

x3�2

3 � x2 dx �18 0 �

2�1�4�3�2

3 � �1�4�2 �2�1�2�3�2

3 � �1�2�2 �2�3�4�3�2

3 � �3�4�2 �12� � 0.172�n � 4�:

Review Exercises for Chapter 5 509

101. � 3 � ln 4�4

1

x � 1x

dx � �4

1�1 �

1x� dx � x � lnx�

4

1

103. ��3

0sec d � lnsec � tan �

�3

0� ln�2 � �3 �

105. Let Then

�e2x � e�2x

e2x � e�2x dx �12

ln�e2x � e�2x� � C

du � 2�e2x � e�2x� dx.

u � e2x � e�2x.�e2x � e�2x

e2x � e�2x dx.

102. �e

1 ln x

x dx � �e

1�ln x�1�1

x� dx � 12

�ln x�2�e

1�

12

104.

� 0 � ln� 1

�2� �

12

ln 2

��4

0tan�

4� x� dx � lncos�

4� x���4

0

106. � e2x

e2x � 1 dx �

12

ln�e2x � 1� � C

107. Let

�12

arctan�e2x� � C

�12� 1

1 � �e2x�2�2e2x� dx

� 1e2x � e�2x dx � � e2x

1 � e4x dx

u � e2x, du � 2e2x dx.

109. Let

� x

�1 � x4 dx �

12� 1

�1 � �x2�2�2x� dx �

12

arcsin x2 � C

u � x2, du � 2x dx.

108. Let

�1

5�3 arctan

5x

�3� C

� 13 � 25x2 dx �

15� 1

��3 �2 � �5x�2�5� dx

u � 5x, du � 5 dx.

110. � 116 � x2 dx �

14

arctan x4

� C

111. Let

� x16 � x2 dx �

12� 1

16 � x2�2x� dx �12

ln�16 � x2� � C

u � 16 � x2, du � 2x dx.

113. Let

�arctan�x�2�4 � x2 dx �

12��arctan

x2��

24 � x2� dx �

14�arctan

x2�

2

� C

u � arctan�x2�, du �

24 � x2 dx.

112. � 4 � x

�4 � x2 dx � 4� 1

�4 � x2 dx �

12��4 � x2��1�2��2x� dx � 4 arcsin

x2

� �4 � x2 � C

114. Let

� arcsin x

�1 � x2 dx �

12

�arcsin x�2 � C

u � arcsin x, du �1

�1 � x2 dx.

510 Chapter 5 Integration

Problem Solving for Chapter 5

1. (a)

(b) by the Second Fundamental Theorem of Calculus.

(c) for

(Note: The exact value of x is e, the base of the natural logarithm function.)

(d) We first show that To see this, let and

Then Now,

� L�x1� � L�x2�.

� �x1

1

1u

du � �x2

1 1u

du

� �1

1�x1

1u

du � �x2

1 1u

du

�using u �t

x1� L�x1x2� � �x1x2

1 1t dt � �x2

1�x1

1u

du

�x1

1 1t dt � �1

1�x1

1ux1

�x1 du� � �1

1�x1

1u

du � �1

1�x1

1t dt.

du �1x1

dt.u �t

x1�x1

1 1t dt � �1

1�x1

1t dt.

�2.718

1 1t dt � 0.999896

x � 2.718L�x� � 1 � �x

1 1t dt

L��1� � 1

L��x� �1x

L�1� � �1

1

1

t dt � 0

116.

Since the area of region A is

the shaded area is �1

0arcsin x dx �

2� 1 � 0.571.

1����2

0sin y dy�,

10.25 0.750.5

π2

π4

y

x

y = arcsin x

117.

y� � 2 �1

2�x�sinh �x� � 2 �

sinh�x

2�x

y � 2x � cosh�x 118.

y� � x� 21 � 4x2� � tanh�1 2x �

2x1 � 4x2 � tanh�1 2x

y � x tanh�1 2x

119. Let

�12

ln�x2 � �x4 � 1 � � C

� x

�x4 � 1 dx �

12� 1

��x2�2 � 1�2x� dx

u � x2, du � 2x dx. 120. Let

�x2�sech x3�2 dx �13��sech x3�2�3x2� dx �

13

tanh x3 � C

u � x3, du � 3x2 dx.

115.

Since when you have Thus,

y � A sin�� km

t�.

sin�� km

t� �yA

C � 0.t � 0,y � 0

arcsin�yA� �� k

m t � C

� dy

�A2 � y2� �� k

m dt

Problem Solving for Chapter 5 511

3.

(a)

x

y

1

1 3

2

−2

−1

S�x� � �x

0sin��t2

2 � dt

(b)

The zeros of correspond to the relative extrema of S�x�.y � sin �x2

2

x

3

y

2

1

1 2 32 3 5 6 7 2 2

(c)

Relative maximum at and

Relative minimum at and

(d)

Points of inflection at x � 1, �3, �5, and �7

S� �x� � cos��x2

2 ���x� � 0 ⇒ �x2

2�

2� n� ⇒ x2 � 1 � 2n ⇒ x � �1 � 2n, n integer

x � �8 � 2.8284x � 2

x � �6 � 2.4495x � �2 � 1.4142

S��x� � sin �x2

2� 0 ⇒ �x2

2� n� ⇒ x2 � 2n ⇒ x � �2n, n integer

2. (a)

(b)

(c)

Since

�Note: sin 4 � �0.7568�

F��x� � sin x2, F��2� � sin 4 � limx→2

G�x�.

� limx→2

G�x�

� limx→2

1

x � 2 �x

2sin t2 dt

F��2� � limx→2

F�x� � F�2�

x � 2

limx→2

G�x� � �0.75

G�x� �1

x � 2 �x

2sin t2 dt

F�x� � �x

2sin t2 dt

x 0 1.0 1.5 1.9 2.0 2.1 2.5 3.0 4.0 5.0

0.0611 0 �0.2769�0.0576�0.0312�0.3743�0.0867�0.0265�0.4945�0.8048F�x�

x 1.9 1.95 1.99 2.01 2.05 2.1

�0.8671�0.8174�0.7697�0.7436�0.6873�0.6106G�x�

512 Chapter 5 Integration

(d)

is a point of inflection, whereas is not. ( is not continuous at )x � 6.f �x � 6x � 2

0 < x < 2

2 < x < 6

6 < x < 8F� �x� � f��x� �

�1,

1,

1,2

6. (a)

(b) v is increasing (positive acceleration) on and

(c) Average acceleration

(d) This integral is the total distance traveled in miles.

(e) One approximation is

(other answers possible)a�0.8� �v�0.9� � v�0.8�

0.9 � 0.8�

50 � 400.1

� 100 mi�hr2.

�1

0v�t� dt �

110

0 � 2�20� � 2�60� � 2�40� � 2�40� � 65� �38510

� 38.5 miles

�v�0.4� � v�0�

0.4 � 0�

60 � 00.4

� 150 mi�hr2

�0.7, 1.0�.�0, 0.4�

0.2 0.4 0.6 0.8 1.0

20

40

60

80

100

v

t

4. Let d be the distance traversed and a be the uniform acceleration.We can assume that and Then

when

The highest speed is The lowest speed is The mean speed is

The time necessary to traverse the distance d at the mean speed is

which is the same as the time calculated above.

t �d

�ad�2��2d

a

12

��2ad � 0� ��ad2

.v � 0.v � a�2da

� �2ad.

t ��2da

.s�t� � d

s�t� �12

at2.

v�t� � at

a�t� � a

s�0� � 0.v�0� � 0

5. (a)

(b)

x

y

1

2 4 5 6 7 8 9

2345

−2−3−4−5

−1

f(0, 0)

(6, 2)(8, 3)

(2, −2)

x 0 1 2 3 4 5 6 7 8

0 314

�2�72

�4�72

�2�12

F�x�

(c)

F is decreasing on and increasingon Therefore, the minimum is at andthe maximum is 3 at x � 8.

x � 4,�4�4, 8�.�0, 4�F��x� � f �x�.

0 ≤ x < 2

2 ≤ x < 6

6 ≤ x ≤ 8��x2�2�

�x2�2��1�4�x2

� 4x

� x

,

� 4,

� 5,

F�x� � �x

0f �t� dt �

0 ≤ x < 2

2 ≤ x < 6

6 ≤ x ≤ 8f �x� �

�x,

x � 4,

1x � 1,

2

Problem Solving for Chapter 5 513

9. Consider Thus,

�12

f �b�2 � f �a�2�.

�12

F�b� � F�a��

� �12

F�x� b

a

�b

a

f �x� f��x� dx � �b

a

12

F��x� dx

F�x� � f �x��2 ⇒ F��x� � 2f �x� f��x�.

11. Consider

The corresponding Riemann Sum using right endpoints is

Thus, limn→�

S�n� � limn→�

15 � 25 � . . . � n5

n6 �16

.

�1n615 � 25 � . . . � n5�. S�n� �

1n��

1n�

5

� �2n�

5

� . . . � �nn�

5

�1

0x5 dx �

x6

6 1

0�

16

.

10. Consider The corresponding

Riemann Sum using right endpoints is

Thus, limn→�

�1 � �2 � . . . � �nn3�2 �

23

.

�1

n3�2 �1 � �2 � . . . � �n�.

S�n� �1n��1

n��2

n� . . . ��n

n

�1

0

�x dx �23

x3�2 1

0�

23

.

7. (a)

Error:

(b)

(Note: Exact answer is )

(c) Let

�2b3

� 2d

p��1�3� � p� 1

�3� � �b

3� d� � �b

3� d�

�2b3

� 2d

�1

�1p�x� dx � �ax4

4�

bx3

3�

cx2

2� dx

1

�1

p�x� � ax3 � bx2 � cx � d.

��2 � 1.5708

�1

�1

11 � x2 dx �

11 � �1�3� �

11 � �1�3� �

32

�1.6829 � 1.6758� � 0.0071

�1

�1cos x dx � sin x

1

�1� 2 sin�1� � 1.6829

� 2 cos� 1�3� � 1.6758

�1

�1cos x dx � cos��

1�3� � cos� 1

�3� 8.

Thus,

Differentiating the other integral,

Thus, the two original integrals have equal derivatives,

Letting we see that C � 0.x � 0,

�x

0f �t��x � t� dt � �x

0��t

0f �v� dv� dt � C.

ddx�

x

0��x

0f �v� dv� dt � �x

0f �v� dv.

ddx�

x

0f �t��x � t� dt � x f �x� � �x

0f �t� dt � x f �x� � �x

0f �t� dt.

� x�x

0f �t� dt � �x

0t f �t� dt

�x

0f �t��x � t� dt � �x

0x f �t� dt � �x

0t f �t� dt

12. (a)

—CONTINUED—

� 227 � 9� � 36

� 2�9x �x3

3 3

0

−4 −2 −1 1

12345678

10

2 4 5

y

x

Area � �3

�3�9 � x2� dx � 2�3

0�9 � x2� dx

514 Chapter 5 Integration

14. (a)

(b)

Hence,

(c)

⇒ �n

i�1i2 �

n�n � 1��2n � 1�6

�n�n � 1��2n � 1�

2

�2n3 � 3n2 � n

2

�2n3 � 6n2 � 6n � 3n2 � 3n � 2n

2

⇒ �n

i�13i2 � n3 � 3n2 � 3n �

3n�n � 1�2

� n

�n � 1�3 � 1 � �n

i�1�3i2 � 3i � 1� � �

n

i�13i2 �

3�n��n � 1�2

� n

�n � 1�3 � �n

i�1�3i2 � 3i � 1� � 1.

� �23 � 13� � �33 � 23� � . . . � ��n � 1�3 � n3�� � �n � 1�3 � 1

�n

i�1�3i2 � 3i � 1� � �

n

i�1�i � 1�3 � i3�

3i2 � 3i � 1 � �i � 1�3 � i3

�1 � i�3 � 1 � 3i � 3i2 � i3 ⇒ �1 � i�3 � i3 � 3i2 � 3i � 1

15. Since

��b

a� f �x�� dx ≤ �b

a

f �x� dx ≤ �b

a� f �x�� dx ⇒ ��b

af �x� dx� ≤ �b

a� f �x�� dx.

��f �x�� ≤ f �x� ≤ �f �x��,

13. By Theorem 5.8,

Similarly,

Thus, On the interval and

Thus, �Note: �1

0

�1 � x4 dx � 1.0894�1 ≤ �1

0

�1 � x4 dx ≤ �2.

b � a � 1.0, 1�, 1 ≤ �1 � x4 ≤ �2m�b � a� ≤ �b

a

f �x� dx ≤ M�b � a�.

m ≤ f �x� ⇒ m�b � a� � �b

a

m dx ≤ �b

a

f �x� dx.

0 < f �x� ≤ M ⇒ �b

a

f �x� dx ≤ �b

a

M dx � M�b � a�.

12. —CONTINUED—

(b)

(c) Let the parabola be given by

Archimedes’ Formula: Area �23�

2ba ��b2� �

43

b3

a

Base �2ba

, height � b2

� 2�b3

a�

13

b3

a �43

b3

a � 2�b2�b

a� �a2

3 �ba�

3

− ba

ba

b2

y

x

� 2�b2x � a2 x3

3 b�a

0 Area � 2�b�a

0�b2 � a2x2� dx

y � b2 � a2x2, a, b > 0.

area �23

bh �23

�6��9� � 36Base � 6, height � 9,

Problem Solving for Chapter 5 515

17. Let

� 0.8109

� 2 ln 3 � 2 ln 2 � 2 ln�32�

� �2 ln u 3

2

� �3

2

2u

du

� �3

2

2�u � 1�u2 � u

du

Area � �4

1

1�x � x

dx � �3

2

2u � 2

�u � 1� � �u2 � 2u � 1� du

dx � �2u � 2� du.

u � 1 � �x, �x � u � 1, x � u2 � 2u � 1, 18. Let

�12

arctan�12�

� �12

arctan�u2�

1

0

� �1

0

duu2 � 4

Area � ���4

0

1sin2 x � 4 cos2 x

dx � ���4

0

sec2 xtan2 x � 4

dx

u � tan x, du � sec2 x dx.

16. (a)

Area B � �12��

6� ��

12� 0.2618

� ��22

��32

��3 � �2

2� 0.1589

Area A � ���4

��6sin y � dy � �cos y

�4

��6

y � f �x� � arcsin x, sin y � x

(c)

(d)

��

12�4�3 � 3� �

12

ln 2 � 0.6818

Area C � ��3

1arctan x dx � ��

3���3� �12

ln 2 � ��

4��1�

� �ln 12

� ln �22

� ln�2 �12

ln 2 A

CB

π4

π3

1 3

y

x

y = arctan x � �ln�cos y�

��3

��4

Area A � ���3

��4tan y dy

tan y � x

Area B � �3

1ln x dx � 3�ln 3� � A � 3 ln 3 � 2 � ln 27 � 2 � 1.2958

1 2 3

AB

ln 3

y

x

y = ln x

ey = x

Area A � �ln 3

0ey dy � ey

ln 3

0� 3 � 1 � 2

(b)

� 0.1346

� ���28

�112� �

�2 � �32

���2

8�

�3 � �22

��

12

��2�2

1�2arcsin x dx � Area�C� � ��

4���22 � � A � B

516 Chapter 5 Integration

19. (a) (i)

−2 2

−1

y

y1

4

y1 � 1 � x

y � ex (ii)

−2 2

−1

y

y2

4

y2 � 1 � x � �x2

2 � y � ex (iii)

−2 2

−1

y y3

4

y3 � 1 � x �x2

2�

x3

6

y � ex

(b) th term is in polynomial:

(c) Conjecture: ex � 1 � x �x2

2!�

x3

3!� . . .

y4 � 1 � x �x2

2!�

x3

3!�

x4

4!xn�n!n

20. (a)

Then,

Thus,

(b) b � 1 ⇒ �1

0

sin xsin�1 � x� � sin x

dx �12

A �b2

.

� �b

01 dx � b.

2A � �b

0

f �x�f �x� � f �b � x� dx � �b

0

f �b � x�f �b � x� � f �x� dx

� �b

0

f �b � x�f �b � x� � f �x� dx

� �b

0

f �b � u�f �b � u� � f �u� du

A � �0

b

f �b � u�

f �b � u� � f �u���du�

Let u � b � x, du � �dx.

Let A � �b

0

f �x�f �x� � f �b � x� dx.

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