MATH 172
FINAL EXAM
Section 504
Solutons
a Ÿ2, 0, 1 1. If
and
b
Ÿ
"1, 2, 1 Spring 1999
P. Yasskin
then
" 3b
2a
a. Ÿ1, 6, 5 b. Ÿ1, "6, 5 c. Ÿ7, 6, "1 d. Ÿ7, "6, "1 correctchoice
e. Ÿ"1, 6, "5 " 3b
2a
2Ÿ2, 0, 1 " 3Ÿ"1, 2, 1 a Ÿ2, 0, 1 2. If
and
a. Ÿ"2, "3, 4 b.
c.
d.
e.
b
Ÿ4, 0, 2 Ÿ3,
Ÿ
"1, 2, 1 "6, "3 then
Ÿ7,
"6, "1 a•
b
correctchoice
Ÿ"2, 3, 4 Ÿ"2, 0, 1 Ÿ4, 0, "1 Ÿ4, 3, "2 a•
b
k
î
F
2
0 1
îŸ0
" 2 " F Ÿ2 " "1 k Ÿ4 " 0 Ÿ
"2, "3, 4 "1 2 1
a Ÿ2, 0, 1 3. If
and
b
Ÿ
"1, 2, 1 are two edges of a triangle, find the area of
the triangle.
a.
b.
c.
d.
e.
A
3
3
2
1 29
2
29
2
29
1 a•
b
2
correctchoice
1 4 9 16
2
1 29
2
1
2 , "1, 1
u
4. If
v
2
is
v Ÿ0, 1,
and
"1 then the angle between
u
and
30°
45°
60°
120°
e. 135°
a.
b.
c.
d.
u v cos 2
correctchoice
|
2 Ÿ0 Ÿ"1 Ÿ1 Ÿ1 Ÿ"1 "2
|u
u v "2 "1
2 135°
||v
|
|u
2 2
2
!
.
5. The series
Ÿ" 1 Ÿx
2 n1
n
n0
" 2 n
211
|v|
2
is the Taylor series about
x
11
2
for
2
(Hint: Just sum the series.)
a. 1
x
correctchoice
b. 2
x
1
x"2
2
d.
x"2
1
e.
2Ÿx " 2 c.
" Ÿx " 2 Geometric series:
ratio !
1
2
"
Ÿx " 2 1"
2
.
n0
Ÿ" 1 Ÿx
2 n1
n
" 2 n
2
2
2
Ÿ" 1 first term 21
1
2 Ÿx " 2 !
.
6. Find the radius of convergence of the series
n0
0
Ÿx
" 2 0
1
2
1
x
Ÿ" 1 Ÿx
2 n1
n
" 2 n.
a. 0
b. 1
c. 2
correctchoice
d. 4
e. 8
Ÿ" 1 Ÿx
2 n1
n
Ratio Test:
L
lim aan1
nv.
n
|x " 2|
2
an
lim
nv.
" 2 n
" 2 n1
a n1
n1
Ÿ" 1 Ÿx
2 n2
2 n1
x"2
lim
n
nv.
2
Ÿx " 2 2
Radius of Convergence 2
Ÿx
n2
" 2 n1
x"2
2
1
2
!
.
7. The series
n1
n!
n1
is
a. Divergent by the n th term Divergence Test
b. Convergent by the Ratio Test
correctchoice
c. Divergent by the Ratio Test
d. Convergent by the Integral Test
e. Divergent by the Integral Test
Ratio Test:
L
an
lim aan1
n
nv.
n1
n!
n2
lim
nv. Ÿn 1 !
a n1
n!
n1
n2
1 !
Ÿn lim
nv.
n2
Ÿn 1 n 1
0
1
Convergent
3
sinŸ2x " 2x 8x
3
lim
xv0
x5
8. Compute
a. ".
b. 0
c. 4
3
2
5!
d.
e. .
correctchoice
3
5
3
5
x" x x "C
sinŸ2x 2x " 8x 32x " C
3!
5!
6
120
3
3
5
3
8x
8x 3
32x
8x
sinŸ2x " 2x 2x "
" C " 2x 8x
3 lim
6
120
3 lim 6
lim
xv0
xv0
xv0
x5
x5
32
8
"
2
lim
x "C .
xv0 6
120
sin x
9. Compute
;
Ÿ2x
3
32x 5 " C
120
x5
" 1 sinŸx 4 " 2x dx
a. 2 cosŸx 4 " 2x C
b. "2 cosŸx 4 " 2x C
c. 1 cosŸx 4 " 2x C
2
d. " 1 cosŸx 4 " 2x C
correctchoice
2
e. cosŸx 4 " 2x C
u
;
x 4 " 2x
Ÿ2x
3
du
Ÿ4x
3
" 1 sinŸx 4 " 2x dx
" 2 dx
1
2
;
1 du Ÿ2x 3 " 1 dx
2
sin u du " 1 cos u C
2
" 1 cosŸx 4 " 2x C
2
3
;
10. Compute
2
0
x2
dx
4 " x2
a. 1
b. 2
c. =
4
d. =
e. 2=
;
x
2
0
correctchoice
2 sin 2
dx 2 cos 2 d2
4 " x 2 4 " 4 sin 2 2 2 1 " sin 2 2 2 cos 2
=/2
4 sin 2 2 2 cos 2 d2 =/2 4 sin 2 2 d2 4 =/2 1 " cosŸ22 d2
x2
dx 2
2 cos 2
0
0
0
4 " x2
=
/2
sinŸ22 = =
2
2 2"
2
2
0
;
;
11. Compute
a.
b.
c.
d.
e.
;
=/2
1
63
2
63
1
21
4
63
2
21
;
1
sin 6 x cos 3 x dx
0
sin 6 x cos 3 x dx
0
.
;
correctchoice
12.
=/2
;
;
1
0
;
=/2
0
sin 6 x 1 " sin 2 x cos x dx
u 6 Ÿ1 " u 2 du
1
dx
x e 2x
1
u7 " u9
7
9
0
1 " 1
7
9
sin x
du
cos x dx
2
63
is
a. Convergent by comparison to
b. Divergent by comparison to
;
;
;
;
.
1
c. Convergent by comparison to
d. Divergent by comparison to
.
1
.
1 dx
e 2x
1 dx
e. 2x
1 dx
x
correctchoice
1
1
1 dx
x
;
.
1 dx which is convergent.
2x
1 e
which is convergent.
For large x, e 2x is larger than x. So we compare to
;
u
1 dx e "2x . 0 e "2 e "2
2x
2
2
"2 1
1 e
1
1
Since
the original series is also convergent.
e 2x
x e 2x
.
4
13.
The basket shown at the right is 9 in tall.
Its horizontal cross sections are circles
whose radius is given by
r
2 y
where y is the height from the bottom.
Find the volume of the basket.
a. 216=
b. 162=
correctchoice
c. 108=
d. 81=
e. 81=
V
14.
;
2
9
0
AŸy dy
;
9
0
= 2 y
2
dy
4=
;
9
y dy
0
4=
y2
2
9
162=
0
A trough filled with water is 3m long and its
end is a 45° right triangle which is 2m high
and 2m wide. Find the work done to pump
the water out of the top.
(> is the density of water and
g is the acceleration of gravity.)
a. 2>g
b. 3>g
c. 4>g
correctchoice
d. 6>g
e. 8>g
Put the origin at the bottom of the trough with y positive upward.
A horizontal slice at height y has volume dV Ÿlength Ÿwidth Ÿheight Ÿ3 Ÿy Ÿdy and weight dF >g dV 3>g y dy. It must be lifted a distance D 2 " y. So
2
2
y3 2
W D dF Ÿ2 " y 3>g y dy 3>g Ÿ2y " y 2 dy 3>g y 2 "
3 0
0
0
8
3>g 4 "
>g¡12 " 8 ¢ 4>g
3
;
;
;
5
;
15. (10 points) Compute
1
x arctan x dx
x 2 arctan x "
2
0
x 2 arctan x " 1
2
2
x arctan x dx.
0
arctan x
dv x dx
2
1
dx v x
du 2
2
1x
u
Integrate by Parts:
;
1
;
1
;
1
0
0
dx
1 x2
1 = " 1 1" =
4
2
2 4
yŸ1 Integrating factor:
x
dy
dx
y
1
x2
Ÿ1
x2
dy
dx
xy
" 1 1 x2
0
dx
1
0
= " 1
2
4
2x 2
1 x2
with the initial
ln 4.
dy
dx
Linear. Put into standard form:
d Ÿxy dx
;
x 2 arctan x " 1 Ÿx " arctan x 2
2
16. (10 points) Solve the differential equation
condition
0
0
1 arctan 1 " 1 Ÿ1 " arctan 1 2
2
x 2 arctan x " 1
2
2
1
1
1"
1
1
x2
dx
2 1 x2
I
e
; P dx
e
1
; x dx
2x
1 x2
xy
Use the initial condition to find C:
ln 4 ln 2 C
C ln 4 " ln 2
1y
x
;
x
2
1 x2
e ln x
1
x
Q
2
1 x2
x
2x dx
1 x2
P
lnŸ1 x 2 C
when
1
y
ln 4 :
ln 2
Substitute back and solve for y:
xy
lnŸ1 x 2 ln 2
y
;
17. (10 points) Approximate
lnŸ1 x 2 ln 2
x
0.1
e "x dx
2
to 7 decimal places.
0
(4 points extra credit:) How do you know it is accurate to 7 decimal places?
ex
;
0.1
0
2
1x x
2
e "x dx X
2
;
0.1
0
x3
6
e "x
C
1 " x2
x 4 dx
2
2
1 " x2
3
x" x
3
x5
10
x4 " x6
2
6
0.1
0
C
.1 " .001
3
.000001
.099 667 666 6
Since this is an alternating series, the partial sum is accurate to within the next term
which gives
0.1
0.1 6
10 "7 X 2.4 • 10 "9 .000 000 002 4
x dx x 7
6
42
42
0
0
which says the error is in is the 8th digit of the answer.
;
6