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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.Previous Question < > Next Question
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