How do you find the area between #y=1/2x^3+2, y=x+1, x=0, x=2#?
1 Answer
Nov 9, 2016
graph{(1/2x^3+2-y)(x+1-y)(x-2)x=0 [-6.67, 9.14, -0.72, 7.18]}
Explanation:
First of all
To explain this call
Then
and
So f has a local minimum in
Then the area is given by