Question

The value of ${\int }_{-2}^2|1-x^2|dx$ is

Answer 1

${\int }_{-2}^2|1-x^2|dx=2{\int }_0^2|1-x^2|dx$
$=2({\int }_0^1|1-x^2|dx+{\int }_1^2|1-x^2|dx)$
$=2\left({\int }_0^1\right(1-x^2)dx+{\int }_1^2(x^2-1\left)dx\right)$
$=2[{(x-\frac{x^3}{3})|}_0^1+{(\frac{x^3}{3}-x)|}_1^2]$
$=2[1-\frac{1}{3}+\frac{8}{3}-2-(\frac{1}{3}-1)]=2\left(2\right)=4$