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

1 Answer
Dec 3, 2017

#(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.