How do you integrate \pi \int _ { 0} ^ { 2} ( \frac { y + 2} { 3} ) ^ { 2} d yπ20(y+23)2dy?

1 Answer
Dec 3, 2017

(56pi)/2756π27

Explanation:

pi int_0^2 ((y+2)/3)^2 dy π20(y+23)2dy
=pi int_0^2 (y+2)^2/9 dy =π20(y+2)29dy
=pi/9 int_0^2 (y+2)^2 dy =π920(y+2)2dy
=pi/9 int_0^2 (y+2)^2 d (y+2) =π920(y+2)2d(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.