Question

How do you evaluate the integral `int x^2lnx dx` from 0 to 1 if it converges?

Answer 1

This problem requires integration by parts.

`int_0^1 x^2 ln(x) dx`, let `u = ln(x) -> du = 1/x dx`

and let `dv = x^2 dx -> v = 1/3 x^3 + C` (although the `C` becomes irrelevant)

Next, we use the formula

`int_a^b u dv = [uv]_a^b - int_a^b v du`

So our integral becomes

`[1/3 x^3 ln(x)]_0^1 - int_0^1 1/3 x^3 * 1/x dx`

`= [1/3 x^3 ln(x)]_0^1 - 1/3 * int_0^1 x^2 dx`

`=[1/3 x^3 ln(x) - 1/3 * 1/3 x^3]_0^1`

`=[1/3 x^3 ln(x) - 1/9 x^3]_0^1`

`=(cancel(1/3(1)^3 ln(1)) - 1/9(1)^3) - cancel((1/3(0)^3 ln(0) - 1/9(0)^3))`

`=-1/9`

The answer is negative because the function `x^2 ln(x)` is only positive for values greater than `x = 1` as seen in the graph below.

Explanation:

The point of using integration by parts is to integrate an expression that's "easier" to integrate first. As in this example, this problem would be very difficult to solve (or even impossible) if we would have chosen a different `u` and `dv`.

For example, if we would have chosen `dv = ln(x)`, then `int ln(x) dx` yields another fairly "unknown" integral which cannot solved without using numerical methods, which defeats the purpose of integration by parts.