# Thread: Using Substitution with Integrals

1. ## Using Substitution with Integrals

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

2. Do you know integration by parts?

If not, read Integration by Parts -- from Wolfram MathWorld.

If so, use it.

3. I read that but still can't get it to work out properly.

4. $4\int r\sqrt{6-r} \, dr$

let $u = 6-r$

$r = 6-u$

$dr = -du$

substitute ...

$4 \int (u-6)\sqrt{u} \, du$

$4 \int u^{\frac{3}{2}} - 6u^{\frac{1}{2}} \, du$

can you finish?

5. ok, but how do you get the 8/5 and the -16?

6. $4 \int u^{\frac{3}{2}} - 6u^{\frac{1}{2}} \, du$

antiderivative ...

$4\left[\frac{2}{5}u^{\frac{5}{2}} - 4u^{\frac{3}{2}}\right] + C$

distribute the constant 4 ...

$\frac{8}{5}u^{\frac{5}{2}} - 16u^{\frac{3}{2}} + C$

now back substitute (6-r) for u.