$\int _{4} ^{5} \dfrac {1} {5x-1} dx$
$\int _{0} ^{1} \dfrac {1} {2x-1} dx$
$\int _{ \infty } ^{ -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx$
$\int _{1} ^{3} \ln (x-1) dx$
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 $
$\int _{5} ^{9} \dfrac {1} {5x+1} dx$
SolutionPart 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.
