469
Answers
701–800
Answers and Explanations
Next, set the original function equal to the average value and solve for c:
5 3
1
4 3
4
3
4
3
2
3
2 3
3
2
2
2
−
=
=
=
±
=
±
=
±
=
c
c
c
c
c
c
Both values fall in the given interval, so both are solutions.
748.
2
π
Using the average value formula with a = 0, b = π
2
, and f
 
(x) = x sin x
gives you
f
x
x dx
x
xdx
avg =
−
=
∫
∫
1
2 0
2
0
2
0
2
π
π
π
π
sin
sin
To integrate this function, you can use integration by parts. Notice that for the indefinite
integral x
xdx
sin
∫
, you can let u = x so that du = dx and let dv = sin x
 
dx so that
v = –cos x
 
dx. Using integration by parts formula gives you
x
xdx
x
x
x dx
x
x
x C
sin
c os
cos
cos
sin
∫
∫
= −
+
= −
+
+
Therefore, to evaluate the definite integral 2
0
2
π
π x
xdx
sin
∫
, you have
2
2
2
2
2
2
0
0
2
0
2
π
π
π
π
π
π
π
π
x
xdx
x
x
x
sin
( cos
sin )
cos
sin
( s
∫
=
−
+
=
−
( ) + ( ) − + i in( ))
( )
0
2 1
2






=
=
π
π