$\int _{4} ^{5} \dfrac {1} {5x-1} dx$
$\int _{0} ^{1} \dfrac {1} {2x-1} dx$
$\int _{ \infty } ^{ -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx$
$\int _{1} ^{3} \ln (x-1) dx$
This
article aims to find
improper integrals. This article uses the concept of
improper integrals; there are
two types of improper integrals. Integrals having
limits of integrations are infinite and are improper. When integrals become
infinite within the limits of integration, integrals are improper.
Improper integrals can be represented as follows:
\[ \int _{ -\infty} ^{ \infty} f(x) dx\]
Expert Answer
An
improper integral is a definite integral
with one or both limits at infinity or an
integrand that approaches infinity at one or more points in the range of
integration.
Part 1
Let’s solve for each integral.
\[\int _{4} ^{5} \dfrac {1} {5x-1} dx\]
since none of the
limits is equal to infinity, we check the
integrand.
As we know that
\[ \dfrac {1}{x} = \ln |x|\]
\[\int _{4} ^{5} \dfrac {1} {5x-1} dx = \dfrac {1} {5} \ln |5x-1| |_ {4} ^{5}\]
\[ = \dfrac {1} {5} ( \ln |24| – \ln |19|)\]
The integrand is not
approaching infinity, so the given
integral is not improper.
Part 2
\[\int _{0} ^{1} \dfrac {1} {2x-1} dx \]
Plugging $x=\dfrac {1} {2}$, the
integral approaches to $\infty$ and $ x = \dfrac {1}{2} $ is also in the range of integration $[ 0 , 1 ]$.
So the
given integral is improper.
Part 3
\[ \int _{ \infty } ^{ -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx \]
As seen clearly, the limits of
integrations are infinity, so by definition, the given
integral is improper.
Part 4
\[ \int _{1} ^{3} \ln (x-1) dx \]
At $x = 1$, the
integral becomes undefined, and this value is also in the
range of integration.
So it is also an
improper fraction.
Numerical Result
$\int _{4} ^{5} \dfrac {1} {5x-1} dx$ is not an
improper integral.
$\int _{0} ^{1} \dfrac {1} {2x-1} dx$ is an
improper integral.
$\int _{ \infty } ^{ -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx$ is an
improper integral.
$\int _{1} ^{3} \ln (x-1) dx$ is an
improper integral.
Example
Which of the following integrals is improper?
$\int _{- \infty} ^{ \infty } \dfrac {1} {6x+1} dx $
$\int _{5} ^{9} \dfrac {1} {5x+1} dx$
Solution
Part a
\[ \int _{- \infty} ^{ \infty } \dfrac {1} { 6x+1} dx \]
As seen clearly, the limits of
integrations are infinity, so by definition, the given
integral is improper.
Part b
Since none of the
limits is equal to infinity, we check the
integrand.
As we know:
\[ \dfrac {1}{x} = \ln |x|\]
\[\int _{5} ^{9} \dfrac {1} {5x+1} dx = \dfrac {1} {5} \ln |5x+1| |_ {5} ^{9}\]
\[ = \dfrac {1} {5} ( \ln |46| – \ln |26|)\]
As the integrand is not
approaching infinity, the given
integral is not improper.
