MATH 152
Exam1
Fall 1997
Version A
Solutions
Part I is multiple choice. There is no partial credit. You may not use a calculator.
Part II is work out. Show all your work. Partial credit will be given. You may use your
calculator.
1
Part I: Multiple Choice (5 points each)
There is no partial credit. You may not use a calculator. You have 1 hour.
1. Compute ;
a. " =
b.
c.
d.
e.
=/4
2
=2 =4
32
1024
1
4
3 correctchoice
4
5
6
u sec x
;
=/4
0
tan x sec 4 x dx
0
du sec x tan x dx
tan x sec 4 x dx ;
2
4
4
=/4
x0
1
4
"
.
u4
4
u 3 du sec 4 x
4
=/4
0
1 " 1 3 (d)
4
4
2. The integral ; x 2 e "x dx
3
0
a. diverges to ".
b. converges to " 1
3
c. converges to 0
d. converges to 1 correctchoice
e. diverges to .
3
1 du x 2 dx
3
When x 0, u 0.
When x ., u ..
.
.
.
3
ulim
" 1 e "u " " 1 e 0
; x 2 e "x dx 1 ; e "u du " 1 e "u
v.
3
3
3
3
0
0
0
u x3
du 3x 2 dx
0 1 1 (d)
3
3
2
3. At time t in days, a lump of Thorium contains M 200e Ÿ"t/40 kg of radioactive
Thorium-234. In other words, the amount of Thorium-234 drops by a factor of 1e every
40 days. Find the average amount of Thorium-234 present in the first 40 days, i.e.
between t 0 and t 40.
a. 8000Ÿe " 1 b. 200 1 " 1
e correctchoice
c. 200Ÿe " 1 d. 4000 1 1
e
e. 100 1 1
e
40
M ave 40
1 ; 200e Ÿ"t/40 dt 200
40 " 0 0
40
e
Ÿ"t/40 "200Ÿe "1 " e 0 200Ÿ1 " e "1 (b)
"1
40
0
4. Find the area between the parabolas y 16 " x 2 and y x 2 " 4x.
a. 38
b. 72 correctchoice
c. 96
d. 102
e. 128
20
15
10
5
-4
-2
0
2
x
4
-5
-10
16 " x 2 x 2 " 4x
0 2x 2 " 4x " 16
x 2 " 2x " 8 0
Ÿx 2 Ÿx " 4 0
4
3
4
4
2
2
2
A ; Ÿ16 " x " Ÿx " 4x dx ; Ÿ16 4x " 2x dx 16x 2x 2 " 2 x
"2
"2
3 "2
3
3
Ÿ" 2 2
4
2
16 • 4 2 • 4 " 2
" 16Ÿ"2 2Ÿ"2 " 2
3
3
128
16
64 32 "
72 (b)
" "32 8 3
3
x "2, 4
3
=
5. Compute ; sin x e cos x dx
a. e "1 " e
b. "e "1
c. e "1 " 1
0
d. 1 " 1
e
e. e " 1
e correctchoice
u = cos x
du " sin x dx
cos x
sin
x
e
dx
" ; e u du "e u "¡e cos x ¢ =0 "e "1 e 1 e " 1e (e)
;
0
6. The area below the parabola y xŸ4 " x and above the x-axis is rotated about the
x-axis. The volume of the solid swept out is given by
4
a. ; 2=y 4 " y dy
0
4
b. ; 2=x 2 Ÿ4 " x dx
0
2
c. 2 ; 2=x 2 Ÿ4 " x dx
0
4
d. ; =x 2 Ÿ4 " x 2 dx correctchoice
0
2
e. 2 ; 2=y 4 " y dy
0
4
2
0
1
2
x
3
4
-2
-4
4
4
4
0
0
0
V ; =y 2 dx ; =¡xŸ4 " x ¢ 2 dx ; =x 2 Ÿ4 " x 2 dx (d)
4
7. The area below the parabola y xŸ4 " x and above the x-axis is rotated about the
y-axis. Find the volume of the solid swept out.
a. 128 = correctchoice
3
b. 64 =
3
c. 512 =
15
496
=
d.
15
e. 108 =
5
4
3
2
1
-4
-2
0
2
4
x
4
4
4
3
4
V ; 2=xh dx ; 2=x¡xŸ4 " x ¢ dx ; 2=¡4x 2 " x 3 ¢ dx 2= 4 x " x
3
4
0
0
0
512= 1 " 1 512 = 128 = (a)
12
3
3
4
4
0
8. Identify which term in the following partial fraction expansion does NOT have the
correct form:
Ÿx
B
2x 5
A C 2D
2
x"1
x2
x 7
" 1 2 Ÿx 2 Ÿx 2 7 Ÿx " 1 where A, B, C and D are constants.
A
a.
x"1
B
b.
2
Ÿx " 1 C
c.
x2
d. 2D correctchoice
x 7
e. None, they all have the correct form.
The quadratic term should have a linear numerator: Dx2 E (d)
x 7
5
9. The integral ;
2
"2
Ÿ
4 " x 2 " x dx gives the area of which of the following regions?
correctchoice
a.
d.
e.
b.
c.
The top is part of the semicircle y 4 " x 2 . The bottom is part of the line y x.
The integral ranges from x "2 where the semicircle meets the x-axis to x 2
where the semicircle meets the line y x.
Note the line must also extend to x "2. (a)
1
10. Compute ; x 2 e x dx
0
a. e
b. e " 1
c. e " 2 correctchoice
d. 2e " 1
e. 2e " 2
Integrate by parts twice: First u x 2 and dv e x dx. So du 2x dx and v e x .
Thus the indefinite integral is
I ; x 2 e x dx x 2 e x " ; 2xe x dx
Second u 2x and dv e x dx. So du 2 dx and v e x . Thus
I x 2 e x " 2xe x " ; 2e x dx x 2 e x " 2xe x 2e x
So the definite integral is
1
1
; x 2 e x dx ¡x 2 e x " 2xe x 2e x ¢ 0 ¡e 1 " 2e 1 2e 1 ¢ " ¡0 2e 0 ¢
0
e " 2e 2e " 2 e " 2 (c)
6
Part II: Work Out (10 points each)
Show all your work. Partial credit will be given.
You may use your calculator but only after 1 hour.
11. Compute ;
2
"2
4 " x 2 dx
Substitute x 2 sin 2 and dx 2 cos 2 d2:
2
1 cosŸ22 4 " x 2 dx ; 4 " 4 sin 2 2 2 cos 2 d2 4 ; cos 2 2 d2 4 ;
d2
I;
2
"2
sinŸ22 2¡2 sinŸ2 cosŸ2 ¢
2 2
2
2
Now x 2 sin 2. So
sin 2 x
cos 2 1 " sin 2 2 1 " x
and
2
4
2 arcsin x
Thus:
2
2
2
x
x
x
I 2¡2 sinŸ2 cosŸ2 ¢ 2 arcsin
1"
2
4 "2
2
=
2¡arcsinŸ1 ¢ " 2¡arcsinŸ"1 ¢ 4 2=
2
12. Compute ;
4
Ÿx
" 1 2 Ÿx 1 dx
We find the partial fraction expansion. The general form is:
4
B
A C
2
x"1
x1
Ÿx " 1 2 Ÿx 1 Ÿx " 1 We clear the denominator:
4 AŸx " 1 Ÿx 1 BŸx 1 CŸx " 1 2
Then we plug in three numbers:
x1:
4 BŸ2 B2
x "1 :
4 CŸ " 2 x0:
4 AŸ"1 B C
2
C1
A "1
So the integral is
2
4
1 dx
dx ; "1 ;
2
x1
x"1
Ÿx " 1 2 Ÿx 1 Ÿx " 1 "
2
ln|x 1| K
" ln|x " 1| x"1
7
13. Compute ; x 5 sinŸx 3 " 1 dx
We first substitute t x 3 " 1. So dt 3x 2 dx and x 3 t 1. Thus,
I ; x 5 sinŸx 3 " 1 dx 1 ;Ÿt 1 sinŸt dt
3
We now integrate by parts with u t 1 and dv sinŸt dt. So du dt and
v " cosŸt Thus,
I 1 "Ÿt 1 cosŸt ; cosŸt dt 1 ¡"Ÿt 1 cosŸt sinŸt ¢ C
3
3
" 1 x 3 cosŸx 3 " 1 1 sinŸx 3 " 1 C
3
3
=
14. It is easy to compute ; sin x dx 2 exactly. However, find the approximate value for
=
; sin x dx
0
using the midpoint rule in a Riemann sum with 4 intervals. Use your
0
calculator to give a decimal value.
x
The area is divided into 4 regions by the vertical lines at = , = , 3= .
4 2
4
The width of each regions is x = and the midpoints are at = , 3= , 5= , and 7= .
8
4
8
8
8
gives
So
the
midpoint
rule
=
; sin x dx X = sin = sin 3= sin 5= sin 7=
4
8
8
8
8
0
X .785¡.383 .923 .923 .383 ¢ X 2.052
8
15. The base of a solid is the triangle with corners Ÿ0, 0 , Ÿ0, 1 and Ÿ1, 0 . The
cross-sections perpendicular to the x-axis are semicircles. Compute the volume of the
solid.
1
At the right is a top view:
1
y
1
y
x
0
x
1
The slice at x has length y 1 " x. So the semicircles have radius r 1 " x .
2
2
= Ÿ1 " 2x x 2 .
So the area of the semicircle is A 1 =r 2 1 = 1 " x
2
2
21
8
1
1
3
=
=
2
= .
x " x2 x
So the volume is V ; AŸx dx ; Ÿ1 " 2x x dx 3 0
24
8
8 0
0
9