Question

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

Answer 1

We have,

$I={\int }_{-1}^2\left|x^3-x\right|dx$

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

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

$I=\frac{1}{4}+\frac{1}{4}+2+\frac{1}{4}=\frac{11}{4}$

Hence, this is the answer.