Name
ID
MATH 152
Sec
Final Exam
Sections 201,202
Spring 2001
Solutions
P. Yasskin
Multiple Choice: (4 points each)
a. e "2
correctchoice
1
e 1/2
e2
b.
c.
d.
e.
.
2n
1 " 1n
lim
nv.
exp nlim
v.
;
=
32
=
16
=
8
=
4
=
2
a.
b.
c.
d.
e.
=/4
0
exp nlim
ln 1 " 1n
v.
2n
exp nlim
v.
"2 e "2
1 " 1n
2. Compute
;
2n
1 " 1n
lim
nv.
1. Compute
=/4
0
ln 1 " 1n
1
2n
1
n2
exp nlim
v.
1 " 1n
"1
2n 2
sin 2 2 cos 2 2 d2
correctchoice
sin 2 2 cos 2 2 d2 1
4
;
=/4
0
sin 2 Ÿ22 d2 1
4
;
=/4
0
sinŸ42 1 " cosŸ42 d2 1 2 "
4
2
8
=/4
=
0
32
3. The region below y 1
x
above the x-axis between x 1 and x . is rotated
about the x-axis. Find the volume of the solid of revolution.
a. =
4
b. =
2
c. =
correctchoice
d. 2=
e. 4=
;
.
1
= 1x
2
dx "=
x
.
1
=
1
;
4. Compute
a.
1
x 2 e "x dx
0
"5e "1
b. 2 " 5e "1
correctchoice
"e "1
c.
d. 2 " e "1
e. e "1 " 2
dv e "x dx
u x2
du 2x dx
v
"e "x
ux
dv e "x dx
du dx
v "e "x
;
1
0
;
;
;
x 2 e "x dx "x 2 e "x 2 x e "x dx
x 2 e "x dx "x 2 e "x 2 "xe "x ;
e "x dx
x 2 e "x dx ¡"x 2 e "x " 2xe "x " 2e "x ¢ 10 ¡"e "1 " 2e "1 " 2e "1 ¢ " ¡"2 ¢ 2 " 5e "x
5. Find the average value of the function fŸx a.
b.
c.
d.
e.
2
6
2
27
2
9
1
9 2
1
27 2
f ave 1
3
f ave 1
3
;
;
Ÿ
1
3/2
x2 9 on the interval
0, 3 ¢.
¡
correctchoice
3
dx
0 Ÿx 2
=/4
9 3/2
0
Let x 3 tan 2. Then dx 3 sec 2 2 d2
3 sec 2 2 d2
1
3/2
2
27
Ÿ9 tan 2 9 ;
=/4
0
sec 2 2 d2 1
3/2
2
27
Ÿsec 2 ;
=/4
0
cos 2 d2 1 sin 2
27
=/4
0
1
27 2
6. Find the angle between the vector u Ÿ2, 1, "2 and the normal to the plane through
Ÿ0, "3, 4 .
P Ÿ3, "4, 12 containing the vectors v Ÿ1, 0, 0 and w
a. arccos "22
b. arccos
c. arccos
d. arccos
e. arccos
39
15
2
3
2
2
15
2
3
correctchoice
v • w
Ÿ1, 0, 0 • Ÿ0, "3, 4 Ÿ0, "4, "3 N
Ÿ2, 1, "2 Ÿ0, "4, "3 2
u N
| |u
N
16 9 5
414 3
cos 2 u N 2
15
| N
|u
2
7. Use the 4 th degree Maclaurin polynomial for e "x
2
to estimate
a. 1 " 1 1
3
;
1
e "x dx.
2
0
5
b. 1 1 1
5
3
1
1
c. 1 "
120
6
1
1
d. 1 10
3
e. 1 " 1 1
10
3
correctchoice
2
4
2
e "x X 1 " x 2 x
et X 1 t t
2
2
1
1
4
3
5
2
e "x dx X
1 " x2 x
dx x " x x
2
3
10
0
0
;
;
!
.
8. The series
n1
n
n 3/2 1
1
0
1" 1 1
3
10
is
!
.
a. convergent by the Comparison Test with
!
.
b. conv. by the Limit Comp. Test with
n1
c. divergent by the Comparison Test with
!
.
d. div. by the Limit Comp. Test with
n1
n1
1
n 1/2
1 but not by the Comp. Test
n 1/2
!
.
n1
1
n 1/2
1 but not by the Comp. Test
n 1/2
correctchoice
e. none of these
!
.
1 is divergent because it is a p-series with p 1 1
1/2
2
n
n1
n
11/2 So the Comp. Test cannot show convergence or divergence.
n
n 3/2 1
n
.
n 3/2 1 lim n 3/2 1 So
n
lim
is divergent by the Limit Comp. Test
3/2
nv.
nv. n 3/2 1
1
1
n
n
1
n 1/2
9. The area below y x 2 , above the x-axis, between x 1 and x 2 is rotated
about the y-axis. Find the volume of the solid of revolution.
!
a. 4=
b. 15=
4
15
=
c.
2
d. 8=
e. 31=
5
;
h y x2
2
2
4
8= " = 15=
2=rh dx 2=x x 2 dx 2= x
2
4 1
2
1
x-integral
V
correctchoice
;
cylinders
rx
3
!
.
10. Compute
n2
n1 " n2
n
n"1
a. 0
b. 1
correctchoice
c. 2
d. 3
e. divergent
!
k
n1 " n2 3 " 4
n
1
2
n"1
n2
3 " k2
1
k
S lim S k lim 3 " k 2 2
kv.
kv.
k
Sk 4 " 5
2
3
C
k " k1
k"2
k"1
k1 " k2
k"1
k
11. The Maclaurin series for sinh x is
sinh x !
.
k0
3
5
7
x 2k1
x x x x C
3!
5!
7!
Ÿ2k 1 !
If you use the 5 th -degree Maclaurin polynomial to approximate sinh x on the interval
1 , 2 , bound the error in the approximation using the Taylor Remainder Inequality
2
M |x " a|n1
|R n Ÿx | t
where M u |f Ÿn1 Ÿc | for all c between x and a.
Ÿn 1 !
HINT: sinh x:
cosh x:
-2
a.
b.
c.
d.
e.
4
15
4
45
4
45
4
15
4
45
-1
1x
2
-2
-1
1x
2
cosh 2
sinh 2
cosh 1
2
1
sinh
2
cosh 2
correctchoice
1 , 2 . The maximum value of |x " a| is 2.
2
f Ÿx sinh x
f Ÿ3 Ÿx cosh x
fŸx sinh x
f U Ÿx cosh x
f Ÿ4 Ÿx sinh x
f Ÿ5 Ÿx cosh x
f Ÿ6 Ÿx sinh x
M u sinh c for c between 0 and 2 and sinh x is increasing. So take M sinh 2.
M |x " a|n1 t sinh 2 2 6 4 sinh 2
|R n Ÿx | t
6!
45
Ÿn 1 !
Here, n 5, a 0, and x is in the interval
UU
4
a, b, c where the line x 2 " t y 3 2t z 4 t
intersects the plane 2x " y 3z 14. Then a b c 12. Find the point
Ÿ
a. 1
b. 3
c. 5
correctchoice
d. 7
e. 9
2Ÿ2 " t " Ÿ3 2t 3Ÿ4 t 13 " t 14 ® t "1
x 2 " t 3 y 3 2t 1 z 4 t 3 ® Ÿa, b, c Ÿ3, 1, 3 ®
abc 7
Work Out (13 points each)
Show all your work. Partial credit will be given. You may not use a calculator.
dy
13. Find the solution of the differential equation x 3
" 2y 4 satisfying the initial
dx
condition yŸ1 3.
2
; P dx
dy
e 1/x
P "23
Ie
" 23 y 43
dx
x
x
x
4
2
d e 1/x 2 y 4 e 1/x 2
1/x 2 dy
1/x 2
1/x 2
" 3e y 3e
e
dx
dx
x
x3
x
2
2
2
4
x 1, y 3
e 1/x y e 1/x dx "2e 1/x C
x3
2
2
2
e 1/x y "2e 1/x 5e
y "2 5e 1"1/x
;
14. The curve y x 2 for 0 t x t
2
e3 "2e C
C 5e
is rotated about the y-axis. Find the surface area
of the resulting surface.
A
A
;
;
2=r ds
2
0
rx
ds 2=x 1 4x 2 dx = Ÿ1 4x 2 3/2
6
1
2
0
dy
dx
2
dx 1 Ÿ2x 2 dx 1 4x 2 dx
= Ÿ9 3/2 " = Ÿ1 3/2 13=
6
6
3
5
15.
A plate in the shape of a semicircle is placed at
the bottom if a tank with the straight edge down.
The radius of the circle is 4 cm and the water
in the tank is 6 cm deep.
What is the force on the plate?
(The density of water is > 1000 kg3 and
m
the acceleration of gravity is g 9.8 m 2 ,
sec
but you may leave your answer in terms of >g.)
Measure y up from the bottom of the tank. The slice at height y has width
w 2x 2 16 " y 2 . The water above this slice has depth h 6 " y. So the force is
F
;
;
4
;
>gh dA >g
;
4
Ÿ
0
6 " y 2 16 " y 2 dy 12>g
;
4
0
16 " y 2 dy " 2>g
;
4
0
y 16 " y 2 dy
16 " y 2 dy Area of a quarter circle of radius 4 1 =4 2 4=
4
3/2 4
2
4
3/2
"Ÿ16 " y 0 16 64
y 16 " y 2 dy 3
3
3
0
0
F 12>gŸ4= " 2>g 64 >g 48= " 128
3
3
0
!
.
16. Find the interval of convergence of the series
n2
x " 3 n
.
nŸlnn 2
Ÿ
Be sure to check the endpoints.
Name or quote the test(s) you use and check out all requirements of the test.
n1
x " 3 n
Ÿx " 3 a n1 2
2
nŸlnn Ÿn 1 ŸlnŸn 1 2
nŸlnn 2
|x " 3|n1
a n1 lim
n
lnn
"
3|lim
L nlim
lim
|x
n
nv. n 1 ln n 1 2 |x " 3|
nv. n 1 nv. lnŸn 1 v. a n
Ÿ
Ÿ Ÿ
|x " 3| 1
Converges on 2 x 4.
Check endpoints:
.
n
Ÿ" 1 At x 2:
converges because it is an alternating, decreasing series and
nŸlnn 2
n2
1
0.
lim
nv. n ln n 2
Ÿ
an Ratio Test:
Ÿ
!
At x 4:
;
!
.
n2
.
1
nŸln n 2
;
Apply the integral test: Let u lnn. Then du 1n dn and
1
1 du "1 "1 . 1 which is finite. So
dn u
2
ln 2
lnn 2
u2
2 nŸlnn converges.
So the interval of convergence is 2 t x t 4.
!
.
n2
1
nŸlnn 2
6