Question

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

Answer 1

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.