Download Name: Instructor: Z 17.(6 pts.) Let g(x) =

17.(6 pts.) Let g(x) =
Z
0
t2 dt. Find g 0 (x).
sin x
(a)
−(cos x)2 cos x
(b)
−(sin x)2 cos x
(c)
(cos x)2 cos x
(d)
−(sin x)2 sin x
(e)
(sin x)2 cos x
MTH 142B FALL 2014: PRACTICE FOR EXAM 1
This is how Monday’s exam will look like. There are 7 multiple choice questions worth 7 points each and 2 partial credit questions carrying 25 points each.
The duration of the exam is 50 minutes. The solutions to the partial credit questions must be written in all detail in order to earn full credit. The answers are
given in the last page. No calculators, books, notes etc. are allowed. There is a
formula sheet on page 7. Please ignore the points written in front of each
question.
18.(6 pts.) Calculate the integral
(a)
(d)
√
5−1
√
5
Z
2
√
0
x
dx
+1
x2
(b)
√
− 5−1
(e)
4
(c)
1−
√
5
Name:
Instructor:
√
23.(6 pts.) Consider the plane region bounded by the graphs of y = x, y = 0 and
x = 1. Rotate this region about the line y = −3 and calculate the volume.
(a)
3π
3
(b)
9π
2
(c)
1
7π
2
10
(d)
15π
2
(e)
27π
2
(e)
Z
√
2
√
− 2
(8 − 2x4 ) dx
Name:
2
MTH 142B FALL 2014: PRACTICE FOR EXAM 1
Instructor:
22.(6 pts.)√The plane region bounded below by the graph of y = x and above by the
graph
y = Differentiate
x is rotated the
about
the line x = 5. Which integral below gives the volume?
3.(7 pts.)
function
(a)
(a)
(b)
(b)
(c)
(c)
(d)
(d)
(e)
(e)
π
Z
1
0
(5 −
√
x)2 − (5 − x)2 dx
(x2 − 1)4
.
f (x) = √
x2 + 1
Z 1
2
4
x(x
−
1)
8
1
√
0
√2 − (5 − x)2 2 dx + 2
πf (x)(5=− x)
x −1 x +1
x2 + 1
0
Z 1
1
(x2 − √
1)4
4
0
−
f (x)(x
=−√5) · ( x − x)dx
2π
x2 + 1 x2 − 1 x2 + 1
0
Z 1
(x2 − 1)4 √ 4
1
0
f (x) (5
= −√x) · (x − x)dx
+
2π
x2 + 1 x2 − 1 x2 + 1
0
Z 1
2
4
√
(x
−
1)
8
1
f 0 (x)(5=−√x) · ( x − x)dx
−
2π
x2 + 1 x2 − 1 x2 + 1
0
1
x(x2 − 1)4
8
0
−
f (x) = √
x2 − 1 x2 + 1
x2 + 1
12
4.(7 pts.) Compute the integral
Z
2e2
2e
(a)
2
(b)
3
2
(c)
1
x dx.
x(ln )2
2
1
2
(d)
3
1
(e)
0
Name:
Instructor:
MTH 142B FALL 2014: PRACTICE FOR EXAM 1
3
5.(7 pts.) Which of the following expressions gives the partial fraction decomposition of
the function
x2 − 2x + 6
f (x) = 3
?
x (x − 3)(x2 + 4)
(a)
(b)
(c)
(d)
(e)
A
C
B
+ 2
+
3
x
x−3 x +4
A
Cx + D
B
+
+
x3 x − 3
x2 + 4
A
B
C
D
E
+ 2+ +
+ 2
3
x
x
x
x−3 x +4
A
B
C
D
Ex + F
+ 2+ +
+ 2
3
x
x
x
x−3
x +4
B
C
D
E
F
A
+
+
+
+
+
x3 x2
x
x−3 x+2 x−2
6.(7 pts.) Find f 0 (x) if
f (x) = xln x .
(a)
2(ln x)xln x
(b)
xln x ln x
(c)
2(ln x)x(ln x)−1
(d)
xln x (ln x + 1)
(e)
x(ln x)−1 ln x
4
Name:
4
MTH 142B FALL 2014: PRACTICE FOR EXAM
1
Instructor:
7.(7 pts.) Calculate the following integral.
Z 1
arctan x
dx .
1 + x2
0
(a)
1
2
(b)
π
8
8.(7 pts.) Evaluate the integral
(c)
Z
π2
32
(d)
ln 2
(e)
π2
8
1
24
(e)
1
4
π/2
sin3 (x) cos5 (x)dx.
0
(a)
0
(b)
π
2
(c)
−
1
24
5
(d)
Name:
MTH 142B FALL 2014: PRACTICE FOR EXAM 1
Instructor:
Partial Credit: You must show all your work to receive credit!
10.(11 pts.) Evaluate the integral
Z
x2 cos(2x)dx.
7
5
6
Name:
MTH 142B FALL 2014: PRACTICE FOR EXAM 1
Instructor:
Partial Credit: You must show all your work to receive credit!
11.(11 pts.) Evaluate:
Z
1 3√
x 9 − x2 dx.
3
8
Name:
MTH 142B FALL 2014: PRACTICE FOR EXAM 1
Instructor:
The following is the list of useful trigonometric formulas:
sin2 x + cos2 x = 1
1 + tan2 x = sec2 x
1
sin2 x = (1 − cos 2x)
2
1
cos2 x = (1 + cos 2x)
2
sin 2x = 2 sin x cos x
1
sin(x − y) + sin(x + y)
2
1
sin x sin y =
cos(x − y) − cos(x + y)
2
1
cos x cos y =
cos(x − y) + cos(x + y)
2
sin x cos y =
The hyperbolic sine and cosine functions are defined to be:
ex + e−x
2
x
e − e−x
sinh x =
2
cosh x =
10
7
8
MTH
³¡
¢1/x2 ´
ln L = lim ln cosh(x)
x→0
¡
¢
ln cosh(x)
= lim
x→0
x2
tanh(x)
= lim
(l’Hospital’s rule)
x→0
2x
142B FALL 2014: PRACTICE
FOR EXAM 1
sech2 (x)
= lim
(l’Hospital’s rule)
x→0
2
Answers
1
= .
2
1
18.(a) 23.(b) 22.(e)
4.(c) 5.(d) 7.(c) 8.(d)
Therefore L = e 2 .
10. Evaluate the integral
Z
x2 cos(2x)dx.
Solution:
Z
x2 cos(2x)dx
Z
1 2
= x sin(2x) − x sin(2x)dx (integration by parts)
2
·
¸
Z
1 2
1
1
= x sin(2x) − − x cos(2x) − − cos(2x)dx
(integration by parts)
2
2
2
1
1
1
= x2 sin(2x) + x cos(2x) − sin(2x) + C
2
2
4
11. Evaluate
Z
1 3p
x 9 − x2 dx.
3
trigonometric
substitution
with x = 3 sin θ and u
4 Solution: Two approaches work:
SOLUTIONS
TO EXAM
1
substitution with u = 9 − x2 . The method of trigonometric substitution is outlined here,
although the latter method may be somewhat easier.
Z
Z
1 3p
2
x 9 − x dx = 81 sin3 θ cos2 θdθ (x = 3 sin θ, dx = 3 cos θdu)
3
Z
= 81(1 − cos2 θ) sin θ cos2 θdθ (u = cos θ, du = − sin θdθ)
Z
= 81(u4 − u2 )du
¡
¢
81 cos5 θ
1p
− 27 cos3 θ + C
cos θ =
9 − x2
5
3
5
2
3
(9 − x ) 2
=
− (9 − x2 ) 2 + C.
15
=
12. Let C(t) be the concentration of a drug in the bloodstream. As the body eliminates
the drug, C(t) decreases at a rate that is proportional to the amount of the drug that is
present at the time. Thus C 0 (t) = kC(t), where k is a constant. The initial concentration
of the drug is 4 mg/ml. After 5 hours, the concentration is 3 mg/ml.