Question

How do you evaluate the integral `int 1/(x-1)^(2/3)dx` from 0 to 2?

Answer 1

To evaluate `int_0^2 1/(x-1)^(2/3)dx`, the integrand, i.e.,

`1/(x-1)^(2/3)` has to be continuous over the interval `[0,2]`.

Here, it is not so at `x=1 in [0,2]`.

Answer 2

The integrand is not defined at `x = 1`, so this is an improper integral.

Explanation:

We need to break this into two improper integrals and try to evaluate each of them. If both integrals converge, then we can add the values to get the integral on `[0,2]`

`int_0^2 1/(x-1)^(2/3) dx = int_0^1 1/(x-1)^(2/3) dx+int_1^2 1/(x-1)^(2/3) dx`

provided that both integrals on the right converge,

`= lim_(brarr1^-)int_0^b 1/(x-1)^(2/3) dx + lim_(ararr1^+)int_a^2 1/(x-1)^(2/3) dx`

provided that both integrals on the right converge.

`lim_(brarr1^-)int_0^b 1/(x-1)^(2/3) dx ={: lim_(brarr1^-) 3(x-1)^(1/3) ]_0^b = 3`

and

`lim_(ararr1^+)int_a^1 1/(x-1)^(2/3) dx ={: lim_(ararr1^+) 3(x-1)^(1/3) ]_a^2 = 3`

We conclude:

`int_0^2 1/(x-1)^(2/3) dx = int_0^1 1/(x-1)^(2/3) dx+int_1^2 1/(x-1)^(2/3) dx`

`= 3+3=6`