MATH 172
EXAM 1
Fall 1999
Section 502
Solutions
P. Yasskin
Multiple Choice: (5 points each)
; x x 2 " 1x dx
2x 7/2 2x "3/2 C
3
7
2x 7/2 " 2x 1/2 C
correctchoice
7
"
3/2
37/2
2x
2x
C
3
3
3
3/2
x x " lnx 2x Ÿx 2 " lnx C
3
3
2x 3/2 x 3 " lnx C
3
3
1. Compute:
a.
b.
c.
d.
e.
; x x 2 " 1x
dx
;
2. Evaluate
2 2=
1=
1 2=
="1
e. 2 =
2
0
;Ÿx 5/2 " x "1/2 dx
4 " x2
1 dx
2x 7/2 " 2x 1/2
7
1
C
by interpretating it as an area.
a.
b.
c.
d.
correctchoice
3
Quarter circle of radius 2
2
on top of rectangle of
width 2 and height 1.
A 1 =Ÿ2 2 2 1 = 2
4
1
0
1
2
1
3. Find the area between the parabola y
x 2 and the line y
2x 3.
a. " 16
3
16
3
32
3
9
88
3
b.
c.
d.
e.
x2
A
correctchoice
;
3
"1
x 2 " 2x " 3
®
2x 3
Ÿ2x 3 " x 2 dx
;
4. Compute
4
1
x2
®
0
Ÿx 3
3
3x " x
3
¡9 "1
1 Ÿx " 3 ®
0
9 " 9¢ " 1 " 3 1
3
x
92" 1
3
"1, 3
32
3
lnx dx
2 x
a. 2 ln 4 " 1
b. 2 ln 4 " 2
correctchoice
c. 2 ln 4 " 4
d. 2 ln 4 " 6
e. ln 2 " 4
u
ln x
dv
Integrate by parts with
du
;
4
1
ln x dx
2 x
1 dx
x
x lnx " ; 1 dx x
2 ln 4 " 4 2 2 ln 4 " 2
v
1 dx
2 x
x
x lnx " 2 x
5. The mass density of a 3 cm bar is >
4
1
4 ln 4 " 2 4
1 x 2 gm
cm
"
1 ln 1 " 2 1
for 0 t x t 3. Find the total
mass of the bar.
a. 4 gm
b. 10 gm
c. 12 gm
correctchoice
d. 18 gm
e. 30 gm
M
3
; >Ÿx dx
0
3
; Ÿ1 x 2 dx
0
3
x x
3
3
39
12
0
2
1 x 2 gm
cm
x-coordinate of the center of mass of the bar.
a. 3
2
b. 2
c. 7
3
d. 33
correctchoice
16
e. 99
4
6. The mass density of a 3 cm bar is >
M1
x
3
3
; x>Ÿx dx
; Ÿx x 3 dx
0
M1
M
7. Compute
0
99
4 12
;
=/2
0
x2
2
x4
4
3
0
9
2
for 0 t x t 3. Find the
81
4
99
4
33
16
sin 4 2 cos 2 d2
a. " 1
b.
c.
d.
e.
;
=/2
0
3
"1
5
1
6
1
5
1
3
correctchoice
sin 4 2 cos 2 d2
8. Compute
;
=/4
0
sin 5 2
5
=/2
0
1
5
tan 3 2 sec 2 2 d2
a. " 1
b. 1
4
4
correctchoice
c. " 1
d. 1
3
3
e. " 1
2
;
=/4
0
tan 3 2 sec 2 2 d2
tan 4 2
4
=/4
0
1
4
3
2
; x 4 " x 2 dx
9. Compute:
0
a.
b.
c.
d.
e.
u
2 2
3
8
correctchoice
3
24
32
3
6
4 " x2
du
2
; x 4 " x 2 dx
"1 ;
2
0
" 1 du
"2x dx
2
2
x0
u du
"1
2
2u 3/2
3
x dx
"1
3
Ÿ4
" x 2 3/2
2
0
0 1 Ÿ4 3/2
3
8
3
1
; Ÿt 2 " t e 2t dt
10. (15 points) Compute
0
1
; Ÿt 2 " t e 2t dt
0
Ÿt
2
Ÿt
2
Ÿt
2
" t 1 e 2t " 1 ;Ÿ2t " 1 e 2t dt
2
2
1 2t 1 ; 2 e 2t dt
Ÿ2t " 1 e "
2
2
2
" t 1 e 2t " 1
2
" t 1 e 2t " 1
2
2
1 Ÿt 2 " t e 2t "
2
0 " 1 Ÿ1 e 2 4
1
"
2
Ÿ2t
u
du
u
Ÿ2t
du
t2 " t
dv
" 1 dt v
2t " 1 dv
2 dt
v
e 2t dt
1 e 2t
2
e 2t dt
1 e 2t
2
" 1 1 e 2t " 1 e 2t
2
2
1 Ÿ2t " 1 e 2t 1 e 2t 1
4
4
0
1 e 2 " 0 " 1 Ÿ" 1 e 0 1 e 0
4
4
4
4
x2 " 1
dx
x
1
You must evaluate any inverse trig functions in the answer.
;
11. (15 points) Compute
x
;
sec 2
dx
x2 " 1
x
2
1
;
2
x1
Ÿ
sec 2
x2 " 1
sec 2 tan 2 d2
dx
2
;
2
x1
tan 2 sec 2 tan 2 d2
sec 2
2 " 1 d2
tan 2 " 2
4 " 1 " arcsec 2 "
because arcsec 2
2
x1
sec 2 2 " 1
;
2
x1
tan 2
tan 2 2 d2
0.
t 0 and t 1.
HINT:
Factor the quantity in the square root.
L
;
dy
dt
3t 5
1
0
2
dx
dt
1 t6, y
2
t 4 between
4t 3
dy
dt
x1
3 " =
3
12. (15 points) Find the arc length of the parametric curve x
dx
dt
2
x 2 " 1 " arcsec x
1 " 1 " arcsec 1
= and arcsec 1
3
2
dt
;
1
9t 10
16t 6 dt
0
1
; t 3 9t 4 16 dt
0
1
1 ;
where
du 36t 3 dt
u du
u 9t 4 16
36 t0
1 2u 3/2 1 1 Ÿ9t 4 16 3/2 1 1 Ÿ25 3/2 " 1 Ÿ16 3/2
54
54
36 3
54
t 0
t 0
125 " 64
54
61
54
13. (10 points) Find the volume of the solid whose base is the triangle with vertices Ÿ0, "1 ,
Ÿ1, 0 and Ÿ0, 1 and whose crosssections perpendicular to the x-axis are semicircles.
1
The top of the triangle is
y
0
-1
x
1
1"x
So the radius of the semicircle is
r"1"x
-1
So the crosssectional area is
1
V
; AŸx dx
0
AŸx 1
2
Ÿ1 " x dx
; =
2
0
"
1 =r 2
2
3
= Ÿ1 " x 3
2
= Ÿ1 " x 2
2
1
0
0
and the volume is
3
= Ÿ1 2 3
=
6
5
x 2 and the line y
the indicated axis. Find the volume of the solid swept out.
a. x-axis
14. (10 points) The area between the parabola y
x is rotated about
This is an x-integral. Use a vertical rectangle.
1
Rotating about the x-axis, gives washers.
upper radius
x lower radius
1
V
2
; =Ÿx 2 " =Ÿx 2 dx
0
3
= x " x
5
3
0
1
x
5
1
0
x2
1
= ; x 2 " x 4 dx
0
= 1 " 1
5
3
=5"3
15
2=
15
b. y-axis
This is again an x-integral. Again use a vertical rectangle.
Rotating about the y-axis, gives cylindrical shells.
radius
x height
x " x2
1
V
; 2=xŸx " x 2 dx
1
0
3
4
2= x " x
3
4
0
1
0
2= ; x 2 " x 3 dx
2= 1 " 1
3
4
2= 4 " 3
12
=
6
6