`int_(-1)^2|x^3-x|dxdot`

Explanation

`f(x)=x^(3)-x=x(x^(2)-1)=x(x-1)(x+1)`
Sign scheme of `f(x)` is as shown in the following figure.

From the sign, shceme, we have
`int_(-1)^(2)|x^(3)-x|dx=int_(-1)^(0)(x^(3)-x)dx+int_(0)^(1)-(x^(3)-x)dx+int_(1)^(2)(x^(3)-x)dx`
`=[(x^(4))/4-(x^(2))/2]_(-1)^(0)+[(x^(2))/2-(x^(4))/4]_(0)^(1)+[(x^(4))/4-(x^(2))/2]_(1)^(2)`
`=-(1/4-1/2)+(1/2-1/4)+(4-2)-(1/4-1/2)`
`=-1/4+1/2+1/2-1/4+2-1/4+1/2=3/2-3/4+2=11/4`