Name
MATH 152H
Exam 1
Sections 201/202 (circle one)
1-13
/52
Spring 2016
14
/15
P. Yasskin
15
/20
16
/15
Total
/102
Solutions
Multiple Choice: (4 points each. No part credit.)
1
1. Compute
x3e
x2 1
dx
0
a. e
b.
c.
d.
e.
correct choice
2
1 e2 1
2
e2
2
1 e 1
2
1 1 e
2
Solution: Use the substitution
1
x3e
x2 1
0
1
2
1 ew
w
2
1
2
dx
w
w
x2
1
dw
Parts:
1 e w dw
u
du
1
1 w
2
e w dw
1 ew
1 w
2
ew
1 dw
2
2x dx
w
1
dv
dw
2 ew
x dx
v
2
1
0
x2
w
1
e w dw
ew
1
2
1 e1
e
2
/2
2. Compute
x cos x dx
0
1
a.
b. 1
c.
2
d. 1
e.
2
2
1
correct choice
Solution: Use integration by parts with
u
x
du
/2
dv
dx v
/2
x cos x dx
x sin x
sin x dx
0
sin
d x sin x
dx
2
cos x
cos
2
sin x
cos 0
x cos x
2
sin x
sin x
/2
cos x
0
0
2
Check:
x sin x
cos x dx
1
x cos x
1
3. Compute
2
x 2 ln x dx
1
a. 8 ln 2
1
3
b. 8 ln 2 7
3
9
c. 4 ln 2 1
d. 4 ln 2
2
e. 4 ln 2
3
correct choice
u
ln x
1 dx
du
x
Solution: Use integration by parts with
x 3 ln x
x 3 1 dx
3
3 x
2
3
x ln x x 3
3
9 1
x 2 ln x dx
2
x 2 ln x dx
1
x 3 ln x
3
d
dx
Check:
x3
9
dv
v
x 2 dx
x3
3
x 3 ln x
x 2 dx
x 3 ln x x 3 C
3
3
3
9
8 ln 2 8
1
8 ln 2 7
3
9
9
3
9
x 2 ln x
x3 1
3 x
x2
3
x 2 ln x
sin 3 d
4. Compute
0
a. 1
b.
c.
d.
e.
3
2
3
1
4
3
2
correct choice
Solution: u
sin 3 d
/3
cos
1
sin 3 d
du
cos 2
cos
0
1
Check:
1
3
d
d
1
cos
sin d
sin d
cos 3
3
1
3
cos 3
3
1
u 2 du
cos
0
u
1 cos 3
3
u3
3
C
cos 0
cos
cos 3
3
C
1 cos 3 0
3
4
3
sin
1 3 cos 2
3
sin
sin 1
cos 2
sin 3
2
/3
5. Compute
sec 3 tan 3 d
0
a. 0
b. 56
15
c. 58
15
d. 128
15
e. 136
15
correct choice
Solution: u
/3
sec
sec 3 tan 3 d
0
du
2
2
u2 u2
u4
1 du
1
25
5
tan 2
sec tan d
sec 2
u5
5
u 2 du
1
23
3
1
5
1
3
u2
1
1
2
u3
3
1
58
15
/4
6. Compute
cos 4 cos 2 d
0
a. 1
b. 1
2
c. 1
3
d. 1
correct choice
6
e. 1
12
Solution: Add the identities:
cos 6
cos 4
2
cos 4 cos 2
sin 4 sin 2
cos 2
cos 4
2
cos 4 cos 2
sin 4 sin 2
cos 6
cos 2
/4
cos 4 cos 2 d
0
1
2
1 sin 3
6
2
1
2
/4
cos 6
0
1 sin
2
2
7. Find the area between, y
1
2
x 2 and y
1
6
sin 6
6
1
2
cos 2 d
1
2 cos 4 cos 2
1 1
2
sin 2
2
1
6
/4
0
3x.
a. 81
2
b. 27
2
c. 9
correct choice
2
d. 1
2
e. 10
3
Solution: The curves intersect when x 2
3
3x
A
0
x 2 dx
3x 2
2
x3
3
3
0
27
2
3x or x
9
0, 3.
9
2
3
8. Find the average value of f x
e 3x on the interval
0, 2 .
a. 1 e 6
b.
c.
d.
e.
6
1 e6 1
6
1 e6
3
1 1 e6
2
e6 1
Solution: f ave
9.
correct choice
2
1
2
e 3x dx
0
1 e 3x
6
2
0
1 e6
6
1
A solid has a base which is a circle of radius 2
and has cross sections perpendicular to the y-axis
which are isosceles right triangles with a leg
on the base. Find the volume of the solid.
a. 64
b.
c.
d.
e.
3
128
3
16
3
32
3
32
3
correct choice
Solution: The width of the base is W y
2x
1 WH
2
y3
2 4y
3
2 4
12 4
2
The area of the cross section is A y
2
2
V
A y dy
2
24
2
y 2 dy
2
4 8
2
y 2 . The height is H
y2 2 4
8
3
y2
32 2
3
24
W
y2 .
64
3
4
10.
The region shown at the right is bounded
y
sin x and the x-axis.
by y
It is rotated about the line y
1.0
0.5
1.
0.0
0
Find the volume swept out.
1
2
3
x
a. 17
4
b. 9
2
c.
2
2
d. 4
2
2
2
e. 4
correct choice
2
Solution: As an x-integral, a slice is vertical and rotates into a washer.
R2
V
r 2 dx
1
0
sin x
2
1
2
dx
2 sin x
0
1
2 sin x
cos 2x
dx
2
0
2 cos
sin 2 x dx
0
2 cos 0
2
x
2
2 cos x
4
1
2
sin 2x
4
0
2
11. The region in Problem 10 is rotated about the the line x
1.
Which formula gives the volume swept out?
1
a.
sin x
2
dx
0
2 x sin x dx
b.
1
x
c.
1 sin x dx
0
2 x
d.
1 sin x dx
correct choice
0
1
e.
sin x
2
1 dx
0
Solution: As an x-integral, a slice is vertical and rotates into a cylinder.
V
2 rh dx
0
2 x
1 sin x dx
0
This integral could be computed using integration by parts.
5
12. A certain spring is at rest when its mass is at x
3.
It requires 24 Joules of work to stretch it from x 3 to x 7 meters.
What is the force required to maintain the mass at 7 meters?
a. 6 Newtons
b. 12 Newtons
correct choice
c. 18 Newtons
d. 24 Newtons
e. 48 Newtons
3 . The work is
1k x
W
F dx
k x 3 dx
2
3
3
So the force constant is k 3. And the force is F
Solution: The force is F
7
kx
7
3
2
7
1 k 16 0
2
3 4 12.
3
kx
8k
24.
13. A 80 kg cable which is 20 meters long hangs down from the top of a building. A 5 kg bag of sand is at
the bottom of the cable. How much work is done to lift the sand and the cable to the top of the
building? Give your answer as a multiple of the acceleration of gravity g.
a. 800g
b. 900g
correct choice
c. 1600g
d. 1700g
e. 2200g
Solution: The work to lift the bag of sand alone is W
FD
5g 20
100g.
The piece of rope at a distance y from the top of length dy
80 kg
weighs dF
g dy 4g dy and is lifted a distance D y.
20 m
So the work is W
20
20
D dF
0
The total work is W
y 4g dy
0
100g
800g
2y 2 g
20
0
800g.
900g.
6
Work Out: (Points indicated. Part credit possible. Show all work.)
15 points
14.
Solution: The curves intersect when x 3
or x 3
x2
or x x
x3
Find the area between the cubic y
1 x
Between
0 or x x 2
2x
x
0 or x
2
2
x2
x 2 and the line y
2x.
2x
0
1, 0, 2
1 and 0, the cubic is above the line.
(Plug in x
1/2.)
Between 0 and 2, the line is above the cubic.
(Plug in x
1.)
So the area is
0
x3
A
2
x2
2x dx
1
x4
4
x 2 dx
0
1
4
0
20 points
15.
x3
2x
1
3
1
4
8
3
4
0
x3 x2
3
3 4 12
12
0
x2
1
32
x4
4
37
12
x3
3
2
0
x4
A water tower is made by rotating the curve y
about the y-axis, where x and y are in meters. If the tower is filled
with water up to height y
25 m, how much work is done to pump
all the water out a faucet at height 30 m?
Assume the acceleration of gravity is g
9. 8 m/sec 2 and
1000 kg/m 3 .
the density of water is
You may give your answer as a multiple of
g .
Solution: The cross section at height y is a circle of radius x
x2
and hence area A
y 1/4
y 1/2 .
A slice of thickness dy has volume dV
A dy
This slice must be lifted a distance h
30
So the work to lift it is dW
g 30
dm g h
y 1/2 dy and mass dm
dV
y 1/2 dy.
y.
y y 1/2 dy.
The total work is
25
W
g 30
25
y y 1/2 dy
g
0
30y 1/2
y 3/2 dy
g 30
0
3
2 5
g 30 2 5
5
3
3. 848 5 10 7 Joules
5
g 4 54
2 54
2 54
2y 3/2
3
g
25
2y 5/2
5
1250
0
g
7
16.
15 points
Consider the curve y
x2.
f x
a. Find the tangent line at the generic point x
Solution: f x
f p
y f tan x
2x.
f p x
p2
p
p. Call it y
2p x
p
2px
b. The region between y
f x and y f tan x for 0
Find the volume swept out. Call it V p .
f tan x .
p2
1 is rotated about the y-axis.
x
Solution: We use an x-integral. There are vertical slices which rotate into cylinders.
Radius r
x. Height h
1
Vp
1
2 rh dx
2
0
f tan x
x x2
2px
x2
p2.
2px
p 2 dx
1
2
0
x4
4
2
f x
3
1
2
p2 x
2
2p x
3
x3
2px 2
p 2 x dx
0
1 p2
2
2
0
2p
3
1
4
c. Find the value of p which minimizes the volume V p .
Solution: V p
V p
2
1 p2
2
0
2
2
3
p
2p
3
1
4
p
2
3
d. Repeat steps (a), (b) and (c) for the general concave up curve y
1
Note: You can’t do the integral
f x
gx .
g x x dx. So leave it unevaluated.
0
(a) Solution: f x
f p
y f tan x
g x .
f p x
(b) Solution: Radius r
g p x
f x
p
f tan x
gx
gp
g p x
p .
1
2 rh dx
2
xgx
0
gp
g p x
p
dx
0
1
2
gp
x. Height h
1
Vp
p
gx x
gp x
g p x2
px
dx
0
1
2
g x x dx
0
1
2
g x x dx
0
2
2
gp x
2
1g p
2
g p
g p
1
3
x3
3
1p
2
2
px
2
1
0
1g p g p 1
1p
1
g p
2
3
2
2
1
1
2
2 g p
p
0
p
2
3
3
provided g p
0 which is true because g x is everywhere concave up.
2
0
p
2 p 2
3
3
(c) V p
2
e. What do you conclude?
Solution: For any concave up curve y f x , if the region between the curve and
its tangent line y f tan x at a point p is rotated about the y-axis, then the volume
2.
swept out is a minimum when p
3
8