Question

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

Answer 1

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

The range in which the function is positive comes out to be $[-1,0]\cup [1,2]$

And it is negative for x belonging to $[0,1]$

$\Rightarrow$ we get ${\int }_{-1}^2|x^3-x|dx$ to be ${\int }_{-1}^0(x^3-x)dx+{\int }_0^1(x-x^3)dx+{\int }_1^2(x^3-x)dx$

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

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

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

$=2.75$