Question

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

Answer 1

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

$\frac{(-1<x<0)x^2-x>0}{\frac{(6\le x<1)x^2-x<0}{(1\le x\le 2)x^2-x>0}}$

$\therefore I\left(x\right)\{\frac{x^3-x}{\frac{x-x^3}{x^2-x}}\frac{-1\le x<0}{\frac{0\le x<1}{1\le x<2}}$

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

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

$=(0-\frac{1}{4})-(0-\frac{1}{2})+(\frac{1}{4}-0)+(\frac{24}{4}-\frac{1}{4})-(\frac{2}{2}-\frac{1}{2})$

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

$=\frac{-3}{4}+\frac{3}{2}+2$

$=\frac{-3+6+8}{4}\Rightarrow \frac{11}{4}$

$\therefore {\int }_{-1}^2|x^3-x|dx=\frac{11}{4}$