Integration
Section CHAPTER 4.1 Section 4.2 Section 4.3 Section 4.4 Section 4.5 Section 4.6
4
Antiderivatives and Indefinite Integration . . . . . . . . . 177 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Riemann Sums and Definite Integrals . . . . . . . . . . . 188 The Fundamental Theorem of Calculus . . . . . . . . . . 192 Integration by Substitution . . . . . . . . . . . . . . . . . 197 Numerical Integration . . . . . . . . . . . . . . . . . . . 204
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
CHAPTER Integration
Section 4.1
4
Antiderivatives and Indefinite Integration
Solutions to Odd-Numbered Exercises
1.
d3 dx x3 dy dt y
C
d 3x dx
3
C
9x
4
9 x4
3.
d 13 x dx 3
4x
C
x2
4
x
2x
2
5.
3t2 t3 C C 3t2
7.
dy dx y
x3
2
d3 t Check: dt
25 x 5
2
C
2
Check:
d 25 x dx 5
C
x3
2
Given 9.
3
Rewrite x1 3 dx
Integrate x4 3 43 x
12
Simplify 34 x 4
3
x dx
C
C
11.
1 xx
dx
x
32
dx
12 1x2 22
C
2 x 1 4x2
C
13.
1 dx 2x3 x2 2 3x
1 x 2
3
dx
C
C
15.
x
3 dx d x2 dx 2
3x C
C x 3
17.
2x Check:
3x 2 dx d2 x dx
x2 x3
x3 C
C 2x 3x 2
Check:
19.
x3 Check:
2 dx d 14 x dx 4
14 x 4 2x
2x C
C x3 2
21.
x3
2
2x d 25 x dx 5
1 dx
2
25 x 5 x2 x
2
x2 C
x x3
2
C 2x 1
Check:
23.
3
x2 dx
x2 3 dx d 35 x dx 5
3
x5 3 53 C x2
3
C
3
35 x 5 x2
3
C
25.
1 dx x3 Check: d dx
x
3
dx
x
2
2 C
C 1 x3
1 2x2
C
Check:
1 2x2
177
178
Chapter 4
Integration
27.
x2
x x
1
dx
2
x3 23 x 3
2
x1
2
x
2
12
dx x3
2
25 x 5 x1
2
23 x 3 x
2
2x1 x2
2
C x x 1
21 x 15
2
3x2
5x
15
C
Check:
d 25 x dx 5
2
2x1
C
2
12
29.
x
1 3x
2 dx x3
3x2 12 x 2 C
x
2 dx 2x 3x2 x C x 1 3x 2 2
31.
y2 y dy Check: d 27 y dy 7
y5 2 dy
2
27 y 7 y5
2
C y2 y
C
2
Check:
d3 x dx
12 x 2
2x
33.
dx Check:
1 dx d x dx
x C
C 1
35.
2 sin x Check: d dx
3 cos x dx 2 cos x
2 cos x 3 sin x C
3 sin x 2 sin x
C 3 cos x
37.
1 Check:
csc t cot t dt d t dt csc t
t C
csc t 1
C csc t cot t
39.
sec2 Check:
sin d tan d
d cos
tan C
cos sec2
C sin
41.
tan2 y
1 dy C
sec2 y dy sec2 y
tan y tan2 y
C 1
43. f x
y
cos x
d Check: tan y dy
3 2
C
2C
3
x
2
2
0 2
3
C
45. f x fx
2 2x
y
5 4
47. f x C
f )x)
2x 2
1 x
x3 3
x2 x3 3
x
y
4
49. C
dy dx y 1
2x 2x 1 x2
2
1, 1, 1 1 dx 1 x 1 x2 CC x C 1
fx
f )x)
f )x)
x3 3
x
3
y
f
3
f )x)
2x
3 2
x
3 2 1 2 3 3 2 1 3 2
x
Answers will vary.
f
Answers will vary.
S ection 4.1
Antiderivatives and Indefinite Integration
179
51.
dy dx y 4 y
cos x, 0, 4 cos x dx sin 0 sin x sin x C 4
53. (a) Answers will vary.
y 5
CC 4
3
x
5
3
(b)
dy dx y 2 2 y
1 x 2 x2 4 42 4 C x2 4 8t3 8t3 4 2 5t
32
1, 4, 2 x 4 C
4
6
8 2
C
x
2
55. f x fx f0 fx
4x, f 0 4x dx 6 2x 2 20 6
2
6 2x 2 C 6
57. h t ht h1 ht 61. f x f4 f0
5, h 1 5 dt 5 11 2t4
4 5t C 11
CC
CC
2t4 x 2 0 x 2 2
59. f x f2 f2 fx f2 fx fx f2 fx
2 5 10 2 dx 4 2x 2x 6 x
2
2x
C1 1
fx f4 x 4 C2 fx fx f0 fx 0
32
dx C1 3
12
2x
12
C1 3
2 x
C1
C1 1
5 C1
2 C1
1 dx
x2
2 x 2x 0 4x1
2
C2 x
10 C2 4
3 dx
4x1
2
3x
C2
C2 3x
0 C2 4x
0 3x
63. (a) h t h0 ht (b) h 6 0
1.5t 0
5 dt C 5t
2
0.75t 2 12 C 12
5t 12
C
0.75t2 0.75 6
56
12
69 cm
180
Chapter 4
Integration
65. f 0 (a) f 4
4. Graph of f is given. 1.0 (f) f is a minimum at x (g)
6 4 2 x 2
3.
(b) No. The slopes of the tangent lines are greater than 2 on 0, 2 . Therefore, f must increase more than 4 units on 0, 4 . (c) No, f 5 < f 4 because f is decreasing on 4, 5 . (d) f is an maximum at x 3.5 because f 3.5 the first derivative test. 0 and
y
2
4
6
8
(e) f is concave upward when f is increasing on ,1 and 5, . f is concave downward on 1, 5 . Points of inflection at x 1, 5. 67. a t vt v0 st s0 st 6 60 32 ft sec2 32 dt C1 32t C2 16t2 60t 6 Position function 60 dt 16t 2 60t C2 32t C1
6
69. From Exercise 68, we have: st 16t2 v0t 0 when t v0 32 v0 32 time to reach
st 32t v0 maximum height. s v02 64 v0 32 v02 32 v02 v0 16 550 35,200
v0 32
2
v0
550
The ball reaches its maximim height when vt 32t 60 32t t s 15 18 16 9.8 9.8 dt v0 9.8t C1 9.8t 4.9t
2
0 60 15 seconds 8
2
187.617 ft sec
15 8
60
15 8
6
62.26 feet 73. From Exercise 71, f t vt 9.8t v0 t v0t v0t C2 s0 f 10 9.8 9.8t 10 10 9.8 7.1 m 10 4.9t2 10t 2. 0.)
71. a t vt v0 ft f0 s0 a vt st s 20
0 (Maximum height when v
C1 v t 9.8t v0 dt
C2 f t 1.6 1.6 dt 1.6t dt 0 1.6t
4.9t 2
75.
v0 s0 s0 s0
1.6t, since the stone was dropped, v0
0.
0.8t2 0.8 20
2
0 320
Thus, the height of the cliff is 320 meters. vt v 20 1.6t 32 m sec
S ection 4.1
Antiderivatives and Indefinite Integration
181
77. x t (a) v t
t3
6t2 xt 3t
2
9t 3t 4t 6t
2
2 12t 3 12
0t5 9 3t 6t 1t 2 3
79. v t xt x1 xt at
1 t
t
12
t>0
2
v t dt 4 2t1 2 vt 21
2t1
C 2
at
vt
CC
(b) v t > 0 when 0 < t < 1 or 3 < t < 5. (c) a t v2 81. (a) v 0 v 13 at vt v0 v 13 6t 31 2 1 0 when t 3 25 80 1000 3600 1000 3600 250 m sec 36 800 m sec 36 2.
2 position function 1 t 2
32
1 acceleration 2t 3 2
25 km hr 80 km hr
83. Truck: v t st
30 30t Let s 0 6 6t Let v 0 3t2 Let s 0 0. 0. 0.
Automobile: a t vt st
a constant acceleration at C at 250 36 250 36
250 vt 36 800 36 550 36 a 13a 13a 550 468 250 t 36
2
At the point where the automobile overtakes the truck: 30t 0 0 (a) s 10 3t2 3t2 3t t 3 10 6 10 30t 10 when t
2
10 sec.
300 ft 60 ft sec 41 mph
275 234 s0 250 13 36 22 ft sec 15
(b) v 10 1.175 m sec2 0 189.58 m
(b)
st s 13
a
t2 2
275 13 234 2
85.
1 mi hr 5280 ft mi 3600 sec hr (a) t V1 ft sec V2 ft sec (c) S1 t S2 t 0 0 0 5
10 10.27 55.73
15 23.47 74.8 0.0416 2 t 2 6.7991t2 2
20 42.53 88
25 66 93.87
30 95.33 95.33
(b) V1 t V2 t
0.1068t2 0.1208t2
0.0416t 6.7991t
0.3679 0.0707
3.67 30.8
V1 t dt V2 t dt
0.1068 3 t 3 0.1208t3 3
0.3679t 0.0707t S2 0 0
In both cases, the constant of integration is 0 because S1 0 S1 30 S2 30 953.5 feet 1970.3 feet
The second car was going faster than the first until the end.
182
Chapter 4
Integration
87. a t vt st
k kt k2 t since v 0 2 s0 0. 160 and k 2 t2 0.7. Since k 2 t2 0.7,
At the time of lift-off, kt t v 1.4 k 1.4k k 1.4 k
1.4 k
160 1602 1.4
1602 k
18,285.714 mi hr2 7.45 ft sec2. 89. True x3 3 91. True x2 2 x2 2
93. False. For example,
x
x dx
x dx
x dx because
C
C1
C2
95. f x
1, 0 x < 2 3x, 2 x 5 x C1, 0x<2 3x2 C2, 2 x 5 2 31 C1 3 C1 2 2:
fx f1
f is continuous: Values must agree at x 4 fx 6 C2 C2 2
0x<2 x 2, 3x 2 2, 2 x 5 2 2 do not agree. Hence f is not differentiable at x 2.
The left and right hand derivatives at x
Section 4.2
5 5
Area
5
1.
i 4 1
2i 1
0 9
1
2
i
i
1 i
1
1
21 1 10 9.
j 1
2 1 17
8
3 158 85
4
5
5
35
4
3.
k
k2
1
1
1 2
1 5
5.
k 1
c 2n ni 1
20
c 2i n 1
2 3
c
c 2i n
19
c
4c 3n 21 ni 1 3i n
2
7.
i
1 3i 1
20
5
j 8
3
11.
13.
20
15.
i 1
2i
2
i
i
1
2
20 21 2
420
17.
i 1
i
i2
i 1
19 20 39 6
2470
S ection 4.2
Area
183
15
15
15
15
19.
i 1
ii
1
i
1 2
i
1
i
1 20
>
2
i3 152 16 4 14,400 12,040
2
i2
i 15 16 2
21. sum seq x i2
i 1
2 3
3, x, 1, 20, 1 20 20
2930 (TI-82) 1 3 20
15 16 31 2 6 2,480 120
1 2 20 6 60
20 21 41 6
2930
23. S s
3 1
4 3
9 2
51
9 2
33 2 25 2
16.5 12.5
25. S s 31 44 1 4 1 1 2 8 1 5 1 6 1 6 1 7 81 4 2 8 3
3 2 3
3 2 2
51 31
11 7
4
1
27. S 4 s4 0 1 4 1 5
11 44 11 44 11 655 11 755
2
11 24 11 24 11 755 11 855 81 lim 4 n 18 lim 2 n n4
1 31 44
0.768
0.518
29. S 5 s5
1
11 855 11 955 2n3 n4 n n2
11 955 11 25 n2 81 1 4
1 7 1 8
1 8 1 9
1 9 1 10
0.746 0.646
11 655
31. lim
n
81 n2 n 1 n4 4 18 n n 1 n2 2 1 n2 S 10
33. lim
n
n2
18 1 2
9
n
35.
i 1
2i
1n 2i n2 i 1 12 10 1.02 1.002 1.0002 1.2
1
1 nn 1 2 n2 2
n
n n
2
Sn
S 100 S 1000 S 10,000
n
37.
k
6k k 1 n3 1
6n 2 k n3 k 1 6 2n2 n2
k 3n 1 6
6 nn n3 3n
1 2n 6 3
1
nn 2 2
1
1 2n2 n2
Sn
S 10 S 100 S 1000 S 10,000
n
1.98 1.9998 1.999998 1.99999998 16i n2 16 n i n2 i 1 16 n n 1 n2 2 n2 n2 n 1 n
39. lim
n i 1
n
lim
n
lim
n
lim
8
8 lim 1
n
8
184
Chapter 4
Integration
n
41. lim
n i
1 i n3 1
1
2
n
lim
1n n3 i
1
i2
1
n
lim
1n n3 n
n
1 n 2n 6 lim 12 6 1 n n
1 3n 1 1 nn 1 n 2 1 n2 1 3 2 lim
n
1 2n3 lim n 6
n
3n2 n3
n
43. lim
n i 1
1
i n
2 n
2 lim
n
1 n
1
i 1
1n i ni 1
2 lim
n
1
n2 n 2n2
21
1 2
3
45. (a)
3
y
(e)
x sn Sn
n
5 1.6 2.4
10 1.8 2.2 2 n 2 n
50 1.96 2.04
100 1.98 2.02 4n i n2i 1
2
1
x
1 3
(f) lim
n i 1
i
1
n
lim lim
1 n 4 n 2
(b)
x
2 n
0
2 n
n
4 nn 1 n2 2 2n n 4n i n2i 1 4 nn 1 n2 2 2n n 1 2 1
Endpoints: 0 <1 (c) Since y sn
i n 1
n
lim
2 2 <2 <...< n n n x is increasing, f mi
n
1
2 2 <n n n
1
2
n
n
lim
i 1
i
2 n
2 n
n
lim lim lim
f xi
on xi
1,
xi .
f xi f
i 1
1
x 2 2 n
n
n
2i n
1,
i
i 1
2 1 n
2 n
n
(d) f Mi Sn
i
f xi on xi
n
xi
n
f xi
1
x
i 1
f
2i 2 nn
n
i
i 1
2 n 0
2 n 1 n
47. y
2x sn
i
3 on 0, 1 .
n
Note: x
n
1 n 3 2
y
f
1
i n 2 n2i
n
1 n i
1
2
i 1
i n 1n
1 n 1 n
2
3 Area 49. y x2 Sn
i 1 n
3
2n 2n2
1
x
lim s n
2 Note: x
n i 1
1
2
3
2 on 0, 1 .
n
1 n
2
y
f 1 n3 i
i n
n
1 n i2 2 7 3
i n nn
2 1 2n 6n3
1 n 1 2 1 2 6 3 n 1 n2 2
3
1
1
x
1 2 3
Area
n
lim S n
S ection 4.2
Area
185
51. y sn
16
n
x 2 on 1, 3 . Note: x f1
i 1
2 n 1 2i n
2
y 18
2i n
2 n 4i2 n2
n
16
i 1
2 n
2n 15 ni 1 2 15n n 30 Area
4i n 1 2n 6 1 4 1 4 n n 70 3 4 n 1 9i n 27 n n n2 1 2n 27 2 1 n 1 2n 6 1 513 4 1 1 9nn 1 n 2 4nn 1 n 2 1 23 1 3 3 n 3i n
3
1
14 12 10 8 6 4 2 x 1 2 3
4 nn n2 1 2n 30
8 n 6n2
n
lim s n
8 3
53. y sn
64
n
x3 on 1, 4 . Note: x f1
i 1
1
y 70
3i n
3 n 27i3 n3
n
64
i 1
3 n
60 50 40 30 20 10 1 x 1 2 3 5 6
3n 63 ni 1 3 63n n 189 Area
27i2 n2
2
27 n2 n 1 n3 4 81 n 4n2 1 189
2
81 n 6n2 81 4 27
27 n 1 2 n 128.25
n
lim s n
55. y
x2
x3 on
1, 1 .
Note: x
2 n
2
y
Again, T n is neither an upper nor a lower sum.
n
Tn
i n 1
f 1
i n 1
1 4i n 10i n 20 n2 10 1 4
2i n
2 n 4i2 n2 16i 2 n2
n
1
i 1
2i n
2
1 8i3 n3 20 n i n2 i 1 1 2 n
2i n 2 n
3
2 n
1
1
1 8i3 n3 32 n3 3 n 4 2 n
6i n
12i 2 n2 4n 1 ni 1 1 2n 6 41
x
1
2
i 1
32 n 2 i n3 i 1 16 n4 1 n2 n2 n 4
16 n 3 i n4 i 1 1
2
4 n n 4 Area
nn 2 1 n 10
1 16 2 3 32 3
nn 1 n2 2 3
n
lim T n
186
Chapter 4
Integration
57. f y Sn
3y, 0 y 2
n
Note: y
n
2 n 2 n
0
n
2 n 2i n 1 n 2 n 6 6 n
4
y
f mi
i 1
y
i 1
f 12 n2
2i n 2 6 n
3
i 1
2
12 n i n2 i 1 Area
n
nn
1
6n
x
2 4 6
lim S n
n
lim 6
6
2
59. f y Sn
y 2, 0 y 3
n
Note: y 3 n
n i 1
3 n
2
0
3 n
6 4 2
y
f
i 1
3i n nn
3i n 1 27 2n
3 n
27 n 2 i n3 i 1 3n 2 9 3 n 1 2 n
10 8 6
27 n3 Area 61. g y Sn
i n 1 n
1 2n 6
n
9 2n2 n2 9 2n2
1
9
27 2n
9 2n2
x 2 2 4 4 6 8 10
lim S n
lim 9
4y2
n
y3, 1 y 3. g1 41 2i n 2i n 4i n 10i n 6 2 n
2
Note: y
y
1 4i2 n2 4i2 n2 10
i
1
2i n 1 8i3 n3 8 3 4
3
2 n 6i n 2 3n n 4 44 3 65. f x Let ci 12i2 n2 8i3 n3 4 nn n2
2 4 2 2 4 x
2n 41 ni 1 2n 3 ni 1 Area 63. f x Let ci x
n n
10 n n 1 n 2
1 2n 6
1
8 n2 n 1 n2 4
2
lim S n
x2 xi
3, 0 x 2, n xi 2
1
tan x, 0 x xi xi 2 , c1 x
i 1
4
,n
4
. 1 ,c 42
4
. , c2
4
1 ,c 21 f ci
i 1
3 ,c 43 ci2 3 3
5 ,c 44 1 2 25 16
7 4 Area
x
n
16
32
3 ,c 32 3 tan ci 16 tan 5 32
5 ,c 32 4
7 32
Area 1 2 69 8 67. f x n
x
i 1
f ci
i 1
1
1 16
3
9 16
3
49 16
3 16
tan
32
tan
3 32
tan
7 32
0.345
x on 0, 4 . 4 5.3838 8 5.3523 12 5.3439 16 5.3403 20 5.3384
Approximate area Exact value is 16 3
S ection 4.2
Area
187
69. f x n
tan
x on 1, 3 . 8 4 2.2223 8 2.2387 12 2.2418 16 2.2430 20 2.2435
Approximate area
71. We can use the line y x bounded by x a and x b. The sum of the areas of these inscribed rectangles is the lower sum.
y
The sum of the areas of these circumscribed rectangles is the upper sum.
y
x a x a b b
We can see that the rectangles do not contain all of the area in the first graph and the rectangles in the second graph cover more than the area of the region. The exact value of the area lies between these two sums. 73. (a)
8 6 4 2 x 1 2 3 4 1 2 3 4
y
(b)
8 6 4 2
y
x
Lower sum: s4 0 (c)
8 6 4 2
4
51 3
6
151 3
46 3
15.333
Upper sum: 4 S4
51 3
6
62 5
2111 15
326 15
21.733
y
(d) In each case, x 4 n. The lower sum uses left endpoints, i 1 4 n . The upper sum uses right endpoints, i 4 n . The Midpoint Rule uses midpoints, i 1 4 n . 2
x 1 2 3 4
Midpoint Rule: M4 22 3 (e) n sn Sn Mn 4 15.333 21.733 19.403
44 5
55 7 8 17.368 20.568 19.201
62 9
6112 315
19.403 100 18.995 19.251 19.125 200 19.06 19.188 19.125
20 18.459 19.739 19.137
(f) s n increases because the lower sum approaches the exact value as n increases. S n decreases because the upper sum approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the exact value, whereas the upper sum is always larger.
188
Chapter 4
Differentiation
75.
y
77. True.
(Theorem 4.2 (2))
4 3 2 1 1 2 3 4 x
b. A 79. f x
6 square units sin x, 0,
y
2
1.0 0.75 0.5 0.25
sin x, the x-axis, x 0 and x 2. Let A2 area of the rectLet A1 area bounded by f x angle bounded by y 1, y 0, x 0, and x 2. Thus, A2 21 1.570796. In this program, the computer is generating N2 pairs of random points in the rectangle whose area is represented by A2. It is keeping track of how many of these points, N1, lie in the region whose area is represented by A1. Since the points are randomly generated, we assume that A1 A2 N1 A1 N2 N1 A. N2 2
f (x) = sin(x)
( , 1) 2
4
2
x
The larger N2 is the better the approximation to A1. 81. Suppose there are n rows in the figure. The stars on the left total 1 n n 1 stars in total, hence 21 1 2 2 ... ... n n nn
1 2
2
...
n, as do the stars on the right. There are
1 1.
nn
83. (a) y
4.09
10
5
x3
0.016x2
2.67x
452.9
(b)
500
(c) Using the integration capability of a graphing utility, you obtain A 76,897.5 ft2.
0 0 350
Section 4.3
1. f x xi lim
i 1
Riemann Sums and Definite Integrals
0, x 1 n2
n 2
x, y 3i2 n2
n
0, x 3 2i n2
n
3, ci 1
3i2 n2
3 2
y
3i
y=
x
n
f ci xi
lim
i 1
3i2 3 2i n2 n2 i
1
1
...
3 n2 3(2)2 . . . 3(n 1)2 3 n2 n2 x
3 3n lim 2i2 n n3 i 1
n
lim
3 3 nn 2 n3 n
1 2n 6 1 2n 3n2 23 1
1 n
nn 2 1 2n2
1
n
lim 3 3 2 3 0
33
3.464
S ection 4.3
Riemann Sums and Definite Integrals
189
3. y
6 on 4, 10 .
n n
Note: x f4
i 1
10 n 6 n
4
n
6 , n 6 6 n
0 as n
n i 1
f ci
i 1 10
xi 6 dx
6i n 36
i
1
36 n
36
4
n
lim 36
5. y
x3 on
n
1, 1 .
n
Note: x f
i 1
1 n 2 n 24 n3 i 1 n
n
1
n
2 , n 1
0 as n 2i n
n 3
f ci
i 1
xi 2 2
1 12 n2 i
n
2i n i
1
i
1
2 n
n
1
i 1
6i n
12i 2 n2
8i 3 n3
2 n
i2
1
16 n4 i 3 n
i3
1
61 2 n 0
42
1 n2
41
2 n
1 n2
2 n
1
x3 dx
1
n
lim
7. y
x2
n
1 on 1, 2 .
n
Note: x i n 1 n
2 n
1
1 , n 1 i n
0 as n
2
n i 1
f ci
i 1
xi
i 1
f1
1 1 n
1 n 1 2 6
n
1
i 1
2i n 1 n2
i2 n2 10 3
1 3 2n
1 n 1 6n2
2
2
2n i n2 i 1 10 3
5
1n2 i n3 i 1 3 2n 1 6n2
2 10 3
1
3 n
x2
1
1 dx
n
lim
n
n
3
9. lim
0 i
3ci
1
10
xi
1
3x
10 dx
11. lim
0 i
ci2
1
4 xi
0
x2
4 dx
on the interval
5
1, 5 .
4
on the interval 0, 3 .
2
13.
0
3 dx
15.
4
4
x dx
17.
2
4
x2 dx
2
19.
0
sin x dx
21.
0
y 3 dy
23. Rectangle A A
0
y
bh
3
34 4 dx 12
5
3 2 1
Rectangle
x
1 2 3 4 5
190
Chapter 4
Integration
25. Triangle A A
0
y
1 bh 2
4
1 44 2 8
4
27. Trapezoid b1 b2 h A 2
Triangle
2
y
5 2 5 dx
9
2
9
6
x dx
2
A
0
x
2 4
2x
14
3
Trapezoid
x
1 2 3
29. Triangle A A
1
y
31. Semicircle
Triangle
y
1 bh 2
1
1 21 2 1 x dx 1
1
1
A
x
12 r 2
3
1 2 9
4
Semicircle
3
2
1
A
3
x2 dx
9 2
x
4 2 2 4
4
4
4
In Exercises 3339,
2
x3 dx
60,
2
x dx
6,
2
dx
2
2
4
4
4
33.
4
x dx
2
x dx
6
35.
2 4
4x dx
4
2
x dx
46
4
24
4 4
4
4
4
37.
2
x
8 dx
2
x dx
8
2
dx
6
82
10
39.
2
13 x 2
3x
2 dx
1 2
x3 dx
2
3
2
x dx 22
6
2
2
dx
1 60 2
7 5 7 6 6
36 f x dx
16
41. (a)
0
f x dx
0 0
f x dx
5 5
f x dx 10
10
3
13
43. (a)
2
fx
g x dx
2
g x dx
2
(b)
5 5
f x dx
0
f x dx 0
5
10
6 6
2 g x dx
8
6
(b)
2
gx
f x dx
2
f x dx
2
(c)
5
f x dx
5
2 3f x dx 3
0
10 2 2
12 4 30
(d)
0
f x dx
3 10
30
6
6
(c)
2
2g x dx
6
2
2
g x dx
6
(d)
2
3f x dx
3
2
f x dx
3 10
45. (a) Quarter circle below x-axis: (b) Triangle: 1 bh 2 (c) Triangle
1 2
1 4
r2
1 4
2
2
42
4
1 2
Semicircle below x-axis: 1 2
21 3 1 20 2
1 2
2
2
1
2
(d) Sum of parts (b) and (c): 4
(e) Sum of absolute values of (b) and (c): 4 (f) Answer to (d) plus 2 10 20: 3 2
2 23
5 2
2
47. The left endpoint approximation will be greater than the actual area: >
49. Because the curve is concave upward, the midpoint approximation will be less than the actual area: <
S ection 4.3
Riemann Sums and Definite Integrals
191
51. f x
1 x 4
53.
4 3 2 1
y
is not integrable on the interval 3, 5 and f has a discontinuity at x 4.
1
2
3
4
x
a. A
5 square units
55.
2
3 2
y
1
1 2
x
1 2
1
3 2
2
1
d.
0 3
2 sin
x dx
1 12 2
1
57.
0
x3 n Ln Mn Rn
2
x dx 4 3.6830 4.3082 3.6830 8 3.9956 4.2076 3.9956 12 4.0707 4.1838 4.0707 16 4.1016 4.1740 4.1016 20 4.1177 4.1690 4.1177
59.
0
sin2 x dx n Ln Mn Rn 4 0.5890 0.7854 0.9817 8 0.6872 0.7854 0.8836 12 0.7199 0.7854 0.8508 16 0.7363 0.7854 0.8345 20 0.7461 0.7854 0.8247
61. True
63. True
65. False
2
x dx
0
2
67. f x x0 x1 c1
x2 0, x1
3x, 0, 8 1, x2 3, x3 7, x4 4, x4 8 f2 10 2 x2 f5 40 4 x3 f8 x4 272 1 8
1, x2 1, c2
4
2, x3 2, c3 x 5, c4 f1 41
f ci
i 1
x1
88 1
192
Chapter 4
Differentiation
69. f x
1, 0,
x is rational is x irrational
0, f ci 1 or f ci 0 in each subinterval since there is not integrable on the interval 0, 1 . As are an infinite number of both rational and irrational numbers in any interval, no matter how small. 71. Let f x
n
x2, 0 x 1, and xi
n
1 n. The appropriate Riemann Sum is
f ci
i 1
xi
i 1
i n 32
2
1 n
1n2 i. n3i 1 ... n2
n
1 lim 3 12 n n
22
lim lim
1 n3 2n2
n 2n 3n 6n2
1n 6 1
1 1 3 1 2n 1 6n2 1 3
n
n
lim
Section 4.4
1. f x 4
0
The Fundamental Theorem of Calculus
5
4 x2 x2 1 1 dx is positive.
5
3. f x
2
x x2 x x2
1 0
5
5
1 dx
5
5
2
5
2
1
1
0
5.
0
2x dx
x2
0
1
0
1
7.
1
x
2 dx
x2 2
0
2x
1
0
1 2
2
5 2
1
9.
1
t2
2 dt
t3 3
1
1
2t
1
1 3
2
1 3 43 t 3
2
10 3
1
1
11.
0 2
2t
1 2 dt
0
4t2
4t
1 dt
2t2
t
0
4 3 1 2 2 3
2
1
1 3
13.
1
3 x2 u
1 dx
3 x
4
2
x
1
3 2
2
3
1
4
15.
1 1
2 du u
3
u1
1
2
2u
12
du
23 u 3 3 4 1 x2 32
4 2
4u1
2 1
4
3
44
2 3
4
2 3
17.
1 1
t
2 dt
34 t 4 1 3
1
1 3
2t
1
2
3 4 23 x 3 3 4
1 2 0
2
4
19.
0 0
x 3
x
dx
x
0
x1
2
dx
11 32 3 5 27 20
2 3
1 18
21.
1
t1
3
t2
3
dt
34 t 4
32
3
35 t 5
0 3 1
0
3
3
23.
0
2x
3 dx
0
3 3x x2
2x dx
32 32
2x
3
3 dx 9 2
split up the integral at the zero x 9 4 0 9 9 9 4 9 2
3 2 2 9 2 9 4 9 2
x2
0
3x
32
S ection 4.4
The Fundamental Theorem of Calculus
193
3
2
3
25.
0
x2
4 dx
0
4 4x 8 23 3
x 2 dx
2
x2 x3 3 4x
4 dx
3 2
x3 3 8 3
2 0
9
12
8 3
8
27.
0
1
sin x dx
x
cos x
0
1
0
1
2
6
6
29.
6
sec2 x dx
tan x
6
3 3
3
3 3
23 3
3
31.
3
4 sec
tan d
4 sec
3
42
42
0
3
33.
0
10,000 t
3
6 dt
10,000
3
t2 2
3
1
6t
0
$135,000
35. A
0
x
x2 dx
x2 2 12 3 5
x3 3
1 0
1 6
37. A
0
3
x
x dx
0
3x1
2
x3
2
dx
2x3
2
25 x 5
3 2 0
xx 10 5
3
2x
0
2
2
39. A
0
cos x dx
sin x
0
1
41. Since y 0 on 0, 2 ,
2 2
A
0
3x2
1 dx
x3
x
0
8
2
10.
43. Since y 0 on 0, 2 ,
2
2
45.
0
x
2 x dx fc 2 c 0 2c 1 1 c
2
x2 2 6 3 3 6
4x3 3 82 3 42 3 42 3 42 3 6
22
2
0
82 3
A
0
x3
x dx
x4 4
x2 2
2
4
0
2
6.
c
2c c
1
1 c c
42 3 6 42 3
2
1
0.4380 or c
1.7908
194
Chapter 4
Integration
4
4
47.
4
2 sec2 x dx fc
2 tan x
4
21
2
1
4
4
4 2 sec2 c sec2 c sec c c
4 8 4 2 2
arcsec arccos
2 13 x 3
0.4817
2 2
49.
1 2 2
2
4
2
x2 dx 8 3
1 4x 4
1 4
8
8 3
8
8 3
8 3
(
2 3
3
,8
3
(
5
(
2 3
3
,8
3
(
Average value 4 x2
3
3 1
8 when x2 3
4
8 or x 3
23 3
1.155.
51.
1 0
0
sin x dx 2 2
1
cos x
0
2
2 2 (0.690, ( 2 (2.451, (
53. If f is continuous on a, b and F x
b
f x on a, b ,
Average value sin x x
2
then
a
f x dx
Fb
Fa.
2
1
3 2
0.690, 2.451
6
2
6
55.
0 6
f x dx
area of region A
6 6
1.5
57.
0
f x dx
0
f x dx
2
f x dx
1.5
5.0
6.5
59.
0
2
f x dx
0
2 dx
0
f x dx 15.5 (b) 1 3
3
12 61. (a) F x F0 Fx k sec2 x k 500
3.5
0
500 sec2 x dx
0
1500 1500
3
tan x
0
500 sec2 x
3
0
826.99 newtons 827 newtons 1 5 0
5
63.
0.1729t
0
0.1552t2
0.0374t3 dt
1 0.08645t2 5
5
0.05073t3
0.00935t4
0
0.5318 liter
S ection 4.4
The Fundamental Theorem of Calculus
195
65. (a)
0
1
(b)
24
10
1
0 0
24
The area above the x-axis equals the area below the x-axis. Thus, the average value is zero. 67. (a) v (b) 8.61
90
The average value of S appears to be g.
10 4t3
0.0782t2
0.208t
0.0952
10 10
70
60
(c)
0
v t dt
8.61 4 t2 2 8 25 2 8
10 4t 4
0.0782t3 3 x2 2
0.208t 2 2
60
0.0952t
0
2476 meters
x
x
69. F x
0
t 4 2 25 2 64 2
5 dt 52 55 58
5t
0
5x
71. F x
10 2 dv 1v 10 x 10 5 8 35 4
x
x
10v
1
2 dv
10 v
x 1
F2 F5 F8
10 1
1 x
F2 F5 F8
10 10 10
1 2 4 5 7 8
x
x
x
73. F x
1
cos d sin 2 sin 5 sin 8
x 3 8
sin
1
sin x
sin 1
75. (a)
0
t d 12 x dx 2
2 dt 2x
t2 2 x
x
2t
0
12 x 2
2x
F2 F5 F8
sin 1 sin 1 sin 1 34 t 4
x 3 8
0.0678 1.8004 0.1479 34 x 4 x1
3 3
(b)
2
77. (a) (b)
t dt
3
3
16
34 x 4
x 3
x
12
79. (a)
x4
sec2 t dt d tan x dx 1
tan t
x4
tan x
1
d 34 x dx 4
x
12
x
x
(b)
sec2 x
x
81. F x
2
t2 x2 2x
2t dt
83. F x
1
t4 x4 1
1 dt
85. F x
0
t cos t dt x cos x
Fx
Fx
Fx
196
Chapter 4
Integration
x
2
87. F x
x
4t
x
1 dt
2
Alternate solution:
x 2
2t
2
t
x
Fx
x 0
4t 4t
x x
1 dt
x 2
2x 8x Fx 8
2 10
2
x
2
2x2
x
1 dt
0 x
4t
2
1 dt 4t 1 dt 8
4t
0
1 dt
0
Fx
sin x
4x
x3
1
4x
2
1
89. F x
0
t dt sin x
1 2 cos
23 t 3 x
2
sin x 0
2 sin x 3
32
91. F x
0
sin t 2 dt sin x3
2
Fx
cos x sin x
Fx
3x2
3x2 sin x 6
Alternate solution
sin x
Fx
0
t dt sin x
x
Fx 93. g x
0
d sin x dx
sin x cos x
x
f t dt 0, g 1 1 ,g 2 2 1, g 3 1 ,g 4 2 0
95. (a) C x
5000 25 5000 25 5000 25
3
0
t1 4 dt 45 t 5
4 x 4 0
g0
y 2 1
3
12 5 x 5
1000 125 $137,000
54 54
12x5
4
f
(b)
g
x 1 2 3 4
C1 C5
1000 125 1000 125 1000 125
12 1 12 5
$214,721 $338,394
1 2
C 10
12 10
g has a relative maximum at x 97. True
2.
1 0 1
99. False;
1
x
2 dx 1
x
2 dx 0
x
2 dx 2
Each of these integrals is infinite. f x has a nonremovable discontinuity at x
1x
x 0.
101. f x
0
1 t
2
x
1
dt
1 1
2 0t
dt
By the Second Fundamental Theorem of Calculus, we have fx 1 1x 1 Since f x
2
1 x2
1 x2 1 x2 1
1 x2 0. 1
1
0, f x must be constant.
S ection 4.5 103. x t xt t3 3t 2 3 t2 6t 2 12t 4t 3t
5
Integration by Substitution
197
9t 9 3 1
2
3t Total distance
x t dt
0 5
3t
0 1
3t 4t 20
1 dt
3 5
3
0
t2 4
3 dt
3
1
t2
4t
3 dt
3
3
t2
4t
3 dt
4
28 units
4
105. Total distance
1 4
x t dt
v t dt
1 4 1
1 t
4 2 1
dt
2t1 22
1
2 units
Section 4.5
f g x g x dx
Integration by Substitution
u gx du g x dx
1. 3. 5.
5x2 x x
2
1
2
10x dx
5x2 x2 tan x
5
1 1
10x dx 2x dx sec2 x dx
1
dx
tan2 x sec2 x dx
7.
1 Check:
2x 4 2 dx d 1 2x dx 5 x2
12
1 5
5
2x
C 21 x2 32 C
32
C 9
2x
4
9.
9 Check:
2x dx x2
32
C 2 3 3 9 2
2 9 3 x2
12
x2
32
C 9 x2 2x
d2 9 dx 3
2x
198
Chapter 4
Integration
11.
x3 x4 Check:
3 2 dx
1 4
3
x4 C
3
2
4x3 dx
2
1 x4 3 4 3 4x3
3
C 3
2
x4 12 x3
3
3
C
d x4 3 dx 12 1 3
3 x4 3 12
x4
13.
x2 x3 Check:
1 4 dx
x3
5
1
4
3x2 dx
1 x3 1 3 5
5
C 1
4
x3 15
1
5
C
d x3 1 dx 15
C
5 x3
1 4 3x2 15
x2 x3
15.
t t2 Check:
2 dt d dt t2
1 2 2 3
t2
32
2 C
12
2t dt 3 2 t2
1 t2 2 2 32 2 3
12
32
C t2 2
t2 3
1 2t
2
32
C
2t
17.
5x 1 Check:
x2 d dx
13
dx
5 2 x2
1
43
x2 C
13
2x dx 15 8 4 1 3
5 2 x2
1
x2 43 2x
43
C 5x 1 x2
15 1 8
13
x2
43
C x2
15 1 8 1 2
13
5x 3 1
19.
x 1 Check: x2 3
dx
1 C
x2
3
2x dx 21 x2
11 2
3
x2 2 2x
2
C x 1 x2
3
1 41 x2
2
C
1 d dx 4 1 x2 1 3 1 31
2
1 4
21.
1
x2 dx x3 2 d dx
1
x3 C
2
3x2 dx 1 3
11 3 11 x3
x3 1
2
1
C 3x2
1 31
2
x3
C
Check:
x3
x2 1 x3
12
23.
x 1 Check: x2 d dx 1 t
3
dx 1
1 2 x2
1
12
x2 C
12
2x dx 1 1 2 x2
1 1 x2 2 12
12
C x 1
4
1
x2
C
2x
x2
25.
1 Check:
1 dt t2 1 1t 4
4
1 C
1 t
3
1 dt t2 1 41 4 1 t
1
3
1t 4 1 t2
C 1 1 t2 1 t
3
d dt
27.
1 dx 2x Check: d dx
1 2 2x
2x C
12
2 dx 1 2x 2
1 2x 1 2 2 12
12
C 1 2x
2x
C
2
S ection 4.5
Integration by Substitution
199
29.
x2
3x x
7
dx
2
x3 2x3
2
2
3x1 14x1
2
7x C
12
dx
25 x 5 3x x
2
2x3
2
14x1
2
C
2 5
x x2
5x
35
C
Check:
d 25 x dx 5 2 dt t d 14 t dt 4
2
x2
7
31.
t2 t Check:
t3 t2 C
2t dt t3
14 t 4 2t
t2 t2 t
C 2 t
33.
9 Check:
y
y dy d 23 y dy 5
2
9y1 15 y
2
y3 C
2
dy
9
23 y 3
2
2
25 y 5
2
2
C 9y1
23 y 5
2
2
15
2
y 9
C y y
d 6y3 dy
25 y 5
C
y3
35. y
4x 4 x dx 4 x2 2 2
4x dx 16 x2 2 16 16 x2
12 12
37. y 2x dx C 1 2
x x2 x2 2x
1 3
2
dx
2
2x 2x 1 1 2x
3 3
2x C C
2 dx
x2 12 x2
1 x2 2 2 x2 dy dx y
1
2x2
4 16
C
3
39. (a)
3
y
(b)
x4 x4 1 2 2 4 3
x2, 2, 2 x2 dx x2 1 4 3 1 4 3
2
1 2
32
4
x2
12
2x dx x2
32
x 2 1 2
C
32
1 4 3 CC 2
C
2, 2 : 2 y
22 x2
2
32
2
2
1
41.
sin x dx
cos x
C
43.
sin 2x dx
1 2
sin 2x 2x dx
1 cos 2x 2
C
45.
1
2
cos
1
d
cos
1
1
2
d
sin
1
C
200
Chapter 4
Integration
47.
sin 2x cos 2x dx sin 2x cos 2x dx sin 2x cos 2x dx
1 2 1 2 1 2
sin 2x 2 cos 2x dx cos 2x
1 sin 2x 2 2
2
C
2
12 sin 2x 4 C1 C2
C
OR C1 OR
2 sin 2x dx 1 2
1 cos 2x 2 2 sin 4x dx
1 cos2 2x 4
2 sin 2x cos 2x dx
1 cos 4x 8
49.
tan4 x sec2 x dx
tan5 x 5
C
15 tan x 5
C
51.
csc2 x dx cot3 x
cot x cot x 2
2
3
csc2 x dx C 1 2 cot2 x C 12 tan x 2 C 1 sec2 x 2 1 C 1 sec2 x 2 C1
53.
cot2 x dx
csc2 x
1 dx
cot x
x
C
55. f x Since f 0 fx
cos
x dx 2 3
2 sin 2 sin 0
x 2
C C, C 3. Thus,
2 sin
x 2
3.
57. u
x
2, x xx
u 2 dx
2, dx u u3 25 u 5
2 2
du 2 u du 2u1 43 u 3
2 2
du C C 2 4 10 C C
2u3 2 3u 15 2 x 15 2 x 15 59. u 1 x, x 1 x dx u, dx 1 u1 23 u 3
2
10
32
2 2
3x 3x
32
du u
2
x2 1
u du
2
2u3 45 u 5
u5
2
du
2
2
2
27 u 7
C C x 8 15 1 C x
2
2u3 2 35 105 2 1 105 2 1 105 x x
42u
32
15u2 42 1 12x
35 15x2
C
32
S ection 4.5
Integration by Substitution
201
61. u
2x
1, x
1 u 2
1 , dx 12 u 1 u 8 1 8
1 du 2 1 u u2 2u1 43 u 3 10u
2
x2 1 dx 2x 1
11 du 2 1 3u 6u1 45 1
2 12
12
2u
2
4 du du C C 10 2x 52 13 1 C C 45 C
u3
2
125 u 85
2
2
2
u1 2 2 3u 60
2x 1 3 2x 60 1 60 1 15 63. u x x 1, x 1 u x x 1 1, dx dx 2x 2x 1 12x2 1 3x2
8x 2x
du u u u u 1 u u x x x where C1 u 2u1 2u 1 2x 2x 1 1 du u 1 u
12
u 1 du C C
1
du
2
2x 1 1 C. 1
1
C C
C1
65. Let u
1
x2 x x2
1
1, du 1 3 dx
2x dx. 1 2
1
x2
1
1
3
2x dx
12 x 8
1
1
4 1
0
67. Let u
2
x3
1, du 1 dx
3x2 dx 2 1 3
2
2x2 x3
1
x3
1
1
12
3x2 dx
2 x3 1 3 32 43 x 9 4 27 9 1
32 2 1 2 32 1
22
12
8 9
2
202
Chapter 4
Integration
69. Let u
4 0
2x 1 2x
1, du 1 dx
2 dx. 1 2
4 4
2x
0
1
12
2 dx
2x
1
0
9
1
2
71. Let u
9 1
1 1 x1 x, x 1, u x
1
x, du
1 dx. 2x
9
x 2
2
dx
2
1
1
x
2
1 2x
dx
2 1 x
9 1
1 2
1
1 2
73. u
2
u, dx 1. When x
0
du 2, u 2
1
When x
2
0.
0
1
2
x dx
u
1
u du
1
u3
2
u1
2
du
25 u 5
2
23 u 3
0 2 1
2 5
2 3
4 15
2
75.
0
cos
2 x dx 3 1, x 0, u
7
3 2 sin x 2 3 u 1, dx 1. When x
2 0
3 2
3 2
33 4
77. u
x
du 7, u
8
When x Area
8. 1
3 3
x3 x
0
1 dx
1 8
u u4
1
u du
3
u1
du
37 u 7
3
34 u 4
8 3 1
384 7
12
3 7
3 4
1209 28
79. A
0
2 sin x
sin 2x dx
2 cos x
1 cos 2x 2 1 dx 2
7
4
0
2
3
81. Area
2
sec2
x dx 2
2
3
2
2
sec2
x 2
2 tan
x 2
2 2
3
2
3
1
4
83.
0
x 2x
3
1
dx
3.333
10 3
85.
3
xx
15
3 dx
28.8
144 5
3
87.
0
5
cos
6
d
7.377
1 1
5 0 0 8
1 1
4
89.
2x 2x
1 2 dx 1 2 dx
1 2
2x 4x2
1 2 2 dx 4x 1 dx C1
1 2x 6 43 x 3 1 . 6
1
3
C1 x
43 x 3 C2
2x2
x
1 6
C1
2x2
They differ by a constant: C2
S ection 4.5
Integration by Substitution
203
91. f x
2
x2 x2 x2 x2
1 is even.
2
93. f x x2 dx 8 3 272 15 2 x5 5 x3 3
2 0 2
x x2 x x2
1 3 is odd. 0
1 dx
2
0
x4 32 5
1 3 dx
2
2
2
2
95.
0
x2 dx
0
x3 3
2 0 2
8 ; the function x2 is an even function. 3 x2 dx
0 2
(a)
2 2
x2 dx x2 dx
0
8 3 8 3
4 4
2
2
(b)
2 0
x2 dx 3x2 dx
2
2
0
x2 dx
2
16 3 8
(c)
x2 dx
0
(d)
3
0
x2 dx
4
4
4
97.
4
x3
6x2
2x
3 dx
4
x3
2x dx
4
6x2
3 dx
0
2
0
6x2
3 dx
2 2x3
3x
0
232
99. Answers will vary. See Guidelines for Making a Change of Variables on page 292.
2
101. f x
x x2
1 2 is odd. Hence,
2
x x2
1 2 dx
0.
103.
dV dt Vt V0 V1
k t t k 1 k 2 1 k 1 C C
2 2
dt
k t 1
C
500,000 400,000 200,000 and
Solving this system yields k C 300,000. Thus, Vt When t
b
200,000 t1 4, V 4
300,000. $340,000.
105.
1 b (a) (b) (c) a
74.50
a
43.75 sin 262.5 262.5 cos cos cos t 6 t 6
t dt 6
3 0 6 3 12 0
1 b a
74.50t 262.5
262.5
cos
t 6
b a
1 74.50t 3 1 74.50t 3 1 74.50t 12
1 223.5 3 1 447 3 1 894 12
102.352 thousand units 223.5 262.5 102.352 thousand units 74.5 thousand units
262.5 262.5
262.5
t 6
204
Chapter 4 1 b (a) (b) a
b
Integration 1 b a 1 cos 60 t 30
1 60
107.
2 sin 60 t
a
cos 120 t
dt
1 120 1 30 0 1
b
sin 120 t
a
1 1 60 1 1 240
0 0
1 cos 60 t 30 1 cos 60 t 30
1 sin 120 t 120 1 sin 120 t 120
60
0 1 240
1 30 1 120 1.382 amps 1 30
4
1.273 amps 1 30
240
0
30 2 22
2 1 1 30 1 cos 60 t 30 1 sin 120 t 120
1 30
5 1 30
(c)
0
30
0
0 amps
109. False 2x 111. True
10 10 10 10
1 2 dx
1 2
2x
1 2 2 dx
1 2x 6
1
3
C
ax3
10
bx2
cx
d dx
10
ax3
cx dx
10
bx2
d dx Even
0
2
0
bx2
d dx
Odd 113. True 4 sin x cos x dx 2 sin 2x dx cos 2x C
115. Let u
b
x fx
a
h, then du
b
dx. When x
h b
a, u
h
a
h. When x
b, u
b
h. Thus,
h dx
a h
f u du
a h
f x dx.
Section 4.6
2
Numerical Integration
x2 dx
0 2
1. Exact: Trapezoidal:
0 2
13 x 3 1 0 4 1 0 6
2 0
8 3 2 4 1 2 1 2
2
2.6667 21
2 2
x2 dx x2 dx
0
2 4
3 2 3 2
2
2
2
2
11 4 8 3
2.7500 2.6667
Simpsons:
21
2
2
2
2
3. Exact:
0 2
x3 dx x3 dx
0 2
x4 4 1 0 4 1 0 6
2
4.000
0
Trapezoidal: Simpsons:
0
2 4
1 2 1 2
3
21
3
3
2 4
3 2 3 2
3
2
3
3
17 4 24 6
4.2500 4.0000
x3 dx
21
3
2
3
S ection 4.6
Numerical Integration
205
2
5. Exact:
0 2
x3 dx x3 dx
0 2
14 x 4 1 0 8 1 0 12
2
4.0000
0
Trapezoidal: Simpsons:
0
2 4
1 4 1 4
3
2
3
2 4 2 2 4
3
2
3
3 4 4 3 4
3
21
3
3
2
3
5 4 4 5 4
3
2
3
6 4 2 6 4
3
2
3
7 4 4 7 4
3
8
3
4.0625 8 4.0000
x3 dx
21
9
7. Exact:
4 9
x dx x dx
4
23 x 3 5 2 16
9 2 4
18 2 37 8
16 3 2
38 3 21 4
12.6667 2 47 8 2 26 4 2 57 8 2 31 4 2 67 8 3
Trapezoidal:
12.6640
9
Simpsons:
4
x dx
5 2 24
4
37 8
21
4
47 8
26
4
57 8
31
4
67 8
3
12.6667
2
9. Exact:
1 2
1 x 1 x 1
2
1
2
dx dx
1 x 11 84 11 84 1 2 32 81 4
2 1
1 3 1 54 8 25 1 54 8 25 1
1 2
2
1 6 2 1 9 2 1 9
0.1667 1 32 1
2
Trapezoidal:
1
2
1 74 1
2
1 9
32 121 1
2
0.1676 1 32 1
2
2
Simpsons:
1
1 x 1
2
dx
11 12 4 11 12 4
4
1 74 1
2
1 9
64 81
64 121
0.1667
2
11. Trapezoidal:
0 2
1 1
0
x3 dx x3 dx
1 1 4 1 1 6
21 41
18 18
22 22
21 41
27 8 27 8
3 3
3.283 3.240
Simpsons:
Graphing utility: 3.241
1
1
13.
0
x1
x dx
0 1
x1
x dx 1 0 8 1 0 12 1 1 4 1 1 4 1 4 1 4 1 1 2 1 1 2 1 2 1 2 3 1 4 3 1 4 3 4 3 4
Trapezoidal:
0
x1
x dx
2
2
2
0.342
1
Simpsons:
0
x1
x dx
4
2
4
0.372
Graphing utility: 0.393
206
Chapter 4
Integration
2
15. Trapezoidal:
0
cos x2 dx
2 8 0.957
cos 0
2 cos
2 4
2
2 cos
2 2
2
2 cos
2 4
2
2
cos
2
2
Simpsons:
0
cos x2 dx
2 12 0.978
cos 0
4 cos
2 4
2
2 cos
2 2
2
4 cos
2 4
2
2
cos
2
Graphing utility: 0.977
1.1
17. Trapezoidal:
1 1.1
sin x2 dx sin x2 dx
1
1 sin 1 80 1 sin 1 120
2 sin 1.025
2
2 sin 1.05
2
2
2 sin 1.075
2
2
sin 1.1
2
2
0.089
2
Simpsons:
4 sin 1.025
2 sin 1.05
4 sin 1.075
sin 1.1
0.089
Graphing utility: 0.089
4
19. Trapezoidal:
0 4
x tan x dx x tan x dx
0
32 48
0 0
2 4
16 16
tan tan
16 16
2 2
2 2 tan 16 16 2 2 tan 16 16
2 4
3 3 tan 16 16 3 3 tan 16 16
4 4
0.194 0.186
Simpsons:
Graphing utility: 0.186 x3 3x2 6x 6 0 20 12 42
3
21. (a)
y
23.
y = f ( x)
fx fx fx fx f
4
x
x
a
b
(a) Trapezoidal: Error
12
0.5 since 2.
The Trapezoidal Rule overestimates the area if the graph of the integrand is concave up.
f x is maximum in 0, 2 when x (b) Simpsons: Error f
4
20 0 180 44
5
0 since
x
0.
25. f x
2 in 1, 3 . x3 1 and f 1 2. 366.
(a) f x is maximum when x Trapezoidal: Error f (b) f
4 4
23 12n2
2 < 0.00001, n2 > 133,333.33, n > 365.15; let n
x
24 in 1, 3 x5 1 and when f
4
x is maximum when x 25 180n4
1
24. 26.
Simpsons: Error
24 < 0.00001, n4 > 426,666.67, n > 25.56; let n
S ection 4.6
Numerical Integration
207
27. f x (a) f x
1
x 1 41 x
32
in 0, 2 . 0 and f 0 1 . 4 130.
f x is maximum when x Trapezoidal: Error (b) f
4
81 12n2 4 in 0, 2
< 0.00001, n2 > 16,666.67, n > 129.10; let n
x
4
16 1
15 x
72
f
x is maximum when x 32 15 180n4 16
0 and f
4
0
15 . 16 12.
Simpsons: Error
< 0.00001, n4 > 16,666.67, n > 11.36; let n
29. f x (a) f x
tan x2 2 sec2 x2 1 4x2 tan x2 in 0, 1 . 1 and f 1
3
f x is maximum when x Trapezoidal: Error (b) f
4
49.5305. 643.
10 12n2 3
49.5305 < 0.00001, n2 > 412,754.17, n > 642.46; let n 32x4 tan x2 1 and f
4
x
4
8 sec2 x2 12x2
36x2 tan2 x2 9184.4734.
48x4 tan3 x2 in 0, 1
f
x is maximum when x
1
Simpsons: Error
1 05 9184.4734 < 0.00001, n4 > 5,102,485.22, n > 47.53; let n 180n4 Cx
5
48.
31. Let f x
Ax3
Bx2
D. Then f 0
4
x
0.
Simpsons: Error
ba 0 180n4
Therefore, Simpsons Rule is exact when approximating the integral of a cubic polynomial.
1
Example:
0
x3 dx
1 0 6
4
1 2
3
1
1 4
This is the exact value of the integral. 33. f x n 4 8 10 12 16 20 2 Ln 12.7771 14.0868 14.3569 14.5386 14.7674 14.9056 3x2 on 0, 4 . Mn 15.3965 15.4480 15.4544 15.4578 15.4613 15.4628 Rn 18.4340 16.9152 16.6197 16.4242 16.1816 16.0370 Tn 15.6055 15.5010 15.4883 15.4814 15.4745 15.4713 Sn 15.4845 15.4662 15.4658 15.4657 15.4657 15.4657
208
Chapter 4
Integration
35. f x n 4 8 10 12 16 20
sin x on 0, 4 . Ln 2.8163 3.1809 3.2478 3.2909 3.3431 3.3734
2
Mn 3.5456 3.5053 3.4990 3.4952 3.4910 3.4888
Rn 3.7256 3.6356 3.6115 3.5940 3.5704 3.5552
Tn 3.2709 3.4083 3.4296 3.4425 3.4568 3.4643
Sn 3.3996 3.4541 3.4624 3.4674 3.4730 3.4759
37. A
0
x cos x dx 14 84 0.701
y
Simpsons Rule: n
2
x cos x dx
0
0 cos 0
4
28
cos
28
2
14
cos
14
4
3 3 cos 28 28
... 2
cos
2
1
1 2
4
2
x
5
39. W
0
100x 125
x 3 dx 12 x3 dx 5 0 3 12 400 400 5 12 15 12 125 125 5 12 15 12
3
Simpsons Rule: n
5
100x 125
0
200
3
10 12 0
125
10 12
3
...
10,233.58 ft
lb
12
41.
0
6 1 x2
dx
Simpsons Rule, n
6
1 0 2 36
6
4 6.0209 3.1416
2 6.0851
4 6.1968
2 6.3640
4 6.6002
6.9282
1 113.098 36 43. Area 1000 125 2 10
2 125
2 120
2 112
2 90
2 90
2 95
2 88
2 75
2 35
89,250 sq m
t
45.
0
sin x dx
2, n
10 2.477.
By trial and error, we obtain t