MAT 254 – Winter Quarter 2002
Test 3
NAME_______________________________________________________
Show work and write clearly.
1. (30 pts.)
a. Find the average value of f ( x)   sin x on [0, ].

ANS: f ave 
1
1
11  2

 sin xdx  cos x 0 

.

 00



b. Find the value c such that f(c) = fave.
ANS: That is, find c such that f (c)   sin c  2 /  . One method is to graph
y1 = –sinx and y2 = –2/ on TI83 and find intersection:
Thus, there are two possible values of c: c  0.6901 and c  2.4515.
c. Sketch the graph of f(x) and construct a rectangle over the interval whose area is the same as
the area under the graph of f(x) over the interval. ANS:
2. (48 pts.) Find the following integrals:
y
a. 
dy ANS: let u = y + 1, then du = dy and y = u – 1. So,
y 1

u 1
u
du 

u
u
du 

1
u
u 2 u 2
2 y  1


C 
3
1
3
2
2
3
1
1
du   u 2 du   u
3
2
 2 y  1
1
2
C
1
2
du

y
y 1
dy =
b.
 tan
3
(5 x) sec 2 (5 x)dx ANS: let u = tan(5x), then du = 5sec2(5x)dx. So,
3
2
 tan (5x) sec (5x)dx =
3
c.
1 3
1 u4
1
u
du

C 
tan 4 (5 x)  C

5
5 4
20
 csc cos(3t )sin( 3t )dt
2
2
ANS: let u = cos(3t), then du = –3sin(3t)dt. When t = 1, u = –0.99
1
3
and when t = 1.5, u = –0.21. So,
 csc
2
2
cos(3t )sin( 3t )dt
0.21

1
1
1
 0.21
csc 2 u du = cot u  0.99 

3
3  0.99
[–4.7 – (–0.66)]/3 = –1.35.
 x2  x  dx ANS: let u = 2 – x2, then du = –2xdx. When x = –2, u = 0 and when x = 0,
0
d.
2 3
 2
 x2  x 
2
u = 2. So,
2 3
0
2
1
1 u4
dx =   u 3 du  
20
2 4
2
 2.
0
3. (22 pts.) Sketch the area between y = x3, y = –x, y = 8. Find the area.
ANS: Here is the sketch of the area:
a = –8 is the intersection point between y = –x and y = 8; b = 2 is the intersection point between y = x3
and y = 8. To find the area between curves, we need to split the area into two parts:

x2
Area =  8  ( x) dx   8  x dx   8 x 
2

8
0
0
2
3
0


x4
   8 x 
4
 8 

(8) 2 
24
 0  8(8) 

8
(
2
)

 0  32  16  4  44

2 
4

2


0