MATH 126 SPRING 2011, QUIZ 6
1. Determine whether each integral is convergent or divergent. Evaluate those
that Rare convergent.
∞
1
a. 1 (3x+1)
2 dx
There is a point of discontinuity at x = 1/3, but this is not in our
region of integration, so we do not need to split up the integral.
Z
∞
Z
a
1
dx
a→∞ 1 (3x + 1)2
1
, du = 3dx
Z
1 a −2
= lim
u du
a→∞ 3 4
1
= lim (−u−1 |a4 )
a→∞ 3
1 1 1
= lim
−
a→∞ 3
4 a
1
=
<∞
12
Therefore
the integral is CONVERGENT.
R∞
b. 2π sin θdθ
1
dx
(3x + 1)2
u = 3x + 1
Z
=
∞
Z
sin θdθ
=
a
lim
a→∞
2π
sin θdθ
2π
lim − cos θ|a2π
=
a→∞
lim cos 2π − cos a
=
a→∞
lim 1 − cos a,
=
c.
lim
a→∞
but
R 2 2lima→∞ cos a does not exist, hence the integral is DIVERGENT.
z ln zdz
0
Z
2
z 2 ln zdz
Z
=
lim+
a→0
0
2
z 2 ln zdz
a
u = ln z
,
dv = z 2
du = 1/zdz
,
v = z 3 /3
Z 2
1 2
1 3
z ln z|2a −
z dz
+
a→0 3
a 3
1
1
= lim+ (8 ln 2 − a3 ln a) − (8 − a3 )
9
a→0 3
=
lim
At this point we must investigate the behavior of a3 ln a as a → 0, so
1
ln a
a−3
1/a
= lim
a→0+ −3a−4
−1 3
= lim+
a = 0.
3
a→0
Hence the integral is equal to
lim a3 ln a
a→0+
=
lim
a→0+
8
8
ln 2 −
3
9
and is therefore CONVERGENT.
2. Use the Comparison Theorem to determine whether the integral is convergent Ror divergent. (Hint: One is convergent and one is divergent.)
∞
1
a. 1 x+e
2x dx
First we note that for x ≥ 1, x + e2x ≥ e2x , so that in particular we
have
1
1
≤ 2x .
0≤
x + e2x
e
Z ∞
Z A
1
e−2x dx
dx
=
lim
2x
A→∞
e
1
1
1
= lim − e−2x |A
1
A→∞
2
1 −2A
= lim − (e
− e−2 )
A→∞
2
= e−2 /2 < ∞.
Therefore, by the Comparison Theorem, the integral is CONVERGENT.
R π/2 1
b. 0 x sin
x dx Again, when in doubt start with the simple expression
x sin x. For 0 < x ≤ π/2, we have 0 < x sin x ≤ x, and so
0≤
1
1
≤
x
x sin x
We then compute
Z
0
π/2
1
dx
x
Z
=
A→0
=
π/2
1
dx
x
lim+ ln(π/2) − ln(A)
lim+
A
A→0
= ∞,
since limA→0+ ln A = −∞. Therefore, by the Comparison Theorem,
the integral is DIVERGENT.