Math 166
Quiz 5
Name:
Directions: This quiz is worth a total of 10 points. To receive full credit, all work must be shown.
Z
1
1. Evaluate the improper integral
−8
dx
. Note that the integrand has a discontinuity at 0.
x1/3
Observe that
Z
1
−8
Z 1
dx
dx
+
1/3
1/3
x
x
−8
0
Z c
Z 1
dx
dx
= lim
+
lim
1/3
1/3
−
+
c→0
c→0
−8 x
c x
c
1
3 2/3 3 2/3 = lim− x + lim− x c→0 2
c→0 2
−8
c
3 2/3
3
2/3
= lim−
c − (−8)
1 − c2/3
+ lim−
c→0 2
c→0 2
9
=−
2
dx
=
x1/3
Z
0
2. Use either the Direct Comparison Test or the Limit Comparison test to determine whether the improper
integral
Z ∞
x dx
√
4−1
x
2
Z ∞
dx
converges. You may use without justification the fact that the integral
diverges when p ≤ 1 and
xp
1
converges when p > 1. Start by giving an inequality if using the Direct Comparison Test or computing
a limit if using the Limit Comparison Test. After, fill in the blanks below to make a concluding
statement.
Solution using direct comparison test: Observe that
√
√
x4 − 1 <
√
x4 = x2 which implies
1
x
x
> 2 = .
x
x
x4 − 1
Therefore, by the Direct Comparison test we conclude that the improper integral
Z ∞
x dx
√
4−1
x
2
diverges because the improper integral
Z
∞
2
dx
x
is known to diverge.
Solution using limit comparison test: Observe that
L=
√ x
4
lim x1 −1
x→∞
x
x2
x→∞
x4 − 1
r
x4
= lim
x→∞
x4 − 1
r
x4
=
lim 4
x→∞ x − 1
= 1.
= lim √
Thus 0 < L < ∞. Therefore, by the Limit Comparison test we conclude that the improper integral
Z ∞
x dx
√
4−1
x
2
diverges because the improper integral
Z
2
is known to diverge.
∞
dx
x