MATH 101 HOMEWORK 6 – SOLUTIONS
1. Evaluate the integrals, or show that they diverge:
(a) Integrate by parts with f (x) = ln x, f 0 (x) = 1/x, g(x) = −x−2 /2, g 0 (x) = x−3 :
Z
∞
1
ln x
ln x
dx = − 2 +
3
x
2x
Z
∞
1
1
1
· 2 dx = 0 +
x 2x
Z
∞
1
dx
1 ∞
1
= − 2 = .
3
2x
4x 1
4
√
√
(b) Integrate by parts with f (x) = ln x, f 0 (x) = 1/x, g(x) = 2 x, g 0 (x) = 1/ x:
Z
0
1
√
ln x
√ = 2 x ln x|10 −
x
Z
1
0
1 √
· 2 xdx = 0 −
x
Note: we used that lim x−2 ln x = 0 and lim
x→∞
x→0
√
Z
1
0
√
2
√ dx = −4 x|10 = −4.
x
x ln x = 0. These are basic facts about the
ln x function that you might already be familiar with; if you are not, you can verify them
using l’Hopital’s Rule.
2. Use the comparison test to decide whether the following improper integrals are convergent. Justify your answer.
(a) Since 2 − sin x ≥ 1 for all x, we have
Z
0
2
2 − sin x
dx ≥
2−x
Z
2
0
dx
=
2−x
Z
0
2
du
=∞
u
(we substituted u = 2 − x). So the integral is divergent.
Z π
Z π/2
dx
dx
√
√
(b)
=2
, by symmetry. For 0 ≤ x ≤ π/2, we have sin x ≥ 2x/π,
sin x
sin x
0
0
so that
Z π/2
Z π/2
dx
dx
√
p
,
≤
sin x
2x/π
0
0
which is convergent. Therefore the first integral is also convergent.
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3. Write out the midpoint approximation Mn to the integral 0 sin(x2 )dx. If we want the
error to be no greater than .001, how large should n be? In this problem, you may use a
calculator.
The midpoint approximation Mn is
1X
Mn =
sin(m2j ),
n j=1
n
1
j
2j − 1
K(b − a)3
1 j−1
K
(
+ )=
. Now, the error is bounded by
=
,
2
2 n
n
2n
24n
24n2
where K is a constant such that |(sin(x2 ))00 | ≤ K on [0, 1]. We have (sin(x2 ))0 = 2x cos(x2 )
and
where mj =
|(sin(x2 ))00 | = | − 4x2 sin(x2 ) + 2 cos(x2 )| ≤ | − 4x2 sin(x2 )| + |2 cos(x2 )| ≤ 4 + 2 = 6,
so we can take K = 6. We need
6
1
2
≤
.001,
n
≥
= 250, n ≥ 15.81.
24n2
4 · .001
Since n should be integer, we take n ≥ 16.
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