Question

Evaluate: ${\int }_{-1}^2|x^3-x|dx.$

Answer 1

$x^3-x=x(x^2-1)=x(x+1)(x-1)$

Thus, the function is positive in the range $(-1,0)\cup (1,\infty )$

The integral thus can be split and written as below:

${\int }_{-1}^2|x^3-x|dx$

$={\int }_{-1}^0(x^3-x)dx+{\int }_0^1(x-x^3)dx+{\int }_1^2(x^3-x)dx$

$={[\frac{x^4}{4}-\frac{x^2}{2}]}_{-1}^0+{[\frac{x^2}{2}-\frac{x^4}{4}]}_0^1+{[\frac{x^4}{4}-\frac{x^2}{2}]}_1^2$

$=0-(\frac{1}{4}-\frac{1}{2})+\frac{1}{2}-\frac{1}{4}+4-2-(\frac{1}{4}-\frac{1}{2})$

$=\frac{1}{4}+\frac{1}{4}+2+\frac{1}{4}$

$=2.75$