Name
Sec
MATH 152 Honors
Final Exam
Sections 201,202
Solutions
ID
Multiple Choice: (6 points each)
1-11
/66
Fall 2007
12
/12
P. Yasskin
13
/15
14
/10
Total
/103
1. A 2 meter bar has linear density ρ = 1 + x 3 kg/m where x is measured from one end.
Find the average density of the bar.
a. 2 kg/m
b. 3 kg/m
correct choice
c. 4. 5 kg/m
d. 5 kg/m
e. 6 kg/m
ρ ave = 1
2
∫ 0 1 + x 3  dx =
2
1 x + x4
4
2
2
0
= 1 2 + 4 = 3
2
2. A 2 meter bar has linear density ρ = 1 + x 3 kg/m where x is measured from one end.
Find the center of mass of the bar.
a. 5 m
b.
c.
d.
e.
M=
7
5 m
6
6 m
5
7 m
5
42 m
5
correct choice
∫ 0 1 + x 3  dx =
M1 =
2
4
x+ x
4
∫ 0 x1 + x 3  dx =
2
2
0
= 2+4 = 6
x2 + x5
5
2
2
0
= 42
5
x̄ = M 1 = 42 = 7
M
5
5⋅6
1
∫ x arctan x dx.
3. Compute
2
a. 3x arctan x − x lnx 2 + 1 + C
b.
c.
d.
e.
2
2
2
3x arctan x + x lnx 2 + 1 + C
2
2
x 2 arctan x + 3x lnx 2 + 1 + C
2
2
2
x arctan x − 1 x − 1 arctan x + C
2
2
2
2
x arctan x − 1 x + 1 arctan x + C
2
2
2
u = arctan x
1 dx
du =
1 + x2
∫
dv = x dx
2
v= x
2
correct choice
∫ x arctan x dx =
x 2 arctan x − 1
2
2
∫
x 2 dx
1 + x2
x 2 + 1 − 1 dx = ∫ 1 − 1 dx = x − arctan x + C
1 + x2
1 + x2
2
∫ x arctan x dx = x2 arctan x − 12 x − arctan x + C
(The second integral can also be done with the trig substitution x = tan θ.)
x 2 dx =
1 + x2
∫
4. Find the arclength of the parametric curve
a. 61
b.
c.
d.
e.
L=
1
dx
dt
u = 16 + 9t 4
L= 1
36
y = 1 t6
2
for 0 ≤ t ≤ 1.
correct choice
54
16
9
11
9
1
9
1
54
∫0
x = t4
∫ 16
25
2
+
dy
dt
2
dt =
du = 36t 3 dt
u du = 1
36
2 u 3/2
3
∫0
1
2
4t 3  + 3t 5  2 dt =
∫0
1
16t 6 + 9t 10 dt =
∫0 t3
1
16 + 9t 4 dt
1 du = t 3 dt
36
25
= 1 25 3/2 − 16 3/2  = 1 125 − 64 = 61
54
54
54
16
2
5. Which term appears in the partial fraction expansion of
a.
b.
c.
d.
e.
−2
x − 1 2
1
x − 1 2
2
x − 1 2
−2
x−1
2
x−1
4x 2 − 2x + 2 ?
x − 1 2 x 2 + 1
correct choice
4x 2 − 2x + 2 = A +
B
+ Cx2 + D
x−1
x +1
x − 1 2 x 2 + 1
x − 1 2
4x 2 − 2x + 2 = Ax − 1x 2 + 1 + Bx 2 + 1 + Cx + Dx − 1 2
x = 1:
4 − 2 + 2 = 0 + B2 + 0
4 = 2B
B=2
6. The base of a solid is the region bounded by the curves y = x 2 , y = −x 2 and x = 2.
The cross sections perpendicular to the x-axis are squares. Find the volume of the solid.
a. 128
b.
c.
d.
e.
5
32
5
16
3
8
3
16
15
correct choice
This is a x-integral. The side of the square is s = 2x 2 . So the area is A = s 2 = 4x 4 .
So the volume is
V=
∫ 0 A dx = ∫ 0 4x 4 dx =
2
2
4x 5
5
2
0
= 128
5
3
dy
= xy 2 + x satisfying the initial condition
dx
7. Find the solution of the differential equation
y0 = 1.
2
a. y = sin x + π
b. y = tan
c. y = sin
d. y = tan
e. y = cos
2
2
2
x + π
2
4
2
x
+1
2
x2 + 1
2
x2
2
Separate variables:
correct choice
dy
= x dx
y +1
2
∫
dy
=
y +1
2
∫ x dx
2
arctan y = x + C
2
arctan 1 = C = π
4
2
2
π
x
arctan y =
+
y = tan x + π
2
4
2
4
Apply the initial conditions x = 0, y = 1:
Substitute back and solve for y:
8. If gx = cosx 2 , what is g 8 0, the 8 th derivative at zero?
HINT: What is the coefficient of x 8 in the Maclaurin series for cosx 2 ?
a. 8 ⋅ 7 ⋅ 6 ⋅ 5
correct choice
1
8⋅7⋅6⋅5
c. 4!
d. 1
4!
e. 1
8!
b.
2
4
On the one hand, the Maclaurin series for cost is cost = 1 − t + t − ⋯.
2!
4!
4
8
So the Maclaurin series for cosx 2  is cosx 2  = 1 − x + x − ⋯.
2!
4!
On the other hand the Maclaurin series for any function gx is
g 8 0 8
gx = g0 + g ′ 0 x + ⋯ +
x +⋯
8!
g 8 0
Since these must be equal, the coefficients of x 8 must be equal:
= 1
8!
4!
8!
So g 8 0 =
= 8 ⋅ 7 ⋅ 6 ⋅ 5 = 1680
4!
4
∞
∑ne
9. Suppose the series
−n 2
9
is approximated by its 9 th partial sum ∑ n e
n=1
−n 2
.
n=1
Use an integral to bound the error in this approximation.
a. 1 e −64
b.
c.
d.
e.
2
1 e −81
2
1 e −100
2
1 e −121
2
1 e −144
2
correct choice
∞
The error is E = ∑ n e
−n 2
.
n=10
So E ≤
∞
∫9 n e
dn = − 1 e
2
−n 2
−n 2
∞
9
= 0−−1 e
2
−9 2
= 1 e −81
2
10. Find the area of the triangle with vertices P = 2, −1, 3, Q = 1, 2, 1 and R = 3, 1, 4.
a. 1
73
2
b. 73
c. 5 3
2
d. 5 3
correct choice
e. −5 3
The edges are
î
PQ × PR =
PQ = Q − P = −1, 3, −2
̂
k̂
−1 3 −2
1
2
PR = R − P = 1, 2, 1
= î3 + 4 − ̂ −1 + 2 + k̂ −2 − 3 = 7, −1, −5
1
A = 1 PQ × PR = 1 7 2 + 1 2 + 5 2 = 1 49 + 1 + 25 = 1 75 = 5 3
2
2
2
2
2
11. If ⃗
u points North East and ⃗v points West, in which direction does ⃗
u × ⃗v point?
a. North West
b. South
c. South East
d. Up
correct choice
e. Down
Hold your right hand with the fingers pointing to the right of forward and the palm facing left.
Then you thumb points up.
5
Work Out: (Points indicated. Part credit possible.)
4
8
12. (12 points) Compute ∫
dx
2 x3 x2 − 4
x = 2 sec θ
dx = 2 sec θ tan θ dθ
x=2 @
sec θ = 1 or θ = 0
∫2
4
x3
8
dx =
x2 − 4
=
π/3
∫0
π/3
∫0
x=4 @
8 ⋅ 2 sec θ tan θ dθ =
8 sec 3 θ 4 sec 2 θ − 4
π/3
∫0
1 + cos 2θ dθ = 1 θ + sin 2θ
2
2
2
π/3
0
sec θ = 2 or θ = π
3
π/3
sec θ dθ = ∫ cos 2 θ dθ
sec 3 θ
0
= 1
2
π + 1 sin 2π
2
3
3
3
= π +
6
8
13. (15 points) The curve y = x 2 is rotated about the y-axis to form a bowl. If the bowl contains
8π cm 3 of water, what is the height of the water in the bowl?
The silce at height y is a circle. So its area is A = πr 2 .
The radius is
r=x=
So the area is
y
and the volume of the slice of thickness dy is
A = πy
dV = πy dy.
So the volume up to height h is
V=
∫ dV = ∫ 0 πy dy = π
h
y2
2
h
0
= π h2
2
We equate the volume to 8π and solve for h:
π h 2 = 8π

h 2 = 16

h = 4 cm
2
6
14. (10 points) This question is designed to teach you about infinite products.
∞
a. (2 pt) Define the partial sum, S k , and the sum, S, of an infinite series
∑ an.
n=0
(Give one sentence including one equation for each.)
∞
th
The k -partial sum of the infinite series
∑ an
is S k =
n=0
∞
∑ an
The sum of the infinite series
k
∑ an.
n=0
is S = lim S k .
k∞
n=0
b. (2 pt) By analogy, define the partial product, P k , and the product, P, of an infinite
∞
product
∏ an.
(Give one sentence including one equation for each.)
n=0
∞
The k th -partial product of the infinite product
∞
∏ an
The product of the infinite product
n=0
∞
c. (4 pt) Compute
∏
n
1+x 2
k
∏ a n is S k =
∏ an.
n=0
n=0
is P = lim P k .
k∞
.
n=0
HINT: Multiply out the first 3 partial products P 0 , P 1 and P 2 . Then find P k and P.
P 0 = 1 + x
P 1 = 1 + x1 + x 2  = 1 + x + x 2 + x 3
P 2 = 1 + x1 + x 2 1 + x 4  = 1 + x + x 2 + x 3 1 + x 4  = 1 + x + x 2 + x 3 + x 4 + x 5 + x 6 + x 7
Pk = 1 + x + ⋯ + x
2 k+1 −1
2 k+1 −1
=
∑
x n (This could be proved by mathematical induction.)
n=0
2 k+1 −1
P = lim P k = lim
k∞
k∞
∑
n=0
∞
x =
n
∑ xn
n=0
d. (2 pt) For which x does this infinite product converge? To what function?
It converges when |x| < 1 to P =
1 .
1−x
7