MATH 101 HOMEWORK 6 – SOLUTIONS
1. For each of the following integrals, use the comparison test to determine whether it is
convergent of divergent.
Z
∞
√
x3 − 1
dx: the integrand looks roughly like x−1/2 for large x, so the integral
2+1
x
1
should be divergent.
We √
now give more details. For all x ≥ 1, we have x2 + 1 ≤ 2x2 . Also,
√
for all x ≥ 2, x3 − 1 ≥ x3 /2. So
(a)
Z
∞
√
x3 − 1
dx =
x2 + 1
1
Z
2
≥
1
Z
∞
(b)
2
√
x3 − 1
dx +
x2 + 1
−x2
(x + 1)e
Z
∞
2
Z
1
dx =
0
Z
2
Z
2
Z
2
∞
+
1
2
√
x3 − 1
dx +
x2 + 1
1
Z
∞
2
dx
= ∞.
4x1/2
∞
; the first integral is finite, so it suffices to check the
1
−x2
second one. For x ≥ 1 we have e
∞
x3
dx =
4x2
+
0
Z
√
Z
≤ e−x , therefore
−x2
(x + 1)e
Z
∞
dx ≤
1
(x2 + 1)e−x dx.
1
The latter integral can be done by parts, see the formula for I2 on page 354 of the textbook.
Using this formula, we get
Z
∞
−1
(x2 + 1)e−x dx = −e−x (x2 + 2x + 2)|∞
,
1 = 5e
1
using that limx→∞ xa e−x = 0 for any x. Therefore the integral is convergent.
Z
Z 1
Z ∞
Z 1
Z ∞
dx
dx
dx
dx
dx
(c)
=
+
≤
+
. The last two
1/3 + x3
1/3
x3
x1/3 + x3
x1/3 + x3
0
0 x
1
0 x
1
integrals are convergent, therefore the original integral is also convergent.
∞
Z
1
2
ex dx.
2. We want to evaluate numerically the integral
0
(a) Write out (but do not try to evaluate) the midpoint and trapezoid approximations M4
and T4 .
1 2
3 2
5 2
7 2
M4 = 0.25(e( 8 ) + e( 8 ) + e( 8 ) + e( 8 ) ),
1
T4 =
1 2
2 2
3 2
0.25 0
(e + 2e( 4 ) + 2e( 4 ) + 2e( 4 ) + e1 ).
2
(b) How large should n be, if we want the error for the midpoint approximation Mn to be
no greater than .001?
If f (x) = ex then f 00 (x) = (4x2 + 2)ex . For 0 ≤ x ≤ 1, 4x2 + 2 is always between 2 and
2
6, and ex is between e0 = 1 and e1 = e, hence we may take K = 6e. We want
2
2
K · (b − a)3
6e · 13
=
≤ .001.
24n2
24n2
6e
, n ≥ 26.06. Since n needs to be integer, we take n ≥ 27
24 · .001
(this is just to be on the safe side, 26 would likely be good enough).
Solve this for n: n2 ≥
2