Find the area encompassed by y=(3-x)x^2 and y=x(4-x) between x=0 and x=2.
This is an integration question.
The limits have been provided.
Firstly find the integral for the first curve
int{x(4-x)}dx
int{4x-x2}dx
2x2-x3/3
Apply the limits x=2,0
[2(2)2-(2)3/3]-[2(0)2-(0)3/3]=
8-8/3=16/3
Secondly, find the integral of the second curve
int{(3-x)x2 }dx
int{3x2-x3}dx
x3-x4/4
Apply the limits x=2,0
[(2)3-(2)4/4]-[(0)3-(0)4/4]=
8-16/4=
8-4=4
Subtract the areas to find the area between them
16/3-4=4/3
