MATH 172
EXAM 2
Spring 1999
Section 504
Solutions
P. Yasskin
7
; Ÿx 2 2x dx
1. Approximate
by using the trapezoid rule with 3 intervals.
1
a. 83
b. 116
c. 162
d. 166
correctchoice
e. 232
7"1 2
3
7
1
2
fŸ1 fŸ3 fŸ5 1 fŸ7 x
; Ÿx 2x dx X
2
2
1
Ÿ3 2Ÿ15 2Ÿ35 Ÿ49 14 166
fŸx x2
2x
x
fŸ1 2
2. Given that the partial fraction expansion for x4 " 12 is
2
; x4 " 12 dx .
x x
correctchoice
2 tan "1 x 1x C
tan "1 x 1x C
2
"
1 x
tan
" 1x C
2
"4x 2 C
2
2
x3
Ÿx 1 None of These
x
x
2fŸ3 2fŸ5 fŸ7 x2 " 1
x4 x2
2
x
2
1
" 12 ,
x
compute
a.
b.
c.
d.
e.
2
; x4 " 12 dx
x x
;
2
x2
1
" 12 dx
x
2 tan "1 x 1x
C
1
;
3. Compute:
2
1
x "1
2
1
a. sin "1 Ÿ2 " =
2
b. = " sin "1 Ÿ2 dx
2
c. ln 2 3
correctchoice
d. ln 2 " 3
e. ln sec 2 tan 2
sec 1 tan 1
xu1
;
2
So:
1
x2 " 1
1
dx
x
;
sec 2
2
x1
dx
sec 2 tan 2 d2
1
sec 2 tan 2 d2
2
sec 2 " 1
2
ln x x 2 " 1
1
;
4. The improper integral
2
1
;
2
x1
sec 2 d2
ln|sec 2 tan 2 |
" ln 1 0
ln 2 3
" lim " 1
xv1
x"1
"1 .
ln 2 3
1
dx
2
Ÿx " 1 diverges to " .
converges to a negative number
converges to 0
converges to a positive number
correctchoice
e. diverges to .
a.
b.
c.
d.
;
2
1
1
dx
2
Ÿx " 1 ".
b. converges and is
c. converges and is
d. converges and is
e. diverges to
"1 t sin x t 1
So ;
.
2
.
2
1
2
1
2
1
2
.
x2 t x2
" 1
2"1
1
;
5. The improper integral
a. diverges to
2
" 1
x"1
.
1
dx
x 1 sin x
2
correctchoice
1 sin x t x 2
.
1
dx t ; 12 dx
x 1 sin x
2 x
2
" 1x
2
.
2
1 u
x2
x2
1
2
1
u 21
1 sin x
x 2
2
dy
dx
6. Which of the following is the direction field of the differential equation
a.
d.
b.
e.
7. Solve the initial value problem
Then find
yŸ2 .
a. 0
b. 1
5
c. 4
5
d. 7
10
correctchoice
e. 1
dy
dx
xy 2 ?
correctchoice
dy
dx
dy
dx
dy
dx
c.
xy 2
0 on both axes
0 when x
0
0 when x
0
with the initial condition
yŸ1 2 .
5
3
; 12 dy ; x dx
y
Initial Condition ( x 1, y 2 ):
"5
5
2
2
2
x
1
1
x
3"
Solve:
"y "3
y
2
2
2
1
Evaluate:
yŸ2 6 " 22
" 1y
Separate variables:
8. The mass density of a 3 ft bar is >
1 C
2
6 " x2
2
1 x 2 lb
ft
x2 C
2
C "3
y
2
6 " x2
for 0 t x t 3. Find the center of
mass of the bar.
a. x 12
b. x 16
33
c. x 33
correctchoice
16
d. x 4
99
e. x 99
4
3
M
; > dx
0
mom
x
3
; Ÿ1 x 2 dx
0
3
3
x x
3
3
; x> dx ; xŸ1 x 2 dx
0
0
mom 99 1 33
M
4 12
16
3
0
3
; x x 3 dx
dx
dt
L
;
4t
1
0
x
x2
2
0
9. Find the arc length of the parametric curve
between t
a. 61
27
b. 125
9
c. 125
27
d. 122
3
250
e.
3
3 27
3
12
2t 2
x4
4
3
9
2
81
4
0
and
y
t3
3
99
4
0 and t 1.
correctchoice
dy
dt
3t 2
2
1
1
dx 2 dy
dt ; 16t 2 9t 4 dt ; t 16 9t 2 dt
dt
dt
0
0
1
1 Ÿ25 3/2 " 1 Ÿ16 3/2 1 Ÿ125 " 64 2 1 Ÿ16 9t 2 3/2
27
27
3 18
27
0
61
27
4
for
is rotated about the x-axis. The area of the
0txt2
y x3
resulting surface may be computed from the integral
10. The curve
2
a. ; =x 1 9x 4 dx
0
2
b. ; 2=x 3 1 9x 4 dx
correctchoice
0
2
c. ; 2=x 1 x 6 dx
0
2
d. ; =x 3 1 x 6 dx
0
2
e. ; 2=x 1 9x 4 dx
0
2
A
; Ÿcircumference ds
2
0
11. (10 points)
0
2x 2 " x 2
xŸx 2 1 0:
2
A
x
1:
3
AŸ2 ŸB C x
"1 :
A
2
5
B
2x 2 " x 2
x3 x
0
; 2=x 3 1 9x 4 dx
0
2x 2 " x 2 .
x3 x
dx
;
1
dx
1
"
x 2 3/2
Ÿ
Bx C
x2 1
2x 2 " x 2
®
4BC
AŸx 2
BC
4B"C
®
1 ŸBx C x
"1
B"C
1
"1
"1
x2
1
Compute:
;
C
2
x
x
AŸ2 Ÿ"B C Ÿ"1 12. (10 points)
0, 1, "1.)
A
x
x
sin 2
2
dx
Find the partial fraction expansion for
(Do not integrate. HINT: Try x
2x 2 " x 2
x3 x
2
dy
dx
; 2=y 1 ;
1
dx.
2 3/2
Ÿ1 " x cos 2 d2
;
1 d2
cos 2 2
1
1 " sin 2 2
; sec 2 2 d2
3/2
cos 2 d2
tan 2 C
;
1
cos 2 d2
cos 3 2
sin 2
cos 2
C
x
1 " x2
C
5
13. (10 points)
Solve the initial value problem
condition
P
Ÿ1 yŸ1 2x
1 x2
x2 dy
dx
2xy
Initial condition ( x
2
Ÿ1 x y x3
2xy
1 x2
I
e
2.
; Pdx
dy
dx
2x dx
1 x2
;
d
dx
3x 2
3
1, y
y
Ÿ1 2 ):
lnŸ1 x 2 x 2 y
Ÿ1 1 2
3x 2
Ÿ1 1C
C
; P dx
x 2 y
3x 2
1 x2
1x 2
e ln
; 3x 2 dx
with the initial
1 x2
x3
C
3
x3 3
1 x2
14. (10 points)
A nuclear power plant went on line at the beginning of the year 1980. It
has produced isotope X at the rate of 10 kg
yr and the half-life of X is 20 yr. (So its
ln
2
decay constant is k . ) The plant stores all of the isotope X it produces. If
20
there was no isotope X at the beginning of 1980, how much isotope X will there be at
(6 points for the equations.)
the beginning of the year 2000?
dX
dt
Let XŸt be the kg of X at time t. Solve:
Separate variables:
;
dX
10 " ln 2 X
20
Use initial conditions ( t
0, X
Solve for X:
ln 10 " ln 2 X
20
0 ):
" 20 ln 10 " ln 2 X
20
ln 2
" ln 2 t ln 10
C
with
XŸ0 " 20 ln 10 " ln 2 X
ln 2
20
0.
tC
" 20 ln 10
ln 2
t " 20 ln 10
ln 2
e ln 10 e "
t ln 2
20
10e "
ln 2
200 1 " e " t 20
ln 2
The minus is needed to give the initial condition. The year 2000 is t
20. So
XŸ20 e ln 10 e "
t ln 2
20
o10e
200 1 " e " 2020ln 2
ln 2
ln 2 X
20
200 Ÿ1 " e "ln 2 ln 2
10 o 10e "
t ln 2
20
t ln 2
20
X
10 " ln 2 X
20
t ln 2
" 20
10 " ln 2 X
20
; dt
10 " ln 2 X
20
20
200 1 " 1
ln 2
2
100 X 144kg
ln 2
6