# How do you test the improper integral int (3x-1)^-5dx from [0,1] and evaluate if possible?

Oct 10, 2017

The integrand function $f \left(x\right) = {\left(3 x - 1\right)}^{- 5}$ is continuous in the interval $\left[0 , 1\right]$ with the exception of the point $x = \frac{1}{3}$.

The improper integral is then defined as:

${\int}_{0}^{1} {\left(3 x - 1\right)}^{- 5} \mathrm{dx} = {\lim}_{x \to {\left(\frac{1}{3}\right)}^{-}} {\int}_{0}^{x} {\left(3 t - 1\right)}^{-} 5 \mathrm{dt} + {\lim}_{x \to {\left(\frac{1}{3}\right)}^{+}} {\int}_{x}^{1} {\left(3 t - 1\right)}^{-} 5 \mathrm{dt}$

As the improper integral is:

$F \left(x\right) = \int {\left(3 t - 1\right)}^{-} 5 \mathrm{dt} = \frac{1}{3} \int {\left(3 t - 1\right)}^{-} 5 d \left(3 t - 1\right) = - \frac{1}{12} {\left(3 x - 1\right)}^{-} 4 + C$

we have:

${\lim}_{x \to {\left(\frac{1}{3}\right)}^{-}} {\int}_{0}^{x} {\left(3 t - 1\right)}^{-} 5 \mathrm{dt} = {\lim}_{x \to {\left(\frac{1}{3}\right)}^{-}} - \frac{1}{12} {\left(3 x - 1\right)}^{-} 4 + \frac{1}{12} = + \infty$

and similarly:

${\lim}_{x \to {\left(\frac{1}{3}\right)}^{+}} {\int}_{x}^{1} {\left(3 t - 1\right)}^{-} 5 \mathrm{dt} = + \infty$

thus the integral does not converge.