Question

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

Answer 1

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

We know that

$x^3-x\ge 0$

$\Rightarrow x(x^2-1)\ge 0$

$\Rightarrow xϵ[-1,0]\cup [1,2]x^3-x\le 0$

$\Rightarrow x(x^2-1)\le 0$

$\Rightarrow xϵ[-\infty ,-1]\cup [0,1]$

Now,

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

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

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

$[\because {\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x\left)dx\right]$

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

$\Rightarrow I=[0-(\frac{(-1{)}^4}{4}-\frac{(-1{)}^2}{2})]-[(\frac{1^4}{4}-\frac{1^2}{2})-0]+[(\frac{2^4}{4}-\frac{2^2}{2})-(\frac{1}{4}-\frac{1}{2})]$

$\Rightarrow I=[0-(-\frac{1}{4})]-[-\frac{1}{4}-0]+\left[\right(4-2)-(-\frac{1}{4})]$

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

$\Rightarrow I=2+\frac{3}{4}=\frac{11}{4}$

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