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

1 Answer
Jul 10, 2016

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.

enter image source here

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.