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

1 Answer
Oct 10, 2017

The integrand function #f(x) = (3x-1)^(-5)# is continuous in the interval #[0,1]# with the exception of the point #x=1/3#.

The improper integral is then defined as:

#int_0^1 (3x-1)^(-5)dx = lim_(x->(1/3)^-) int_0^x (3t-1)^-5dt + lim_(x->(1/3)^+) int_x^1 (3t-1)^-5dt #

As the improper integral is:

#F(x) = int (3t-1)^-5dt = 1/3 int (3t-1)^-5d(3t-1) = -1/12 (3x-1)^-4 +C#

we have:

#lim_(x->(1/3)^-) int_0^x (3t-1)^-5dt =lim_(x->(1/3)^-) -1/12 (3x-1)^-4 +1/12 = +oo#

and similarly:

#lim_(x->(1/3)^+) int_x^1 (3t-1)^-5dt = +oo#

thus the integral does not converge.