PRACTICE PROBLEMS: SET 7
MATH 101: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. Determine with proof whether the following integrals are divergent
or convergent.
(a)
∫ ∞ 2
x +2
dx
x4 + x
2
(b)
∫ ∞
x
dx
2+1
x
1
(c)
∫ ∞
sin(x)
dx
3+1
x
0
Problem 2. Determine with proof whether the following integral is divergent or
convergent. If it is convergent, compute its value.
∫ 3
2
(x − 1)− 5 dx
1
1
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Problem 1. In each case we first guess what is the correct answer, and
then try to replace the function we integrate with a simpler function
which would be useful for us to prove our guess is indeed correct.
(a) The order of magnitude for the given fraction is x12 (by taking the∫quotient
∞
of the leading terms of both numerator and denominator). Since 2 x12 dx
is convergent, we suspect the integral is convergent; so we prove our guess
is indeed correct. In order to do this we have to compare our function with
a larger function whose integral would still be convergent.
We note that for x ≥ 2 we have
x4 + x > x4
and thus
x2 + 2
x2 + 2
1
2
<
= 2 + 4.
4
4
x +x
x
x
x
Therefore
∫ ∞
2
x2 + 2
dx
x4 + x
∫
<
=
=
=
=
=
=
∞
1
2
+ 4 dx
2
x
x
∫2 ∞
∫ ∞
1
2
dx +
dx
2
4
x
x
2
2
∫ t
∫ t
1
2
lim
dx + lim
dx
t→∞ 2 x2
t→∞ 2 x4
(
)
(
)
1 t
2
lim −
| + lim − 3 |t2
t→∞
x 2 t→∞
3x
)
)
(
(
2
1 1
1
lim
−
+ lim
− 3
t→∞ 2
t→∞ 12
t
3t
1
1
+
2 12
7
,
12
which proves that indeed the given integral is convergent. Note that the
original function we integrated was positive and therefore we were allowed
to use the usual comparative test for convergence/divergence of integrals.
(b) The order of magnitude of the given fraction is x1 , whose integral on the
interval [1, ∞] is divergent. So, we will compare our function to a suitable
function which would prove the divergence of our integral. We need this
time to choose a function smaller than x2x+1 ; so x1 is not the right choice
since
x
1
< .
2
x +1
x
However we note that
x2 ≥ 1 on the interval [1, ∞];
PRACTICE PROBLEMS: SET 7
3
therefore
x2
x
x
1
≥ 2
=
.
2
+1
x +x
2x
Then
∫
∞
1
x2
x
dx
+1
∫
∞
1
dx
2x
1
∫ t
1
= lim
dx
t→∞ 1 2x
(
)
ln(x) t
= lim
|1
t→∞
2
ln(t)
= lim
,
t→∞ 2
≥
which diverges to infinity, thus proving that indeed our integral is divergent.
Note that the original function we integrated was positive and therefore we
were allowed to use the usual comparative test for convergence/divergence
of integrals.
(c) We know that
| sin(x)| ≤ 1 for all x.
Thus the integral behaves roughly as
∫
0
∞
1
dx
x3 + 1
which is convergent. However, in order to use the comparative test for
convergence/divergence of integrals we have to take absolute value of the
integral and only then apply our test. In this problem there is also an extra
complication since we cannot simply drop the +1 from the denominator
since then we would have a problem integrating near x = 0. Therefore we
will split the integral as follows. (Note that we need to take first absolute
4
MATH 101: PROF. DRAGOS GHIOCA
values since we want to exploit the inequality | sin(x)| ≤ 1.)
∫ ∞
sin(x) dx
x3 + 1 0
∫ ∞
sin(x) ≤
x3 + 1 dx
0
∫ ∞
1
≤
dx
3+1
x
0
∫ 1
∫ ∞
1
1
=
dx +
dx
3+1
3+1
x
x
0
1
∫ 1
∫ ∞
1
1
<
dx +
dx
3
1
x
0
1
∫ t
1
= 1 + lim
dx
t→∞ 1 x3
(
)
1
= 1 + lim − 2 |t1
t→∞
2x
)
(
1
1
= 1 + lim
− 2
t→∞ 2
2t
1
= 1+
2
3
=
;
2
therefore the integral is indeed convergent. Note that in the above estimations we used that
1 ≤ 1 + x3 on [0, 1]
and
x3 < x3 + 1 on [1, ∞].
Problem 2. Using the easy substitution u = x − 1, we immediately compute
∫
3
2
5
(x − 1)− 5 dx = (x − 1) 5 + C .
3
Therefore, we compute
∫
2
(x − 1)− 5 dx
∫ 3
2
= lim
(x − 1)− 5 dx
t→1 t
(
)
3
5
5
(x − 1)
|3t
= lim
t→1 3
3
3
5 · 25
5
− · (t − 1) 5
t→1
3
3
3
5 · 25
.
=
3
=
lim
This proves the integral is actually convergent and it equals
5
√
5
8
3 .