$\displaystyle{\int_0^4} x \sqrt{36-x^2}dx$
This is a pretty basic question but I have something very wrong with my idea of U substitution. So what I do is set u = $ 36-x^2 $ which then gives me $\displaystyle du =-2xdx$ So what I do next is divide by -2 and get –$\displaystyle \frac{1}{2}du = dx$ So then you need to change your bounds which would I think give you
$\displaystyle-\frac{1}{2} {\int_{36}^{20}} \sqrt{u}$ du if you plug the original bounds back into the equation for u. So I get the integral to be $\displaystyle\frac{1}{3}u^{3/2}$ after integrating. I did flip the bounds so 36 is on top and 20 is on bottom and $\displaystyle \frac{1}{2}$ becomes positive. I evaluate on the bounds and I am getting $\displaystyle \frac{216}{3}-\frac{40\sqrt{5}}{3}$ which I know to be incorrect. Where am I going wrong?
The correct answer based on my book is $\displaystyle \frac{21\sqrt{21}-5\sqrt{5}}{3}$
Best Answer
Guessing: A study guide or a computerized grader is expecting to see $72 - \frac{40\sqrt{5}}{3}$. But $\frac{216}{3} = 72$, so these are equivalent.