# MATH 172.504 NAME Examination 2 SIGNATURE ```MATH 172.504
Examination 2
March 8, 2012
NAME
SIGNATURE
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
This exam consists of 8 problems, numbered 1–8. For partial credit you must present your
The use of calculators is not permitted on this exam.
The point value for each question is shown next to each question.
CHECK THIS EXAMINATION BOOKLET BEFORE
YOU START. THERE SHOULD BE 8 PROBLEMS ON
6 PAGES (INCLUDING THIS ONE).
Do not mark in the box below.
1–5
6
7
8
Total
Points
Possible Credit
30
40
12
12
94
NAME
MATH 172
Examination 2
Page 2
Multiple Choice: [6 points each] In each of Problems 1–5, circle the best answer.
Z
1.
π
x cos(x) dx.
Evaluate
0
(A) 2
(B) 1
(C) 0
(D) −1
(E) −2
2.
After trigonometric substitution, the integral
Z
√
x4 x2 − 4 dx
becomes
Z
(A) 16
Z
(B) 32
Z
(C) 64
Z
(D) 32
Z
(E) 64
tan4 θ sec3 θ dθ
tan θ sec4 θ dθ
tan2 θ sec5 θ dθ
sin4 θ cos θ dθ
sin4 θ cos2 θ dθ
March 8, 2012
NAME
MATH 172
Z
3.
Evaluate
1
∞
(x2
Examination 2
Page 3
4x
dx.
+ 1)2
(A) 0
(B) 1
(C) 2
(D) 4
(E) The integral diverges.
4.
Consider the following improper integrals:
Z 1
Z 0
1
I.
dx
II.
ex dx
2
x
−1
−∞
Z
III.
1
∞
1
√ dx
x
Which of these integrals converge?
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
March 8, 2012
NAME
5.
MATH 172
Examination 2
Page 4
If we use Simpson’s rule to approximate
Z 2
f (x) dx
0
with 4 equal subintervals, we obtain which expression below?
6.
(A)
1
f (0) + 4f ( 12 ) + 2f (1) + 4f ( 23 ) + f (2)
6
(B)
1 f (0) + 4f ( 12 ) + 2f (1) + 4f ( 32 ) + f (2)
12
(C)
1 f (0) + 2f ( 12 ) + 4f (1) + 2f ( 23 ) + f (2)
12
(D)
1
f (0) + 2f ( 12 ) + 2f (1) + 2f ( 32 ) + f (2)
8
(E)
1
f (0) + 2f ( 12 ) + 2f (1) + 2f ( 23 ) + f (2)
2
[40 points] Evaluate the following integrals. (This problem continues on the next
page.)
Z
(a)
sin(x)e2x dx
March 8, 2012
NAME
MATH 172
Examination 2
Page 5
Z
(b)
sin(7x) sin(3x) dx
Z
3x2
√
dx
1 − 9x2
Z
x2 − x + 3
dx
x3 + x
(c)
(d)
March 8, 2012
NAME
7.
MATH 172
Examination 2
Page 6
[12 points] Find all solutions of the differential equation
y
dy
&middot;
− sin3 (x) = 0.
4
cos (x) dx
8.
[12 points] Solve the initial value problem,
x
dy
− y = x ln x,
dx
y(e) = 2e.
March 8, 2012
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