Name
ID
MATH 172
EXAM 2
Fall 1998
Section 502
Solutions
P. Yasskin
Multiple Choice: (5 points each)
4
; lnx dx
1
x
1. Compute
a. 4 ln 4
b. 4 ln 4 " 2
c. 4 ln 4 " 4
correctchoice
d. 4 ln 4 " 12
e. 2 ln 2 " 8
Integration by parts:
1 dx
x
u lnx
dv du 1x dx
v2 x
4 2 x
4
4
4
4
; ln x dx ¡2 x lnx ¢ 1 " ; x dx ¡2 x lnx ¢ 1 " ; 2 dx
1
1
1
x
x
4
4
¡2 x lnx ¢ 1 " ¡4 x ¢ 1 2 4 ln 4 " 2 1 ln 1 " 4 4 4 1 4 ln 4 " 4
6
; x 3 dx
2. Approximate the integral
by using the Midpoint Rule with 3 rectangles
0
with equal widths
a. 162
b. 306
correctchoice
c. 324
d. 360
e. 648
200
With fŸx x 3 we have
150
6
; x 3 dx X 2fŸ1 2fŸ3 2fŸ5 0
100
X 2Ÿ1 27 125 306
50
0
1
2
3
4
5
6
1
3. Compute
;
=/2
0
sin 4 2 cos 2 d2
a. " 1
3
b. " 1
c. 1
6
d. 1
5
e. 1
3
5
correctchoice
u sin 2
du cos 2 d2
;
=/2
0
sin 4 2 cos 2 d2 ;
=/2
2 0
May be skipped.
4. Compute
;
=/4
0
u5
5
u 4 du =/2
2 0
=/2
sin 5 2
5
1
5
0
May be skipped.
tan 3 2 sec 2 2 d2
a. " 1
b. 1
4
4
correctchoice
c. " 1
d. 1
3
3
e. " 1
2
u tan 2
du sec 2 2 d2
;
=/4
0
tan 3 2 sec 2 2 d2 ;
=/4
2 0
May be skipped.
5. Compute
a. ".
;
.
e
u 3 du u4
4
=/4
2 0
tan 4 2
4
=/4
0
May be skipped.
1
dx
xŸlnx 2
b. " 1
e
1
c. e
d. 1
e. .
.
correctchoice
.
1
1 du lim " 1
dx ;
u
bv.
e xŸlnx 2
xe u 2
e
1
6. Compute
dx
;
2
1 xŸlnx ;
b
xe
lim " 1
bv.
ln x
u lnx
du 1x dx
lim " 1 1
1
bv.
ln e
lnb
b
e
a. ".
b. "1
c. 1
e
d. 1
e. .
correctchoice
e
1
1 du lim " 1
dx ;
;
u
2
av1 x1 u 2
1 xŸln x e
e
xa
lim " 1
av1
ln x
e
a
lim " 1 1
av1
ln e
lna
.
2
7. Which of the following is the direction field of the differential equation
a.
b.
No,
dy
x 2 y?
dx
y U 0 in Quadrant II
correctchoice
c.
No,
y U 0 in Quadrant I
d.
No,
y U 0 in Quadrant III
e.
No,
y U 0 in Quadrant I
3
1 " x2
x Ÿ1 x 2 8. Given that the partial fraction expansion of
1 " x2
12 " 2 2 ,
1x
x Ÿ1 x 2 x
a. " 1 C
2x
4x
C
b. " 23 2 2
x
Ÿ1 x 2
compute:
2
;
is
1 " x2
dx
x Ÿ1 x 2 2
c. ln x 2 " lnŸ1 x 2 C
d. ln x 2 " 2 lnŸ1 x 2 C
e. " 1
x " 2 arctan x C
;
1"x
x 2 Ÿ1 x 2 2
correctchoice
dx ; 12 " 2 2 dx " 1x " 2 arctan x C
x
1x
9. Solve the differential equation
dy
xy 2 y 2
dx
with the initial condition
yŸ0 2.
Then find
yŸ1 .
a. "1
correctchoice
b. 0
c. 1
d. " 2
3
e. 3
2
dy
dy
dy
Separate variables:
y 2 Ÿx 1 Ÿx 1 dx
; 2 ;Ÿx 1 dx
dx
y
y2
2
2
x
1
1
Initial Cond: x 0, y 2
xC
" C
" 1y x x " 1
"y 2
2
2
2
2
x
1
1
1
1
Solve for y: y "
So: yŸ1 "1
y
"x
2
2
2
"1 "1 1
"x "x 1
2
2
2
2
y
dy
with the initial condition
10. Solve the differential equation
" x 2x 2
yŸ1 3.
dx
Then find
yŸ2 .
a. 2
b. 4
c. 6
d. 8
Linear equation: PŸx " 1x
e. 12
correctchoice
QŸx 2x 2
1
"
dx
"1
; Pdx
e ; x e "ln x e ln x x "1 1x
Ie
1 dy " 1 y 2x
d 1 y 2x
Multiply equation by I 1x :
x dx
dx x
x2
3 1C
Initial Cond:
x 1, y 3
Solve: 1x y ; 2x dx x 2 C
1 y x2 2
y x 3 2x
So: yŸ2 2 3 2 2 12
x
Integrating factor:
C2
4
1
; Ÿt 2 " t e 2t dt
11. (15 points) Compute
0
1
Step 1: u t 2 " t
; Ÿt " t e dt
2
2t
0
Ÿt
2
" t 1 e 2t " ; 2t " 1 e 2t dt
2
Ÿt 2 " t 1 e 2t
2
Ÿt 2 " t 1 e 2t
2
" 2 " 1 e2 4
x2 " 1
x
2
1
;
;
;
=/3
0
=/3
0
=/3
0
du 2t " 1 v 1 e 2t
2
Step 2: u 2t " 1 dv e 2t dt
2
du dt
v 1 e 2t
2
2
0
1
1
2t
1
"
e 2t ; e 2t dt
"
2
4
0
1
2t
1
1
"
2t
2t
e e
"
4
4
0
1 e 2 " " "1 1 " 1
2
4
4
4
;
12. (10 points) Compute
;
1
2
1
x2 " 1
x
dx
x sec 2
dx
sec 2 2 " 1
sec 2
dx sec 2 tan 2 d2
sec 2 tan 2 d2
Change limits: x 1 when 2 0
tan 2 2 d2
sec 2 2 " 1 d2
tan 2 " 2
dv e 2t dt
x 2 when sec 2 2
=/3
or cos 2 1 or 2 =
2
3
0
tan = " = 3 " =
3
3
3
OR substitute back:
tan 2 " 2
2
x1
sec 2 2 " 1 " 2
4 " 1 " arcsec 2 "
2
x1
1 " 1 " arcsec 1
x 2 " 1 " arcsec x
3 " =
3
2
x1
5
13. (10 points) Find the partial fraction expansion for
(Do not integrate.)
Hints:
Try:
x 1,
x0
and
Ÿx
4x " 6
.
" 1 2 Ÿx 2 1 d .
dx
4x " 6
A
B
Cx2 D
2 2
2
x
"
1
Ÿ
Ÿx 1 Ÿx " 1 Ÿx 1 Ÿx " 1 4x " 6 AŸx " 1 Ÿx 2 1 BŸx 2 1 ŸCx D Ÿx " 1 2
x1:
" 2 BŸ2 x0:
" 6 AŸ"1 B D "A " 1 D
D A"5
d :
4 A¡Ÿx 2 1 Ÿx " 1 Ÿ2x ¢ BŸ2x CŸx " 1 2 ŸCx D 2Ÿx " 1 dx
x1:
4 AŸ2 BŸ2 2A " 2
2A 6
D A"5 3"5
x0:
4 AŸ1 C D2Ÿ"1 A C " 2D
C 4 " A 2D 4 " 3 2Ÿ"2 So:
Ÿx
B "1
A3
D "2
C "3
4x " 6
3
"2
"1 2 "3x
2
Ÿx " 1 " 1 2 Ÿx 2 1 Ÿx 1 Ÿx " 1 14. (15 points) A tank initially contains
of sugar water with
of dissolved
12L
8gm
sugar. Sugar water that contains
of sugar per liter is poured into the tank at
2gm
the rate of
The solution is kept thoroughly mixed and drains from the tank
3L/min.
at the rate of
How much sugar is in the tank
3L/min.
(a) after
(b) after
and
(c) asymptotically after a large time?
t min
16min
Let
be the grams of sugar in the tank at time t.
s Ÿt sŸt gm
gm
ds 6 " 1 s
ds 2
3 L "
3 L
L
4
dt
12 L min
dt
min
IN
OUT
; dt
" 4 ln 6 " 1 s t C
; ds
1
4
6" s
4
InitialCond:
s Ÿ0 8
t 0, s 8
" 4 ln 6 " 1 8 C
4
Plug back and solve:
"4 ln 6 " 1 s t " 4 ln 4
ln 6 " 1 s
4
4
1
1
t/4ln 4
t/4
t/4
"
"
"
6" s e
4e
s 6 " 4e
4
4
(a)
(b)
(c)
C "4 ln 4
" t ln 4
4
sŸt 24 " 16e "t/4
sŸ16 24 " 16e "4
lim sŸt 24
tv.
6