hey guys, I'm trying to teach myself how to integrate with u substitution but I really don't get this:

$\displaystyle \int_{0}^{1}x^2\sqrt{x^3-1} dx $

$\displaystyle u = x^3 - 1$

$\displaystyle \frac{du}{dx} = 3x^2$

$\displaystyle du = 3x^2 dx$

$\displaystyle \frac{1}{3}\int_{0}^{1} u^\frac{1}{2} dx$

$\displaystyle = [\frac{1}{3}.\frac{2}{3}u^\frac{3}{2}]$

substituting in

$\displaystyle = \frac{2}{9}\sqrt{(x^3 - 1)^3}$

But when I substitute x = 1 and x = 0 in, it yields a negative number within the square root sign. btw, no complex numbers involved at this stage.

or have I done something wrong?

Thanks