How do you find the area between #y=1/2x^3+2, y=x+1, x=0, x=2#?

Question

4986 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T10:11:52

graph{(1/2x^3+2-y)(x+1-y)(x-2)x=0 [-6.67, 9.14, -0.72, 7.18]}

#A=2#

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#