Additional review problems for Exam 2, solutions
12. The integrand becomes infinite at x = 1. Therefore you have to examine separately
whether
Z 2
1
dx
3
0 (x − 2)
converges, and whether
Z 3
2
1
dx
(x − 2)3
converges. Neither converges, since the power of x − 2 in the denominator is 3. For convergence, that power would have to be less than 1. Explicitly:
s
Z 2
Z s
1
1
1
=
dx = lim
dx = lim
3
3
2
s→2−
s→2−
(x
−
2)
(x
−
2)
(−2)(x
−
2)
0
0
0
1
1
= −∞
+
lim −
2
s→2−
2(s − 2)
8
and
Z 3
2
3
1
1
=
dx
=
lim
3
s→2+ (−2)(x − 2)2 s
(x − 2)3
s
1
1
lim − +
= ∞.
s→2+
2 2(s − 2)2
1
dx = lim
s→2+
(x − 2)3
Z s
1
For 03 (x−2)
3 dx to converge, both integrals would have to converge. In fact, neither of the
two integrals converges.
R
13. If r(t) is the rate at time t (measured in hours past the beginning of the day), the average
rate is
Z
1 24
r(t)dt.
24 0
Using the trapezoidal method, we approximate this by
1 2+4
4+6
6+4
4+2
2+6
6+4
102
×4+
×2+
×6+
×2+
×6+
×4 =
= 4.25.
24
2
2
2
2
2
2
24
14. (a)
L=
Z 1q
(2t)2 + (3t 2 )2 dt
Z 1
=
−1
−1
Z 1 p
p
2
|t| 4 + 9t dt = 2
t 4 + 9t 2 dt.
0
(b) Substitute u = 4 + 9t 2 :
L=2
Z 13
√ du
4
1
u =
18 9
Z 13
1/2
u
4
1 2 3/2 13
2 √
du = · u =
13 13 − 8 .
9 3
27
4
√
15. If n terms are included, the error
is
at
most
±1/
√
√ n + 1. Therefore
√ if 10,000 terms are
included, the error is at most ±1/ 10, 001, and 1/ 10, 001 < 1/ 10, 000 = 0.01.
16. The trick is to write
n1/n = eln(n
1/n )
= e(1/n) ln n .
Now by l’Hospital’s rule,
ln n
= 0,
n→∞ n
lim
therefore
lim n1/n = e0 = 1.
n→∞