
Integration 11-31
= =
−
∫
=
−
sin
.
1
2
0
1
1
2
1
2
p
pdx
x
Example 11.29 Examine the improper integral
1
3
2
5
x
dx
−
∫
.
Solution: The integrand 1/x
3
is not continuous at x = 0. So, we split the given integral up to that point.
That is,
1 1 1
3
2
5
3
2
0
3
0
5
x
dx
x
dx
x
dx
− −
∫
=
∫
+
∫
.
Now, we evaluate each of these integrals as h ® 0.
1 1
1
2
3
2
0
0
3
2
0
0
2
2
x
dx
x
dx
x
h
h
h
h
h
−
→
−
+
→
−
→
∫
=
∫
= −
=
lim
lim
lim
00
2
1
2
1
8
− +
= −∞
h
.
Hence, the integral is not convergent.
Thus, we need not find
1
3
0
5
x
dx
ò
. Therefore,
1
3
2
5
x
dx
−
∫
is divergent.
11.3.3 Tests for Convergence of Infinite Integrals
Sometimes the fundamental theorem of Calculus proves to be ineffective in evaluating the infinite
integrals. ...