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]}

A=2

Explanation:

First of all
1/2x^3+2>x+1, AA0<=x<=2
To explain this call f(x)=1/2x^3-x+1

Then f^'(x)=3/2x^2-1 so f^'(x)=0 only when x=+-sqrt(2/3)

and f^('')(x)=3x>0, AA 0 < x <=2

So f has a local minimum in x=sqrt(2/3) and this minimum is
f(sqrt(2/3))=1-2/3sqrt(2/3)>0

Then the area is given by

A=int_0^2(1/2x^3-x+1)dx=[x^4/8-x^2/2+x]_0^2=16/8-4/2+2=2