Can someone please show me how to do this problem? I have the answer but do not understand the steps in order to arrive at it.
Integral 4r square root(6-r) dr
Answer: 8/5*(6-r)^5/2 - 16(6-r)^3/2 +C
Do you know integration by parts?
If not, read Integration by Parts -- from Wolfram MathWorld.
If so, use it.
$\displaystyle 4\int r\sqrt{6-r} \, dr$
let $\displaystyle u = 6-r$
$\displaystyle r = 6-u$
$\displaystyle dr = -du$
substitute ...
$\displaystyle 4 \int (u-6)\sqrt{u} \, du$
$\displaystyle 4 \int u^{\frac{3}{2}} - 6u^{\frac{1}{2}} \, du$
can you finish?
$\displaystyle 4 \int u^{\frac{3}{2}} - 6u^{\frac{1}{2}} \, du$
antiderivative ...
$\displaystyle 4\left[\frac{2}{5}u^{\frac{5}{2}} - 4u^{\frac{3}{2}}\right] + C$
distribute the constant 4 ...
$\displaystyle \frac{8}{5}u^{\frac{5}{2}} - 16u^{\frac{3}{2}} + C$
now back substitute (6-r) for u.