Calculus II
Practice Problems 6: Answers
Determine whether or not the integral converges. If it does, try to find its value (you may not be able to do
this in some cases).
∞
dx
1.
2
2 x ln x
dx x. We get
du
1
1
1
u 1 ln 2 ln a ln 2
u
Answer. This integral converges. We make the substitution u
a
2
dx
x ln x
2
ln a
ln a
ln 2
2
ln 2
lnx du
∞.
as a
10
2.
1
dx
x lnx
Answer. Since ln 1 0, the problem is at the lower limit of integration. Nevertheless, this integral converges.
We make the same substitution u ln x and get (taking a 0 and small)
10
a
as a
3.
1
dx
x lnx
ln 10
ln a
du
u1 2
2u1
2
2 ln 10
ln 2
ln 10
2
2
ln a
ln 10
1.
∞
ln 5x
dx
2
5 x
Answer. Since ln x
x for x large enough, the integrand is less than 5x 3 2 , so by comparison (propoosition
8.6), our integral converges. We can now proceed to evaluate it. We make the substitution u ln 5x du
dx x, and x e u 5:
a
1
ln 5x
dx
2
5 x
which converges to 5 as a
∞
e
ln 5a
5
0
u
ln 5a 5 e u e 5 5a u
udu
ln 5a
0
u
∞
dx
1 x2
1
5a
1
∞.
Answer. This integral converges, by the comparison text. For, 1
4.
3 2
1 1
1
x2
1
x2
3 2
x 1
2 3 2
x2 , so
and the integral of the latter is finite ( π ). Now, we can find the value by a trigonometric substitution. Let
x tanu dx sec2 udu
1 x2 sec u. (Draw the triangle corresponding to these substitutions!). This
gives us
dx
sec2 udu
x
cos udu sin u C
C
2
3
2
sec3 u
1 x
1 x2
Thus
x
b
a
b
a
1 x 1 b 1 a 1 b 1 a
Now the limit of this, as a and b go to infinity is (as we saw in problem 10 of practice set 5) 1+1 =2.
b
a
π 2
5.
1
0
dx
cos x
dx
1 x2
b
3 2
a
2
2
2
2
2
Answer. The improperness here is at x 0 (cos 0
1).There is no easy comparison to make, so we calculate
the integral. To do so we use some trigonometric identities:
1
1 cos x
Thus
1
Now we calculate
π 2
a
If we let a
dx
1 cos x
dx
cos x
1
cos x
cos x
csc2 xdx
1
1
cos x
cos2 x
cotx csc xdx
cota cot π 2
csc π 2
csc a
1 cos x
x sin
2
cot x
csc x
1 cota csc a
0 , the term in parentheses becomes infinite. Thus the integral diverges.
1 x 1
1
1
cos x 1
dx
6.
3 2
0
Answer. Make the change of variable u
1 x du
dx. Then the integral becomes
we saw (see display (7) of the Notes), this integral diverges.
1 2
7.
0
dx
x1 x
Answer. In this range 1
x 1 2, so
u 8. Find the area under the curve y
x 2 x 1 , above the x-axis and to the right of the line x
Answer. First, we find the indefinite integral. We find the partial fraction representation
1
1
x x 1 x
dx
dx
dx
x 1
ln x x
x 1
x
x
dx
x 1
a 1
2 1
a 1
ln ln ln ln x x
x
a
2
a ∞, the first term goes to zero, so the answer is ln2.
1
x2
2
As a
a
2
2
1
0
3 2 du.
1
2
x 1 x x so by comparison the integral converges.
Now,
a
2
ln2
2.
But