Question

How do you find the integral of `int (dx / x^(2/3))` from 1 to -15?

Answer 1

`int_1^-15 (dx / x^(2/3)) = int_1^-15 x^(-2/3) dx` is an improper integral. The integrand is not defined at `0`. Early in the treatment of integration, improper integrals are not (yet) defined..

When improper integral have been defined, we do the following:
To (attempt to) evaluate the integral, we have to split it at `0`

`int_1^-15 x^(-2/3) dx = int_1^0 x^(-2/3) dx +int_0^-15 x^(-2/3) dx`

`= lim_(brarr0^+) int_1^b x^(-2/3) dx + lim_(ararr0^-) int_a^-15 x^(-2/3) dx`

Note that `int x^(-2/3) dx = 3x^(1/3) = 3root(3)x`,

So for the first integral we get:

`int_1^0 x^(-2/3) dx = lim_(brarr0^+) int_1^b x^(-2/3) dx`

`= lim_(brarr0^+) [3root(3)x]_1^b`

`= lim_(brarr0^+) [3root(3)b-3root(3)1] = 0 - 3 = -3`

And for the second integral we get:

`int_0^-15 x^(-2/3) dx = lim_(ararr0^-) int_a^-15 x^(-2/3) dx`

`= lim_(ararr0^-) [3root(3)x]_a^-15`

`= lim_(ararr0^-) [3root(3)(-15)- 3root(3)a] = -3root(3)(15) - 0=-3root(3)(15)`

Both improper integrals converge, so we get:

`int_1^-15 x^(-2/3) dx = int_1^0 x^(-2/3) dx +int_0^-15 x^(-2/3) dx`

`= (-3) + (-3root(3)15)`

`= -3 -3root(3)15)`

Note on limits of integration

`int_a^b f(x) dx` is read "the integral from a to b"

Recall that `int_a^b f(x) dx = -int_b^a f(x) dx`.

So if the intended question was `int_-15^1 x^(-2/3) dx` `" "`(which is the integral from -15 to 1),

then the answer should be

`int_-15^1 x^(-2/3) dx = 3+3root(3)15`.

And the details of the solution should be changed appropriately.