Q #5!
Math 106-C (Salomone)
February 27, 2009
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Score (25 points possible):
Problem 1. (10 points) Determine whether each of the following improper integrals converges or diverges. You do
not need to calculate their value, but you should provide a rigorous argument. (Write a sentence or two.)
(a) (4 points) !2
0
√1+x2
1−x2dx
The integrand has a vertical asymptote at x=1. In fact, partial fraction decomposition shows
1
1−x2=1/2
1−x+1/2
1+x.
Therefore, we should focus on the integral
!2
0
1/2√1+x2
1−xdx.
But
√1+x2≥1
#####
√1+x2
1−x#####≥#####
1
1−x#####
and both "1
0
1
1−xdx and "2
1
1
1−xdx diverge (for the same reason as "1
0
1
xdx does). Therefore, this integral
diverges.
(b) (3 points) !∞
2
dx
x4+x+12
A simpler one; as x→∞, the behavior of the integrand will be characterized by its highest-order term.
Since this ”looks like” "∞
2
1
x4dx, we suspect the integral will converge.
Rigorously: if we wish to show this converges, we need to show that it’s less than something convergent. This
means we need to show that the denominator is bigger than something. But for all x≥0, we have
x4+x+12 ≥x4
!∞
2
1
x4+x+12 ≤!∞
2
1
x4
!∞
2
dx
x4+x+12 ≤(converges)
Therefore, this integral converges.
(c) (3 points) !∞
1
5−z+3z2
9−9z−17z2dz
No tricky comparison necessary here: the integrand doesn’t go to zero!
lim
z→∞
5−z+3z2
9−9z−17z2=−3
17
Since the integrand does not approach zero, the integral diverges.
