Question

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

Answer 1

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

$I=2[{(x-\frac{1}{3}{x}^3)}_0^1+{(\frac{1}{3}{x}^3-x)}_1^2]=2\left[(1-\frac{1}{3}-0+\frac{8}{3}-2-\frac{1}{3}+1)\right]=4$