MATH 152
Exam1
Spring 2000
Test Form A
Solutions
Part I is multiple choice. There is no partial credit.
Part II is work out. Show all your work. Partial credit will be given.
You may not use a calculator.
Formulas:
sinŸA sinŸB 1 cosŸA " B " 1 cosŸA B 2
2
1
sinŸA " B 1 sinŸA B sinŸA cosŸB 2
2
cos A cos B 1 cosŸA " B 1 cosŸA B 2
2
1
fŸx 0 4fŸx 1 2fŸx 2 4fŸx 3 2fŸx 4 C 2fŸx n"2 4fŸx n"1 fŸx n Sn 3
KŸb " a 3
where K u |f Ÿx | for a t x t b
|E T | t
12n 2
KŸb " a 5
where K u f Ÿ4 Ÿx for a t x t b
|E S | t
180n 4
; sec 2 d2 ln|sec 2 tan 2 | C
x
UU
; csc 2 d2 ln|csc 2 " cot 2 | C
; ln x dx x lnx " x C
1
Part I: Multiple Choice (5 points each)
There is no partial credit. You may not use a calculator.
1. Compute ;
1/2
xe 2x dx.
0
a. 1 e " 1
b.
c.
d.
e.
2
4
1 e2
4
1 Ÿe 2 " 1 4
1
correctchoice
4
3
4
ux
dv e 2x dx
Integrate by parts with
to get
du dx
v 1 e 2x
2
1/2
1/2
x e 2x " 1 e 2x
; xe 2x dx 2x e 2x " 12 ; e 2x dx
2
4
0
0
1/2
0
1 e1 " 1 e1 "
4
4
0 e0 " 1 e0
2
4
1
4
=
2. Compute ; cos 2 Ÿx dx.
0
a. 1
b. =
2
c. 2
d. = " 1
2
e. =
correctchoice
=
=
; cos 2 Ÿx dx ; 1 cos 2x dx 1 x sin 2x
2
2
2
0
0
=
0
1 = sin 2=
2
2
" 1 0 sin 0
2
2
=
2
3. Find the average value of the function y xe x on the interval 0 t x t 2.
2
a. 1 Ÿe 4 " 1 2
b. 1 Ÿe 4 " 1 4
c. e
d. 1 e 4
2
e. e 4
correctchoice
Use the substitution u x 2 . So du 2x dx and x dx 1 du:
2
2
2
2
2
1
1
1
1
y ave 1 e4 " 1 e0 1 e4 " 1
ex
; xe x dx ; e u du e u 2 0
4
4
4
4
4
4
4
0
2
5
4. Use the Trapezoid Rule with n 2 to approximate ; 1
x dx.
1
a. 14
15
b. 23
15
c. 28
15
d. 46
15
e. 76
45
correctchoice
x 5"1 2
fŸx 1x
2
1 fŸ1 fŸ3 1 fŸ5 x 1 1 1 1 1 2 2 15 10 3
2 5
2
2
2
3
30
T2 28
15
5
5. If you use the Trapezoid Rule with n 8 to estimate ; 1
x dx you obtain the approximation
1
T 8 1.628968. (Don’t compute this.) Use the Trapezoid Error formula
KŸb " a 3
where K u |f Ÿx | for a t x t b
|E T | t
12n 2
to find an upper bound for the error in this approximation.
a. |E T | t 1
correctchoice
6
b. |E T | t 1
60
c. |E T | t 1
72
d. |E T | t 1
300
e. |E T | t 1
750
Max at x 1 :
fŸx 1x
f Ÿx "12
f Ÿx 23
K f Ÿ1 2 2
1
x
x
3
3
KŸb " a 2 5 " 1 Ÿ
2 64 1
|E T | t
2
12 64
6
12n
12Ÿ8 2
UU
U
6. Compute ;
a.
=/4
0
UU
UU
tan 2 Ÿx sec 4 Ÿx dx
2
2
" ln 1 "
2
2
2
ln 1 " 2
2
2
8
c.
correctchoice
15
d. 4
35
e. 4
b.
Let u tan x. Then du sec 2 x dx and sec 2 x tan 2 x 1 u 2 1. So
;
=/4
0
tan 2 Ÿx sec 4 Ÿx dx ;
u5 u3
5
3
=/4
0
tan 2 Ÿx sec 2 Ÿx sec 2 Ÿx dx ; u 2 Ÿu 2 1 du ;Ÿu 4 u 2 du
tan 5 x tan 3 x
5
3
=/4
0
tan 5 =
4
5
tan 3 =
4
3
"0 1 1 8
5
3
15
3
7.
Find the volume of the solid obtained
by rotating the region bounded by
y x 3 " x and y 0 for x u 0
about the y-axis.
a.
b.
c.
d.
e.
4 =
15
1=
2
2 =
15
8 =
105
16 =
105
correctchoice
Use an x-integral and cylinders. The radius is r x. The height is h 0 " Ÿx 3 " x x " x 3 .
1
1
1
1
3
5
2= 1 " 1 4 =
V ; 2=rh dx ; 2=xŸx " x 3 dx 2= ; Ÿx 2 " x 4 dx 2= x " x
5
5 0
15
3
3
0
0
0
2
8.
Which integral gives the volume
1
of the solid obtained by rotating
the region bounded by
0
x 2 y 2 2 and y 1
-1
about the x-axis?
-2
a. ;
b. ;
1
"1
=
2
" 2
Ÿ2
" x 2 2 " 1 dx
2 " x2 " 1
=
2
dx
1
c. ; 2=xŸ1 " x 2 dx
0
1
2 " x 2 " 1 dx
d. ; 2=x
e. ;
0
1
"1
=Ÿ1 " x 2 dx
correctchoice
Use an x-integral and washers. The outer radius is R 2 " x 2 . The inner radius is r 1.
The two curves intersect when x 2 1 2 2 or x o1. So these are the limits.
V;
1
"1
=ŸR 2 " r 2 dx ;
1
"1
=
2
2 " x 2 " 1 2 dx ;
1
"1
=Ÿ2 " x 2 " 1 dx ;
1
"1
=Ÿ1 " x 2 dx
4
9.
Find the area of the region
bounded by
x 3y 2 ,
x " 2y "2,
y "2 and y 3.
a. 8
b. 24
c. 40
correctchoice
d. 56
e. 64
Use a y-integral.
A;
3
"2
Ÿright " Ÿleft dy ;
3
"2
2
Ÿ3y " Ÿ2y " 2 dy ;
3
"2
Ÿ3y
2
" 2y 2 dy
¡y 3 " y 2 2y ¢ "3 2 ¡27 " 9 6 ¢ " ¡"8 " 4 " 4 ¢ 40
10. Which of the following definite integrals equals ;
a. 4 ;
b. 2 ;
c. 2 ;
d. 4 ;
e. 4 ;
=/2
0
=/2
0
=/4
0
=/4
0
=/4
0
3
x 2 " 2x 5 dx?
1
sin 2 d2
sin 2 2 d2
tan 2 d2
sec 2 d2
sec 3 2 d2
correctchoice
Complete the square:
;
3
1
x 2 " 2x 5 dx ;
3
1
Ÿx
" 1 2 4 dx
Use the trig substitution: x " 1 2 tan 2. Then dx 2 sec 2 2 d2. So
;
3
1
x 2 " 2x 5 dx ;
3
1
4 tan 2 2 4 2 sec 2 2 d2 4 ;
3
1
tan 2 2 1 sec 2 2 d2 4 ;
=/4
0
sec 3 2 d2
5
Part II: Work Out (10 points each)
Show all your work. Partial credit will be given.
You may not use a calculator.
11.
An object has a base that is
2
the triangle with vertices
Ÿ0, 0 ,
Ÿ1, 0 ,
and
Ÿ0, 2 .
The crosssections perpendicular
y
to the x-axis are semicircles.
Find the volume of the object.
You must use an integral.
0
x
1
The upper edge of the triangle is y 2 " 2x. This is the diameter of the semicircles. So the
radius is r 1 " x and the area is
AŸx 1 =r 2 1 =Ÿ1 " x 2 1 =Ÿx " 1 2
2
2
2
So the volume is
3 1
1
1
"1 3
Ÿx " 1 2
1
1
=Ÿx " 1 dx =
0"=Ÿ =
V ; AŸx dx ;
2
3
6
6
0
0 2
0
12. Compute ;
x
2
1
dx.
x2 " 4
Use the trig substitution x 2 sec 2. Then dx 2 sec 2 tan 2 d2 and
;
x
2
1
1
sec 2 tan 2
dx ;
2 sec 2 tan 2 d2 1 ;
d2
4
2
2
2
2
4 sec 2 4 sec 2 " 4
x "4
sec 2 sec 2 2 " 1
x2 " 4
C
x
where we have used a triangle whose hypotenuse is x, adjacent side is 2 and opposite side is
x2 " 4 .
1 ; 1 d2 1 ; cos 2 d2 1 sin 2 C 1
4
4
4
4 sec 2
6
13.
A tank has the shape of a circular cone with height 20 m,
radius 8 m and vertex down. It has a spout which extends
5 m above the top of the tank. The tank is initially full of
water. Set up the integral that gives the work required to
pump all the water out of the spout. Do not evaluate the
integral.
water density 1000
kg
m3
g 9.8 m 2
sec
We choose to measure y up from the vertex of the cone. The slice at height y with thickness
dy is a circle of radius r. So its volume is dV =r 2 dy. The radius r satisfies the similar triangle
relation yr 8 . So r 2 y and the volume is dV = 4 y 2 dy. Hence the weight (force) is
5
20
25
dF >g dV >g= 4 y 2 dy. This slice at height y must be lifted to 25 m. So it is lifted a distance
25
D 25 " y. So the work is
20
20
0
0
W ; D dF ;
Ÿ25
20
" y >g= 4 y 2 dy 4 1000 9.8= ; Ÿ25 " y y 2 dy
25
25
0
Other answers are possible if you put the origin at different heights.
e
14. Compute ; x lnx dx.
1
Integrate by parts with
e
; x lnx dx 1
u lnx
du 1x dx
x2
2
e2
2
dv x dx
2
v x
2
e
2
lnx " ; x 1x dx
2
1
2
lne " e
" 1 ln 1 "
2
4
Then
e
e
2
2
x 2 lnx " ; x dx
x lnx " x
2
2
2
4 1
1
2
2
2
1 e " e
" "1 e 1
4
4
2
4
4
4
7
15.
a. Find the partial fraction expansion for x3 " 2 .
x x
Factor the denominator and write the general expansion:
x " 2 x " 2 A Bx C
x3 x
xŸx 2 1 x2 1
x
Clear the denominator:
x " 2 AŸx 2 1 ŸBx C Ÿx Plug in numbers and solve for A, B and C:
x 0:
®
" 2 AŸ1 ®
A "2
x 1:
®
" 1 AŸ2 ŸB C Ÿ1 "4 B C
®
x "1:
®
" 3 AŸ2 Ÿ"B C Ÿ"1 "4 B " C
So:
A "2
B2
C1
and
x " 2 "2 2x 1
x3 x
x2 1
x
®
BC 3
B"C 1
b. Then compute ; x3 " 2 dx.
x x
; x3 " 2 dx ; "2 2x2 1 dx ; "2 dx ; 22x dx ; 2 1 dx
x 1
x x
x 1
x 1
x
x
"2 ln|x| ln|x 2 1| arctan x C
8