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C H A P T E R 4
Integration
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177
1. d
dx3
x3 C ddx3x3 C 9x4
9
x4 3.
d
dx1
3x3 4x C x2 4 x 2x 2
5.
Check: d
dtt3 C 3t2
y t3 C
dy
dt 3t2 7.
Check: d
dx2
5x52 C x32
y 2
5x52 C
dy
dxx32
Given Rewrite Integrate Simplify
9.3
4x 43 C
x43
43 Cx13dx3xdx
11. 2
x C
x12
12 Cx32dx 1
xxdx
13. 1
4x2 C
1
2x2
2 C1
2x3dx 12x3dx
15.
Check: d
dxx2
2 3x C x 3
x 3dx x2
2 3x C 17.
Check: d
dxx2 x3 C 2x 3x2
2x 3x2dx x2 x3 C
19.
Check: d
dx1
4x 4 2x C x3 2
x3 2dx 14x4 2x C 21.Check:
d
dx2
5x52 x2 x C x32 2x 1
x32 2x 1dx 25x52 x2 x C
23.
Check: d
dx3
5x53 C x23 3x2
3
x2
dx x
23
dx
x53
53 C3
5x53
C 25.
Check: d
dx1
2x2 C 1x3
1
x3dx x3
dx x2
2 C 1
2x2 C
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178 Chapter 4 Integration
27.
Check: x2 x 1
x
d
dx2
5x52
2
3x32 2x12 C x32 x12 x12
2
15x123x2 5x 15 C
2
5x52
2
3x32 2x12 Cx2 x 1
xdx x32 x12 x12dx
29.
Check:
x 13x 2
d
dxx3 1
2x2 2x C 3x2 x 2
x3 1
2x2 2x C
x 13x 2dx 3x2 x 2dx 31.Check:
d
dy2
7y72 C y52 y2y
y2ydy y52dy 27y72 C
33.
Check: d
dxx C 1
dx 1 dx x C 35.Check:
d
dx2 cosx 3 sinx C 2 sinx 3 cosx
2 sinx 3 cosxdx 2 cosx 3 sinx C
37.
Check: d
dtt csc t C 1 csc tcot t
1 csc tcot tdt t csc t C 39.Check:
d
dtan cos C sec2 sin
sec2 sin d tan cos C
41.
Check: d
dytany C sec2
y tan2
y 1
tan2y 1dy sec2ydy tany C 43.
2
2
3
3
2 2
x
3C
2C
0C
y
fx cosx
45.
Answers will vary.
5
4
3
32123
y
x
2x)x)
22x) )xf
ff
fx 2x C
fx 2 47.
Answers will vary.
3
4
3
2
231
2
x
y3
3
3
xxx)f
xx3
x)f )3
)
f
fx x x3
3 C
fx 1 x2 49.
y
x
2
x
1
1 12 1 C C 1
y 2x 1dx x2 x C
dy
dx 2x 1, 1, 1
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Section 4.1 Antiderivatives and Indefinite Integration 179
51.
y
sinx
4
4 sin 0 C C 4
y cosxdx sinx C
dy
dx cosx, 0, 4
53. (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
6dy
dx
1
2x 1, 4, 2
x
y
3 5
3
5
55.
fx 2x2 6
f0 6 202 CC 6
fx 4xdx 2x2 Cfx 4x,f0 6 57.
ht 2t4 5t 11
h1 4 2 5 CC 11
ht 8t3 5dt 2t4 5t Cht 8t3 5, h1 4
59.
fx x2 x 4
f2 6 C2 10 C
2 4
fx 2x 1dx x2 x C2fx 2x 1
f2 4 C1 5 C
1 1
fx 2 dx 2x C1f2 10
f2 5
fx 2 61.
fx 4x12 3x 4x 3x
f0 0 0 C2 0 C
2 0
fx 2x12 3dx 4x12 3x C2fx
2
x 3
f4 2
2 C
1 2 C
1 3
fx x32dx 2x12 C1 2x
C1
f0 0
f4 2
fx x32
63. (a)
(b) h6 0.7562 56 12 69 cm
ht 0.75t2 5t 12
h0 0 0 C 12 C 12
ht 1.5t 5dt 0.75t2 5t C
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180 Chapter 4 Integration
65. Graph of is given.
(a)
(b) No. The slopes of the tangent lines are greater than 2
on Therefore,fmust increase more than 4 units
on
(c) No, becausefis decreasing on
(d) fis an maximum at because and
the first derivative test.
(e) fis concave upward when is increasing on
and fis concave downward on Points
of inflection atx 1, 5.
1, 5.5,., 1f
f3.5 0x 3.5
4, 5.f5 < f4
0, 4.0, 2.
f4 1.0
ff0 4.
(f ) is a minimum at
(g)
x
2
2
4
4
6
6 82
6
y
x 3.f
67.
The ball reaches its maximim height when
s15
18
1615
8
2
6015
8
6 62.26 feet
t15
8seconds
32t 60
vt 32t 60 0
st 16t2 60t 6 Position function
s0 6 C2
st 32t 60dt 16t2 60t C
2
v0 60 C1
vt 32 dt 32t C1at 32 ftsec2 69. From Exercise 68, we have:
when time to reach
maximum height.
v0 187.617 ftsec
v02 35,200
v02
64
v02
32 550
s v032 16v0
322
v0 v032 550
tv0
32st 32t v
0 0
st 16t2 v0t
71.
f0 s0 C
2ft 4.9t2 v
0t s
0
ft 9.8t v0dt 4.9t2 v0t C2v0 v
0 C
1 vt 9.8t v
0
vt 9.8 dt 9.8t C1at 9.8 73. From Exercise 71,
(Maximum height when )
f 109.8 7.1 m
t10
9.8
9.8t 10
v0.v t 9.8t 10 0
ft 4.9t2 10t 2.
75.
since the stone was dropped,
Thus, the height of the cliff is 320 meters.
v20 32 msec
vt 1.6t
s0 320
s20 0 0.8202 s0 0
st 1.6tdt 0.8t2 s0v0 0.vt 1.6 dt 1.6t v0 1.6t,
a 1.6
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Section 4.1 Antiderivatives and Indefinite Integration 181
77.
(a)
(b) when or(c) when
v2 311 3
t 2.at 6t 2 03 < t < 5.0 < t < 1vt > 0
at vt 6t 12 6t 2
3t2 4t 3 3t 1t 3
vt xt 3t2 12t 9
0 t 5xt t3 6t2 9t 2 79.
at vt 1
2t32
1
2t32acceleration
xt 2t12 2 position function
x1 4 21 C C 2
xt vtdt 2t12 Ct > 0vt
1
t t12
81. (a)
(b)
s13 275
234132
2
250
3613 189.58 m
st at2
2
250
36t s0 0
a 550
468
275
234 1.175 msec2
550
36 13a
v13 800
36 13a
250
36
v0 250
36 vt at250
36
vt at C
at a constant acceleration
v13 80 kmhr 80 1000
3600
800
36msec
v0 25 kmhr 25 1000
3600
250
36msec 83. Truck:
Automobile:
At the point where the automobile overtakes the truck:
when sec.
(a)
(b) v10 610 60 ftsec 41 mph
s10 3102 300 ft
t 100 3tt 10
0 3t2 30t
30t 3t2
st 3t2Let s0 0.
vt 6tLet v0 0.
at 6
st 30tLet s0 0.
vt 30
85.
(a) (b)
(c)
The second car was going faster than the first until the end.
S230 1970.3 feet
S130 953.5 feet
In both cases, the constant of integration is 0 because S10 S
20 0
S2t V2tdt
0.1208t3
3 6.7991t2
2 0.0707t
S1t V1tdt 0.10683 t3 0.04162 t2 0.3679t
V2t 0.1208t2 6.7991t 0.0707
V1
t 0.1068t2 0.0416t 0.3679
1 mihr5280 ftmi
3600 sechr
22
15ftsec
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 95.33V2ftsec
V1ftsec
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87.
since
At the time of lift-off, and Since
7.45 ftsec2.
18,285.714 mihr2
1.4k 1602 k1602
1.4
v1.4k k1.4k 160t
1.4
k
k2t2 0.7,k2t2 0.7.kt 160
v0 s0 0.st k
2t2
vt kt
at k
89. True 91. True
93. False. For example, becausex3
3 C x
2
2 C
1x2
2 C
2x xdx xdx xdx
95.
is continuous: Values must agree at
The left and right hand derivatives at do not agree. Hence is not differentiable atx 2.fx 2
fx x 2,
3x2 2,
2
0 x < 2
2 x 5
4 6 C2C2 2
x 2:f
f1 31 C1 3C
1 2
fx x C
1,
3x2 C
22,
0 x < 2
2 x 5
fx 1,3x,0 x < 2
2 x 5
182 Chapter 4 Integration
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Section 4.2 Area 183
19.
12,040
14,400 2,480 120
152162
4 2
151631
6
1516
2
15
i1
ii 12 15
i1
i 3 215
i1
i 2 15
i1
i 21. sum seq (TI-82)
202141
6 60 2930
20
i1
i2 3 2020 1220 1
6 320
2 3,x, 1, 20, 1 2930x
23.
s 1 3 4 921 252 12.5
S 3 4 92 51 332 16.5 25.
s 2 2 31 7
S 3 3 51 11
27.
s4 014 1
41
4 1
21
4 3
41
4 1 2 3
8 0.518
S4 141
4 1
21
4 3
41
4 11
4 1 2 3 2
8 0.768
29.
s5 1
651
5 1
751
5 1
851
5 1
951
5 1
21
5 1
6
1
7
1
8
1
9
1
10 0.646
S5 115 16515 17515 18515 19515 15 16 17 18 19 0.746
31. limn
81n4n2n 12
4 81
4limn
n4 2n3 n2
n4 81
41
81
4
33. limn
18n2nn 1
2 18
2limn
n2 n
n2 18
21 9
39. 8 limn
1 1n 8 limn8n2 n
n2 limn16
n2nn 1
2 limnn
i1
16in2 limn16
n2
n
i1
i
35.
S10,000 1.0002
S1000 1.002
S100 1.02
S10 12
10 1.2
n
i1
2i 1n2
1n2
n
i1
2i 1 1n22nn 12 n n 2n Sn
37.
S10,000 1.99999998
S1000 1.999998
S100 1.9998
S10 1.98
6n22n
2 3n
1
3n
36 1n22n2 2 Sn
n
k1
6kk 1
n3
6
n3
n
k1
k2 k 6
n3nn 12n 1
6
nn 1
2
>
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184 Chapter 4 Integration
41.
limn
162 3n 1n2
1 1
3 lim
n
1
62n3 3n2 n
n3
limn
1
n3n 1n2n 1
6 limnn
i1
1
n3i 12 lim
n
1
n3n1
i1
i 2
43. 2 limn1 n2 n
2n2 21 1
2 3 2 limn1
nn 1
nnn 1
2 limnn
i11 i
n2
n 2 limn1
nn
i11
1
n
n
i1i
45. (a)
(b)
Endpoints:
(c) Since is increasing, on
(d) on
Sn
n
i1
fxix
n
i1
f2i
n 2
n
n
i1i2
n2
n
xi1
,xifM
i fx
i
n
i1
f2i 2n
2
n n
i1
i 12n2
n
sn n
i1
fx
i1x
xi1
,xi.fm
i fx
i1y x
0 < 12n< 22
n< . . .< n 12
n< n2
n 2
x 2 0
n
2
n
x31
3
2
1
y (e)
(f)
limn
2n 1
n 2
limn
4n2nn 1
2
limn
n
i1
i2n2
n limn4
n2n
i1
i
limn
2n 1n 4
n 2
limn
4
n2nn 1
2 n
limn
n
i1
i 12n2
n limn4
n2n
i1
i 1
x 5 10 50 100
1.6 1.8 1.96 1.98
2.4 2.2 2.04 2.02Sn
sn
47. on
Area limn
sn 2
3 2
n2n
i1
i 3 2n 1n
2n2 2
1
n
sn n
i1
f in
1
n n
i1
2 in 31
n
321x
2
1
yNote: x 1 0n 1
n0, 1.y 2x 3
49. on
Area limn
Sn 7
3
1n3n
i1
i 2 2 nn 12n 16n3 2 1
62 3
n
1
n2 2
Sn n
i1
f in1
n n
i1
in2
21n
x2 3
3
1
1
y
Note: x
1
n0, 1.y x2 2
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Section 4.2 Area 185
51. on
Area limn
sn 30 8
3 4
70
3 23
1
3
30 8
6n2n 12n 1
4
nn 1
2
n15n 4n2
nn 12n 1
6
4
nnn 1
2
2
n
n
i1
15 4i2
n2
4i
n
sn n
i1
f1 2in 2
n n
i1
16 1 2in 2
2n
x
y
11
2
4
6
8
10
12
14
18
2 3
1, 3. Note:x 2ny 16 x2
53. on
Area limn
sn 189 81
4 27
27
2
513
4 128.25
189 81
4n2n 12
81
6n2n 12n 1
27
2n 1
n
3
n63n 27
n3n2n 12
4
27
n2nn 12n 1
6
9
nnn 1
2
3
n
n
i1
63 27i3
n3
27i2
n2
9i
n
sn n
i1
f1 3in 3
n n
i1
64 1 3in 3
3n
x5 6
30
y
40
50
60
70
20
10
1 21 3
1, 4. Note:x 4 1n 3
ny 64 x3
55. on
Again, is neither an upper nor a lower sum.
Area limn
Tn 4 10 32
3 4
2
3
4 101 1
n 16
32 3
n
1
n2 41 2
n
1
n2
4
nn
20
n2
nn 1
2
32
n3
nn 12n 1
6
16
n4
n2n 12
4
n
i1
2 10in 16i 2
n2
8i3
n3 2
n 4
nn
i1
1 20
n2n
i1
i 32
n3n
i1
i 2 16
n4n
i1
i 3
n
i1
1 4in 4i2
n2 1 6i
n
12i 2
n2
8i3
n3 2
n
Tn n
i1
f1 2in
2
n n
i1
1 2in 2
1 2in 3
2n
Tn
x1
2
1
1
yNote: x 1 1n 2
n1, 1.y x2 x3
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186 Chapter 4 Integration
57.
Area limn
Sn limn
6 6n 6
6n 1
n 6
6
n
12
n2
n
i1
i 12n2 nn 1
2
n
i1
f2in
2
n n
i1
32in 2
nSn n
i1
fm
iy
64x
4
2
2
2
yNote:y 2 0n 2
n0 y 2fy 3y,
59.
Area limn
Sn limn
9 272n
9
2n2 9
9
n22n2 3n 1
2 9 27
2n
9
2n2
27
n3
nn 12n 1
6
n
i1
3in 2
3n 27
n3
n
i1
i2Sn n
i1
f3in
3
nx
2 4 6 8 10
2
4
6
2
4
yNote:y 3 0n 3
n0 y 3fy y2,
61.
Area limn
Sn 6 10 8
3 4
44
3
2
n
n
i1
3 10in 4i2
n2
8i3
n3 2
n3n 10
nnn 1
2
4
n2nn 12n 1
6
8
n2n2n 12
4
2
n
n
i1
41 4in 4i2
n2 1 6i
n
12i2
n2
8i3
n3
n
i1
41 2in 2
1 2in 3
2n
Sn n
i1
g1 2in 2
n
x
6
y
8
10
2
2
4
24
gy 4y2 y3, 1 y 3. Note: y 3 1n 2
n
63.
Let
69
8
1
21
16 3 9
16 3 25
16 3 49
16 3
Area n
i1
fcix
4
i1
ci
2 312
c4
7
4c3
5
4,c
2
3
4,c
1
1
4,x
1
2,
ci
xix
i1
2.
n 40 x 2,fx x2 3, 65.
Let
16tan 32 tan 332 tan 532 tan 732 0.345
Area n
i1
fcix
4
i1
tan ci16
c4
7
32c3
5
32,c
2
3
32,c
1
32,x
16,
ci
xix
i1
2.
n 40 x
4,fx tanx,
67. on
Exact value is 163
0, 4.fx x
n 4 8 12 16 20
5.3838 5.3523 5.3439 5.3403 5.3384Approximate area
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Section 4.2 Area 187
69. on 1, 3.fx tanx8n 4 8 12 16 20
2.2223 2.2387 2.2418 2.2430 2.2435Approximate area
71. We can use the line bounded by and
The sum of the areas of these inscribed rectangles is the
lower sum.
x
y
a b
x b.x ay x The sum of the areas of these circumscribed rectangles is the
upper sum.
We can see that the rectangles do not contain all of the area in
the 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
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 19.125Mn
Sn
sn
73. (a)
Lower sum:
(c)
Midpoint Rule:
(e)
(f) increases because the lower sum approaches the exact value as n increases. 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.
Snsn
M4 223 4
45 5
57 6
29
6112315 19.403
x
1
2
2 3
4
4
6
8
y
s4 0 4 513 6 15
13
463 15.333
x
1
2
2 3
4
4
6
8
y (b)
Upper sum:
(d) In each case, The lower sum uses left end-
points, The upper sum uses right endpoints,
The Midpoint Rule uses midpoints,i 124n.i4n.i 14n.
x 4n.
S4 4 513 6 6
25 21
1115
32615 21.733
x
1
2
2 3
4
4
6
8
y
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188 Chapter 4 Differentiation
Section 4.3 Riemann Sums and Definite Integrals
75.
b. square unitsA 6
x1 2 3 4
1
2
3
4
y 77. True. (Theorem 4.2 (2))
79.
Let area bounded by thex-axis, and Let area of the rect-
angle bounded by and Thus,
In this program, the computer is generating pairs of random points in the rectangle whose
area is represented by It is keeping track of how many of these points, lie in the region
whose area is represented by Since the points are randomly generated, we assume that
The larger is the better the approximation toA1.N
2
A1
A2
N
1
N2
A1
N1
N2
A2.
A1.
N1,A
2.
N2
A2 21 1.570796.x 2.x 0,y 0,y 1,
A2x 2.x 0fx sinx,A
1
x
0.25
0.5
0.75
1.0
4
2
2( ), 1f x x( ) sin( )=
y0, 2fx sinx,
81. Suppose there are n rows in the figure. The stars on the left total as do the stars on the right. There are
stars in total, hence
1 2 . . . n 12nn 1.
21 2 . . . n nn 1
nn 11 2 . . . n,
83. (a)
(c) Using the integration capability of a graphing utility,
you obtain
A 76,897.5 ft2.
y 4.09 105x3 0.016x2 2.67x 452.9 (b)
0
0 350
500
1.
3323 0 23 3.464
limn
33n 12n 13n2 n 1
2n2
limn
33
n32nn 12n 1
6
nn 1
2
limn
33
n3 n
i1
2i2 i
limn
n
i1
fcix
i lim
n
n
i1
3i2
n2
3
n22i 1
xi
3i2
n2
3i 12
n2
3
n22i 1
3n2
3(2)2
n23(n1)2
n2. . .
. . .
3
1
2
3
y
x
y= x
fx x,y 0,x 0,x 3, ci
3i2
n2
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Section 4.3 Riemann Sums and Definite Integrals 189
3. on
104
6 dx limn
36 36
n
i1
66n n
i1
36
n 36
n
i1
fcix
i
n
i1
f4 6in 6
n
Note: x 10 4n 6
n, 0 as n4, 10.y 6
5. on
1
1
x3dx limn
2
n
0
2 61 1n 42 3
n
1
n2 41 2
n
1
n2 2
n
2 12
n2n
i1
i 24
n3
n
i1
i2 16
n4
n
i1
i3
n
i1
fc
ix
i
n
i1
f1 2in 2
n n
i11 2in
3
2n n
i11 6in
12i 2
n2
8i3
n32
n
Note:x 1 1n 2
n, 0 as n1, 1.y x3
7. on
21
x2 1dx limn
103 3
2n
1
6n2 10
3
2 2
n2
n
i1
i 1
n3
n
i1
i2 2 1 1n 1
62 3
n
1
n2 10
3
3
2n
1
6n2
n
i1
fc
ix
i
n
i1
f1 in1
n n
i11 in
2
11n n
i11 2in
i2
n2 11n
Note:x 2 1n 1
n, 0 as n1, 2.y x2 1
9.
on the interval 1, 5.
lim0
n
i1
3ci 10x
i 5
1
3x 10dx 11.
on the interval 0, 3.
lim0
n
i1
ci2 4 x
i 3
0
x2 4 dx
13.50
3 dx 15.44
4 xdx 17.22
4 x2dx
19.0
sinxdx 21.20
y3dy 23. Rectangle
x5
5
3
2
42 31
1
Rectangle
y
A 3
0
4 dx 12
A bh 34
8/12/2019 Problemas Impares - Copia
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190 Chapter 4 Integration
25. Triangle
A 40
xdx 8
A 1
2bh
1
244
x
Triangle
4
2
42
y27. Trapezoid
A 20
2x 5dx 14
A b1 b
2
2 h 5 92 2
x
9
6
321
3
Trapezoid
y
29. Triangle
A 11
1 xdx 1
A 1
2bh
1
221
1 Triangle
11x
y 31. Semicircle
A 33
9 x2dx 9
2
A 1
2r2
1
232
x
4 Semicircle
4224
y
33.24
xdx 42
xdx 6 35.42
4xdx 442
xdx 46 24
In Exercises 3339,42
x3dx 60, 42
xdx 6, 42
dx 2
37.42
x 8dx 42
xdx 842
dx 6 82 10 39.
1
260 36 22 16
42
12x3 3x 2dx 1
24
2
x3 dx 342
xdx 242
dx
41. (a)
(b)
(c)
(d)50
3fxdx 350
fxdx 310 30
55
fxdx 0
0
5
fxdx
5
0
fxdx 10
70
fxdx 50
fxdx 75
fxdx 10 3 13 43. (a)
(b)
(c)
(d)62
3fxdx 362
fxdx 310 30
62
2gxdx 262
gxdx 22 4
2 10 12
6
2
g x fxdx 6
2
gxdx 6
2
fxdx
10 2 8
62
fx gxdx 62
fxdx 62
gxdx
45. (a) Quarter circle belowx-axis:
(b) Triangle:
(c) Triangle Semicircle belowx-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 2210 20:
4 1 2 5 2
4 1 2 3 2
1
2
21 1
222 1 2
12bh
12 42 4
14r
2 142
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:
8/12/2019 Problemas Impares - Copia
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Section 4.3 Riemann Sums and Definite Integrals 191
51.
is not integrable on the interval andfhas a
discontinuity atx 4.
3, 5
fx 1
x 4
53.
a. square unitsA 5
x1 2 3 4
1
2
3
4
y
55.
d.1
0
2 sin x dx1
212 1
y
x1
1
1 2
2
2
12
32
32
57.30
x3 xdx
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 4.1177Rn
Mn
Ln
n
59.2
0
sin2xdx
4 8 12 16 20
0.5890 0.6872 0.7199 0.7363 0.7461
0.7854 0.7854 0.7854 0.7854 0.7854
0.9817 0.8836 0.8508 0.8345 0.8247Rn
Mn
Ln
n
61. True 63. True 65. False
2
0
xdx 2
67.
41 102 404 881 272
4
i1
fc
ix f1x
1f2x
2f5x
3f8x
4
c4 8c
3 5,c
2 2,c
1 1,
x4 1x
3 4,x
2 2,x
1 1,
x4 8x
3 7,x
2 3,x
1 1,x
0 0,
0, 8fx x2 3x,
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192 Chapter 4 Differentiation
Section 4.4 The Fundamental Theorem of Calculus
69.
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.
fci 0fc
i 10,0, 1.
fx 1,0,xis rational
xis irrational
71. Let and The appropriate Riemann Sum is
limn
2n2 3n 1
6n2 lim
n
13 1
2n
1
6n2 1
3
limn
1
n312 22 32 . . . n2 lim
n
1
n3
n2n 1n 1
6
n
i1
fcix
i
n
i1
in2 1
n
1
n3n
i1
i 2.
xi 1n.fx x2, 0 x 1,
1.
is positive.
0
4
x2 1dx
2
5 5
fx 4
x2 1 3.
2
2
xx2 1 dx 0
5
5 5
fx xx2 1
5.10
2xdx x21
0
1 0 1 7.01
x 2dx x2
2 2x
0
1
0 12 2 5
2
9.11
t2 2dt t3
3 2t
1
1
13 2 1
3 2 103
11.
1
0
2t 12dt
1
0
4t2 4t 1dt 43
t3 2t2 t1
0
4
3
2 1 1
3
13.21
3x2 1dx 3
xx
2
1
32 2 3 1 1
2
15.41
u 2
udu 4
1
u12 2u12du 23u32 4u124
1
2343 44 23 4
2
3
17.11
3t 2dt 34t43 2t1
1
34 2 3
4 2 4
19.
1
0
x x
3
dx 1
3
1
0
x x12dx 1
3
x2
2
2
3
x32
1
0
1
3
1
2
2
3
1
18
21.01
t13 t23dt 34t43 3
5t53
0
1
0 34 3
5 27
20
23. split up the integral at the zero
3x x232
0
x2 3x3
32 92
9
4 0 9 9 9
4
9
2 29
2
9
4 9
2
x 3
23
0
2x 3dx 320
3 2xdx 332
2x 3dx
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Section 4.4 The Fundamental Theorem of Calculus 193
25.
23
3
8 83 9 12 8
3 8
4x x3
3 2
0
x3
3 4x
3
2
30
x2 4dx 20
4 x2dx 32
x2 4dx
27.0
1 sinxdx x cosx
0
1 0 1 2
29.66
sec2xdx tanx6
63
3 33
23
3
31.33
4 sec tan d 4 sec 3
3 42 42 0
33.30
10,000t 6dt 10,000t2
2 6t
3
0
$135,000 35. A 10
x x2dx x2
2
x3
3 1
0
1
6
37. A 30
3 xxdx 30
3x12 x32dx 2x32 25x523
0
xx5 10 2x3
0
123
5
39. A 20
cosxdx sinx2
0
1 41. Since on
A 20
3x2 1dx x3 x2
0
8 2 10.
0, 2,y 0
43. Since on
A 20
x3 xdx x4
4
x2
2 2
0
4 2 6.
0, 2,y 0 45.
c 0.4380 or c 1.7908
c 1 6 423 2
c 1 6 423
c 12 6 42
3
c 2c 1 3 42
3 1
c 2c 3 42
3
fc2 0 6 82
3
20
x 2xdx x2
2
4x32
3 2
0
2 82
3
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194 Chapter 4 Integration
47.
arccos
2 0.4817
c arcsec 2
sec c 2
sec2c 4
2 sec2c 8
fc4
4 4
44
2 sec2xdx 2 tanx4
4 21 21 4
49.
Average value
when orx 23
3 1.155.x2 4
8
34 x2
8
3
8
3
5
1
3 3
2 3
3 3
8,( ( 2 33 38,( (1
2 2
2
2
4 x2dx 1
44x 1
3x3
2
2
1
48 8
3 8 8
3 8
3
51.
x 0.690, 2.451
sinx 2
1
2
2
0.690,
2.451,
((
((
2
2
3
2
Average value 2
1
0
0
sinxdx 1cosx
0
2
53. If is continuous on and
then ba
fxdx Fb Fa.
Fx fxon a, b,a, bf
55.20
fxdx area of regionA 1.5 57.60
fxdx 20
fxdx 62
fxdx 1.5 5.0 6.5
59.
12 3.5 15.5
60
2 fxdx 60
2 dx 60
fxdx
61. (a)
Fx 500 sec2x
F0 k 500
Fx ksec2x (b)
newtons
newtons 827
826.99
1500
3 0
1
3 03
0
500 sec2xdx 1500
tanx3
0
63.1
5 05
0
0.1729t 0.1552t2 0.0374t3dt1
50.08645t2 0.05073t3 0.00935t45
0
0.5318 liter
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Section 4.4 The Fundamental Theorem of Calculus 195
65. (a)
The area above the -axis equals the area below the
-axis. Thus, the average value is zero.x
x
24
1
0
1
67. (a)
(b)
(c) meters600
vtdt 8.61 104t4
4
0.0782t3
3
0.208t2
2 0.0952t
60
0
2476
70
10
10
90
v 8.61 104t3 0.0782t2 0.208t 0.0952
(b)
The average value of appears to be g.S
24
0
0
10
69.
F8 64
2 58 8
F5 25
2 55
25
2
F2 4
2 52 8
Fx x0
t 5dt t2
2 5t
x
0
x2
2 5x 71.
F8 1078 35
4
F5 1045 8
F2 1012 5
10
x 10 101 1x
Fx x1
10
v2dv x
1
10v2 dv 10
v x
1
73.
F8 sin 8 sin 1 0.1479
F5 sin 5 sin 1 1.8004
F2 sin 2 sin 1 0.0678
Fx x1
cos d sin x
1
sinx sin 1 75. (a)
(b) d
dx1
2x2 2x x 2
x0
t 2dt t2
2 2t
x
0
1
2x2 2x
77. (a)
(b) d
dx3
4x43 12 x13 3x
x8
3tdt 34t43x
8
3
4x43 16
3
4x43 12 79. (a)
(b) d
dxtanx 1 sec2x
xx4
sec2tdt tan tx
x4 tanx 1
81.
Fx x2 2x
Fx x2
t2 2tdt 83.
Fx x4 1
Fx x1
t4 1 dt 85.
Fx xcosx
Fx x0
tcos tdt
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196 Chapter 4 Integration
87.
Fx 8
8x 10
2x 22 x 2 2x2 x
2t2 tx2
x
Fx x2x
4t 1dt Alternate solution:
Fx 4x 1 4x 2 1 8
x
0
4t 1dt x2
0
4t 1dt
0x
4t 1dt x20
4t 1dt
Fx x2x
4t 1dt
89.
Alternate solution
Fx sinx ddx sinx sinxcosx
Fx sin x0
tdt
Fx sinx12 cosx cosxsinx
Fx sin x0
tdt 23t32
sin x
0
2
3sinx32 91.
Fx sinx32 3x2 3x2sinx 6
Fx x30
sin t2dt
93.
has a relative maximum at x 2.g
x
y
1
1 2 3 4
2
2
1
f g
g0 0, g11
2, g2 1, g3
1
2, g4 0
gx x0
ftdt 95. (a)
(b)
C10 1000125 121054 $338,394
C5 1000125 12554 $214,721
C1 1000125 121 $137,000
500025 125x54 1000125 12x54
500025 345t54x
0
Cx 500025 3x0
t14dt
97. True 99. False;
Each of these integrals is infinite.
has a nonremovable discontinuity atx 0.
fx x2
11
x2dx 01
x2dx 10
x2dx
101.
By the Second Fundamental Theorem of Calculus, we have
Since must be constant.fxfx 0,
1
1 x2
1
x2 1 0.
fx 1
1x2 11
x2 1
x2 1
fx
1x
0
1
t2
1
dt
x
0
1
t2
1
dt
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Section 4.5 Integration by Substitution 197
103.
28 units
4 4 20
310
t2 4t 3dt 331
t2 4t 3dt 353
t2 4t 3dt
50
3t 3t 1dt
Total distance
5
0
xt
dt
3t 3t 1
3t2 4t 3
xt 3t2 12t 9
xt t3 6t2 9t 2
105.
22 1 2 units
2t124
1
41
1
tdt
4
1
vtdt
Total distance 41
xtdt
Section 4.5 Integration by Substitution
du gxdxu gxfgxgxdx1. 10x dx
3. 2x dx
5. tanx sec2xdxtan2xsec2xdxx2 1 x
x2 1dx
5x2 15x2 1210xdx
7.
Check: d
dx1 2x5
5 C 21 2x4
1 2x42 dx 1 2x5
5 C
9.
Check: d
dx2
39 x232 C 23
3
29 x2122x 9 x22x
9 x2122xdx 9 x23232
C2
39 x232 C
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198 Chapter 4 Integration
11.
Check: d
dxx4 33
12 C 3x
4 32
12 4x3 x4 32x3
x3x4 32dx 14x4 324x3dx 1
4x4 33
3 C
x4 33
12 C
13.
Check: d
dxx3 15
15 C 5x
3 143x2
15 x2x3 14
x2x3 14dx
1
3x3 143x2dx
1
3x3 15
5 Cx3 15
15 C
15.
Check: d
dtt2 232
3 C 32t
2 2122t
3 t2 212t
tt2 2 dt 12t2 2122tdt 1
2t2 232
32 C
t2 232
3 C
17.
Check: d
dx15
81 x243 C 158
4
31 x2132x 5x1 x213 5x 31 x2
5x1 x213dx 521 x2132xdx 5
2
1 x243
43 C
15
81 x243 C
19.
Check: d
dx1
41 x22 C 1421 x232x
x
1 x23
x1 x23dx 121 x232xdx 121 x22
2 C
1
41 x22 C
21.
Check: ddx 131 x3 C 1311 x323x2 x
2
1 x32
x21 x32dx 131 x323x2dx 131 x31
1 C 1
31 x3 C
23.
Check: d
dx1 x212 C
1
21 x2122x
x
1 x2
x1 x2
dx 1
21 x2122xdx 1
21 x212
12 C 1 x2 C
25.
Check:
d
dt
1 1t4
4
C
1
441
1
t3
1
t2
1
t21
1
t3
1 1t3
1t2dt 1 1t3
1t2dt 1 1t4
4 C
27.
Check: d
dx2x C 1
22x122
1
2x
12x
dx 1
22x122 dx 1
22x12
12 C2x C
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Section 4.5 Integration by Substitution 199
29.
Check: d
dx2
5x52 2x32 14x12 C x
2 3x 7
x
x2 3x 7x
dx x32 3x12 7x12dx 25x52 2x32 14x12 C
2
5xx2 5x 35 C
31.
Check: d
dt1
4t4 t2 C t3 2t t2t 2t
t2t
2
tdt
t3 2tdt
1
4t4 t2 C
33.
Check: d
dy2
5y3215 y C ddy6y32
2
5y52 C 9y12 y32 9 yy
9 yydy 9y12 y32dy 923y32 25y52 C 25y3215 y C
35.
2x2 416 x2 C
4x2
2 216 x212
12 C
4xdx 216 x2122xdxy
4x
4x
16 x2
dx 37.
1
2x2 2x 3 C
1
2x2 2x 31
1 C
1
2x2 2x 322x 2dx
y
x 1
x2
2x 32dx
39. (a)
x
y
2 2
1
3
(b)
2 2
1
2
y 1
34 x232 2
2 1
34 2232 C C 22, 2:
1
2
2
34 x232 C
1
34 x232 C
y
x4
x
2
dx
1
2
4
x2
12
2xdx
dy
dxx4 x2, 2, 2
41. sin xdx cos x C 43. sin 2xdx 12sin 2x2xdx 12cos 2x C
45. 12
cos1
d cos 1
1
2d sin1
C
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200 Chapter 4 Integration
47. OR
OR
sin 2xcos 2xdx
1
2
2 sin 2xcos 2xdx
1
2
sin 4xdx
1
8cos 4x C
2
sin 2xcos 2xdx 12cos 2x2 sin 2xdx 1
2cos 2x2
2 C
1
1
4cos22x C
1
sin 2xcos 2xdx 12sin 2x2 cos 2xdx 1
2sin 2x2
2 C
1
4sin22x C
49. tan4xsec2xdx tan5x5 C
1
5tan5x C
51.
cotx2
2 C
1
2 cot2x C
1
2tan2x C
1
2sec2x 1 C
1
2sec2x C
1
csc2xcot3x
dx cotx3csc2xdx
53. cot2xdx csc2x 1dx cotx x C 55.Since Thus,
fx 2 sinx
2 3.
C 3.f0 3 2 sin 0 C,
fx cosx2
dx 2 sinx
2 C
57.
2
15x 2323x 4 C
2
15x 2323x 2 10 C
2u32
15 3u 10 C
2
5u52
4
3u32 C
u32 2u12duxx 2 dx u 2udu
dx dux u 2,u x 2,
59.
2
1051 x3215x2 12x 8 C
2
1051 x3235 421 x 151 x2 C
2u32
10535 42u 15u2 C
23u32 4
5u52
2
7u72 C
u12 2u32 u52dux21 xdx 1 u2udu
dx dux 1 u,u 1 x,
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Section 4.5 Integration by Substitution 201
61.
1
152x 13x2 2x 13 C
1
602x 112x2 8x 52 C
2x 1
60 32x 12 102x 1 45 C
u12
603u2 10u 45 C
1
82
5u52
4
3u32 6u12 C
1
8u32 2u12 3u12du
1
8u12u2 2u 1 4du
x2 12x 1
dx 12u 12 1u
1
2du
dx 1
2dux
1
2u 1,u 2x 1,
63.
where C1 1 C.
x 2x 1 C1
x 2x 1 1 C
x 1 2x 1 C
u 2u C
u
2u12
C
1 u12du u 1u 1
uu 1du
xx 1 x 1
dx u 1u u
du
dx dux u 1,u x 1,
65. Let
11
xx2 13dx 1
211
x2 132xdx 18x2 141
1
0
du 2xdx.u x2 1,
67. Let
4
927 22 12 8
92
4
9x3 1322
1
23x3 132
32 2
1
21
2x2x3 1 dx 2 1
321
x3 1123x2dx
u x3 1, du 3x2dx
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202 Chapter 4 Integration
69. Let
40
1
2x 1dx
1
240
2x 1122dx 2x 14
0
9 1 2
du 2 dx.u 2x 1,
71. Let
91
1
x1 x2dx 29
1
1 x2 12x
dx 21 x
9
1
1
2 1
1
2
du 1
2x
dx.u 1 x,
73.
When When
01
u32 u12du 25u52 2
3u32
0
1
25 2
3 4
1521
x 12 xdx 01
2 u 1udu
u 0.x 2,u 1.x 1,
dx dux 2 u,u 2 x,
75. 20
cos23xdx 3
2sin23x
2
0
3
23
2 33
4
77.
When When
81
u43 u13du 37u73 3
4u43
8
1
3847 12 3
7
3
4 1209
28
Area 70
x3x 1 dx 81
u 13udu
u 8.x 7,u 1.x 0,
dx dux u 1,u x 1,
79. A 0
2 sinx sin 2xdx 2 cosx 12cos 2x
0
4
81. Area 232
sec2x2dx 223
2sec2x2
1
2dx 2 tanx
223
2 23 1
83.
1
1 5
3
40
x
2x 1dx 3.333
10
3 85.
0
0 8
15
73
xx 3 dx 28.8 144
5 87.
1
1 4
30
cos 6d 7.377
89.
They differ by a constant: C2 C
1
1
6.
2x 12dx 4x2 4x 1dx 43x3 2x2 x C
2
2x 12dx 122x 122 dx 1
62x 13 C
1
4
3x3 2x2 x
1
6 C
1
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Section 4.5 Integration by Substitution 203
91. is even.
2325 8
3 272
15
22
x2x2 1dx 220
x4 x2dx 2x5
5
x3
3 2
0
fx x2x2 1 93. is odd.
22
xx2 13dx 0
fx xx2 13
95. the function is an even function.
(a)
(c)20
x2dx 20
x2dx 8
3
02
x2dx 20
x2dx 8
3
x220
x2dx x3
3 2
0
8
3;
(b)
(d) 02
3x2dx 320
x2dx 8
22
x2dx 220
x2dx 16
3
97. 0 240
6x2 3dx 22x3 3x4
0
23244
x3 6x2 2x 3dx 44
x3 2xdx 44
6x2 3dx
99. Answers will vary. See Guidelines for Making a Change of Variables on page 292.
101. is odd. Hence, 22
xx2 12dx 0.fx xx2 12 103.
Solving this system yields and
. Thus,
When V4 $340,000.t 4,
Vt 200,000
t 1 300,000.
C 300,000
k 200,000
V1 1
2k C 400,000
V0 k C 500,000
Vt kt 12dt kt 1 CdV
dt
k
t 12
105.
(a) thousand units
(b) thousand units
(c) thousand units11274.50t 262.5 cos t6
12
0
112894 262.5 262.5 74.5
1
374.50t262.5
cos
t
66
3
1
3447 262.5
223.5 102.352
1
374.50t262.5
cos
t
63
0
1
3223.5 262.5
102.352
1
b aba
74.50 43.75 sin t6dt1
b a74.50t262.5
cos
t
6b
a
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204 Chapter 4 Integration
109. False
2x 12dx 122x 122 dx 1
62x 13 C
111. True
Odd Even10
10
ax3 bx2 cx ddx
10
10
ax3 cxdx
10
10
bx2 ddx 0 2
10
0
bx2 ddx
113. True
4sinxcosxdx 2sin 2xdx cos 2x C
115. Let then When When Thus,
ba
fx hdx bhah
fudu bhah
fxdx.
u b h.x b,u a h.x a,du dx.u x h,
107.
(a) amps
(b)
amps
(c) amps1
130 01
30cos60t
1
120sin120t
130
0
30 130 1
30 0
2
5 22 1.382
1
1240 01
30cos60t
1
120sin120t
1240
0
240 1302
1
120 1
30
1
160 01
30cos60t
1
120sin120t
160
0
60 130 0 1
30 4
1.273
1
b aba
2 sin60t cos120tdt1
b a1
30cos60t
1
120 sin120t
b
a
Section 4.6 Numerical Integration
1. Exact:
Trapezoidal:
Simpsons: 20
x2dx1
60 41
22
212 4322
22 83 2.6667
20
x2dx1
40 21
22
212 2322
22 114 2.7500
20
x2dx 13x32
0
8
3 2.6667
3. Exact:
Trapezoidal:
Simpsons: 20
x3dx1
60 41
23
213 4323
23 246 4.0000
20
x3dx1
40 21
23
213 2323
23 174 4.2500
2
0
x3dx x4
4 2
0
4.000
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Section 4.6 Numerical Integration 205
5. Exact:
Trapezoidal:
Simpsons: 20
x3dx1
120 41
43
2243
4343
213 4543
2643
4743
8 4.0000
20
x3dx1
80 21
43
2243
2343
213 2543
2643
2743
8 4.0625
20
x3dx 14x42
0
4.0000
7. Exact:
Trapezoidal:
Simpsons: 94
xdx5
242 4378 21 4478 26 4578 31 4678 3 12.6667 12.6640
94
xdx5
162 2378 2214 2478 2264 2578 2314 2678 3
94
xdx 23x329
4
18 16
3
38
3 12.6667
9. Exact:
Trapezoidal:
Simpsons:
1
121
4
64
81
8
25
64
121
1
9 0.1667
21
1
x 12dx
1
121
4 4 154 12 2
1
32 12 41
74 12 1
9
1
81
4
32
81
8
25
32
121
1
9 0.1676
21
1
x 12dx
1
81
4 2 154 12 2
1
32 12 21
74 12 1
9
2
1
1x 12
dx 1x 12
1
13 1
2 1
6 0.1667
11. Trapezoidal:
Simpsons:
Graphing utility: 3.241
20
1 x3dx1
61 41 18 22 41 278 3 3.240
2
0
1 x3dx1
41 21 18 22 21 278 3 3.283
13.
Trapezoidal:
Simpsons:
Graphing utility: 0.393
10
x1 xdx1
120 4141 14 2121 12 4341 34 0.372
10
x1 xdx1
8
0 2141
1
4 21
21 1
2 23
41 3
4
0.342
10
x1 xdx 10
x1 xdx
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206 Chapter 4 Integration
15. Trapezoidal:
Simpsons:
Graphing utility: 0.977
0.978
2
0
cosx2dx2
12 cos 0 4 cos24 2
2 cos22 2
4 cos24 2
cos22
0.957
2
0
cosx2dx2
8 cos 0 2 cos24 2
2 cos22 2
2 cos24 2
cos22
17. Trapezoidal:
Simpsons:
Graphing utility: 0.089
1.11
sinx2dx1
120sin1 4 sin1.0252 2 sin1.052 4 sin1.0752 sin1.12 0.089
1.11
sinx2dx1
80sin1 2 sin1.0252 2 sin1.052 2 sin1.0752 sin1.12 0.089
19. Trapezoidal:
Simpsons:
Graphing utility: 0.186
40
xtanxdx
480 4
16tan
16 22
16tan2
16 43
16tan3
16
4 0.186
4
0xtanxdx
320 2
16tan
16 2
2
16tan2
16 2
3
16tan3
16
4 0.194
21. (a)
The Trapezoidal Rule overestimates the area if the graph
of the integrand is concave up.
xa b
y f x= ( )
y 23.
(a) Trapezoidal: since
is maximum in when
(b) Simpsons: since
f4x 0.
Error 2 05
180440 0
x 2.0, 2fx
Error 2 03
1242 12 0.5
f4
x
0
fx 6
fx 6x
fx 3x2
fx x3
25. in .
(a) is maximum when and
Trapezoidal: let
in
(b) is maximum when and when
Simpsons: let n 26.n > 25.56;n4 > 426,666.67,Error 25
180n424 < 0.00001,
f41 24.x 1f4x
1, 3f4x 24
x5
n 366.n > 365.15;n2 > 133,333.33,Error 23
12n22 < 0.00001,
f1
2.x 1
fx
1, 3fx 2
x3
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Section 4.6 Numerical Integration 207
27.
(a) in .
is maximum when and
Trapezoidal: Error < 0.00001, > 16,666.67,n > 129.10; let
(b) in
is maximum when and
Simpsons: Error < 0.00001, > 16,666.67, n > 11.36; let n 12.n432
180n415
16f40
15
16.x 0f4x
0, 2f4x 15
161 x72
n 130.n2
8
12n21
4
f0 1
4.x 0fx
0, 2fx 1
41 x32
fx 1 x
29.
(a) in .
is maximum when and
Trapezoidal: Error < 0.00001, > 412,754.17, n > 642.46; let
(b) in
is maximum when and
Simpsons: Error < 0.00001, > 5,102,485.22,n > 47.53; let n 48.n41 05
180n4 9184.4734
f41 9184.4734.x 1f4x
0, 1f4x 8 sec2x212x2 3 32x4tanx2 36x2tan2x2 48x4tan3x2
n 643.n21 03
12n2 49.5305
f1
49.5305.x 1
fx
0, 1fx 2 sec2x21 4x2tanx2
fx tanx2
31. Let Then
Simpsons: Error
Therefore, Simpsons Rule is exact when approximating the integral of a cubic polynomial.
Example:
This is the exact value of the integral.
10
x3dx 1
60 41
23
1 14
b a5
180n4 0 0
f4x 0.fx Ax3 Bx2 Cx D.
33. on .0, 4fx 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
SnTnRnMnLn
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208 Chapter 4 Integration
35. on .0, 4fx sinx
n
4 2.8163 3.5456 3.7256 3.2709 3.3996
8 3.1809 3.5053 3.6356 3.4083 3.4541
10 3.2478 3.4990 3.6115 3.4296 3.4624
12 3.2909 3.4952 3.5940 3.4425 3.4674
16 3.3431 3.4910 3.5704 3.4568 3.4730
20 3.3734 3.4888 3.5552 3.4643 3.4759
SnTnRnMnLn
37.
Simpsons Rule:
x
4 2
1
1
2
y
0.701
20
xcosxdx
840 cos 0 4
28cos
28 214cos
14 4328cos
3
28 . . . 2 cos
2
n 14
A 20
xcosxdx
39.
Simpsons Rule:
4001512125 15
123
. . . 0 10,233.58 ft lb
50
100x125 x3dx5
3120 400 512125 5123
2001012125 10
123
n 12
W
5
0 100x125
x
3
dx
41. Simpsons Rule,
1
36113.098 3.1416
1
2
0
36 6 46.0209 26.0851 46.1968 26.3640 46.6002 6.9282
n 6120
6
1 x2dx
43. Area1000
210125 2125 2120 2112 290 290 295 288 275 235 89,250 sq m
45 10t i d 2
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