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C H A P T E R 6Integration

Section 6.1 Antiderivatives and Indefinite Integration . . . . . . . . . 477

Section 6.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486

Section 6.3 Riemann Sums and Definite Integrals . . . . . . . . . . . 498

Section 6.4 The Fundamental Theorem of Calculus . . . . . . . . . . 505

Section 6.5 Integration by Substitution . . . . . . . . . . . . . . . . . 514

Section 6.6 Numerical Integration . . . . . . . . . . . . . . . . . . . 525

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541

C H A P T E R 6Integration

Section 6.1 Antiderivatives and Indefinite Integration

477

1.ddx�

3x3 � C� �

ddx

�3x�3 � C� � �9x�4 ��9x 4 2.

ddx�x 4 �

1x

� C� � 4x3 �1x2

3.ddx�

13

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

5.

Check:ddt

�t 3 � C� � 3t 2

y � t 3 � C

dydt

� 3t 2

7.

Check:ddx�

25

x5�2 � C � x3�2

y �25

x5�2 � C

dydx

� x3�2

Given Rewrite Integrate Simplify

9.34

x 4�3 � Cx 4�3

4�3� Cx1�3 dx3�x dx

11. �2

�x� C

x�1�2

�1�2� Cx�3�2 dx 1

x�x dx

13. �1

4x2 � C12�

x�2

�2� � C12x�3 dx 1

2x3 dx

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�

6.

Check:dd�

��� � C� � �

r � �� � C

drd�

� �

8.

Check:ddx�

�1x2 � C � 2x�3

y �2x�2

�2� C �

�1x2 � C

dydx

� 2x�3

10. �1x

� Cx�1

�1� Cx�2 dx 1

x2 dx

12.14

x 4 �32

x2 � Cx 4

4� 3�x2

2 � � C�x3 � 3x� dxx�x2 � 3� dx

14.�19x

� C19�

x�1

�1� � C19x�2 dx 1

�3x�2 dx

478 Chapter 6 Integration

16.

Check:ddx�5x �

x2

2� C � 5 � x

�5 � x� dx � 5x �x2

2� C15.

Check:ddx�

x2

2� 3x � C � x � 3

�x � 3�dx �x2

2� 3x � C

17.

Check:ddx

�x2 � x3 � C� � 2x � 3x2

�2x � 3x2�dx � x2 � x3 � 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

19.

Check:ddx�

14

x 4 � 2x � C� � x3 � 2

�x3 � 2� dx �14

x 4 � 2x � C 20.

Check:ddx�

x 4

4� 2x2 � 2x � C � x3 � 4x � 2

�x 3 � 4x � 2� dx �x 4

4� 2x2 � 2x � C

21.

Check:ddx�

25

x5�2 � x2 � x � C� � x3�2 � 2x � 1

�x3�2 � 2x � 1� dx �25

x5�2 � x2 � x � C

22.

Check: � �x �1

2�x

ddx�

23

x3�2 � x1�2 � C� � x1�2 �12

x�1�2

�23

x3�2 � x1�2 � C�x3�2

3�2�

12�

x1�2

1�2� � C��x �1

2�x� dx � �x1�2 �12

x�1�2� dx

23.

Check:ddx�

35

x5�3 � C� � x2�3 � 3�x2

3�x2 dx � x2�3 dx �x5�3

5�3� C �

35

x5�3 � C 24.

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

26.

Check:ddx��

13x3 � C� �

1x4

1x4 dx � x�4 dx �

x�3

�3� C � �

13x3 � C25.

Check:ddx ��

12x2 � C� �

1x3

1x3 dx � x�3 dx �

x�2

�2� C � �

12x2 � C

27.

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 � Cx2 � x � 1

�x dx � �x3�2 � x1�2 � x�1�2� dx

28.

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

Section 6.1 Antiderivatives and Indefinite Integration 479

29.

Check:

� �x � 1��3x � 2�

ddx�x3 �

12

x2 � 2x � C� � 3x2 � x � 2

� x3 �12

x2 � 2x � C

�x � 1��3x � 2� dx � �3x2 � x � 2� dx

31.

Check:ddy�

27

y7�2 � C� � y5�2 � y2�y

y2�y dy � y5�2 dy �27

y7�2 � C

33.

Check:ddx

�x � C� � 1

dx � 1 dx � x � C

35. y

x2 33

3

44

4

55

5

6

6

−6

−−−

−5

C = 0

C = 3

C = −2

30.

Check:

� �2t2 � 1�2

ddt �

45

t 5 �43

t 3 � t � C� � 4t 4 � 4t 2 � 1

�45

t 5 �43

t 3 � t � C

�2t 2 � 1�2 dt � �4t 4 � 4t 2 � 1� dt

32.

Check:ddt�

13

t3 �34

t 4 � C� � t2 � 3t3 � �1 � 3t�t2

�1 � 3t�t2 dt � �t2 � 3t3� dt �13

t3 �34

t4 � C

34.

Check:ddt

�3t � C� � 3

3 dt � 3t � C

36.

x4 6 8−2

−2

−4

2

4

6

C = 0

C = 3

C = 2−

y

f �x� � �x

38.

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

37.

Answers will vary.

5

4

3

32123

y

x

2x)x)

22x) )xf

ff ′

f �x� � 2x � C

f��x� � 2

39.

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 40.

x−2−4

−2

−4

4

f x( ) = −

f x( ) 1= − +

1

1

x

x

f ′

y

f �x� � �1x

� C

f��x� �1x2

480 Chapter 6 Integration

41.

y � x2 � x � 1

1 � �1�2 � �1� � C ⇒ C � 1

y � �2x � 1� dx � x2 � x � C

dydx

� 2x � 1, �1, 1� 42.

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�

43.

y � x3 � x � 2

2 � 03 � 0 � C ⇒ C � 2

y � �3x2 � 1� dx � x3 � x � C

dydx

� 3x2 � 1 44.

y �1x

� 2, x > 0

3 �11

� C ⇒ C � 2

y � �x�2 dx �1x

� C

dydx

� �1x2 � �x�2, �1,3�

45. (a) Answers will vary.

(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�

x

y

−3 5

−3

5

46. (a)

(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�

x− 4

−5

5

4

y

47.

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 48.

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

49.

h�t� � 2t 4 � 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 50.

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

Section 6.1 Antiderivatives and Indefinite Integration 481

51.

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 52.

f �x� �1

12x4 � 6x � 3

f �0� � 0 � 0 � C2 � 3 ⇒ C2 � 3

f �x� � �13

x3 � 6� dx �1

12x 4 � 6x � C2

f��x� �13

x3 � 6

f��0� � 0 � C1 � 6 ⇒ C1 � 6

f��x� � x 2 dx �13

x3 � C1

f �0� � 3

f��0� � 6

f ��x� � x2

53. (a)

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

h�t� � 0.75t 2 � 5t � 12

h�0� � 0 � 0 � C � 12 ⇒ C � 12

h�t� � �1.5t � 5� dt � 0.75t2 � 5t � C 54.

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

kt3�2 � C

dPdt

� k�t, 0 ≤ t ≤ 10

55. 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 has a relative minimum at and at both these values of x, and is negative to

the left and positive to the right of each.

(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 �f � � 0x � 6.25.x � �0.75

�4, 5�.f �5� < f �4�

�0, 2�.�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 �

56. Since is negative on is decreasing on Since is positive on is increasing on has a relative minimum at Since is positive on f is increasing on ���, ��.

���, ��,f��0, 0�.f��0, ��.f��0, ��,f �

x1 2 3−2−3

1

2

3

−2

−3f

f ′

f ′′

y���, 0�.f����, 0�,f �

482 Chapter 6 Integration

57.

The ball reaches its maximim height when

s�158 � � �16�15

8 �2

� 60�158 � � 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 58.

f �t� � �16t2 � v0t � s0

f �0� � 0 � 0 � C2 � s0 ⇒ C2 � s0

f �t� � s�t� � ��32t � v0� dt � �16t 2 � 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

59. From Exercise 58, we have:

when time to reach

maximum 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� � �16t 2 � v0t

60.

(a)

Choosing the positive value,

(b)

� �16�17 � �65.970 ft�sec

v�1 � �172 � � �32�1 � �17

2 � � 16

v�t� � s��t� � �32t � 16

t �1 � �17

2� 2.562 seconds.

t �1 ± �17

2

�16�t 2 � t � 4� � 0

s�t� � �16t 2 � 16t � 64 � 0

s0 � 64 ft

v0 � 16 ft�sec

61.

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 62. From Exercise 61, (Using thecanyon floor as position 0.)

t2 �18004.9

⇒ t � �367.35 � 19.2 sec

4.9t2 � 1800

f �t� � 0 � �4.9t2 � 1800

f �t� � �4.9t2 � 1800.

Section 6.1 Antiderivatives and Indefinite Integration 483

63. From Exercise 61,

(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� � �4.9t 2 � 10t � 2. 64. From Exercise 61, 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 � �4.9t 2 � v0t � 2,

f �t� � �4.9t 2 � v0t � 2.

65.

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 � �0.8t 2 � s0

v0 � 0.

v�t� � �1.6 dt � �1.6t � v0 � �1.6t,

a � �1.6 66.

When

v2 � v02 � 2GM�1

y�

1R�

v2 �2GM

y� v0

2 �2GM

R

12

v 2 �GM

y�

12

v02 �

GMR

C �12

v02 �

GMR

12

v02 �

GMR

� C

y � R, v � v0.

12

v 2 �GM

y� C

v dv � �GM 1y2 dy

67.

(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�t 2 � 4t � 3� � 3�t � 1��t � 3�

v�t� � x��t� � 3t 2 � 12t � 9

0 ≤ t ≤ 5x�t� � t 3 � 6t 2 � 9t � 2 68.

(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� � t 3 � 7t 2 � 15t � 9

0 ≤ t ≤ 5x�t� � �t � 1��t � 3�2

484 Chapter 6 Integration

69.

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 70. (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

71.

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

t 2 � 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 mph to30 mph, 1.333 seconds to reduce the speed from 30 mphto 15 mph, and 1.333 seconds to reduce the speed from 15 mph to 0 mph. Each time, less distance is needed toreach the next speed reduction.

0 20 40 60 80 100 120 140

73.33 117.33 132x

45 mph = 66 ft/sec 30 mph = 44 ft/sec 15 mph = 22 ft/sec 0 mph

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

72. Truck:

Automobile:

At the point where the automobile overtakes the truck:

when sec.t � 10 0 � 3t�t � 10�

0 � 3t 2 � 30t

30t � 3t 2

s�t� � 3t 2 �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

Section 6.1 Antiderivatives and Indefinite Integration 485

73. (a)

(b)

s�6� � 197.9 feet

�Note: Assume s�0� � 0 is initial position�

s�t� � v�t�dt �0.6139t 4

4�

5.525t 3

3�

0.0492t 2

2� 65.9881t

v � 0.6139t 3 � 5.525t 2 � 0.0492t � 65.9881

74.

(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 � �0.1208t 3

3�

6.7991t 2

2� 0.0707t

s1�t� � v1�t� dt �0.1068

3 t 3 �

0.04162

t 2 � 0.3679t

v2�t� � �0.1208t 2 � 6.7991t � 0.0707

v1�t� � 0.1068t 2 � 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�

75. True 76. True 77. True 78. True

79. False. For example, because x3

3� C �x2

2� C1��x2

2� C2�.x x dx x dx x dx

80. False. has an infinite number of antiderivatives, each differing by a constant.f

81.

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 82.

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

486 Chapter 6 Integration

Section 6.2 Area

2. �6

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

4. �5

j�3 1j

�13

�14

�15

�4760

6. �4

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

1. �5

i�1�2i � 1� � 2�

5

i�1i � �

5

i�11 � 2�1 � 2 � 3 � 4 � 5� � 5 � 35

3. �4

k�0

1k2 � 1

� 1 �12

�15

�1

10�

117

�15885

5. �4

k�1c � c � c � c � c � 4c

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

4�2

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�01 � � i

n�2

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�

17.

� �19�20��39�6 � 2470

�20

i�1�i � 1�2 � �

19

i�1i 2 18.

� �10�11��21�6 � 10 � 375

�10

i�1�i 2 � 1� � �

10

i�1i 2 � �

10

i�11

19.

� 12,040

� 14,400 � 2,480 � 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�i 2 � 1� � �

10

i�1i 3 � �

10

i�1i

21. sum seq (TI-82)

��20��21��41�

6� 60 � 2930

�20

i�1�i 2 � 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�i 3 � 2i� �

�15� 2�15 � 1� 2

4� 2

15�15 � 1�2

3 � 2x, x, 1, 15, 1� � 14,160�x >

Section 6.2 Area 487

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

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

26.

s � �2 � 1 �23

�12

�13 �

92

� 4.5

S � �5 � 2 � 1 �23

�12 �

556

28.

s�8� � �0 � 2� 14

� �14

� 2�14

� �12

� 2�14

� . . . � �74

� 2�14

� 5.685

�14�16 �

12

�22

�32

� 1 �52

�62

�72

� 2� � 6.038

� �54

� 2�14

� �32

� 2�14

� �74

� 2�14

� �2 � 2�14

S�8� � �14

� 2�14

� �12

� 2�14

� �34

� 2�14

� �1 � 2�14

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

5�2

�15� �1 � �3

5�2

�15� �1 � �4

5�2

�15� � 0 � 0.659

�15�1 �

245

�21

5�

165

�95 � 0.859

S�5� � 1�15� �1 � �1

5�2

�15� �1 � �2

5�2

�15� �1 � �3

5�2

�15� �1 � �4

5�2

�15�

31.

�43

�2� �83

limn→�

�� 43n3��2n3 � 3n2 � n� �

43

limn→�

�2n3 � 3n2 � nn3 32. lim

n→� �8

3�

4n

�4

3n2� �83

� 0 � 0 �83

33.

�814

�1� �814

limn→�

��81n4�n2�n � 1�2

4 �814

limn→�

�n4 � 2n3 � n2

n4 34.

�646

�2� �643

limn→�

��64n3�n�n � 1��2n � 1�

6 �646

limn→�

�2n3 � 3n2 � nn3

488 Chapter 6 Integration

35. limn→�

��18n2�n�n � 1�

2 �182

limn→�

�n2 � nn2 �

182

�1� � 9

37.

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. limn→�

�� 1n2�n�n � 1�

2 �12

limn→�

�n2 � nn2 �

12

�1� �12

38.

S�10,000� � 2.0005

S�1000� � 2.005

S�100� � 2.05

S�10� �2510

� 2.5

�n

j�1

4 j � 3n2 �

1n2 �

n

j�1�4j � 3� �

1n2�4n�n � 1�

2� 3n �

2n � 5n

� S�n�

39.

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

40.

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

4i 2�i � 1�n4 �

4n4 �

n

i�1�i3 � i2� �

4n4�n2�n � 1�2

4�

n�n � 1��2n � 1�6

41. � 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

Section 6.2 Area 489

42. � 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

43.

� 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

44.

� 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

45. � 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

46.

� 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

47. (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

490 Chapter 6 Integration

47. —CONTINUED—

(e) (f)

� limn→�

2�n � 1�

n� 2

� limn→�

� 4n2�n�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�

48. (a)

(c) Since is increasing, on

(d) on

(e)

(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

n�n�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�

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

Section 6.2 Area 491

49. on

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

50. on

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

51. on

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

n�2

� 2�1n�

x2 3

3

1

1

y�Note: �x �1n��0, 1�.y � x2 � 2

52. on

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

l�1l 2 �

3n

�n

l�11

S�n� � �n

l�1 f �3i

n ��3n� � �

n

l�1��3i

n �2

� 1�3n�

y

x1 2

2

4

6

8

10

12

3

�0, 3�. �Note: �x �3 � 0

n�

3n�y � x2 � 1

492 Chapter 6 Integration

53. on

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 �

4i 2

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 �2n�y � 16 � x2

54. on 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

n�2

�1n�

x

2

3

−2 2

−1

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

55. on

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 �

27i 3

n3 �27i 2

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�

3n�y � 64 � x3

56. on

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

Section 6.2 Area 493

57. on

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

�4i 2

n2 � � ��1 �6in

�12i 2

n2 �8i 3

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

58. on

Area � limn→�

s�n� � 2 �52

�43

�14

�7

12

� 2 �52

�5

2n�

43

�2n

�2

3n3 �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

59.

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 ��2n�S�n� � �

n

i�1 f �mi � �y

64x

4

2

2

2

y�Note: �y �2 � 0

n�

2n�0 ≤ y ≤ 2f �y� � 3y,

60.

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�

494 Chapter 6 Integration

61.

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�

3n�0 ≤ y ≤ 3f �y� � y2,

62.

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

�i 2

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�

63.

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�1�4�1 �

4in

�4i2

n2 � �1 �6in

�12i 2

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�

Section 6.2 Area 495

64.

Area � limn →�

S�n� � 2 �14

� 1 �32

�194

� 2 ��n � 1�2

4n2 �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

2

1

2

3

4

5

4 6 8 10

y

x

S�n� � �n

i�1h�1 �

in��

1n�

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

65.

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, 66.

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,

67.

Let

� 0.6730

�14

�1�8 � 3�8 � 5�8 � 7�8 �

Area � � n

i�1f �ci� �x � �

4

i�1

ci � 1�14�

c4 �158

c3 �138

,c2 �118

,c1 �98

,�x �14

,

ci �xi � xi�1

2.

n � 4 �1, 2�,f �x� � x � 1, 68.

�12� 1

116 � 1

�1

916 � 1

�1

2516 � 1

�1

4916 � 1 � 1.1088

Area � �n

l�1

f �ci ��x � �4

l�1

1

ci2 � 1 �

12�

c4 �74

c3 �54

,c2 �34

,c1 �14

,�x �12

,

Let ci �xi � xi�1

2

n � 4�0, 2�,f �x� �1

x2 � 1,

69. on

�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

496 Chapter 6 Integration

70. �0, 2�.f �x� � x3�2 � 2,

n 4 8 12 16 20

6.2435 6.2577 6.2605 6.2615 6.2619Approximate area

72. on �2, 6�.f �x� �8

x2 � 1

n 4 8 12 16 20

2.3397 2.3755 2.3824 2.3848 2.3860Approximate area

71. f �x� �5x

x2 � 1, �1, 3�

n 4 8 12 16 20

Approximate Area 4.0272 4.0246 4.0241 4.0239 4.0238

73. 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 area lies between these two sums.

x

y

a b

74. See the Definition of Area. Page 404.

75. (a) In each case, The lower sum uses left endpoints, The upper sum uses right endpoints,The Midpoint Rule uses midpoints,

(b)

(c) 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�

�i �12��4�n�.

�i��4�n�.�i � 1��4�n�.�x � 4�n.

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 6.2 Area 497

76.

(Note: exact answer is 12.)

0 ≤ x ≤ 8f �x� � 3x,

n 10 20 50 100 200

10.998 11.519 11.816 11.910 11.956

12.598 12.319 12.136 12.070 12.036

12.040 12.016 12.005 12.002 12.001M�n�

S�n�

s�n�

77.

b. square unitsA � 6

x1 2 3 4

1

2

3

4

y 78.

a. square unitsA � 3

1 2

1

2

3

4

3 4x

y

79. True. (Theorem 6.2 (2)) 80. True. (Theorem 6.3)

81. (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. Answers will vary. 83. Suppose there are n rows and columns in the figure.The stars on the left total as do thestars 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

84. (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�1i 3 �

n2�n � 1�2

4

Section 6.3 Riemann Sums and Definite Integrals

498 Chapter 6 Integration

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

2.

� limn→�

1n4�3n4

4�

n3

2�

n2

4 � � limn→�

�34

�1

2n�

14n2� �

34

� limn→�

1n4�3n4 � 6n3 � 3n2

4�

2n3 � 3n2 � n2

�n2 � n

2 �

� limn→�

1n4�3�n2�n � 1�2

4 � 3�n�n � 1��2n � 1�6 �

n�n � 1�2 �

� limn→�

1n4 �

n

i�1�3i 3 � 3i 2 � i�

limn→�

�n

i�1 f �ci� �xi � 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� � 3�x,

3. on

10

4 6 dx � lim

n→� 36 � 36

� �n

i�16�6

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

Section 6.3 Riemann Sums and Definite Integrals 499

5. on

1

�1x3 dx � lim

n→� 2n

� 0

� �2 � 6�1 �1n � 4�2 �

3n

�1n2 � 4�1 �

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

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�1�1 �

4in

�4i2

n2

� �n

i�13�1 �

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 �

16�2 �

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

500 Chapter 6 Integration

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�3�1 �

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

10.

on the interval �0, 4 .

lim���→0

�n

i�16ci�4 � ci�2 �xi � 4

06x�4 � x�2 dx

12.

on the interval �1, 3 .

lim���→0

�n

i�1� 3

ci2 �xi � 3

1

3x2 dx

18. 1

�1

1x2 � 1

dx

9.

on the interval ��1, 5 .

lim���→0

�n

i�1�3ci � 10� �xi � 5

�1�3x � 10� dx

17. 2

�2�4 � x2� dx 19. 2

0

�x � 1 dx

11.

on the interval �0, 3 .

lim���→0

�n

i�1

�ci2 � 4 �xi � 3

0

�x2 � 4 dx

13. 5

03 dx 14. 2

0�4 � 2x� dx 15. 4

�4�4 � �x�� dx 16. 2

0x2 dx

20. 1

�1 �x2 � 1�3 dx

22. 2

0�y � 2�2 dy21. 2

0y3 dy

23. Rectangle

x5

5

3

2

42 31

1

Rectangle

y

A � 3

04 dx � 12

A � bh � 3�4�

25. Triangle

x

Triangle

4

2

42

y

A � 4

0x dx � 8

A �12

bh �12

�4��4�

24. Rectangle

x

1

2

3

5

−a a

Rectangle

y

A � a

�a

4 dx � 8a

A � bh � 2�4��a�

Section 6.3 Riemann Sums and Definite Integrals 501

26. Triangle

x1 2 3 4

−1

1

2

3

Triangle

y

A � 4

0

x2

dx � 4

A �12

bh �12

�4��2�

27. Trapezoid

x

9

6

321

3

Trapezoid

y

A � 2

0�2x � 5� dx � 14

A �b1 � b2

2h � �5 � 9

2 2

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

29. Triangle

A � 1

�1�1 � �x�� dx � 1

A �12

bh �12

�2��1� 1 Triangle

11x

y 30. Triangle

A � a

�a

�a � �x�� dx � a2

A �12

bh �12

�2a�a

x

Triangle

−a a

a

y

31. Semicircle

A � 3

�3

�9 � x2 dx �9�

2

A �12

�r 2 �12

��3�2

x

4 Semicircle

4224

y 32. Semicircle

A � r

�r

�r 2 � x 2 dx �12

�r 2

A �12

�r2

x−r

−r

r

r Semicircle

y

In Exercises 33–40, 4

2x3 dx � 60, 4

2x dx � 6, 4

2dx � 2

37. 4

2�x � 8� dx � 4

2x dx � 84

2dx � 6 � 8�2� � �10

33. 2

4x dx � �4

2x dx � �6

35. 4

24x dx � 44

2x dx � 4�6� � 24

39.

�12

�60� � 3�6� � 2�2� � 16

4

2�1

2x3 � 3x � 2 dx �

12

4

2x3 dx � 34

2x dx � 24

2dx

34. 2

2x3 dx � 0

36. 4

215 dx � 154

2dx � 15�2� � 30

38. 4

2�x3 � 4� dx � 4

2x3 dx � 44

2dx � 60 � 4�2� � 68

40.

� 6�2� � 2�6� � 60 � �36

4

2�6 � 2x � x3� dx � 64

2dx � 24

2x dx � 4

2x3 dx

502 Chapter 6 Integration

41. (a)

(b)

(c)

(d) 5

03f �x� dx � 35

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 � 36

2f �x� dx � 3�10� � 30

6

22g�x� dx � 26

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

42. (a)

(b)

(c)

(d) 6

3�5f �x� dx � �56

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 � 31

0f �x� dx � 3�5� � 15

1

�13f �x� dx � 31

�1f �x� dx � 3�0� � 0

1

0f �x� dx � 0

�1 f �x� dx � 5 � ��5� � 10

0

�1f �x� dx � 1

�1f �x� dx � 1

0f �x� dx � 0 � 5 � �5

45. (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 � ��

46. (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

47. (a)

(c) � f even�5

�5f �x� dx � 25

0f �x� dx � 2�4� � 8

5

0� f �x� � 2 dx � 5

0f �x� dx � 5

02 dx � 4 � 10 � 14 (b)

(d) � f odd�5

�5f �x� dx � 0

�Let u � x � 2.�3

�2f �x � 2� dx � 5

0f �x� dx � 4

Section 6.3 Riemann Sums and Definite Integrals 503

48.

8

0 f �x� dx � 4�4� � 4�4� �

12

�4��4� � 40

4 8

4

8(8, 8)

x

y

f �x� � �4,x,

x < 4 x ≥ 4

50. The right endpoint approximation will be less than theactual area: <

52. is integrable on but is not contin-uous on There is discontinuity at To seethat

is integrable, sketch a graph of the region bounded byand the x-axis for You see that

the integral equals 0.

x1 2−2

−2

1

2

y

�1 ≤ x ≤ 1.f �x� � �x��x

1

�1

�x�x

dx

x � 0.��1, 1 .��1, 1 ,f �x� � �x��x

54.

d. 3

1 x3 � 1

x2 dx � 412

x

y

1

1

2

3

2 3

49. The left endpoint approximation will be greater than theactual area: >

51.

is not integrable on the interval and f has adiscontinuity at x � 4.

�3, 5

f �x� �1

x � 4

53.

a. square unitsA � 5

x1 2 3 4

1

2

3

4

y

55. 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

56. 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

504 Chapter 6 Integration

57. True 58. False

1

0x�x dx � �1

0x dx�1

0

�x dx59. True

60. True 61. False

2

0��x� dx � �2

62. False

4

�2x dx � 6

63.

� �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,

64.

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

65.

The limit

does not exist. This does not contradict Theorem 6.4because f is not continuous on

1 2

1

2

x

f(x) = 1x

y

�0, 1 .

lim���→0 �

n

i�1f �ci ��xi

f �x� � �0,1x

,

x � 0 0 < x ≤ 1

66. 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 67. To find use a geometric approach.

Thus,

2

0�x� dx � 1�2 � 1� � 1.

x1 2 3−1

−1

1

2

3

y

�20 �x� dx,

Section 6.4 The Fundamental Theorem of Calculus

Section 6.4 The Fundamental Theorem of Calculus 505

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

7. �0

�1�x � 2� dx � �x2

2� 2x�

0

�1� 0 � �1

2� 2� � �

52

9. �1

�1�t 2 � 2� dt � �t 3

3� 2t�

1

�1� �1

3� 2� � ��

13

� 2� � �103

2.

�5

1 3�x � 3 dx � 0

1

−2

2

5

f �x� � 3�x � 3

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

8. �5

2��3v � 4� dv � ��

32

v2 � 4v�5

2� ��

752

� 20� � ��6 � 8� � �392

10. � 38 �3

1

�3x2 � 5x � 4� dx � �x3 �5x2

2� 4x�

3

1� �27 �

45

2� 12� � �1 �

5

2� 4�

11. �1

0�2t � 1�2 dt � �1

0�4t 2 � 4t � 1� dt � �4

3t 3 � 2t 2 � t�

1

0�

43

� 2 � 1 �13

12. �1

�1�t 3 � 9t� dt � �1

4t 4 �

92

t 2�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

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�u12 � 2u�12� du � �2

3u32 � 4u12�

4

1� �2

3��4�3

� 4�4� � �23

� 4� �23

16. �3

�3v13 dv � �3

4v43�

3

�3�

34

�3�3 �4 � �3��3 �4� � 0

506 Chapter 6 Integration

17. �1

�1�3�t � 2� dt � �3

4t 43 � 2t�

1

�1� �3

4� 2� � �3

4� 2� � �4

19. �1

0

x � �x3

dx �13�

1

0�x � x12� dx �

13�

x2

2�

23

x32�1

0�

13�

12

�23� � �

118

21. �0

�1�t13 � t23� dt � �3

4t 43 �

35

t 53�0

�1� 0 � �3

4�

35� � �

2720

18. �8

1�2

x dx � �2�8

1x�12 dx � ��2�2�x12�

8

1� �2�2x�

8

1� 8 � 2�2

20. �2

0�2 � t��t dt � �2

0�2t12 � t32� dt � �4

3t32 �

25

t52�2

0� �t�t

15�20 � 6t��

2

0�

2�215

�20 � 12� �16�2

15

22.

�12�

35

x53 �38

x83��1

�8� �x53

80�24 � 15x��

�1

�8� �

180

�39� �3280

�144� �456980

��1

�8

x � x2

2 3�x dx �

12�

�1

�8�x23 � x53� dx

23. split up the integral at the zero

� �3x � x2�32

0� �x2 � 3x�

3

32� �9

2�

94� � 0 � �9 � 9� � �9

4�

92� � 2�9

2�

94� �

92

x �32���3

0�2x � 3� dx � �32

0�3 � 2x� dx � �3

32�2x � 3� dx

24.

� 4 � 16 � 18 �92

�132

� �92

�12� � ��24 � 8� � �18 �

92��

� �x2

2 �3

1� �6x �

x2

2 �4

3

� �3

1x dx � �4

3�6 � x� dx

�4

1�3 � �x � 3�� dx � �3

13 � �x � 3�� dx � �4

33 � �x � 3�� dx

25. �233

� �8 �83� � �9 � 12� � �8

3� 8�� �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

Section 6.4 The Fundamental Theorem of Calculus 507

27. A � �1

0�x � x2� dx � �x2

2�

x3

3 �1

0�

16

28. A � �3

�1 ��x2 � 2x � 3� dx � ��

x3

3� x2 � 3x�

3

�1� ��9 � 9 � 9� � �1

3� 1 � 3� �

323

29. A � �1

�1 �1 � x 4� dx � �x �

15

x5�1

�1�

85

30. A � �2

1

1x2 dx � �

1x�

2

1� �

12

� ��1� �12

31.

� 6

� 213 34

4 � 413

� 213�34

x 43�4

0

A � �4

0 �2x�13 dx � 213�4

0 x13 dx 32.

�12�3

5

� �x�x5

�10 � 2x��3

0

� �2x32 �25

x52�3

0

A � �3

0�3 � x��x dx � �3

0�3x12 � x32� dx

33. Since on

A � �2

0�3x2 � 1� dx � �x3 � x�

2

0� 8 � 2 � 10.

0, 2�,y ≥ 0

35. Since on

A � �2

0�x3 � x� dx � �x 4

4�

x2

2 �2

0� 4 � 2 � 6.

0, 2�,y ≥ 0

37.

c � 213 1.2599

c3 � 2

f �c��2 � 0� � 4

�2

0 x3 dx �

x 4

4 �

2

0� 4

34. Since on

� 8 �34

�16� � 20

Area � �8

0�1 � x13� dx � �x �

34

x 43�8

0

0. 8�,y ≥ 0

36. Since on

A � �3

0�3x � x2� dx � �3

2x2 �

x3

3 �3

0�

92

.

0, 3�,y ≥ 0

38.

c � 3�92

1.6510

c3 �92

9c3 � 2

f �c��3 � 1� � 4

�3

1

9x3 dx � ��

92x2�

3

1� �

12

�92

� 4

39.

c � 1, 3

�c � 1��c � 3� � 0

c2 � 4c � 3 � 0

�c2 � 4c � 3

f �c��3 � 0� � 9

�3

0��x2 � 4x� dx � ��

x3

3� 2x2�

3

0� �9 � 18 � 9

508 Chapter 6 Integration

40.

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�

4x32

3 �2

0� 2 �

8�23

41.

Average value

when or

5

−1

−3 3

2 33 3

8,−( ( 2 33 3

8,( (

x � ±2�3

3 ±1.155.x2 � 4 �

83

4 � x2 �83

�83

�83

14��8 �

83� � ��8 �

83��

12 � ��2��

2

�2�4 � x2� dx �

14�4x �

13

x3�2

�2

43.

�x 0.179, 2.488�x �43

±2�3

3

�x �6 ± �36 � 24

6� 1 ±

�33

3x � 6x12 � 2 � 0

x � 2x12 � �23

�14�8 �

43

�8�� � �23

14 � 0�

4

0�x � 2x12� dx �

14�

x2

2�

43

x 32�4

042.

Average value:

when

or x � �3 1.732

12�x2 � 1� � 16x24�x2 � 1�x2 �

163

�163

� 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

44.

Take x � 3 � �3 1.268.

x � 3 ± �3

x � 3 � ±�3

�x � 3�2 � 3

1

�x � 3�2 �13

12 � 0

�2

0 �x � 3��2 dx �

12�

�1x � 3�

2

0�

12

�1 �13� �

13

47. 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

45. The distance traveled is The area under the curve from is approximately

squares �30� 540 ft.��180 ≤ t ≤ 8

�80 v�t� dt.

46. The distance traveled is The area under the curve from is approximately

squares �5� � 145 ft.��290 ≤ t ≤ 5

�50 v�t� dt.

Section 6.4 The Fundamental Theorem of Calculus 509

48. (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) Average value

(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

��7

1f �x� dx

7 � 1�

86

�43

51. �6

0� f �x�� dx � ��2

0f �x� dx ��6

2f �x� dx � 1.5 � 5.0 � 6.5

53.

� 12 � 3.5 � 15.5

�6

02 � f �x�� dx � �6

02 dx � �6

0f �x� dx

55.1

5 � 0�5

0�0.1729t � 0.1522t2 � 0.0374t3� dt

15�0.08645t2 � 0.05073t3 � 0.00935t4�

5

0 0.5318 liter

52. �2

0�2 f �x� dx � �2�2

0f �x� dx � �2��1.5� � 3.0

54. Average value �16�

6

0f �x� dx �

16

�3.5� � 0.5833

49. �2

0f �x� dx � ��area of region A� � �1.5 50.

� 3.5 � ��1.5� � 5.0

�6

2f �x� dx � �area of region B� � �6

0f �x� dx � �2

0f �x� dx

56.1

R � 0�R

0k�R2

� r2� dr �kR�R2r �

r3

3�R

0�

2kR2

3

57. (a)

(b)

(c) �60

0v�t� dt � ��8.61 � 10�4t 4

4�

0.0782t3

3�

0.208t2

2� 0.0952t�

60

0 2472 meters

70

−10

−10

90

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

510 Chapter 6 Integration

58. (a) histogram

t2

21

4

43

6

8

10

12

14

16

18

65 8 97

N

(b) customers

(c) Using a graphing utility, you obtain

(d)

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

Hence, per minute.3017240 12.6

��7

3N�t� dt��60� �50.28��60� 3017.

−2

−2 10

16

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

� 49806 � 7 � 9 � 12 � 15 � 14 � 11 � 7 � 2�60 � �83�60

59.

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

60.

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

�x 4

4� x2 � 2x � 4� �x4

4� x2 � 2x� � �4 � 4 � 4� F�x� � �x

2�t3 � 2t � 2� dt � �t 4

4� t2 � 2t�

x

2

61.

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

162.

F�8� �1936

�32�2

3

F�5� �383

�10�5

3

F�2� �22

2�

23

232 �16

�136

�43�2

�x2

2�

23

x32 �16

� �x2

2�

23

x32� � �12

�23�

F�x� � �x

1� y � �y � dy � �y2

2�

23

y32�x

1

(e)

The estimated number of customers is�85.162��60� 5110.

�9

0N�t� dt 85.162

Section 6.4 The Fundamental Theorem of Calculus 511

63.

(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 64.

(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

65. (a)

(b)ddx�

12

x2 � 2x� � x � 2

�x

0�t � 2� dt � �t2

2� 2t�

x

0�

12

x2 � 2x

67. (a)

(b)ddx�

34

x43 � 12� � x13 � 3�x

�x

8

3�t dt � �34

t43�x

8�

34

�x43 � 16� �34

x43 � 12

66. (a)

(b)ddx�

14

x4 �12

x2� � x3 � x � x�x2 � 1�

�14

x4 �12

x2 �x2

4�x2 � 2�

�x

0t�t 2 � 1� dt � �x

0�t3 � t� dt � �1

4t 4 �

12

t 2�x

0

68. (a)

(b)ddx�

23

x32 �163 � � x12 � �x

�x

4

�t dt � �23

t32�x

4�

23

x32 �163

�23

�x32 � 8�

70. (a)

(b) F��x� �25�

52�x32 � x32

F�x� � �x

0t32 dt �

25

t52�x

0�

25

x5269. (a)

(b) F��x� �ddx�1 �

1x� �

1x2

F�x� � �x

1 1t2 dt � �

1t�

x

1� 1 �

1x

71.

F��x� � x2 � 2x

F�x� � �x

�2�t 2 � 2t� dt

73.

F��x� � �x4 � 1

F�x� � �x

�1

�t4 � 1 dt

72.

F��x� � 4�x

F�x� � �x

1

4�t dt

74.

F��x� �x2

x2 � 1

F�x� � �x

1

t2

t2 � 1 dt

512 Chapter 6 Integration

75.

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

� �2t 2 � t�x�2

xF�x� � �x�2

x

�4t � 1� dt 76.

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

t 3 dt � �t 4

4�x

�x� 0

77.

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

F�x� � �x2

2t �3 dt � � t �2

�2�x2

2� ��

12t 2�

x2

2�

�12x4 �

18

⇒ F��x� � 2x�5

78.

� 3�1 � 27x3

F��x� � �1 � �3x�3 �3�

F�x� � �3x

0

�1 � t3 dt 79.

has a relative maximum at x � 2.g

x

y

1

1 2 3 4

2

−2

−1

f g

g�0� � 0, g�1� 12

, g�2� 1, g�3� 12

, g�4� � 0

g�x� � �x

0f �t� dt

80. (a) (b)

(c) Minimum of g at

(d) Maximum at Relative maximum at

(e) On g increases at a rate of

(f) Zeros of g: x � 3, x � 8.

124

� 3.6, 10�

�2, 2�.�10, 6�.

�6, �6�.2

2 4 6 10 12−2

−4

−6

4

6

y

x

x 1 2 3 4 5 6 7 8 9 10

1 2 0 0 3 6�3�6�4�2g�x�

81. (a)

� 5000�25 �125

x54� � 1000�125 � 12x54�

� 5000�25 � 3�45

t 54�x

0�

C�x� � 5000�25 � 3�x

0t14 dt� (b)

C�10� � 1000�125 � 12�10�54� $338,394

C�5� � 1000�125 � 12�5�54� $214,721

C�1� � 1000�125 � 12�1�� � $137,000

Section 6.4 The Fundamental Theorem of Calculus 513

82. (a)

Horizontal asymptote: y � 4

limt→�

g�t� � 4

g�t� � 4 �4t2

(b)

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

limx→�

A�x� � limx→�

�4x �4x

� 8� � �

�4x2 � 8x � 4

x�

4�x � 1�2

x

� �4t �4t�

x

1� 4x �

4x

� 8

A�x� � �x

1�4 �

4t2� dt

83. True 84. True 85. The function is not

continuous at You cannotsimply integrate from to 1.�1

x � 0.

f �x� �1x2

86. 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� 87.

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

�1x�2 � 1��1x2� �

1x2 � 1

f �x� � �1x

0

1t2 � 1

dt � �x

0

1t2 � 1

dt

88.

(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.

89.

� 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�� dt

Total 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

514 Chapter 6 Integration

90.

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 91.

� 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 6.5 Integration by Substitution

du � g�x dxu � gx�f �g�x��g��x� dx

1. 10x dx

2.

3. 2x dx

4. 3x2 dxx3 � 1�x2�x3 � 1 dx

x2 � 1� x

�x2 � 1 dx

3x2 dxx3 � 3��x3 � 3� 3x2 dx

5x2 � 1��5x2 � 1�2�10x� dx

5.

Check:ddx�

�1 � 2x�5

5� C� � 2�1 � 2x�4

��1 � 2x�4 2 dx ��1 � 2x�5

5� C

6.

Check:ddx�

�x2 � 9�4

4� C� �

4�x2 � 9�3

4�2x� � �x2 � 9�3�2x�

��x2 � 9�3�2x� dx ��x2 � 9�4

4� C

8.

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

7.

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

Section 6.5 Integration by Substitution 515

9.

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

10.

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

12.

Check:ddx�

�4x2 � 3�4

32� C� �

4�4x2 � 3�3�8x�32

� x�4x2 � 3�3

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

18�

�4x2 � 3�4

4 � � C ��4x2 � 3�4

32� C

14.

Check:ddt�

16

�t4 � 5�3�2 � C� �16

�32

�t4 � 5�1�2�4t3� � �t4 � 5�1�2�t3�

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

14

�t4 � 5 �3�2

3�2� C �

16

�t4 � 5�3�2 � C

16.

Check:ddu�

2�u3 � 2�3�2

9� C� �

29

� 32

�u3 � 2�1�2�3u2� � �u3 � 2�1�2�u2�

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

13

�u3 � 2�3�2

3�2� C �

2�u3 � 2�3�2

9� C

11.

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

13.

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

15.

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

17.

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

516 Chapter 6 Integration

18.

Check:ddx�

�1

4�1 � x4�� C� �

14

�1 � x4��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

19.

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

20.

Check:ddx�

13�16 � x3� � C� �

13

��1��16 � x3��2��3x2� �x2

�16 � x3�2

� x2

�16 � x3�2 dx � �13��16 � x3��2��3x2� dx � �

13�

�16 � x3��1

�1 � � C �1

3�16 � x3� � C

21.

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

22.

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

23.

Check:ddt��

�1 � �1�t� 4

4� C� � �

14

�4��1 �1t �

3

��1t2� �

1t2�1 �

1t �

3

��1 �1t �

3

�1t2� dt � ���1 �

1t �

3

��1t2� dt � �

�1 � �1�t� 4

4� C

24.

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

25.

Check:ddx

��2x � C �12

�2x��1�2�2� �1

�2x

� 1

�2x dx �

12��2x��1�2 2 dx �

12�

�2x�1�2

1�2 � � C � �2x � C

26.

Check:ddx

��x � C �1

2�x

� 1

2�x dx �

12�x�1�2 dx �

12�

x1�2

1�2� � C � �x � C

Section 6.5 Integration by Substitution 517

27.

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

28.

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

t 5�2 � C �2

15t 3�2�5 � 6t� � C

29.

Check:ddt�

14

t 4 � t2 � C� � t3 � 2t � t2�t �2t �

�t2�t �2t � dt � ��t3 � 2t� dt �

14

t 4 � t2 � C

30.

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

31.

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 � 9�23

y3�2� �25

y5�2 � C �25

y3�2�15 � y� � C

32.

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

33.

� 2x2 � 4�16 � x2 � C

� 4�x2

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 34.

�203�1 � x3 � C

�103 ��1 � x3�1�2

1�2 � � C

�103 ��1 � x3��1�2�3x2� dx

y � � 10x2

�1 � x3 dx

35.

� �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 36.

� �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

518 Chapter 6 Integration

37. (a)

(b)

−3 3

−1

3

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�

x

y

−2 2

−1

3

38.

(a)

Slope field of with two solutions

shown,one through and one though

(b)

−1

6

7

−6

y � �x2 � 1 � 2

3 � �02 � 1 � C ⇒ C � 2

�12

�x2 � 1�1�2

1�2� C � �x2 � 1 � C

�dy � � x�x2 � 1�1�2 dx �

12

��x2 � 1��1�2�2x� dx

�0, 0�.�0, 3�

y� �x

�x2 � 1

y

4 62−4 −2x

−2

−4

6

4

2

8

dydx

�x

�x2 � 1

39.

�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, 40.

�1

15�2x � 1�3�2�3x � 1� � C

�1

30�2x � 1�3�2�6x � 2� � C

�1

30�2x � 1�3�2�3�2x � 1� � 5 � C

�u3�2

30�3u � 5� � C

�14�

25

u5�2 �23

u3�2� � C

�14��u3�2

� u1�2� du

�x�2x � 1 dx � �12

�u � 1��u 12

du

dx �12

dux �12

�u � 1�,u � 2x � 1,

Section 6.5 Integration by Substitution 519

41.

� �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,

42.

� �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,

43.

�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,

520 Chapter 6 Integration

44. 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. 45.

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,

46.

�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, 47. 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,

48. Let

�19

��64 � 8�3 � ��8 � 8�3 � 41,472

� �13

�x3 � 8�3

3 �4

�2

�4

�2x2�x3 � 8�2 dx �

13�

4

�2�x3 � 8�2�3x2� dx

u � x3 � 8, du � 3x2 dx. 49. Let

�49

�27 � 2�2 � 12 �89�2

�49��x3 � 1�3�2�

2

1

� �23

�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

50. Let

�1

0x�1 � x2 dx � �

12�

1

0�1 � x2�1�2��2x� dx � ��

13

�1 � x2�3�2�1

0� 0 �

13

�13

du � �2x dx.u � 1 � x2,

51. 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,

Section 6.5 Integration by Substitution 521

52. 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,

53. 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,

54. 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,

55.

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,

56. 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�.

57.

�2�243�

5� 10�9� �

9365

� �2u5

5�

10u3

3 �3

0

� �3

0�2u4 � 10u2� du

�14

5 x�x � 5 dx � �3

0 �u2 � 5� u �2u du�

dx � 2u du.x � u2 � 5,Let u � �x � 5, 58. Let

�1

0

1�x � 1

dx � �2

1u�1�2 du � 2�u�

2

1� 2�2 � 2

du � dx.u � x � 1,

59.

When When

� �8

1�u4�3 � u1�3� du � �3

7u7�3 �

34

u4�3�8

1� �384

7� 12� � �3

7�

34� �

120928

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,

522 Chapter 6 Integration

60.

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,

62.

00

2

20

�2

0x3�x � 2 dx � 7.581

64.

1 50

50

�5

1x2�x � 1 dx � 67.505

61.

−1

−1 5

3

�4

0

x

�2x � 1 dx � 3.333 �

103

63.

00 8

15

�7

3x�x � 3 dx � 28.8 �

1445

65.

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�2 2 dx �

16

�2x � 1�3 � C1 �43

x3 � 2x2 � x �16

� C1

66.

�16

�x2 � 1�3 � C1

�16

�x6 � 3x4 � 3x2 � 1� � C1 �16

�16

�x6 � 3x4 � 3x2� � C1

�x6

6�

x4

2�

x2

2� C1

�x�x2 � 1�2 dx � ��x5 � 2x3 � x� dx

�12

�x2 � 1�3

3� C �

16

�x2 � 1�3 � C

�x�x2 � 1�2 dx �12��x2 � 1�2 �2x� dx 67. 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�

Section 6.5 Integration by Substitution 523

71. 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

68.

Even function

� 2�27 � 9 � 36

� 2�9x �x3

3 �3

0

�3

�3�9 � x2� dx � 2�3

0�9 � x2� dx 69. is odd.

�2

�2x�x2 � 1�3 dx � 0

f �x� � x�x2 � 1�3 70.

Odd function

�3

�3 x�9 � x2 dx � 0

72. (a)

(b) Odd function

(c) Odd function

(d)

� 2�x4

4�

x2

2 �3

0� 2�81

4�

92� �

992

�3

�3 �x��x2 � 1� dx � 2�3

0 x�x2 � 1� dx � 2�3

0�x3 � x� dx

�5

�5

x�x2 � 1

dx � 0

�2

�2 x3�x2 � 1� dx � 0

� 2�15

�13� �

1615

�1

�1 x2�x2 � 1� dx � 2�1

0 �x4 � x2� dx � 2�x5

5�

x3

3 �1

0

73. � 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

74.

� 2�325

� 10� �1645

� 2�x5

5� 5x�

2

0

� 2�2

0�x4 � 5� dx � 0

�2

�2�x4 � 3x � 5� dx � �2

�2�x4 � 5� dx � �2

�2� � 3x� dx

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

12�u3 du.du � �2x dxu � 5 � x2,

76. is odd. Hence, �2

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

524 Chapter 6 Integration

77.

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 78.

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

79. 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,

(b) Let

� �1

0xb�1 � x�adx

� �1

0ub�1 � u�adu

�1

0xa�1 � x�bdx � �0

1�1 � u�aub��du�

x � 0 ⇒ u � 1, x � 1 ⇒ u � 0

u � 1 � x, du � �dx, x � 1 � u.80. (a) Let

� �1

0x5�1 � x�2dx

� �1

0u5�1 � u�2du

�1

0x2�1 � x�5dx � �0

1�1 � u�2u5��du�

x � 0 ⇒ u � 1, x � 1 ⇒ u � 0

u � 1 � x, du � �dx, x � 1 � u.

81. False

��2x � 1�2 dx �12��2x � 1�2 2 dx �

16

�2x � 1�3 � C

82. False

�x�x2 � 1� dx �12��x2 � 1��2x� dx �

14

�x2 � 1�2 � C

83. 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

84. Let

� �cb

ca

f �x�dx

� �cb

ca

f �u�du

c�b

a

f �cx�dx � c�cb

ca

f �u�duc

u � cx, du � c dx: 85. Let be an odd function and let Then,

Thus,

2�a

�a

f �x� dx � 0 ⇒ �a

�a

f �x� dx � 0

� ��a

�a

f �u� du

� ��a

a

f �u� du

�a

�a

f �x� dx � ��a

a

f ��u���du�

u � �x.f �x�

Section 6.6 Numerical Integration

Section 6.6 Numerical Integration 525

1. Exact:

Trapezoidal:

Simpson’s: �2

0x2 dx �

16�0 � 4�1

2�2

� 2�1�2 � 4�32�

2

� �2�2� �83

� 2.6667

�2

0x2 dx �

14�0 � 2�1

2�2

� 2�1�2 � 2�32�

2

� �2�2� �114

� 2.7500

�2

0x2 dx � �1

3x3�

2

0�

83

� 2.6667

3. Exact:

Trapezoidal:

Simpson’s: �2

0x3 dx �

16�0 � 4�1

2�3

� 2�1�3 � 4�32�

3

� �2�3� �246

� 4.0000

�2

0x3 dx �

14�0 � 2�1

2�3

� 2�1�3 � 2�32�

3

� �2�3� �174

� 4.2500

�2

0x3 dx � �x4

4 �2

0� 4.000

5. Exact:

Trapezoidal:

Simpson’s: �2

0x3 dx �

112�0 � 4�1

4�3

� 2�24�

3

� 4�34�

3

� 2�1�3 � 4�54�

3

� 2�64�

3

� 4�74�

3

� 8� � 4.0000

�2

0x3 dx �

18�0 � 2�1

4�3

� 2�24�

3

� 2�34�

3

� 2�1�3 � 2�54�

3

� 2�64�

3

� 2�74�

3

� 8� � 4.0625

�2

0x3 dx � �1

4x4�

2

0� 4.0000

2. Exact:

Trapezoidal:

Simpson’s: �1

0�x2

2� 1� dx �

112�1 � 4��14�2

2� 1� � 2��12�2

2� 1� � 4��34�2

2� 1� � �12

2� 1�� �

76

� 1.1667

�1

0�x2

2� 1� dx �

18�1 � 2��14�2

2� 1� � 2��12�2

2� 1� � 2��34�2

2� 1� � �12

2� 1� �

7564

� 1.1719

�1

0�x2

2� 1� dx � �x3

6� x�

1

0�

76

� 1.1667

4. Exact:

Trapezoidal:

Simpson’s: �2

1

1x2 dx �

112�1 � 4�4

5�2

� 2�46�

2

� 4�47�

2

�14� � 0.5004

�2

1

1x2 dx �

18�1 � 2�4

5�2

� 2�46�

2

� 2�47�

2

�14� � 0.5090

�2

1

1x2 dx � ��1

x �2

1� 0.5000

6. Exact:

Trapezoidal:

Simpson’s: �8

0

3x dx �13

�0 � 4 � 2 32 � 4 33 � 2 34 � 4 35 � 2 36 � 4 37 � 2� � 11.8632

�8

0

3x dx �12

�0 � 2 � 2 32 � 2 33 � 2 34 � 2 35 � 2 36 � 2 37 � 2� � 11.7296

�8

0

3x dx � �34

x43�8

0� 12.0000

526 Chapter 6 Integration

7. Exact:

Trapezoidal:

Simpson’s: �9

4

x dx �5

24�2 � 4378

� 21 � 4478

� 26 � 4578

� 31 � 4678

� 3� � 12.6667

� 12.6640

�9

4

x dx �5

16�2 � 2378

� 2214

� 2478

� 2264

� 2578

� 2314

� 2678

� 3��9

4

x dx � �23

x32�9

4� 18 �

163

�383

� 12.6667

9. Exact:

Trapezoidal:

Simpson’s:

�1

12�14

�6481

�8

25�

64121

�19� � 0.1667

�2

1

1�x � 1�2 dx �

112�

14

� 4� 1��54� � 1�2� � 2� 1

��32� � 1�2� � 4� 1��74� � 1�2� �

19�

�18�

14

�3281

�8

25�

32121

�19� � 0.1676

�2

1

1�x � 1�2 dx �

18�

14

� 2� 1��54� � 1�2� � 2� 1

��32� � 1�2� � 2� 1��74� � 1�2� �

19�

�2

1

1�x � 1�2 dx � ��

1x � 1�

2

1� �

13

�12

�16

� 0.1667

11.

Trapezoidal Rule:

Simpson’s Rule:

Graphing utility: 1.6094

1.6222

1.6833

n � 4�4

0

1x � 1

dx,

8. Exact:

Trapezoidal:

Simpson’s: �3

1�4 � x2� dx �

16�3 � 4�4 �

94� � 0 � 4�4 �

254 � � 5� � �0.6667

�3

1�4 � x2� dx �

14 3 � 2�4 � �3

2�2

� � 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

0xx2 � 1 dx �

16�0 � 4�1

2��12�2 � 1 � 2�1�12 � 1 � 4�32��32�2 � 1 � 222 � 1� � 3.392

�2

0xx2 � 1 dx �

14�0 � 2�1

2��12�2 � 1 � 2�1�12 � 1 � 2�32��32�2 � 1 � 222 � 1� � 3.457

�2

0xx2 � 1 dx �

13��x2 � 1�32�

2

0�

13

�532 � 1� � 3.393

12. Trapezoidal:

Simpson’s:

Graphing utility: 1.402

�2

0

1

1 � x3 dx �

16�1 � 4� 1

1 � �12�3� � 2� 1

1 � 13� � 4� 1

1 � �32�3� �13� � 1.405

�2

0

1

1 � x3 dx �

14�1 � 2� 1

1 � �12�3� � 2� 1

1 � 13� � 2� 1

1 � �32�3� �13� � 1.397

Section 6.6 Numerical Integration 527

13. Trapezoidal:

Simpson’s:

Graphing utility: 3.2413

�2

0

1 � x3 dx �16

�1 � 41 � �18� � 22 � 41 � �278� � 3� � 3.2396

�2

0

1 � x3 dx �14

�1 � 21 � �18� � 22 � 21 � �278� � 3� � 3.2833

15.

Trapezoidal:

Simpson’s:

Graphing utility: 0.3927

�1

0

x�1 � x� dx �1

12�0 � 414�1 �

14� � 21

2�1 �12� � 43

4�1 �34� � � 0.3720

�1

0

x�1 � x� dx �18�0 � 21

4�1 �14� � 21

2�1 �12� � 23

4�1 �34� � � 0.3415

�1

0

x1 � x dx � �1

0

x�1 � x� dx

14.

Trapezoidal Rule: 9.3741

Simpson’s Rule: 9.3004

Graphing Utility: 9.2936

n � 4�4

0

x2 � 1 dx,

16.

Trapezoidal Rule: 0.775

Simpson’s Rule: 0.7833

Graphing Utility: 0.7854

n � 2�1

0

1x2 � 1

dx, 17.

Trapezoidal Rule:

Simpson’s Rule:

Graphing utility: 2.2143

2.2103

2.2077

n � 8�2

�2

1x2 � 1

dx

18.

Trapezoidal Rule: 0.4830

Simpson’s Rule: 0.4082

Graphing Utility: 0.3771

n � 4�1

�1xx � 1 dx, 19.

Trapezoidal Rule:

Simpson’s Rule:

Graphing utility: 2.5326

2.4385

2.3521

n � 6�7

1

x � 1x

dx,

20.

Trapezoidal Rule:

Simpson’s Rule:

Graphing Utility: �3.8303

�3.8331

�3.8693

n � 6�6

3

1

1 � x � 1 dx, 21. The Trapezoidal Rule

overestimates the area if the graph of the integrand is concave up.

xa b

y f x= ( )

y

22. Trapezoidal: Linear polynomials

Simpson’s: Quadratic polynomials

23.

The error is 0 for both rules.

f��x� � 0

f��x� � 2

f �x� � 2x � 3

528 Chapter 6 Integration

24.

(a) Trapezoidal:

since is a maximum of 6 at

(b) Simpson’s:

since is a maximum of 120 at x � 2.� f �4��x��

Error ≤ �4 � 2�5

180�44� �120� �1

12

x � 2.� f � �x��

Error ≤ �4 � 2�3

12�42� �6� �14

,

f �4��x� � 120�x � 1��6

f���x� � �24�x � 1��5

f ��x� � 6�x � 1��4

f��x� � �2�x � 1��3

f �x� � �x � 1��2 25.

(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

26.

f �4��x� �24

�x � 1�5

f���x� ��6

�x � 1�4

f ��x� �2

�x � 1�3

f��x� ��1

�x � 1�2

f �x� �1

x � 1(a) Trapezoidal:

since is maximum in when

(b) Simpson’s:

since is maximum in when x � 0.�0, 1�f �4��x�

Error ≤ �1 � 0�5

180�44� �24� �1

1920� 0.0005

x � 0.�0, 1�f��x�

Error ≤ �1 � 0�2

12�42� �2� �1

96� 0.01

27. in .

(a) is maximum when and

Trapezoidal: let

in

(b) is maximum when and when

Simpson’s: let n � 26.n > 25.56;n4 > 426,666.67,Error ≤ 25

180n4�24� < 0.00001,

� f �4��1�� � 24.x � 1� f �4��x��

�1, 3�f �4��x� �24x5

n � 366.n > 365.15;n2 > 133,333.33,Error ≤ 23

12n2�2� < 0.00001,

� f��1�� � 2.x � 1� f��x��

�1, 3�f��x� �2x3

28. in .

(a) is maximum when and

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

in

(b) is maximum when and

Simpson’s: Error ≤ > 13,333.33, n > 10.75; let (In Simpson’s Rule n must be even.)n � 12.n41180n4�24� < 0.00001,

� f �4��0�� � 24.x � 0� f �4��x��

�0, 1�f �4��x� �24

�1 � x�5

n � 130.n2112n2�2�

� f��0�� � 2.x � 0� f ��x��

�0, 1�f��x� �2

�1 � x�3

Section 6.6 Numerical Integration 529

29.

(a)

is maximum when

Trapezoidal: Error Let

(b)

is maximum when and

Simpson’s Rule: Error Let (must be even)n � 40n > 38.2; n4 > 2,133,333.3,≤32

180n4 �120� < 0.00001,

�f �4��1�� � 120.x � 1� f �4��x��f �4��x� � 120x�6f����x� � �24x�5,

n � 633 n > 632.5; n2 > 400,000,≤8

12n2 �6� < 0.00001,

x � 1 and � f��1�� � 6.� f��x��f��x� � �2x�3, f��x� � 6x�4

�1, 3�f �x� � x�2,

30.

(a)

is maximum when and

Trapezoidal: Error

Let n � 46.

n > 45.6

n2 > 2083.3

≤1

12n2 �14� < 0.00001

� f� �0�� �14

x � 0� f��x��

f��x� � ��x � 2��2, f��x� � 2�x � 2��3 �2

�x � 2�3

�0, 1�f �x� � �x � 2��1,

(b)

is maximum when and

Simpson’s Rule: Error

Let (must be even)n � 6.

n > 4.5

n4 > 416.67

≤1

180n4 �34� < 0.00001

� f �4��0�� �2432

�34

.x � 0� f �4��x��

f �4��x� � 24�x � 2��5f����x� � �6�x � 2��4,

31.

(a) in

is maximum when and

Trapezoidal: let

(b) in

is maximum when and

Simpson’s: let n � 12.n > 11.36;n4 > 16,666.67,Error ≤32

180n4�1516� < 0.00001,

� f �4��0�� �1516

.x � 0� f �4��x��

�0, 2�f �4��x� ��15

16�1 � x�72

n � 130.n > 129.10;n2 > 16,666.67,Error ≤8

12n2�14� < 0.00001,

� f � �0�� �14

.x � 0� f � �x��

�0, 2�.f� �x� � �1

4�1 � x�32

f �x� � 1 � x

32.

(a) in

is maximum when and

Trapezoidal: let

(b) in

is maximum when and

Simpson’s: let (In Simpson’s Rule n must be even.)n � 12.n > 10.530;n4 > 12,290.81,Error ≤32

180n4�5681� < 0.00001,

� f �4��0�� �5681

.x � 0� f �4��x��

�0, 2�f �4��x� � �56

81�x � 1�103

n � 122.n > 121.72;n2 > 14,814.81,Error ≤8

12n4�29� < 0.00001,

�f � �0�� �29

.x � 0� f��x��

�0, 2�.f � �x� � �2

9�x � 1�43

f �x� � �x � 1�23

530 Chapter 6 Integration

33. Let Then

Simpson’s: Error ≤

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

Example:

This is the exact value of the integral.

�1

0x3 dx �

16�0 � 4�1

2�3

� 1� �14

�b � a�5

180n4 �0� � 0

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

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

35. 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�

36. 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�

37.

Simpson’s Rule:

� 400�1512�125 � �15

12�3

� . . . � 0� � 10,233.58 ft � lb

�5

0100x125 � x3 dx �

53�12��0 � 400� 5

12�125 � � 512�

3

� 200�1012�125 � �10

12�3

n � 12

W � �5

0100x125 � x3 dx

Section 6.6 Numerical Integration 531

38. (a) Trapezoidal:

(b) 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

(c) Using a graphing utility,

Integrating, �2

0y dx � 12.53

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

39. 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�12

0

6

1 � x2 dx

41. 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

43.

Use a computer algebra system to determine t such that

to find t � 6.366.

t30� 1

0 � 1� 4� 1

t10

� 1� � 2� 12t10

� 1� � 4� 13t10

� 1� � . . . � 4� 19t10

� 1� �1

t � 1� � 2

n � 10f �x� �1

x � 1,

40. Simpson’s Rule:

� 3.14159

� � 4�1

0

11 � x2 dx �

43�6��1 �

41 � �16�2 �

21 � �26�2 �

41 � �36�2 �

21 � �4�6�2 �

41 � �56�2 �

12�

n � 6

42.

� 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�

44. 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

Review Exercises for Chapter 6

532 Chapter 6 Integration

1.

x

f

f

y

3. ��2x2 � x � 1� dx �23

x3 �12

x2 � x � C

5. �x3 � 1x2 dx � ��x �

1x2� dx �

12

x2 �1x

� C

7.

�37

x7�3 �94

x4�3 � C

� 3�x �x � 3� dx � ��x4�3 � 3x1�3� dx

2.

x

f ′

f

y

4.

� 23�3x

dx �23��3x��1�3�3� dx � �3x�2�3 � C

du � 3 dx

u � 3x

6.

�12

x2 � 2x �1x

� C

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

8.

�x5

5�

5x4

2�

25x3

3� C

� � x4 � 10x3 � 25x2 dx

� x2�x � 5�2 dx � � x2�x2 � 10x � 25� dx

9.

When

y � 2 � x2

C � 2

y � �1 � C � 1

x � �1:

f �x� � ��2x dx � �x2 � C

��1, 1�f��x� � �2x, 10.

Since the slope of the tangent line at is 3, it followsthat when

f �x� � �x � 1�3

f �2� � 1 � C2 � 1 when C2 � 0.

f �x� � �3�x � 1�2 dx � �x � 1�3 � C2

f��x� � 3�x � 1�2

C1 � 0.f��2� � 3 � C1 � 3�2, 1�

f��x� � �6�x � 1� dx � 3�x � 1�2 � C1

f ��x� � 6�x � 1�

Review Exercises for Chapter 6 533

13.

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 14.

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

11. (a)

(b)

1

8

−7

−4

y � x2 � 4x � 2

�2 � 16 � 16 � C ⇒ C � �2

y � ��2x � 4�dx � x 2 � 4x � C

dydx

� 2x � 4, �4, �2�

−2 6

−6

2

x

y 12. (a)

(b)

y �16

x 3 � x 2 � 2

2 �16

�63� � 6 2 � C ⇒ C � 2

y � ��12

x2 � 2x�dx �16

x3 � x2 � C

dydx

�12

x2 � 2x, �6, 2�

x

y

7

6

−2

(6, 2)

15.

(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 16.

(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

534 Chapter 6 Integration

17. 8

i�1

14i

�1

4�1� �1

4�2� � . . . �1

4�8� 18. 12

i�1

i � 22i

�1 � 22�1� �

2 � 22�2� �

3 � 22�3� � . . . �

12 � 22�12�

19. n

i�1�3

n��i � 1

n �2�

3n �

1 � 1n �2

�3n �

2 � 1n �2

� . . . �3n �

n � 1n �2

20. n

i�13i2 �

�i � 1�2

n � � 32 �4n� � 62 �

9n� � . . . � 3n2 �

�n � 1�2

n �

21. 10

i�13i � 3�10�11�

2 � � 165 22. 20

i�1�4i � 1� � 4

20�21�2

� 20 � 820

23.

� 2870 � 420 � 20 � 3310

�20�21��41�

6� 2

20�21�2

� 20

20

i�1�i � 1�2 �

20

i�1�i2 � 2i � 1� 24.

� 6084 � 78 � 6006

��122��132�

4�

12�13�2

12

i�1i�i2 � 1� �

12

i�1�i 3 � i�

25. (a) (b) (c) 10

i�1�4i � 2�

n

i�1i3

10

i�1�2i � 1�

26.

(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,

27.

9.0385 < Area of Region < 13.0385

� 9.0385

s�n� � s�4� �12

10�1�2�2 � 1

�10

1 � 1�

10�3�2�2 � 1

�10

22 � 1� � 13.0385

S�n� � S�4� �12

101

�10

�1�2�2 � 1�

10�1�2 � 1

�10

�3�2�2 � 1�

y �10

x2 � 1, �x �

1

2, n � 4

Review Exercises for Chapter 6 535

28.

� 14.5

s�4� � 1�9 �14

�9�� � �9 �14

�16�� � �9 �14

�25�� � �9 � 9�� � 22.5

S�4� � 1�9 �14

�4�� � �9 �14

�9�� � �9 �14

�16�� � �9 �14

�25���

y � 9 �1

4x2, �x � 1, n � 4

29.

� limn→�

24 � 8 n � 1

n � � 24 � 8 � 16

� limn→�

4n6n �

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

30. right endpoints

� limn→�

43

�n � 1��2n � 1�

n2 � 6� �83

� 6 �263

� 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

2

y

x

y � x2 � 3, �x �2n

31.

� 3 � 18 � 9 � 12

� limn→�

3 � 18 n � 1

n�

92

�n � 1��2n � 1�

n2 �

� limn→�

3nn �

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

536 Chapter 6 Integration

32.

� 4 � 6 � 4 � 1 � 15

� limn→�

4nn �

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�18 �

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

33.

� 18 �92

� 9� �272

� limn→�

3n6n �

3n

n�n � 1�

2�

9n2

n�n � 1��2n � 1�6 �

� limn→�

3n

n

i�16 �

3in

�9i2

n2 �

� limn→�

3n

n

i�110 �

15in

� 4 � 12in

�9i2

n2 �

Area � limn→�

n

i�15�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

34. (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

35. lim���→�

n

i�1�2ci � 3� �xi � �6

4�2x � 3� dx

36. lim���→�

n

i�13ci�9 � ci

2� �xi � �3

13x�9 � x2� dx

Review Exercises for Chapter 6 537

37.

�12

�5��5� �252

Area �12

�base��height�

x9

y

6

9

12

3

−3

3−3 6

Triangle

38.

(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

39. (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

40. (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

41. (c)�734�8

1� 3�x � 1� dx � 3

4x4�3 � x�

8

1� 3

4�16� � 8� � 3

4� 1�

43. �4

0�2 � x� dx � 2x �

x2

2 �4

0� 8 �

162

� 16

45. �1

�1�4t3 � 2t� dt � t4 � t2�

1

�1� 0

47. �9

4x�x dx � �9

4x3�2 dx � 2

5x 5�2�

9

4�

25

���9 �5� ��4 �5

�25

�243 � 32� �4225

42. (d)�3

1

12x3 dx � 12x�2

�2 �3

1� �6

x2 �3

1�

�69

� 6 �163

,

44. �1

�1�t2 � 2� dt � t3

3� 2t�

1

�1�

143

46.

�5215

� �325

�163

� 10� � ��325

�163

� 10�

�2

�2�x4 � 2x2 � 5� dx � x5

5�

2x3

3� 5x�

2

�2

538 Chapter 6 Integration

48. �2

1� 1

x2 �1x3� dx � �2

1�x�2 � x�3� dx � �

1x

�1

2x2�2

1� ��

12

�18� � ��1 �

12� �

18

49.

−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 50.

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

51.

8

6

4

x842

2

6

y

�643

�543

�103

� �643

� 36� � �9 � 27�

�4

3�x2 � 9� dx � x3

3� 9x�

4

352.

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

53.

x

1

1

1

y

�12

�14

�14�1

0�x � x3� dx � x2

2�

x4

4 �1

054.

x1

1

y

�23

�25

�4

15

� 23

x3�2 �25

x5�2�1

0

�1

0

�x�1 � x� dx � �1

0�x1�2 � x3�2� dx

55. � 8�3 � 1� � 16� 4x1�2

�1�2��9

1 Area � �9

1

4�x

dx

2−2−2

2

4

6

8

10

4 6 8 10x

y

Review Exercises for Chapter 6 539

57. Average value

x �254

�x �52

2

5�

1�x 1

2 4 6 8 10

2

2255,4

x

y1

9 � 4 �9

4

1�x

dx � 1

5 2�x�

9

4�

25

�3 � 2� �25

56.

−2 2

−2

−1

1

2

x

y

Area � �1

0 �x � x5� dx �

x2

2�

x6

6 �1

0�

13

58.

x � 3�2

x3 � 2

x1 2

2

4

6

8

( 2 , 2)3

y12 � 0�

2

0x3 dx � x4

8 �2

0� 2

59. F��x� � x2�1 � x3 60. F��x� � x2 � 3x � 2

61. �x7

7�

35

x5 � x3 � x � C��x2 � 1�3 dx � ��x6 � 3x4 � 3x2 � 1� dx

62.

�x3

3� 2x �

1x

� C

��x �1x�

2

dx � ��x2 � 2 � x�2� dx 63. �x�x2 � 1�3 dx �12��x2 � 1�3 �2x� dx �

�x2 � 1�4

8� C

64. ,

�x2�x3 � 3 dx �13��x3 � 3�1�2 3x2 dx �

29

�x3 � 3�3�2 � C

du � 3x2 dxu � x3 � 3

65. ,

� x2

�x3 � 3 dx � ��x3 � 3��1�2 x2 dx �

13��x3 � 3��1�2 3x2 dx �

23

�x3 � 3�1�2 � C

du � 3x2 dxu � x3 � 3

66.

� �19�25 � 9x2 � C� �

118

2�25 � 9x2 � C � x

�25 � 9x2 dx � �

118�

�18x�25 � 9x2

dx

du � �18x dxu � 25 � 9x2,

540 Chapter 6 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 �

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.

�2�x � 5�7�2

7� 4�x � 5�5�2 �

503

�x � 5�3�2 � C

� ��2u6 � 20u4 � 50u2� du �x2�x � 5 dx � ��u2 � 5�2 u�2u� du

dx � 2u dux � u2 � 5,u � �x � 5,

70.

�25

�x � 5�5�2 �103

�x � 5�3�2 � C� ��u 3�2�5u1�2� du �x�x � 5 dx � ��u � 5��u du

u � x � 5, x � u � 5, dx � du

71. �2

�1x�x2 � 4� dx �

12�

2

�1�x2 � 4��2x� dx �

12

�x2 � 4�2

2 �2

�1�

14

�0 � 9 � �94

73. �3

0

1

�1 � x dx � �3

0�1 � x��1�2 dx � 2�1 � x�1�2�

3

0� 4 � 2 � 2

75.

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,

72. �1

0x2�x3 � 1�3 dx �

13�

1

0�x3 � 1�3�3x2� dx �

112�x3 � 1�4�

1

0�

112

�16 � 1� �54

74. �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�

76.

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

�1x2�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,

Problem Solving for Chapter 6 541

77.

(a) 2005 corresponds to

dollarsC �15,000

M �1.20s � 0.02s2�16

15�

27,300M

t � 15.

�15,000

M �1.20s � 0.02s2�t�1

t

C �15,000

M �t�1

t

p ds �15,000

M �t�1

t

�1.20 � 0.04s� ds

(b) 2007 corresponds to

dollarsC �15,000

M �1.20s � 0.02s2�18

17�

28,500M

t � 17.

78. 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�:

80. (a)

(b)

� 5.417 �13�4 � 4�2� � 2�1� � 4�1

2� �14�

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

R < I < T < L

79. 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�:

Problem Solving for Chapter 6

1. (a)

(b) by the Second Fundamental Theorem

of Calculus.

(c) for

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

e,x

�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 (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.

542 Chapter 6 Integration

2. (a) F�x� � �x

2 �1 � t3 dt

x 0 1.0 1.5 1.9 2.0 2.1 2.5 3.0 4.0 5.0

0 0.310 1.763 4.100 10.742 20.355�0.290�1.265�2.130�3.241F�x�

(b)

(c)

Since F��x� � �1 � x3, F��2� � �1 � 8 � 3 � limx→2

G�x�.

� limx→2

G�x�� limx→2

1

x � 2�x

2 �1 � t3 dt F��2� � lim

x→2 F�x� � F�2�

x � 2

limx→2

G�x� � 3

G�x� �1

x � 2�x

2 �1 � t3 dt

x 1.9 1.95 1.99 2.01 2.1

2.9011 2.9503 2.9900 3.0100 3.1011G�x�

3. (a)

(b)

Average value

(c) Average value �1

b � a �b

a

f��x� dx �f �b� � f �a�

b � a

�13�x2�

3

0� 3

�1

3 � 0 �3

0 f��x� dx �

13

�3

0 2x dx

f��x� � 2x

Slope �9 � 03 � 0

� 3 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)

—CONTINUED—

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�

(b)

Problem Solving for Chapter 6 543

5. —CONTINUED—

(c)

F is decreasing on and increasing onTherefore, the minimum is at and the

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

(d)

is a point of inflection, whereas is not.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 (positiveacceleration) on and �0.7, 1.0�.

�0, 0.4�

0.2 0.4 0.6 0.8 1.0

20

40

60

80

100

y

x

(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

7. (a)

Exact: 2.7974

Error:

(c) Let

p��1�3� � p� 1

�3� � �b3

� d� � �b3

� d� �2b3

� 2d

�1

�1 p�x� dx � �ax4

4�

bx3

3�

cx2

2� dx�

1

�1�

2b3

� 2d

p�x� dx � ax3 � bx2 � cx � d.

� 0.0007

� 1.1927 � 1.6054 � 2.7981

�1

�1

�x � 2 dx � f ��1�3� � f � 1

�3� (b)

�Note: Exact answer is �

2� 1.5708.�

�1

�1

11 � x2 dx �

1

1 � �13�

�1

1 �13

�32

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 � �t

0��t

0f �v� dv� dt � C.

ddx�

x

0��t

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

0f �t��x � t� dt � �x

0x f �t� dt � �x

0t f �t� dt � x�x

0f �t� dt � �x

0t f �t� dt

544 Chapter 6 Integration

10. Consider The corresponding

Riemann Sum using right-hand 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

.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

.

12. (a)

(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�ba� �

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� � 36.Base � 6, height � 9.

� 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

Problem Solving for Chapter 6 545

13. By Theorem 6.8,

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 ≤ �2

m�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�.

14. (a)

by the formula

(b)

(c) by part (b)

by part (a).

Hence, 13 � 23 � . . . � n3 � �n�n � 1�2 �

2

.

A � �n�n � 1�2 �

2

A � A1 � A2 � . . . � An � 13 � 23 � . . . � n3

�k2

2 2k � k3

� k k�k � 1�

2� k

k�k � 1�2

Ak � k 1 � 2 � . . . � k � k1 � 2 � . . . � �k � 1�

1 � 2 � . . . � n �n�n � 1�

2

� �n�n � 1�2 �

2

A � S2 � �1 � 2 � . . . � n�2

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