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8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler
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C H A P T E R 4
Integration
Section 4.1 Antiderivatives and Indefinite Integration . . . . . . . . . 450
Section 4.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
Section 4.3 Riemann Sums and Definite Integrals . . . . . . . . . . . 462
Section 4.4 The Fundamental Theorem of Calculus . . . . . . . . . . 466
Section 4.5 Integration by Substitution . . . . . . . . . . . . . . . . . 472
Section 4.6 Numerical Integration . . . . . . . . . . . . . . . . . . . 479
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
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C H A P T E R 4
Integration
Section 4.1 Antiderivatives and Indefinite Integration
Solutions to Even-Numbered Exercises
450
2.d
dxx4 1
x C 4x3 1x2 4.
x12 x32 x2 1
x32
d
dx2x2 3
3x C ddx
2
3x32 2x12 C
6.
Check:d
d
C
r Cdr
d 8.
Check: ddx
1
x2 C 2x
3
y 2x2
2 C
1
x2 c
dy
dx 2x3
Given Rewrite Integrate Simplify
10. 1
x C
x1
1 Cx2dx 1
x2dx
12.1
4x 4
3
2x2 C
x4
4 3x
2
2 Cx3 3xdxxx2 3dx
14.1
9x C
1
9
x1
1 C
1
9x2dx
1
3x2
dx
16.
Check:d
dx5x x2
2 C 5 x
5 xdx 5x x22 C 18.Check:
d
dxx 4 2x 3x C 4x 3 6x 2 1
4x3 6x2 1dx x 4 2x3 x C
20.
Check:d
dxx4
4 2x2 2x C x3 4x 2
x3 4x 2dx x 44 2x2 2x C
22.
Check: x 1
2x
d
dx2
3x32 x12 C x12 12x12
2
3x32 x12 C
x32
32
1
2x12
12 Cx 12xdx x12 1
2x12dx
8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler
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Section 4.1 Antiderivatives and Indefinite Integration 451
24.
Check:d
dx4
7x74 x C x34 1 4x3 1
4x3 1dx x34 1dx 47x74 x C 26.Check:
d
dx1
3x3 C 1x4
1x4
dx x4dx x33
C 1
3x3 C
28.
Check:
x2 2x 3
x4
d
dx1
x
1
x2
1
x3 C x2 2x3 3x4
1
x
1
x2
1
x3 C
x1
1
2x2
2
3x3
3 C
x
2
2x 3x4
dx x2 2x3 3x4dx 30.
Check:
2t2 12
d
dt4
5t5
4
3t3 t C 4t4 4t2 1
4
5t5
4
3t3 t C
2t2 12dt 4t
4 4t2 1dt
32.
Check:d
dt1
3t3
3
4t4 C t2 3t3 1 3tt2
1 3tt2dt t2 3t3dt 13t3 34t4 C 34.Check:
d
dt3t C 3
3 dt 3t C
36.
Check:d
dt1
3t3 cos t C t2 sin t
t2 sin tdt 13t3 cos t C 38.Check:
d
d1
33 tan C 2 sec2
2 sec2d 133 tan C
40.
Check:
secytany secy
d
dysecy tany C secy tany sec2y
secy tany Csecytany
secydy
secy tany
sec
2
ydy 42.
cosx
1 cos2x
Check:d
dxcscx C cscx cotx
1
sinx
cosx
sinx
cscx cotxdx cscx C
cosx
1 cos2xdx
cosx
sin2xdx
1
sinxcosx
sinxdx
44.
x4 6 82
2
4
2
4
6
C= 0
C= 3
C= 2
y
fx x 46.
x
y
4
6
8
24 2 4
f x( ) 2= +x
2
2
f x( ) =x
2
2
f
fx x2
2
C
fx x 48.
x24
2
4
4
f x( ) =
f x( ) 1= +
1
1
x
x
f
y
fx 1
x C
fx 1
x2
8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler
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452 Chapter 4 Integration
50.
y
x
2
2x
1
2 32 23 C C 1
y 2x 1dx x2 2x C
dy
dx 2x 1 2x 2, 3, 2 52.
y 1
x 2, x > 0
3 1
1 C C 2
y x2dx 1x C
dy
dx
1
x2 x2, 1,3
54. (a)
x 4
5
5
4
y
56.
gx 2x3 1
g0 1 203 CC 1
gx 6x2 dx 2x3 Cgx 6x2, g0 1
(b)
y x3
3x
7
3
C7
3
3 1
3 1 C
3 13
3 1 C
y x3
3x C 4 4
5
( 1, 3)
5dy
dxx2 1, 1, 3
58.
fs 3s2 2s4 23
f2 3 322 224 C 12 32 CC 23
fs 6s 8s3ds 3s 2 2s4 Cfs 6s 8s3, f2 3
60.
fx 1
12x4 6x 3
f0 0 0 C2 3 C2 3
fx
1
3
x3 6
dx
1
12
x4 6x C2
fx 13x3 6
f0 0 C1 6 C1 6
fx x2dx 13x3 C1f0 3
f0 6
fx x2 62.
fx sinx 2x 6
f0 0 0 C2 6 C2 6
fx cosx 2dx sinx 2x C2fx cosx 2
f0 1 C1 1 C1 2
fx sinxdx cosx C1f0 6
f0 1
fx sinx
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Section 4.1 Antiderivatives and Indefinite Integration 453
64.
P7 100732 500 2352 bacteria
Pt 2
3150t32 500 100t32 500
P1 23
k 500 600 k 150
P0 0 C 500 C 500
Pt kt12dt 23kt32 C
dP
dt kt, 0 t 10 66. Since is negative on is decreasing on
Since is positive on is increasing
on has a relative minimum at Since is
positive on fis increasing on
x1 2 323
1
2
3
2
3f
f
f
y
,.,,f0, 0.f0,.
f0,,f, 0.f, 0,f
68.
ft 16t2 v0t s0
f0 0 0 C2 s0 C2 s0
ft st 32t v0dt 16t2 v0t C2ft 32t v0
f0 0 C1 v0 C1 v0
ft vt
32 dt 32t C
1
f0 s0
f0 v0
ft at 32 ftsec2 70.
(a)
Choosing the positive value,
(b)
1617 65.970 ftsec
v1 172 321 17
2 16
vt st 32t 16
t1 17
2 2.562 seconds.
t1 17
2
16t2 t 4 0
st 16t2 16t 64 0
s0 64 ft
v0 16 ftsec
72. From Exercise 71, (Using the
canyon floor as position 0.)
t2 1600
4.9 t 326.53 18.1 sec
4.9t2 1600
ft 0 4.9t2 1600
ft 4.9t2 1600. 74. From Exercise 71, If
then
for this tvalue. Hence, and we solve
v02 3880.8 v0 62.3 msec.
4.9 v02 9.82 198
4.9 v02 9.8 v0
2 9.82 198
4.9 v0
2
9.82 v0
2
9.8 198
4.9 v09.82
v0 v09.8 2 200
t v09.8
vt 9.8t v0 0
ft 200 4.9t2 v0t 2,
ft 4.9t2 v0t 2.
8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler
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454 Chapter 4 Integration
76.
When
v2 v02 2GM1y
1
R
v2 2GM
y v0
2
2GM
R
1
2v2
GM
y
1
2v0
2
GM
R
C1
2v0
2
GM
R
1
2v02 GM
R C
y R, v v0.
1
2v2
GM
y C
vdv GM 1y2
dy 78.
(a)
(b) when and
(c) when
v73 37
3 57
3 3 223
4
3
t7
3.at 6t 14 0
3 < t < 5.0 < t 0
at vt 6t 14
vt xt 3t2 14t 15 3t 5t 3
t3 7t2 15t 9
0 t 5xt t 1t 32
80. (a)
(b) for k 0, 1, 2, . . .t k,vt 0 sin t
ft cos t 4
f0 3 cos0 C2 1 C2 C2 4
ft vtdt sin tdt cos t C2vt atdt cos tdt sin t C1 sin t since v0 0at cos t
82.
st 8.25t2 66t
vt 16.5t 66
at 16.5
132 when a 33
2 16.5.
s66a
a2
66a
2
6666a
at 66 0 when t66
a.
vt 0 after car moves 132 ft.
st a
2t2 66t Let s0 0.
vt at 66
at a
15 mph 22 ftsec
30 mph 44 ftsec
v0 45 mph 66 ftsec (a)
(b)
(c)
It takes 1.333 seconds to reduce the speed from 45 mph to
30 mph, 1.333 seconds to reduce the speed from 30 mph
to 15 mph, and 1.333 seconds to reduce the speed from
15 mph to 0 mph. Each time, less distance is needed to
reach the next speed reduction.
0
73.33 117.33
132
feet feet
45mph
=66ft/s
ec
30mph
=44ft/s
ec
15mph
=22ft/s
ec
0mph
s 4416.5 117.33 ft
t44
16.5 2.667
16.5t 66 22
s 2216.5 73.33 ftt
22
16.5 1.333
16.5t 66 44
8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler
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Section 4.1 Antiderivatives and Indefinite Integration 455
84. No, car 2 will be ahead of car 1. If and are the respective velocities, then 300
v2tdt > 30
0v1tdt.v2tv1t
86. (a) (b)
s6 196.1 feet
Note: Assume s0 0 is initial position
st vtdt 0.6139t44 5.525t3
3
0.0492t2
2 65.9881tv 0.6139t3 5.525t2 0.0492t 65.9881
88. Let the aircrafts be located 10 and 17 miles away from the airport, as indicated in the figure.
(a) When aircraft A lands at time you have
kA 50
tA 5012510 625SAt S1t
625
2t2 150t 10
tA 10
125.
125tA 10
1
250
tAt2A 150tA 10
sAtA 1
2
kAt2A 150tA 10 0
vAtA kAtA 150 100 kA 50
tA
tA
sAt 1
2kAt
2 150t 10 sB
1
2kBt
2 250t 17
10 170
A BirportvAt kAt 150 vB kBt 250
Similarly, when aircraft B lands at time you have
(b)
0
0
0.1
20
sA
sB
kB
135
tB 135
365
34
49,275
34S
B
t S2t 49,275
68t2 250t 17
tB 34
365.
365
2tB 17
1
2135
tBt2B 250tB 17
sBtB 1
2kBt
2B 250tB 17 0
vBtB
kBtB
250
115
kB
135
tB
tB
(c)
Yes, for t > 0.0505.d < 3
0
0
0.1
20
d3
d sBt sAt
90. True 92. True
8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler
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456 Chapter 4 Integration
Section 4.2 Area
96.
Thus, for some constant k. Since,
and
Therefore,
[Note that and satisfy these
properties.]
cx cosxsx sinx
sx2 cx2 1.
k 1.c0 1,s0 0
sx2 cx2 k
0
2sxcx 2cxsx
d
dxsx2 cx2 2sxsx 2cxcx94. False. has an infinite number of antiderivatives, each
differing by a constant.
f
2. 6
k3
kk 2 31 42 53 64 50 4. 5
j3
1
j
1
3
1
4
1
5
47
60
6. 4
i1
i 12 i 13 0 8 1 27 4 64 9 125 238
8. 15
i1
5
1 i10.
4
j11 j4
2
12. 2nn
i11 2in 1
2
14. 1nn1
i01 in
2
16.
215162 45 19515
i1
2i 3 215
i1
i 315 18.
1011216 10 37510
i1
i2 1 10
i1
i2 10
i1
1
20.
102112
4 10112 3080
10
i1
ii2 1 10
i1
i3 10
i1
i 22. sum seq (TI-82)
152162
4 1516 14,160
15
i1
i3 2i 15 215 1 2
4 2
1515 1
2
3 2x,x, 1, 15, 1 14,160x
24.
s 4 4 2 01 10
S 5 5 4 21 16 26.
s 2 1 23 1
2
1
3 9
2 4.5
S 5 2 1 23 1
2 55
6
28.
s8 0 21
4 14 214 12 214 . . . 74 214 5.685
1
416 1
22
23
2 1
5
26
27
2 2 6.038
54 214 32 214 74 214 2 214
S8 14 214 12 214 34 214 1 214
>
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Section 4.2 Area 457
30.
s5 1 152
15 1 2
52
15 1 3
52
15 1 4
52
15 0 0.659
1
51 24
521
516
59
5 0.859
S5 115 1 1
52
15 1 2
52
15 1 3
52
15 1 4
52
15
32. limn
64n3nn 12n 1
6 64
6limn
2n3 3n2 n
n3 64
62
64
3
34. limn
1n2nn 1
2 1
2limn
n2 n
n2 1
21
1
2
36.
S10,000 2.0005
S1000 2.005
S100 2.05
S10 25
10 2.5
n
j1
4j 3
n2
1
n2
n
j1
4j 3 1
n24nn 1
2 3n 2n 5n Sn
38.
S10,000 1.000066657
S1000 1.00066567
S100 1.006566
S10 1.056
1
3n
33n3 2n2 3n 2 Sn
1
3n33n3 6n2 3n 4n2 6n 2
4
n3n3 2n2 n
4
2n2 3n 1
6
n
i1
4i2i 1
n4
4
n4
n
i1
i3 i2 4
n4n2n 12
4
nn 12n 1
6
40. limn
4
21 1
n 2 limn4
n2nn 1
2 limnn
i12in
2
n limn4
n2n
i1
i
42.
21 2 43 26
3
2 limn
1 2 2n 4
3
2
n
2
3n2
limn
2
n3n3 4nnn 1
2 4nn 12n 1
6
limn
2
n3n
i1
n2 4nn
i1
i 4 n
i1
i 2
limn
n
i11 2in
2
2n limn2
n3n
i1
n 2i2
8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler
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458 Chapter 4 Integration
44.
2 limn
10 13n 4
n2 20
2 limn
1 3 3n 4 6
n
2
n2 2
4
n
2
n2
2 limn
1
n4n4 6n2nn 12 12n
nn 12n 1
6 8n2n 12
4
2 limn
1
n4
n
i1
n3 6n2i 12ni2 8i 3
limn
n
i11 2in
3
2n 2 limn1
n4n
i1
n 2i3
46. (a)
(c) Since is increasing, on
(d) on
(e)
Sn n
i1fxix
n
i1f1 i2n
2
n n
i11 i2n
2
n
[xi1,xifMi fxi
n
i1
f1 i 12n2
n n
i11 i 12n
2
n
sn n
i1fxi1x
xi1,xi.fmi fxi1y 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 4.02Sn
sn
(b)
Endpoints:
1 < 1 12n < 1 22n < . . . < 1 n 12n < 1 n2n
1 < 1 2
n< 1
4
n< . . . < 1
2n
n 3
x 3 1
n
2
n
(f )
limn
2 2n 1n limn4 2n 4
limn
n
i11 i2n
2
n limn2
nn 2
nnn 1
2
limn
2 2n 2n 4
n limn4 2
n 4
limn
n
i11 i 12n
2
n limn2
nn 2
nnn 1
2 n
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Section 4.2 Area 459
48. on
Area limn
Sn 6 27
2
39
2
6 27
n2n 1n
2 6 27
2 1 1
n
n
i132 3in 4
3
n 18 33
n2
n
i1
i 12
Sn n
i1
f2 3in 3
n
x4 6 8 10 1 2
2
4
6
8
10
12
yNote: x 5 2n 3
n2, 5.y 3x 4
50. on
Area limn
Sn 9
22 3 12
27
n3nn 12n 1
6
3
nn
9
22n2 3n 1
n2 3
27
n3
n
l1
l 23
n
n
l1
1
Sn n
l1f3in
3
n n
l13in
2
13n
y
x1 2
2
4
6
8
10
12
3
0, 3. Note:x 3 0n 3
ny x2 1
52. on Find area of region over the interval
Area 4
3
1
2Area lim
n
sn 1 1
3
2
3
1 1
n3
n
i1
i 2 1 nn 12n 1
6n3 1
1
62 3
n
1
n2
sn n
i1f in
1
n n
i11 in
2
1n
x
2
3
2 2
1
yNote: x 1n0, 1.1, 1.y 1 x2
54. on
Sincey both increases and decreases on is neither an upper nor lower sum.
Area limn
Tn 1 1
4
3
4
1 1
n
1
4
2
4n
1
4n2
2
n2
n
i1
i 1
n4
n
i1
i3 nn 1
n2
1
n4n2n 12
4
Tn n
i1f in
1
n n
i12 in
i
n3
1n
Tn0, 1,
x0.5 1.0 2.0
0.5
1.0
1.5
2.0
yNote: x 1 0n 1
n0, 1.y 2x x3
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460 Chapter 4 Integration
56. on
Area limn
sn 2 5
2
4
3
1
4
7
12
2 52
52n
43
2n
23n3
14
12n
14n2
n
i12 5in
4i 2
n2
i 3
n31
n 2 5
n2
n
i1
i 4
n3
n
i1
i2 1
n4
n
i1
i3
sn n
i1f1 in
1
n n
i11 in
2
1 in3
1n
x1 212
1
1
2
3
yNote: x 0 1n 1
n1, 0.y x2 x3
58.
y
x1
1
2
3
4
5
2 3 4 5
Area limn
Sn 2 1 3
2
nn 1
nnn 1
2 2 n 1
n
n
i1
1
22 2i
n 2
n 2
n
n
i11 in
Sn n
i1
g2 2in 2
n
gy 1
2y, 2 y 4. Note:y 4 2n
2
n 60.
1
1
2
3
5
2 3 4 5
y
x
Area limn
Sn 3 1 1
3
11
3
3 n 1
n
n 12n 1
6
1
n3n 2n
nn 1
2
1
n2
nn 12n 1
6
1
n
n
i1
3 2in i2
n2
1
n
n
i1
4 4in 1 2i
n
i2
n2
1
n
n
i1
41 in 1 i
n2
Sn n
i1
f1 in1
n
fy 4y y2, 1 y 2. Note:y 2 1n 1
n
62.
Area limn
Sn 2 1
4 1
3
2
19
4
2 n 12
n24
1
2n 12n 1
n2
3n 1
2n
1
n2n 1
n3
n2n 12
4
3
n2nn 12n 1
6
3
nnn 1
2
1
n
n
i12 i
3
n3
3i2
n2
3i
n
n
i11 in
3
11n
2
1
2
3
4
5
4 6 8 10
y
x
Sn n
i1
h1 in1
n
hy y3 1, 1 y 2 Note:y 1n
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Section 4.2 Area 461
64.
Let
53
14 2 9
4 6 254 10
49
4 14
Area
n
i1fci
x
4
i1ci2
4ci1
c4 7
2c3
5
2,c2
3
2,c1
1
2,x 1,
ci xi xi1
2.
n 40 x 4,fx x2 4x, 66.
Let
8sin
16 sin
3
16 sin
5
16 sin
7
16 1.006
Area
n
i1fcix
4
i1sin ci
8
c4 7
16c3
5
16,c2
3
16,c1
16,x
8,
ci xi xi1
2.
n 40 x
2,fx sinx,
68. on 2, 6.fx 8
x2 1
n 4 8 12 16 20
2.3397 2.3755 2.3824 2.3848 2.3860Approximate area
70. on 0, 2.fx cosx
n 4 8 12 16 20
1.1041 1.1053 1.1055 1.1056 1.1056Approximate area
72. See the Definition of Area. Page 259.
74.
(Note: exact answer is 12.)
0 x 8fx 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 12.001Mn
Sn
sn
76.
a. square unitsA 3
x1 2 3 4
1
2
3
4
y 78. True. (Theorem 4.3)
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462 Chapter 4 Integration
Section 4.3 Riemann Sums and Definite Integrals
2.
limn
1
n43n4
4
n3
2
n2
4 limn3
4
1
2n
1
4n2 3
4
limn
1
n43n4 6n3 3n2
4
2n3 3n2 n
2
n2 n
2
limn
1
n43n2n 12
4 3nn 12n 1
6 nn 1
2
limn
1
n4
n
i1
3i 3 3i 2 i
limn
n
i1
fcixi limn
n
i1
3i
3
n3
3i 2 3i 1
n3
xi i 3
n3
i 13
n3
3i 2 3i 1
n3
ci i 3
n3x 1,x 0,y 0,fx 3x,
80. (a)
(b)
(c)
Let As
limn
An limx0
r2sinxx r21 r2x0.n,x 2n.
An n12r2 sin2
n r2n
2sin
2
n r2sin2n2n
A 1
2
bh 1
2
rrsin 1
2
r2 sin
h rsin
sin h
r
r
r
h
2
n
82. (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 1k 2
kk 1 2k 1
k1
i1
2i k
i1
2i 2k 1
k
i1
2i kk 1.
n k:
n 1: 2 11 1 2
n
i1
2i nn 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 12
4 k 22
k 12
4k2 4k 1
k2k 12
4 k 13
k1
i1
i3 k
i1
i3 k 13
k
i1
i3 k2k 12
4
n k:
13 121 12
4
4
4
1
n 1
n
i1
i3 n2n 12
4
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Section 4.3 Riemann Sums and Definite Integrals 463
4. on
3
2
xdx limn
52 252n 52
10 25n2nn 1
2 10
25
2 1 1
n 5
2
25
2n
n
i1fcixi
n
i1
f2 5in 5
n n
i12 5in
5
n 10 25
n2
n
i1
i
Note:x 3 2n 5
n, 0 as n2, 3.y x
6. on
6 12 8 26
31
3x2 dx limn
6 12n 1n 4n 12n 1
n2
6 12n 1
n 4
n 12n 1
n2
6
nn 4
nnn 1
2
4
n2nn 12n 1
6
6
nn
i11 4in
4i2
n2
n
i1
31 2in 2
2n
n
i1fcixi
n
i1f1 2in
2
n
1, 3. Note: x 3 1n 2
n, 0 as ny 3x2
8. on
15 27 27 15
2
1
3x2 2dx limn
15 27
n 1
n
27
2
n 12n 1
n2
15 27n 1
n
27
2n 12n 1
n2
3
n3n 18
nnn 1
2
27
n2nn 12n 1
6 2n
3
n
n
i131 6in
9i2
n2 2
3
nn
i131 3in
2
2
n
i1fcixi
n
i1f1 3in 3n
1, 2. Note:x 2 1n 3
n; 0 as ny 3x2 2
10.
on the interval 0, 4.
lim0
n
i1
6ci4 ci2xi 4
0
6x4 x2dx 12.
on the interval 1, 3.
lim0
n
i1 3ci2xi
3
1
3
x2dx
14. 20
4 2xdx 16. 20
x2dx 18. 11
1
x2 1dx 20. 4
0
tanxdx
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464 Chapter 4 Integration
22. 20
y 22dy 24. Rectangle
x
1
2
3
5
a a
Rectangle
y
A aa
4 dx 8a
A bh 24a
26. Triangle
x1 2 3 4
1
1
2
3
Triangle
y
A 40
x
2dx 4
A 1
2bh
1
242
28. Triangle
2
2
4
6
8
10
4 6 8 10
y
x
A 8
0
8 xdx 32
A 1
2bh
1
288 32
30. Triangle
x
Triangle
a a
a
y
A
a
aa xdx a
2
A 1
2bh
1
22aa
32. Semicircle
xr
r
r
r Semicircle
y
A
r
rr2 x2dx 12
r2
A 1
2r2
In Exercises 3440,
4
2x
3
dx
60,
4
2xdx
6,
4
2dx
2.
34. 22
x3 dx 0 36. 42
15 dx 1542
dx 152 30
38. 42
x3 4dx 42
x3 dx 442
dx 60 42 68 40.
62 26 60 36
42
6 2x x3dx 642
dx 242
xdx 42
x3 dx
42. (a)
(b)
(c)
(d) 63
5fxdx 563
fxdx 51 5
33
fxdx 0
3
6
fxdx 6
3
fxdx 1 1
60
fxdx 30
fxdx 63
fxdx 4 1 3 44. (a)
(b)
(c)
(d) 10
3fxdx 310
fxdx 35 15
11
3fxdx 311
fxdx 30 0
1
0
fxdx 0
1
fxdx 5 5 10
01
fxdx 11
fxdx 10
fxdx 0 5 5
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Section 4.3 Riemann Sums and Definite Integrals 465
46. (a)
(c) f even55
fxdx 250
fxdx 24 8
50
fx 2dx 50
fxdx 50
2 dx 4 10 14 (b)
(d) f odd55
fxdx 0
Let u x 2.32
fx 2dx 50
fxdx 4
48. The right endpoint approximation will be less than the
actual area:
52. is integrable on but is not contin-
uous on There is discontinuity at To see
that
is integrable, sketch a graph of the region bounded by
and thex-axis for You see that
the integral equals 0.
x1 22
2
1
2
y
1 x 1.fx xx
11
xx
dx
x 0.1, 1.1, 1,fx xx 54.
b. square unitsA 43
x1 1314 42
1
2
3
4
y
56.
c. Area 27.
y
x1
1
2
3
4
5
6
7
8
9
2 3 4 5 6 7 8 9
58. 30
5
x2 1dx
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 5.9072Rn
Mn
Ln
n
60. 30
x sinxdx
4 8 12 16 20
2.8186 2.9985 3.0434 3.0631 3.0740
3.1784 3.1277 3.1185 3.1152 3.1138
3.1361 3.1573 3.1493 3.1425 3.1375Rn
Mn
Ln
n
62. False
10
xxdx 1
0
xdx1
0
xdx64. True 66. False
42
xdx 6
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466 Chapter 4 Integration
68.
12
4 3
2
12 3
2 2
3 1 0.7084
i1fcixi f6x1 f
3x2 f2
3 x3 f3
2 x4c4
3
2c3
2
3 ,c2
3 ,c1
6 ,
x4 x3 2
3,x2
12,x1
4,
x4 2x3 ,x2
3,x1
4,x0 0,
0, 2fx sinx,
70. To find use a geometric approach.
Thus,
20
xdx 12 1 1.
x1 2 31
1
1
2
3
y
2
0 xdx,
Section 4.4 The Fundamental Theorem of Calculus
2.
0
cosxdx 0 0
2
2
fx cosx
6. 72
3 dv 3v7
2
37 32 15
8. 52
3v 4dv 32 v2 4v5
2
752 20 6 8 39
2
10.
38
3
1
3x2 5x 4dx x3 5x2
2 4x
3
1
27 452 12 1 5
2 4
12. 11
t3 9tdt 14t4 9
2t2
1
1
14 9
2 1
4
9
2 0
14. 12u 1u2du
u2
2
1
u1
2
12 1 2 1
2 2
4.
is negative.22
x2 xdx 2 2
5
5
fx
x
2
x
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Section 4.4 The Fundamental Theorem of Calculus 467
16. 33
v13dv 34v433
3
3
433 4 33 4 0
18. 812xdx 2
8
1
x12dx 22x128
1
22x8
1
8 22
20.
2
0
2 ttdt
2
0
2t12 t32dt
4
3
t32 2
5
t52
2
0
tt
15
20 6t
2
0
22
15
20 12 162
15
22.
1
23
5x53
3
8x83
1
8
x53
8024 15x
1
8
1
8039
32
80144
4569
80
18
x x2
2 3xdx
1
21
8
x23 x53dx
24.
4 16 18 9
2
13
2
92 1
2 24 8 18 9
2
x2
2
3
1
6x
x2
2
4
3
31
xdx 43
6 xdx
41
3 1x 31dx 31
3 x 3dx 43
3 x 3dx
26.
4
3 0
4
3
4
3 0 4
1
3 2 3
9 18 9
1
3 2 3
64
3 32 12
9 18 9
x3
3 2x2 3x
1
0
x3
3 2x2 3x
3
1
x3
3 2x2 3x
4
3
40x2 4x 3dx
1
0
x2 4x 3dx 31
x2 4x 3dx 43
x2 4x 3dx
28. 40
1 sin2
cos2d 4
0
d 4
0
4
30. 24
2 csc2xdx 2x cotx2
4 0 2 1
2 1
2
2
32. 22
2t cos tdt t2 sin t2
2
2
4 1
2
4 1 2
34. P 2
2
0
sin d 2cos 2
0
2
0 1
2
63.7%
36. A 11
1 x4dx x 15x51
1
8
538. A 2
1
1
x2dx 1x
2
1
1
2 1
1
2
40. A 0
x sinxdx x2
2 cosx
0
2
2 2
2 4
2
split up the integral at the zeros
x 1, 3
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468 Chapter 4 Integration
44. Since on
A
3
0 3x
x2
dx
3
2x2
x3
3 3
0
9
2.
0, 3,y 0 46.
c 392 1.6510
c3 9
2
9
c3 2
fc3
1
4
31
9
x3dx 92x2
3
1
1
2
9
2 4
48.
c 0.5971
cos c 33
2
fc
3
3 3
33
cosxdx sinx3
3 3 50.
2
3
1
3
16
3
1
3 13
1
4x2 1
x2dx 23
1
1 x2dx 2x 1x3
1
42. Since on
Area 80
1 x13dx x 34x 438
0
8 3
416 20
0. 8,y 0
52.
x 0.881
cosx 2
Average value 2
0.5
0 2.71
0.881, 2((
1.51
2 02
0
cosxdx 2sinx2
0
2
54. (a)
x1 2 3 4 5 6 7
1
2
3
4
A
A A A
1
2 3 4
y
8
1
23 1
1
21 2
1
22 1 31
A1 A2 A3 A4
71
fxdx Sum of the areas (b) Average value
(c)
Average value
x1 2 3 4 5 6 7
1
2
34
5
6
7
y
20
6
10
3
A 8 62 20
71
fxdx
7 1
8
6
4
3
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Section 4.4 The Fundamental Theorem of Calculus 469
56.
3.5 1.5 5.0
62
fxdx area or regionB 60
fxdx 20
fxdx 58. 20
2fxdx 220
fxdx 21.5 3.0
60. Average value 1
66
0
fxdx 1
63.5 0.5833 62.
1
R 0R0
kR2 r
2dr
k
RR2r
r3
3R
0
2kR
2
3
64.
(a)
Average profit
(b)
(c) The definite integral yields a better approximation.
1
66.5
0.5
5t 30dt1
652
3t32 30t
6.5
5.0
954.061
6 159.010
1
6155 157.071 158.660 160 161.180 162.247
954.158
6 159.026
P 5t 30
t 1 2 3 4 5 6
P 155 157.071 158.660 160 161.180 162.247
66. (a)
(b)
1 5
20
100
R 2.33t4 14.67t3 3.67t2 70.67t (c)
181.957
4
0
Rtdt 2.33t5
5 14.67t
4
4 3.67t
3
3 70.67t
2
2 4
0
68. (a) histogram
t
2
21
4
43
68
10
12
14
16
18
65 8 97
N
(b) customers
(c) Using a graphing utility, you obtain
(d)
(e)
The estimated number of customers is
(f) Between 3 P.M. and 7 P.M., the number of customers is approximately
Hence, per minute.3017240 12.6
7
3
Ntdt60 50.2860 3017.85.16260 5110.
90
Ntdt 85.162
2
2 10
16
Nt 0.084175t3 0.63492t2 0.79052 4.10317.
49806 7 9 12 15 14 11 7 260 8360
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470 Chapter 4 Integration
70.
F8 84
4 64 16 4 1068
F5 625
4 25 10 4 167.25
Note: F2 2
2
t3 2t 2dt 0F2 4 4 4 4 0
x4
4x2 2x 4
x4
4x2 2x 4 4 4
Fx x2
t3 2t 2dt t4
4 t2 2t
x
2
72.
F8 1
64
1
4
15
64
F5
1
25
1
4
21
100 0.21
F2 1
4
1
4 0
Fx x2
2
t3dt x
2
2t3 dt1
t2x
2
1
x2
1
4
74.
F8 1 cos 8 1.1455
F5 1 cos 5 0.7163
F2 1 cos 2 1.4161
Fx x0
sin d cos x
0
cosx cos 0 1 cosx
76. (a)
(b)d
dx1
4x4
1
2x2 x3 x xx2 1
x
0 tt2
1dt
x
0 t3
tdt
1
4t4
1
2t2
x
0
1
4x4
1
2x2
x2
4 x2
2
78. (a)
(b)d
dx2
3x32
16
3 x12 x
x4
tdt 23t32x
4
2
3x32
16
3
2
3x32 8 80. (a)
(b)d
dxsecx 2 secx tanx
x3
sec ttan tdt sec tx
3 secx 2
82.
Fx x2
x2 1
Fx x1
t2
t2 1dt 84.
Fx 4x
Fx x1
4tdt 86.
Fx sec3x
Fx x0
sec3tdt
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Section 4.4 The Fundamental Theorem of Calculus 471
88.
Fx 0
Fx xx
t3dt t4
4x
x 0 Alternate solution
Fx x31 x3 0
x
0t3
dt x
0t3
dt
0x
t3dt x0
t3dt
Fx xx
t3dt
90.
Alternate solution: Fx x232x 2x5
Fx x2
2
t3 dt t2
2x2
2
12t2x2
2
1
2x4
1
8Fx 2x5
92.
Fx sin x222x 2x sinx 4
Fx x2
0
sin 2d
94. (a) (b)
(c) Minimum ofg at
(d) Minimum at Relative maximum at
(e) On g increases at a rate of
(f) Zeros ofg: x 3,x 8.
12
4 3.6, 10
2, 2.10, 6.
6, 6.
2
2 4 6 10 122
4
6
4
6
y
x
x 1 2 3 4 5 6 7 8 9 10
1 2 0 0 3 63642gx
96. (a)
Horizontal asymptote: y 4
limt
gt 4
gt 4 4t2
(b)
The graph of does not have a horizontal asymptote.Ax
limx
Ax limx
4x 4x 8 0 8
4x2 8x 4
x
4x 12
x
4t 4tx
1
4x 4
x 8
Ax x
1 4 4t2dt
98. True 100. Let be an antiderivative of Then,
.fvxvx fuxux
Fvxvx Fuxux
d
dxvx
uxftdt ddxFvx Fux
vx
uxftdt Ft
vx
ux Fvx Fux
ft.Ft
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Section 4.5 Integration by Substitution
du gxdxu gxfgxgxdx
2.
4. 2x 2 dx
6. sinx cosx dxcosxsin2x dxsec 2x tan 2xdx
3x2dxx3 1x2x3 1 dx
12.
Check:d
dxx3 55
15 C 5x
3 543x2
15 x3 54x2
x2x3 54dx 13x3 543x2dx 13x3 55
5 C
x3 55
15 C
14.
Check:d
dx4x2 34
32 C 44x
2 338x
32x4x2 33
x4x2 33dx 184x2 338xdx 184x2 34
4 C4x2 34
32 C
102.
(a)
(c)
(d) G0 0 f0 0
0
ftdt 0
Gx x fx x0
ftdt
G0 00s
s
0
ftdtds 0Gx x
0s
s
0
ftdtds(b) Let
G0 000
ftdt 0
Gx Fx xx
0
ftdt
Gx x0
Fsds
Fs ss0
ftdt.
104.
Using a graphing utility,
Total distance 50xtdt 27.37 units
xt 3t2 14t 15
xt t 1t 32 t3 7t2 15t 9
472 Chapter 4 Integration
10.
Check:d
dx3
41 2x243 C 34
4
31 2x2134x 1 2x2134x
1 2x2134xdx 341 2x243 C
8.
Check:d
dxx2 94
4 C 4x
2 93
42x x2 932x
x2 932xdx x2 944 C
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16.
Check:d
dt1
6t4 532 C 16
3
2t4 5124t3 t4 512t3
t3t4 5 dt 14t4 5124t3dt 14t4 5 32
32 C
1
6t4 532 C
18.
Check:d
du2u3 232
9 C 29
3
2u3 2123u2 u3 212u2
u2u3 2 du
1
3u3 2123u2du
1
3
u3 232
32 C
2u3 232
9
C
20.
Check:d
dx1
41 x4 C 141 x424x3
x3
1 x42
x31 x42dx 141 x424x3dx 141 x41 C 141 x4 C
22.
Check:d
dx1
316 x3 C 13116 x323x2
x2
16 x32
x216 x32dx 1316 x323x2dx 1316 x31
1 C1
316 x3 C
24.
Check:d
dx1 x4
2 C 12 121 x4124x3 x
3
1 x4
x31 x4
dx 1
41 x4124x3dx 1
41 x412
12 C
1 x4
2 C
26.
Check:
d
dx1
3x3
1
9x1
C x2
1
9x2
x2
1
3x2
x2 13x2dx x2 19x2dx x3
3
1
9x1
1 Cx3
3
1
9x C
3x4 1
9x C
28.
Check:d
dxx C
1
2x
12x
dx 1
2x12dx 12x12
12 Cx C
30.
Check:d
dt2
3t32
4
5t52 C t12 2t32 t 2t
2
t
t 2t2t
dt t12 2t32dt 23t32 45t52 C 215t325 6t C
32.
Check:d
dt1
12t4
1
4t C 13t3
1
4t2
t33 14t2dt 13t3 14t2dt 13t4
4 1
4t1
1 C1
12t4
1
4t C
Section 4.5 Integration by Substitution 473
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34.
Check:d
dy4y2
714 y32 C ddy24y2
2
7y72 C 16y 2y52 2y8 y32
2y8 y32dy 28y y52dy 24y2 27y72 C 4y2
714 y32 C
36.
20
31 x3 C
10
3 1 x312
12 C
10
3 1 x3123x2dxy
10x2
1 x3dx 38.
x2 8x 1 C
1
2x2 8x 112
12 C
1
2x2 8x 1122x 8dxy
x 4
x2
8x 1
dx
42. 4x3 sinx4dx sinx44x3dx cosx4 C 44. cos 6xdx 16cos 6x6dx 16 sin 6x C
46. x sinx2dx 12sinx22xdx 12 cosx2 C
48. sec1 x tan1 xdx sec1 x tan1 x1dx sec1 x C
50. tanx sec2xdx tanx3232 C 23tanx32 C
52. sinxcos3xdx cosx3sinxdx cosx2
2 C
1
2 cos2x C
1
2sec2x C
54. csc2x2dx 2csc2x212dx 2 cotx2 C 56.Since Thus
fx sec x 1.
C 1.f13 1 sec3 C,
fx sec x tan xdx sec x C
40. (a)
x55
2
3
3
y (b)
y 1
2sinx2 1
1 1
2sin0 C C 10, 1:
1
2sin x2 C
y x cosx2dx 12cosx22xdx
dy
dxx cosx2, 0, 1
474 Chapter 4 Integration
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58.
1
152x 1323x 1 C
1
302x 1326x 2 C
1
302x 13232x 1 5C
u32
303u 5 C
1
42
5u52
2
3u32 C
1
4u32 u12dux2x 1 dx 12u 1u12du
dx 1
2dux
1
2u 1,u 2x 1, 60.
2
52 x32x 3C
2
52x325 2 xC
2u32
55 uC
2u32 2
5u52 C
3u12 u32dux 12 xdx 3 uudu
dx dux 2 u,u 2 x,
62. Let
2
3x 42x 13 C
2
3
x 42x 4 21 C
2
3u122u 21 C
4
3u32 14u12 C
2u12 7u12du 2x 1x 4
dx 2u 4 1u
du
u x 4,x u 4, du dx. 64.
3
7t 443t 3 C
3
7
t 443t 4 7 C
3u43
7u 7 C
3
7u73 3u43 C
u43 4u13dut3t 4 dt u 4u13du
dt dut u 4,u t 4,
66. Let
1
964 83 8 83 41,472
42
x2x3 82dx 1
34
2
x3 823x2dx 13x3 83
3 4
2
u x3 8, du 3x2dx.
68. Let
1
0
x1 x2dx 1
21
0
1 x2122xdx 131 x2321
0
0 1
3
1
3
du 2xdx.u 1 x2,
70. Let
20
x
1 2x2dx
1
42
0
1 2x2124xdx 121 2x22
0
3
2
1
2 1
du 4xdx.u 1 2x2,
Section 4.5 Integration by Substitution 475
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74. Let
When When
16
3
1
42
327 23 23 2
1
42
3u32 2u12
9
1
51
x
2x 1dx 9
1
12u 1
u1
2du
1
491
u12 u12du
x 5, u 9.x 1, u 1.
u 2x 1, du 2 dx,x 1
2u 1.
76.
2
3
x cosxdx x2
2 sinx
2
3
2
8 1
2
183
2 52
72
2 3
2
78.
When When
80
u73 4u43 4u13du 310u103 12
7u73 3u43
8
0
4752
35Area 6
2
x23x 2 dx 80
u 223udu
u 8.x 6,u 0.x 2,
dx dux u 2,u x 2,
80. A 0
sinx cos 2xdx cosx 12 sin 2x
0
2
82. Let
Area 412
csc 2x cot 2xdx 1
24
12csc 2x cot 2x2dx 12 csc 2x
4
12
1
2
du
2 dx.u
2x,
84.
0
0
2
20
20
x3x 2 dx 7.581 86.
1 5
0
50
51
x2x 1 dx 67.505 88.
0
1
2
2
20
sin 2xdx 1.0
90.
They differ by a constant: .C2 C1 1
2
cos2x
2 C2
1 sin2x
2 C2
sin2x
2
1
2 C2
sinx cosxdx cosx1sinxdx cos2x2 C2sinx cosxdx sinx1cosxdx sin2x2 C1
72. Let
20
x34 x2dx 1
22
0
4 x2132xdx 384 x2432
0
3
8843 443 6
3
234 3.619
du 2xdx.u 4 x2,
476 Chapter 4 Integration
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92. is even.
2
3
2sin3x
3 2
0
22
sin2x cosxdx 20
sin2xcosxdx
fx sin2x cosx 94. is odd.
22
sinx cosxdx 0
fx sinx cosx
96. (a) since sinx is symmetric to the origin.
(b) since cosx is symmetric to they-axis.
(c)
(d) since and hence, is symmetric to the origin.sinx cosx sinx cosx22
sinx cosxdx 0
22
cosxdx 220
cosxdx 2 sinx2
0
2
44
cosxdx 240
cosxdx 2 sinx4
0
2
44
sinxdx 0
98.
sin 3x cos 3xdx
sin 3xdx
cos 3xdx 0 2
0
cos 3xdx 23 sin 3x
0
0
100. If then and x5 x23dx 125 x232xdx 12u3du.du 2xdxu 5 x2,
102.
Thus, When
$250,000.
Q50 t 50,Qt 2100 t3.
Q0 k
31003 2,000,000 k 6
Qt k
3100 t3
Q100 C 0
Qt k100 t2dt k3100 t3 CdQ
dt k100 t2 104.
(a)
Relative minimum: or June
Relative maximum: or January
(b) inches
(c) inches1
312
9
Rtdt1
313 4.33
120
Rtdt 37.47
0.4, 5.5
6.4, 0.7
0 12
0
8
R 3.121 2.399 sin0.524t 1.377
106. (a)
Maximum flow: at .
is a relative maximum.18.861, 61.178t 9.36R 61.713
0 24
0
70 (b) (5 thousand of gallons)Volume
24
0
Rtdt
1272
Section 4.5 Integration by Substitution 477
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108. (a)
0 9.4
4
f
g
4
(e)
The graph ofh is that ofg shifted 2 units downward.
2
0
fxdx t
2fxdx 2 ht.
gt t0
fxdx
0 9.4
4
4
(b) g is nonnegative because the graph offis positive at the
beginning, and generally has more positive sections than
negative ones.
(d) No, some zeros off, like do not correspond toan extrema ofg. The graph ofg continues to increase
after becausefremains above thex-axis.x 2
x 2,(c) The points on g that correspond to the extrema offarepoints of inflection ofg.
110. False
xx2 12dx 12x2 12xdx 14x2 12 C
112. True
ba
sinxdx cosxb
a cos b cos a cosb 2 cos a b2
a
sinxdx
114. False
sin2 2x cos 2xdx 12sin 2x22 cos 2xdx 12sin 2x3
3 C
1
6sin3 2x C
116. Becausefis odd, Then
Let in the first integral.
When When
a0
fudu a0
fxdx 0
1a
fxdx a0
fudu a0
fxdx
x a, u a.x 0, u 0.
x u, dx du
a0
fxdx a0
fxdx.
aa
fxdx 0a
fxdx a0
fxdx
fx fx.
478 Chapter 4 Integration
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Section 4.6 Numerical Integration
Section 4.6 Numerical Integration 479
2. Exact:
Trapezoidal:
Simpsons: 10x
2
2 1dx 1121 4
142
2 1 212
2
2 1 434
2
2 1 1
2
2 1 76 1.1667
10x
2
2 1dx 181 2
142
2 1 212
2
2 1 234
2
2 1 1
2
2 1 7564 1.1719
10x
2
2 1dx x
3
6x
1
0
7
6 1.1667
4. Exact:
Trapezoidal:
Simpsons: 21
1
x2dx
1
121 44
52
2462
4472
1
4 0.5004
21
1
x2dx
1
81 24
52
2462
2472
1
4 0.5090
21
1
x2dx 1x
2
1
0.5000
6. Exact:
Trapezoidal:
Simpsons: 80
3xdx 1
30 4 2 32 4 33 2 34 4 35 2 36 4 37 2 11.8632
80
3xdx 1
20 2 2 32 2 33 2 34 2 35 2 36 2 37 2 11.7296
8
0
3xdx 34x438
0
12.0000
8. Exact:
Trapezoidal:
Simpsons: 3
1
4 x2dx 1
63 44 9
4 0 44 25
4 5 0.6667
31
4 x2dx 1
43 24 3
22
20 24 522
5 0.750031
4 x2dx 4x x3
3 3
1
3 11
3
2
3 0.6667
10. Exact:
Trapezoidal:
Simpsons: 20
xx2 1 dx 1
60 41
2122 1 2112 1 43
2322 1 222 1 3.392
20
xx2 1 dx 1
40 21
2122 1 2112 1 23
2322 1 222 1 3.457
20
xx2 1 dx 1
3x2 1322
0
1
3532 1 3.393
12. Trapezoidal:
Simpsons:
Graphing utility: 1.402
20
1
1 x3dx
1
61 41
1 123 21
1 13 4 11 323
1
3 1.405
2
0
11 x3
dx 141 2 11 123 2 11 13 2 11 323 13 1.397
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480 Chapter 4 Integration
14. Trapezoidal:
Simpsons:
Graphing utility: 1.458
2x sinxdx
242 458 sin58 234 sin34 478 sin78 0 1.4582
x sinxdx
162 1 258 sin58 234 sin34 278 sin78 0 1.430
16. Trapezoidal:
Simpsons:
Graphing utility: 0.256
0.257
40
tanx2dx 4
12 tan 0 4 tan44 2
2 tan42 2
4 tan344 2
tan42
0.2714
0
tanx2dx 48 tan 0 2 tan
44
2
2 tan42 2
2 tan344 2
tan
42
18. Trapezoidal:
Simpsons:
Graphing utility: 1.910
2
0
1 cos2xdx
242 41 cos28 21 cos24 41 cos238 1 1.910
20
1 cos2xdx
162 21 cos28 21 cos24 21 cos238 1 1.910
20. Trapezoidal:
Simpsons:
Graphing utility: 1.852
0
sinx
xdx
121 4 sin4
4
2 sin22
4 sin34
34 0 1.852
0
sinx
xdx
81 2 sin4
4
2 sin22
2 sin34
34 0 1.836
22. Trapezoidal: Linear polynomials
Simpsons: Quadratic polynomials
24.
(a) Trapezoidal: since
is maximum in when
(b) Simpsons:
since is maximum in whenx 0.0, 1f4x
Error 1 05
1804424
1
1920 0.0005
x 0.0, 1fx
Error 1 02
12422
1
96 0.01
f4x 24x 15
fx 6
x 14
fx 2
x 13
fx 1
x 12
fx 1
x 1
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Section 4.6 Numerical Integration 481
26. in .
(a) is maximum when and
Trapezoidal: Error < 0.00001, > 16,666.67, n > 129.10; let
in
(b) is maximum when and
Simpsons: Error > 13,333.33, n > 10.75; let (In Simpsons Rule n must be even.)n 12.n41
180n424 < 0.00001,
f40 24.x 0f
4x
0, 1f4x 24
1 x5
n 130.n21
12n22
f0 2.x 0fx
0, 1fx 2
1 x3
28.
(a) in .
is maximum when and
Trapezoidal: Error < 0.00001, > 14,814.81, n > 121.72; let
(b) in
is maximum when and
Simpsons: Error < 0.00001, > 12,290.81, n > 10.53; let (In Simpsons Rule n must
be even.)
n 12.n432
180n456
81
f40
56
81.x 0f
4x
0, 2f4x 56
81x 1103
n 122.n28
12n4
2
9
f0 2
9.x 0fx
0, 2fx 2
9x 143
fx x 123
30.
(a) in .
is maximum when and
Trapezoidal: Error < 0.00001, > 19,044.17, n > 138.00; let
(b) in
is maximum when and
Simpsons: Error < 0.00001, > 15,793.61, n > 11.21; let n 12.n41 05
180n428.4285
f40.852 28.4285.x 0.852f
4x
0, 1f4x 16x4 12 sinx2 48x2 cosx2
n 139.n21 03
12n22.2853
f1 2.2853.x
1fx
0, 1fx 22x2 sinx2 cosx2
fx sinx2
32. The program will vary depending upon the computer or programmable calculator that you use.
34. on .0, 1fx 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
SnTnRnMnLn
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482 Chapter 4 Integration
36. on .1, 2fx sinx
x
n
4 0.7070 0.6597 0.6103 0.6586 0.6593
8 0.6833 0.6594 0.6350 0.6592 0.6593
10 0.6786 0.6594 0.6399 0.6592 0.6593
12 0.6754 0.6594 0.6431 0.6593 0.6593
16 0.6714 0.6594 0.6472 0.6593 0.6593
20 0.6690 0.6593 0.6496 0.6593 0.6593
SnTnRnMnLn
38. Simpsons Rule:
17.476
83201 23 sin2d
3
6 1 23 sin2 0 41 23 sin2 16 21 23 sin28 . . . 1 23 sin22
n 8
40. (a) Trapezoidal:
Simpsons:
20
fxdx 2
384.32 44.36 24.58 45.79 26.14 47.25 27.64 48.08 8.14 12.592
20
fxdx 2
284.32 24.36 24.58 25.79 26.14 27.25 27.64 28.08 8.14 12.518
(b) Using a graphing utility,
Integrating,20
ydx 12.53
y 1.3727x3 4.0092x2 0.6202x 4.2844
42. Simpsons Rule:
3.14159
410
1
1 x2dx
4
361 4
1 162
2
1 262
4
1 362
2
1 462
4
1 562
1
2
n 6
44.
7435 sq m
Area 120
21275 281 284 276 267 268 269 272 268 256 242 223 0
46. The quadratic polynomial
passes through the three points.
px x x2x x3
x1 x2x1 x3y1
x x1x x3
x2 x1x2 x3y2
x x1x x2
x3 x1x3 x2y3