Name
Sec
MATH 152
ID
Final Exam
Sections 513 - 515
Solutions
1-11
/66
Fall 2007
12
/12
P. Yasskin
13
/15
14
/10
Total
/103
Multiple Choice: (6 points each)
1. A plate is bounded by the curves y = x 2 , y = −x 2 and x = 2 measured in meters.
Find the total mass of the plate if the surface density is ρ = 3 kg/m 2 .
a. 8
3
b. 16
3
32
c.
3
d. 8
e. 16
correct choice
2
2
3
M = ρ ∫ f − g dx = 3 ∫ x 2 − −x 2  dx = 3 2x
3
0
0
2
0
= 16
2. A plate is bounded by the curves y = x 2 , y = −x 2 and x = 2 measured in meters.
Find the center of mass of the plate if the surface density is ρ = 3 kg/m 2 .
a.
b.
c.
d.
e.
3 ,0
4
3, 1
4 10
3, 6
4 5
3 ,0
2
3, 6
2 5
correct choice
2
2
4
M y = ρ ∫ x f − g dx = 3 ∫ xx 2 − −x 2  dx = 3 2x
4
0
0
My
x̄ =
= 24 = 3
ȳ = 0 by symmetry.
M
16
2
2
0
= 24
1
3. Compute
∫ x 2 e 2x dx.
2
a. x e 2x − x e 2x − 1 e 2x + C
b.
c.
d.
e.
2
x2
2
x2
2
x2
2
x2
2
e 2x −
e 2x −
e 2x −
e 2x −
2
x e 2x +
2
x e 2x −
2
x e 2x +
2
x e 2x −
4
u = x2
4
1 e 2x + C
4
1 e 2x + C
2
1 e 2x + C
2
1 e 2x + C
4
correct choice
dv = e 2x dx
v = 1 e 2x
2
du = 2x dx
u=x
∫ x 2 e 2x dx =
∫ x 2 e 2x dx =
x 2 e 2x −
2
2
= x e 2x −
2
dv = e 2x dx
v = 1 e 2x
2
du = dx
x 2 e 2x − ∫ x e 2x dx
2
x e 2x − 1 ∫ e 2x dx
2
2
x e 2x + 1 e 2x + C
2
4
x = 2t 2
y = t3
for 0 ≤ t ≤ 1 is rotated about the y-axis.
Which integral gives the area of the surface swept out?
4. The parametric curve
a. 2π
∫0 t
b. 2π
∫0 t3
16 + 9t 2 dt
c. 4π
∫0 t3
16 + 9t 2 dt
d. 2π
∫0 t4
16 + 9t 2 dt
e. 4π
∫0 t4
16 + 9t 2 dt
1
16 + 9t 2 dt
1
1
1
1
A=
∫ 2πr ds = ∫ 0 2πx
1
correct choice
dx
dt
2
+
dy
dt
2
dt =
∫ 0 2π2t 2
1
2
4t 2 + 3t 2  dt
= 4π ∫ t 2 16t 2 + 9t 4 dt = 4π ∫ t 3 16 + 9t 2 dt
1
1
0
0
2
5. Which term appears in the partial fraction expansion of
a.
b.
c.
d.
e.
−2
x − 2 2
1
x − 2 2
2
x − 2 2
−2
x − 2
1
x − 2
3x 2 − 4x − 20 ?
x − 2 2 x 2 + 4
correct choice
3x 2 − 4x − 20 =
A
B
+
+ Cx2 + D
x − 2
x + 4
x − 2 2 x 2 + 4
x − 2 2
3x 2 − 4x − 20 = Ax − 2x 2 + 4 + Bx 2 + 4 + Cx + Dx − 2 2
x = 2:
12 − 8 − 20 = 0 + B8 + 0
− 16 = 8B
6. Find the solution of the differential equation x
y1 = 2.
B = −2
dy
= 2y + x 3 satisfying the initial condition
dx
a. y = x 3 − 7 x 2
4
b. y = x + x 2
3
c. y = 2x − x
3
correct choice
2
9 + 1 x3
5
5x 2
e. y = 5 2 + 1 x 3
3
3x
d. y =
Put the equation in standard linear form:
∫ − 2x
dy
− 2x y = x 2
dx
dx
= e −2 ln x = x −2 = 12
x
dy
1
Multiply by the integrating factor:
− 23 y = 1
x 2 dx
x
d
1 y =1
1 y = ∫ 1 dx = x + C
Rewrite and integrate:
dx x 2
x2
Apply the initial conditions x = 1, y = 2:
2 = 1+C
C=1
1 y = x+1
Substitute back and solve for y:
y = x3 + x2
x2
The integrating factor is:
I=e
3
7. The region in the first quadrant bounded by the curves y = x 2 , y = 0 and x = 2 is rotated
about the y-axis. Find the volume of the solid swept out.
a. 32 π
5
b. 64 π
5
c. 8π
correct choice
d. 4π
e. 2π
This is a x-integral. A vertical slice rotates into a cylinder of radius r = x and height h = x 2 .
So the volume is
V=
∫ 0 2πrh dx = ∫ 0 2π x x 2 dx = 2π x4
2
2
4
2
0
= 8π
1, 3
by its 3 rd degree Taylor
2 2
polynomial centered at x = 1, namely T 3 x = x − 1 − 1 x − 1 2 + 1 x − 1 3 , then the
2
3
Taylor Remainder Theorem says the error |R 3 x| is less than
8. If you approximate fx = lnx on the interval
Taylor Remainder Theorem:
If T n x is the n th degree Taylor polynomial for fx centered at x = a
then there is a number c between a and x such that the remainder is
f n+1 c
R n x =
x − a n+1
n + 1!
a. 1
b. 1
2
c. 1
4
d. 1
8
e. 0
correct choice
Here a = 1, and n = 3, and |R 3 x| =
|f 4 c|
|x − 1| 4
4!
1, 3 .
2 2
4
f x = −6x −4
We need the largest values of |f 4 c| and |x − 1| 4 for c and x in
f ′ x = 1x = x −1
f ′′ x = −x −2
f ′′′ x = 2x −3
The max of |f 4 c| = 64 occurs at c = 1 and is M = 6 ⋅ 2 4 .
2
c
The max of |x − 1| 4 occurs at 1 or 3 and is 14 .
2
2
2
4
6
⋅
2
1
6
1
So |R n x| ≤
=
=
4! 2 4
24
4
4
9. Compute lim
x0
sinx 2  − x 2
.
3
e x − 1 − x3
a. − 1
b.
c.
d.
e.
6
1
6
2
3
−1
3
1
3
correct choice
3
5
6
10
sint = t − t + t − ⋯
sinx 2  = x 2 − x + x − ⋯
3!
5!
3!
5!
2
3
6
9
t
t
x
x3
3
t
e = 1+t+
+
+⋯
e
= 1+x +
+ x +⋯
2!
3!
2!
3!
6
10
4
x
x
1
x
−
+
−⋯
−
+
−⋯
sinx 2  − x 2
3!
5!
3!
5!
lim x 3
=
lim
=
lim
= − 2! = − 1
6
9
3
3
x0 e
x0
x0
3!
3
x
x
1
x
−1−x
+
+⋯
+
+⋯
2!
3!
3!
2!
10. Find the volume of the parallelepiped with edge vectors
⃗
a = −2, 2, 1,
⃗
b = 3, 2, 4
and
⃗c = 1, −2, 3
a. −26
b. 26
c.
26
d. −46
correct choice
e. 46
V=
−2
2
1
3
2
4
1
−2 3
= |−26 + 8 − 29 − 4 + 1−6 − 2| = |−28 − 10 − 8| = |−46| = 46
11. If ⃗
u points Down and ⃗v points North West, in which direction does ⃗
u × ⃗v point?
a. North East
correct choice
b. South
c. South West
d. South East
e. Up
Hold your right hand with the fingers pointing down and the palm facing left of forward. Then
you thumb points right of forward.
5
Work Out: (Points indicated. Part credit possible.)
1
x2
12. (12 points) Compute ∫
dx
0
4 − x2
x = 2 sin θ
x=0 @
∫0
1
dx = 2 cos θ dθ
or θ = π
6
π/6
π/6
2
4 sin θ 2 cos θ dθ = ∫ 4 sin 2 θ dθ = 4 ∫ 1 − cos 2θ dθ
2
0
0
4 − 4 sin 2 θ
sin θ = 0 or θ = 0
x2
dx =
4 − x2
π/6
∫0
= 2 θ − sin 2θ
2
π/6
0
x=1 @
= 2 π − 1 sin π
6
3
2
sin θ = 1
2
3
= π −
3
2
13. (15 points) A cone of radius 4 and height 6 and vertex down is filled with water up to
height 2. Find the work done to pump the water out the top. Give your answer as a multiple
of ρg.
The silce at height y is a circle. So its area is A = πr 2 .
r
4
The radius is found using similar triangles:
y = 6
2
A = π 2y
= 4 πy 2
3
9
and the volume of the slice of thickness dy is
or
r = 2y
3
So the area is
This slice is lifted a distance
W=
D = 6 − y.
∫ ρg D dV = ∫ 0 ρg 6 − y 49 πy 2 dy =
2
y4
= 4 ρgπ 2y 3 −
9
4
2
0
dV = 4 πy 2 dy.
9
So the work is
4 ρgπ ∫ 26y 2 − y 3  dy
9
0
= 4 ρgπ16 − 4 = 16 ρgπ
9
3
6
14. (10 points) Start from the Maclaurin series:
a. (6 pt) Find the Maclaurin series for
1 =
1+x
∞
∑−1 n x n
n=0
1
.
1 + x 2
HINT: Differentiate both sides of the given series.
d
dx
1
1+x
−1
=
1 + x 2
d
dx
∞
∞
∑−1 x
=
n n
n=0
∑−1 n n x n−1
n=0
∞
1
= − ∑−1 n n x n−1
1 + x 2
n=0
∞
=
∑−1 k k + 1 x k
n=0
n=1
k=0
∑
n=1
n=1
∞
∑−1 n−1 n x n−1 =
b. (4 pt) Compute
∑
∞
∑−1 n−1 n x n−1 =
∞
∞
You can stop here.
−1 n−1 n
=
2 n−1
∞
−1 n−1 n
.
2 n−1
∑−1 n−1 n
n=1
1
2
n−1
=
1
1+ 1
2
2
=
1
3
2
2
= 4
9
7