1. ## Substitution

Can someone please explain to me how the final answer has -16?
Here is my work

Integral 3r square root(8-r)

u= 8-r

dr= -du

3 integral (u-8) square root u du

3 integral u^3/2 - 8^1/2 du

3[2/5u^5/2 -6u^3/2] du + C

6/5(8-r)^5/2-?(8-r)^3/2 + C

The question mark represents the 16. Not sure how to get it and I prob messed up earlier in the problem.

2. Originally Posted by McDiesel
Can someone please explain to me how the final answer has -16?
Here is my work

Integral 3r square root(8-r)

u= 8-r

dr= -du

3 integral (u-8) square root u du

3 integral u^3/2 - 8u^1/2 du

3[2/5u^5/2 -8(2/3)u^3/2] du + C

6/5(8-r)^5/2-16(8-r)^3/2 + C
Does that clear it up?

3. Yes, thank you