Question

How do you integrate `\pi \int _ { 0} ^ { 2} ( \frac { y + 2} { 3} ) ^ { 2} d y`?

Answer 1

`(56pi)/27`

Explanation:

`pi int_0^2 ((y+2)/3)^2 dy`
`=pi int_0^2 (y+2)^2/9 dy`
`=pi/9 int_0^2 (y+2)^2 dy`
`=pi/9 int_0^2 (y+2)^2 d (y+2)`

Used part 2 of Fundamental Theorem Calculus

`F'(x)=f(x)`

`int_a^b f(x)dx=F(b)-F(a)`

as the following:
`=pi/9 [(y+2)^3/3 ]_0^2`
`=pi/9 [(2+2)^3/3-(0+2)^3/3]`
`=pi/9 [(4)^3/3-(2)^3/3]`
`=pi/9 (64/3-8/3)`
`=pi/9 *56/3`
`=(56pi)/27`

Here is the answer :)
Hope it can help you. Feel free to ask me if you don't understand.