## Troubleshooting: Evaluating an Trigonometric Integral Algebraically

Hehe…we are always amused by calculus students who manage to mis-integrate a function to the point of yielding nonsensical answers. To illustrate what we mean by this, here is Steve, who — at the point of this writing at least — still haven’t got a clue where his computations went awry:

###### How do you evaluate $\displaystyle \int_{-5}^5 x – \sqrt{25-x^2} \, dx$ algebraically? When I take the areas geometrically, I get $\displaystyle -\frac{25\pi}{2}$, as $\displaystyle \int_{-5}^5 x \, dx=0$ and $\displaystyle \int_{-5}^5 \sqrt{25-x^2} \, dx$ is the area of a half-circle with radius $5$. But when I try to evaluate it algebraically, I don’t seem to get the same answer: $\displaystyle \int^5_{-5}\left(x-\sqrt{25-x^2}\right)\,dx=\left[\frac{x^2}{2}-\frac{2}{3}(25-x^2)^{3/2}+C\right]^{5}_{-5} \dots$

SteveOK…What’s going on here? Well, keep reading! More