Originally Posted by

**pflo** I don't understand the answer to the following definite integral. Can anyone help me explain why it is the way it is?

$\displaystyle \int_0^2{\frac{1}{^3\sqrt{t-1}}}dt$

I think the easiest way to see my point on this is to use u-substitution (though it isn't necessary): $\displaystyle u=t-1$

$\displaystyle \int_{-1}^1{u^{\frac{-1}{3}}}du$

$\displaystyle (\frac{3u^{\frac{2}{3}}}{2})|_{-1}^1$

$\displaystyle \frac{3(1)^{\frac{2}{3}}}{2} - \frac{3(-1)^{\frac{2}{3}}}{2}$

And I imagine going from the last step to the next one is my problem....

$\displaystyle \frac{3*1}{2} - \frac{3*1}{2} = 0$

But the answer cannot be 0! If you think about this being an area-under-the-curve problem, then the area is infinite because of the asymptote at the discontinuity within the limits of integration.

My suspicion is that $\displaystyle (-1)^{\frac{2}{3}}$ is somehow imaginary, but if it is I don't know why.

Can anyone explain this in reasonably understandable terms?