Question

what is the answer of integration of -1 to 2 , |x^3-x| dx ??

Answer 1

$$\begin{gathered} LetI={\int }_{-1}^2\left|x^3-x\right|dx \\ Letf\left(x\right)=\left|x^3-x\right| \end{gathered}$$

When $x\in [-1,0),f\left(x\right)=-\left(x^3-x\right)=x-x^3$
When $x\in (0,2],f\left(x\right)=x^3-x$

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

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

$$\begin{gathered} \Rightarrow I=\left(0-0\right)-\left(\frac{1}{2}-\frac{1}{4}\right)+\left(4-2\right)-\left(0-0\right) \\ \Rightarrow I=2-\frac{1}{4}=\frac{7}{4} \end{gathered}$$