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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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