139)
ex
1 + 2exdx
139)
A)
ln 1 + 2ex+ C
B)
e2x + C
C)
1
2 ln 1 + 2ex+ C
D)
(1 + 2ex)2+ C
E)
none of these
140)
1
0
x3exdx; n = 4
Express your answer to four decimal places.
140)
A)
1.2392
B)
4.9568
C)
0.6196
D)
0.9593
141)
A money market fund has a continuous flow of money at a rate of 0.09x +700 dollars for 10 years.
Find the present value of this flow if interest is earned at 4% compounded continuously.
141)
A)
$4440.66
B)
$5772.86
C)
$4810.72
D)
$29,339.64
142)
eax sin bx dx
142)
A)
eax a sin bx – b cos bx
a2+b2+ C
B)
eax(sin bx – cos bx) + C
C)
1
2b eax cos2 bx + C
D)
eax(sin bx + cos bx) + C
143)
x3ln 7x dx
143)
A)
1
4x4 ln 7x –1
20 x5+ C
B)
ln 7x –1
4 x4+ C
C)
1
4x4 ln 7x +1
16 x4+ C
D)
1
4 x4 ln 7x –1
16 x4+ C
144)
x
(7x2+ 3)5 dx
144)
A)
–7
3(7x2+ 3)4+ C
B)
–1
56(7x2+ 3)4+ C
C)
–1
14(7x2+ 3)6+ C
D)
–7
3(7x2+ 3)6+ C
145)
(x + 1)ex dx
145)
A)
xex+ C
B)
(x + 1)ex+ C
C)
xex + 1 + C
D)
(x + 2)ex+ C
146)
5x4 dx
(8 +x5)5
146)
A)
1
6(8 +x5)6+ C
B)
–1
4(8 +x5)4+ C
C)
–1
6(8 +x5)6+ C
D)
–5x4
(8 +x5)4+ C
147)
(3x +5) e–2x dx
147)
A)
–6x e–2x –22 e–2x + C
B)
–3
2x e–2x –13
4e–2x + C
C)
3
2x e–2x +13
4e–2x + C
D)
–3
2 x e–2x –e–2x + C
148)
e2x sin 3x dx
148)
A)
1
13 e2x(2 sin 3x – 3 cos 3x) + C
B)
1
13 e2x(sin 3x – 3 cos 3x) + C
C)
–e2x(sin 3x – 3 cos 3x) + C
D)
e2x(sin 3x – cos 3x) + C
149)
An investment is expected to produce a uniform continuous rate of money flow of $500 per year for
10 years. Find the present value at 6% compounded continuously.
149)
A)
$4573.43
B)
$3759.90
C)
$12,906.76
D)
$6850.99
150)
x6x7+3 dx
150)
A)
2
21 (x7+3)3/2 + C
B)
2
21 x7(x7+3)3/2 + C
C)
2
3(x7+3)3/2 + C
D)
1
14 x7+3
+ C
151)
9x ln x dx
151)
A)
9
2x2 ln x –x2
4+ C
B)
9
2x ln x –9
4x + C
C)
9
2x2 ln x –9
4x2+ C
D)
x2
2 ln x –x2
4+ C
152)
1
dx
x
152)
A)
1
B)
divergent
C)
–1
D)
0
153)
–8 x –5; n = 6
153)
A)
x = 0.5; x1= – 7.5, x2= – 7.25, x3= – 6.75, x4= – 6.25, x5= – 5.75, x6= – 5
B)
x = 0.25; x1= – 7.75, x2= – 7.25, x3= – 6.75, x4= – 6.25, x5= – 5.75, x6= – 5.25
C)
x = 0.5; x1= – 7.75, x2= – 7.25, x3= – 6.75, x4= – 6.25, x5= – 5.75, x6= – 5.25
D)
x = 0.5; x1= – 7.5, x2= – 7, x3= – 6.5, x4= – 6, x5= – 5.5, x6= – 5
154)
12 sin x cos x dx
154)
A)
–8(cos x)3/2 + C
B)
–18(cos x)3/2 + C
C)
18(cos x)3/2 + C
D)
8(cos x)3/2 + C
155)
The following data give the marginal cost for different levels of production at Zipperty–Doo–Dah
Inc. Here x represents the number of zippers produced and C'(x) is in dollars per zipper.
Approximate the total in going from a production level of 50 zippers to 90 zippers.
x50 60 70 80 90
C'(x) 4.5 5.0 5.3 5.8 6.5
155)
A)
$216.00
B)
$43.20
C)
$432.00
D)
$135.50
156)
2
1
–e1/x
x2 dx
156)
A)
e1/2
B)
e1/2 – e
C)
e
D)
e –e2
157)
e1/x
x2dx
157)
A)
xe1/x + C
B)
–xe1/x + C
C)
–e1/x + C
D)
e1/x + C
E)
none of these
158)
4
1
3t dt
158)
A)
51
2
B)
24
C)
45
2
D)
765
4
159)
4
0
xe– x dx
159)
A)
–5e–4+ 1
B)
–5e–4
C)
–5e–4– 1
D)
–3e–4+ 1
160)
x2e–xdx
160)
A)
–(2x +2)e–x+ C
B)
(x2+ 2x)e–x+ C
C)
(x2+ 2x +2)e–x+ C
D)
–(x2+ 2x +2)e–x+ C
161)
sin cos2
1 +cos3

161)
A)
sin2 cos2 + C
B)
–1
3ln 1 +cos3 + C
C)
1
2(sin cos2)2+ C
D)
1
2(1 +cos3)2+ C
E)
none of these
162)
The rate of a continuous money flow is 1000e–.4 dollars per year for 10 years. Find the present
value if interest is earned at 3% compounded continuously.
162)
A)
$7191.64
B)
$21,379.79
C)
$28,767.90
D)
$14,482.18
163)
 
Consider
1/2
0
f(x)dx, where f(x) =e–2x. Find a number A such that f(x) A for all x
satisfying 0 x 1
2. Use this A to obtain a bound on the error of using Simpson’s rule with n =5 to
approximate the definite integral.
163)
A)
1.1111 × 10–6
B)
6.9444 × 10–6
C)
2.7778 × 10–7
D)
1.6667 × 10–3
164)

Consider
2
0
f(x)dx, where f(x) =1
20 x5+2x2. Find a number A such that f(x) A for all x
satisfying 0 x 2. Use this A to obtain a bound on the error of using the midpoint rule with n =8
to approximate the definite integral.
164)
A)
0.0208
B)
0.0078
C)
0.3542
D)
0.0625
165)
7
0
xexdx
165)
A)
8e7+ 1
B)
6e7
C)
6e7+ 1
D)
6e7– 1
166)
3(2x + 5)3 dx
166)
A)
1
4(2x + 5)4+ C
B)
1
2(2x + 5)4+ C
C)
3
4(2x + 5)4+ C
D)
3
8(2x + 5)4+ C