This is a definite integral problem
First we need to find the intersections of these 2 functions by setting them equal to each other.
#|x|=x^2-2#
This has to be split up into 2 equations.
#x=x^2-2# AND #-x=x^2-2#
FIRST
#x=x^2-2#
#x^2-x-2=0#
Factor
#(x-2)*(x+1)=0#
#x-2=0#
#x=2#
#x+1=0#
#x=-1#
SECOND
#-x=x^2-2#
#x^2+x-2=0#
Factor
#(x+2)*(x-1)=0#
#(x+2)=0#
#x=-2#
#(x-1)=0#
#x=1#
The interval is from #[-2,2]#
The function #|x|# is greater than the function #x^2-2# over the interval [-2,2].
#int_(-2)^(2)|x|-(x^2-2)dx=int_(-2)^(2)|x|-x^2+2dx#
#=int_(-2)^(2)|x|dx-int_(-2)^(2)x^2dx+int_(-2)^(2)2dx#
Note: |x| is an even function and symmetric about the #y#-axis so make this function easier to manage we can assume x>0 and double the area found by multiplying it by 2 and changing the boundaries from 0 to 2.
#=2*int_0^(2)xdx-int_(-2)^(2)x^2dx+int_(-2)^(2)2dx#
#=2*[x^2/2]_0^2 -1* [x^3/3]_(-2)^2+1*[2x]_(-2)^2#
#=2*[(2)^2/2-(0)^2/2] -1* [(2)^3/3-(-2)^3/3]+1*[2(2)-2(-2)]#
#=2*[2-0] -1* [8/3+8/3]+1*[4+4]#
#=2*[2] - [16/3]+[8]#
#=[4] - [16/3]+[8]#
#=[12/3] - [16/3]+[24/3]#
#= [-4/3]+[24/3]#
#= [20/3]#
#= 6 2/3 or 6.6667#
