Part II: The Answers
470
749.
4 1
2
16
1
4
2
1
π
π
π
ln
tan
+ +
−
Using the average value formula with a = 0, b = π
4
, and f x
x
x
( ) = +
+
1
1
2
gives you
f
x
x
dx
avg =
−
+
+
∫
1
4
0
1
1
2
0
4
π
π
To evaluate this integral, begin by splitting up the fraction:
4
1
1
1
2
0
4
2
0
4
π
π
π
x
x
dx
x
dx
+
+
+
∫
∫
For the first integral, use a substitution, letting u = x
2
+ 1 so that du = 2x
dx, or
1
2
du x dx
=
. You can also find the new limits of integration for the first integral by
noting that if x = π
4
, then u = ( ) + = +
π
π
4
1 16
1
2
2
, and if x = 0, then u = 0
2
+ 1 = 1. With these
new values, you have
4
1
1
1
4
1
2
1
1
1
2
0
4
2
0
4
1
16
1
2
2
π
π
π
π
π
x
x
dx
x
dx
u
du
x
dx
+
+
+
=
( ) + +
∫
∫
∫
+
0 0
4
1
16
1
1
0
4
2
4 1
2
4 1
2
16
1
2
π
π
π
π
π
π
∫
=
+
=
+ +
+
−
ln
tan
ln
u
x
t tan
−
1
4
π
Answers
701–800
470
749.
4 1
2
16
1
4
2
1
π
π
π
ln
tan
+ +
−
Using the average value formula with a = 0, b = π
4
, and f x
x
x
( ) = +
+
1
1
2
gives you
f
x
x
dx
avg =
−
+
+
∫
1
4
0
1
1
2
0
4
π
π
To evaluate this integral, begin by splitting up the fraction:
4
1
1
1
2
0
4
2
0
4
π
π
π
x
x
dx
x
dx
+
+
+
∫
∫
For the first integral, use a substitution, letting u = x
2
+ 1 so that du = 2x
dx, or
1
2
du x dx
=
. You can also find the new limits of integration for the first integral by
noting that if x = π
4
, then u = ( ) + = +
π
π
4
1 16
1
2
2
, and if x = 0, then u = 0
2
+ 1 = 1. With these
new values, you have
4
1
1
1
4
1
2
1
1
1
2
0
4
2
0
4
1
16
1
2
2
π
π
π
π
π
x
x
dx
x
dx
u
du
x
dx
+
+
+
=
( ) + +
∫
∫
∫
+
0 0
4
1
16
1
1
0
4
2
4 1
2
4 1
2
16
1
2
π
π
π
π
π
π
∫
=
+
=
+ +
+
−
ln
tan
ln
u
x
t tan
−
1
4
π
Answers
701–800