Question 1006409
{{{f(x)-g(x)=x^3-x-3x=x^3-4x}}} 
{{{g(x)-f(x)=-(f(x)-g(x))=-x^3+4x}}}
The integrals are
{{{int((f(x)-g(x)), dx, -2, 0 )=int((x^3-4x), dx, -2, 0 )=(matrix(3,1,"",x^4/4-2x^2,""))}}}{{{drawing(30,90,-1,-0.6,-1,1,
line(-0.95,-0.95,-0.95,0.95),locate(-0.65,0.1,"="),
locate(-0.9,0.93,0),locate(-0.9,-0.73,-2)
)}}}{{{(0^4/4-2*0^2)-((-2)^4/4-2(-2)^2)=0-(16/4-2*4)=-(4-8)=-(-4)=4}}}
and
{{{int((f(x)-g(x)), dx, 0, 2 )=int((-x^3+4x), dx, -2, 0 )=(matrix(3,1,"",-x^4/4+2x^2,""))}}}{{{drawing(30,90,-1,-0.6,-1,1,
line(-0.95,-0.95,-0.95,0.95),locate(-0.65,0.1,"="),
locate(-0.9,0.93,0),locate(-0.9,-0.73,-2)
)}}}{{{(-2^4/4+2*2^2)-(-0^4/4+2*0^2)=(-16/4+2*4)-0=-4+8=4}}}
Both integrals are the same, but that should be no surprise,
because {{{f(x)}}} is an odd function,
meaning that {{{f(-x)=-f(x) for all values of {{{x}}} ,
and that makes its graph symmetrical with respect to the origin,
meaning that its graph rotated {{{180^o}}} falls on to of the original graph.
The same can be said of {{{g(x)}}} ,
and of course, the same can be said of {{{f(x)-g(x)}}} .
{{{graph(300,300,-3,3,-8,8,x^3-x,3x)}}}