Student (Print)
Section
Last, First Middle
Student (Sign)
Student ID
Instructor
MATH 152,
Fall 2007
Common Exam 1
Test Form A
Solutions
Instructions:
You may not use notes, books, calculator or computer.
For Dept use Only:
1-10
/50
11
/10
12
/10
after 90 minutes and may not be returned.
13
/10
Part II is work out. Show all your work. Partial credit will be given.
14
/10
15
/10
Part I is multiple choice. There is no partial credit.
Mark the Scantron with a #2 pencil. For your own records,
also circle your choices in this exam. Scantrons will be collected
THE AGGIE CODE OF HONOR:
An Aggie does not lie, cheat or steal, or tolerate those who do.
TOTAL
1
Part I: Multiple Choice (5 points each)
There is no partial credit.
1
∫ 0 xe
1. Compute
x 2 +1
dx
a. 1 e
b.
c.
d.
e.
2
1 (e − 1)
2
1 e2
2
1 (e 2 − 1)
2
1 (e 2 − e)
2
correct choice
u = x2 + 1
Use the substitution
1
∫ 0 xe x +1
2
dx = 1
2
2. Compute
π/2
∫0
a. π − 1
b.
c.
d.
e.
2
∫ 1 e u du =
π/2
2
1 eu
2
1
1 du = x dx
2
= 1 (e 2 − e)
2
x cos x dx
correct choice
2
1+ π
2
π
2
1−π
π−1
Use integration by parts with
∫0
du = 2x dx
x cos x dx = x sin x − ∫ sin x dx
=
u=x
dv = cos x dx
du = dx v = sin x
π/2
0
π sin π + cos π
2
2
2
= x sin x + cos x
π/2
0
− (cos 0) = π − 1
2
3. Find the area below the parabola, y = 3x − x 2 , above the x-axis.
a. 1
b.
c.
d.
e.
2
9
2
27
2
81
2
10
3
correct choice
The parabola intersects the x-axis when 3x − x 2 = 0 or x = 0, 3.
A=
3
∫0
(3x − x 2 ) dx =
3x 2 − x 3
2
3
3
0
= 27 − 9 = 9
2
2
2
4. Find the average value of f(x) = e 3x on the interval [0, 2].
a. 1 (e 6 − 1)
b.
c.
d.
e.
3
1 e6
3
1 (e 6 − 1)
6
1 e6
6
(e 6 − 1)
f ave = 1
2
5.
correct choice
2
∫ 0 e 3x dx =
2
1 e 3x
6
0
= 1 (e 6 − 1)
6
The region shown at the right is bounded
above by y = sin x and below by the x-axis.
It is rotated about the x-axis.
Find the volume swept out.
2
a. π
2
b. 2π 2
y
1.0
0.5
0.0
0
1
2
3
x
correct choice
c. 2π
d. π
2
π
e.
4
As an x-integral, a slice is vertical and rotates into a disk.
π
π
π 1 − cos(2x)
sin(2x)
V = ∫ πR 2 dx = ∫ π(sin x) 2 dx = π ∫
dx = π x −
0
0
0
2
2
2
π
0
2
= π
2
6. The region in Problem 5 is rotated about the the line x = −1.
Which formula gives the volume swept out?
a.
b.
c.
d.
e.
π
∫ 0 π (1 + sin x) 2 − 1
π
∫ 0 2π(x + 1) sin x dx
π
∫ 0 π(x − 1) sin x dx
π
∫ −1 2πx sin x dx
π
∫ 0 π(1 + sin x) 2 dx
dx
correct choice
As an x-integral, a slice is vertical and rotates into a cylinder.
V=
π
π
∫ 0 2πrh dx = ∫ 0 2π(x + 1) sin x dx
This integral can be computed using integration by parts.
3
7. The region bounded by the curves x = 1, y = 1 and y = 4
x
is rotated about the x-axis.
Find the volume swept out.
a.
b.
c.
d.
e.
π(8 ln 4 − 15)
π(15 − 8 ln 4)
12π
9π
correct choice
8π
y
5
4
3
2
1
0
0
1
2
3
4
5
x
As an x-integral, a slice is vertical and rotates into a washer.
2
4
4
4
V = ∫ (πR 2 − πr 2 ) dx = ∫ π 4x
− π(1) 2 dx = π ∫ 162 − 1 dx = π −16
x −x
1
1
1
x
8.
4
1
= 9π
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. 32
b.
c.
d.
e.
3
64
3
128
3
16 π
3
32 π
3
correct choice
The width of the base is W(y) = 2 4 − y 2 . The height is H = W
The area of the cross section is A(y) = 1 WH = 1 2 4 − y 2 2 4 − y 2 = 2(4 − y 2 ).
2
2
2
3
2
2
y
V = ∫ A(y) dy = ∫ 2(4 − y 2 ) dy = 2 4y −
= 4 8 − 8 = 32 2 = 64
−2
−2
3 −2
3
3
3
4
9. A certain spring is at rest when its mass is at x = 0.
It requires 24 Joules of work to stretch it from x = 0 to x = 4 meters.
What is the force required to maintain the mass at 4 meters?
a.
b.
c.
d.
e.
48 Newtons
18 Newtons
12 Newtons
6 Newtons
24 Newtons
correct choice
1 kx 2 4 = 8k = 24.
2
0
So the force constant is k = 3. And the force is F = kx = 3 ⋅ 4 = 12.
The force is F = kx. The work is W =
4
4
∫ 0 F dx = ∫ 0 kx dx =
2
10. Find the partial fraction expansion for f(x) = 5x 3+ x + 12 .
x + 4x
a. 1 + 3x2 − 2
b.
c.
d.
e.
x
2
x
1
x
2
x
3
x
+
+
+
+
x +4
x−3
x2 + 4
2x + 3
x2 + 4
3x + 1
x2 + 4
2x + 1
x2 + 4
correct choice
2
2
f(x) = 5x 3+ x + 12 = 5x +2 x + 12 = A + Bx2 + C
x + 4x
x(x + 4)
x +4
x
Clear the denominator:
5x 2 + x + 12 = A(x 2 + 4) + x(Bx + C) = (A + B)x 2 + Cx + 4A
Equate coefficients:
A+B = 5
Solve:
A=3
Substitute back:
C=1
4A = 12
B=2
C=1
2
f(x) = 5x 3+ x + 12 = 3 + 2x2 + 1
x + 4x
x +4
x
5
Part II: Work Out (10 points each)
Show all your work. Partial credit will be given.
11. Compute
a. (5 points)
u = sin θ
∫ cos 3 θ dθ
du = cos θ dθ
∫ cos 3 θ dθ = ∫(1 − sin 2 θ) cos θ dθ = ∫(1 − u 2 ) du =
Check:
b. (5 points)
d sin θ − sin 3 θ
3
dθ
Check:
3
+ C = sin θ − sin θ + C
3
= cos θ − 1 3 sin 2 θ(cos θ) = cos θ(1 − sin 2 θ) = cos 3 θ
3
∫ x 3 ln x dx
Use integration by parts with
∫ x 3 ln x dx =
3
u− u
3
u = ln x
du = 1x dx
dv = x 3 dx
4
v= x
4
x 4 ln x − x 4 1 dx = x 4 ln x − x 3 dx = x 4 ln x − x 4 + C
∫ 4 x
∫ 4
4
4
4
16
d
dx
x 4 ln x − x 4
4
16
4
3
= x 3 ln x + x 1x − x = x 3 ln x
4
4
6
12. Find the area between the cubic y = x 3 − x 2 and the line y = 2x.
The curves intersect when x 3 − x 2 = 2x or x 3 − x 2 − 2x = 0 or x(x 2 − x − 2) = 0
or x(x + 1)(x − 2) = 0 or x = −1, 0, 2
Between −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
A=
0
= (0) −
13.
0
3
4
x4 − x3 − x2
+ x2 + x − x
4
3
3
4
−1
1 − −1 − 1 + 4 + 8 − 4 − (0) = −3 − 4 + 12 + 32 = 37
4
3
3
12
12
2
∫ −1 (x 3 − x 2 − 2x) dx + ∫ 0 (2x − x 3 + x 2 ) dx =
2
0
A water tower is made by rotating the curve y = x 4 about the
y-axis, where x and y are in meters. If the tower is filled with water
(of density ρ = 1000 kg/m 3 ) 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 .
You may give your answer as a multiple of ρg.
The cross section at height y is a circle of radius x = y 1/4 an hence area A = πx 2 = πy 1/2 .
A slice of thickness dy has volume dV = A dy = πy 1/2 dy and mass dm = ρ dV = ρπy 1/2 dy.
This slice must be lifted a distance h = 30 − y.
So the work to lift it is dW = dm g h = ρπg(30 − y)y 1/2 dy.
The total work is
W=
25
∫0
25
ρπg(30 − y)y 1/2 dy = πρg ∫ (30y 1/2 − y 3/2 ) dy = πρg 30
0
= πρg 30 2 ⋅ 5 − 2 ⋅ 5
5
3
7
= 3. 848 5 × 10 Joules
3
5
2y 3/2
2y 5/2
−
5
3
25
0
= πρg(4 ⋅ 5 4 − 2 ⋅ 5 4 ) = 2 ⋅ 5 4 πρg = 1250πρg
7
14. Compute
1
∫0
x2
dx
(4 − x 2 ) 3/2
Let x = 2 sin θ. Then dx = 2 cos θ dθ and 4 − x 2 = 4 − 4 sin 2 θ = 4 cos 2 θ
Change limits: x = 0 @ θ = 0
and
x=1 @
sin θ = 1
or
2
2
1
π/6
sin 2 θ 2 cos θ dθ = π/6 tan 2 θ dθ = π/6(sec 2 θ − 1) dθ
∫ 0 (4 −xx 2 ) 3/2 dx = ∫ 0 84cos
∫0
∫0
3
θ
π/6
= tan θ − θ 0 = 1 − π
6
3
15. Compute
∫
4
∫0
x − 4 dx =
x 2 + 16
x − 4 dx
x 2 + 16
∫
x
dx + ∫ 2 −4 dx
x 2 + 16
x + 16
Integral 1:
u = x 2 + 16
∫
x
dx = 1
2
x + 16
2
θ= π
6
1 du = x dx
2
1 du = 1 ln|u| + C = 1 ln(x 2 + 16) + C
u
2
2
du = 2x dx
∫
x = 4u
dx = 4du
−16
dx = ∫
du = − arctan(u) + C = − arctan x
∫ x 2 −4
4
+ 16
16u 2 + 16
So ∫ x2 − 4 dx = 1 ln(x 2 + 16) − arctan x + C
2
4
x + 16
4
So ∫ x2 − 4 dx = 1 ln 32 − arctan(1) − 1 ln 16 − arctan(0)
0 x + 16
2
2
π
1
=
ln 2 −
2
4
Integral 2:
+C
8