The value of ${\int }_{-2}^3|1-x^2|dx$ is

The function $|1-x^2|$ changes behavior based on the sign of $1-x^2.$
$1-x^2=0$ at $x=-1$ and $x=1.$ For $x\in [-2,-1],1-x^2\le 0,$ so $|1-x^2|[cit{e}_{s}tart]=x^2-1$.For $x\in [-1,1],1-x^2\ge 0,$ so $|1-x^2|[cit{e}_{s}tart]=1-x^2$.For $x\in [1,3],1-x^2\le 0,$ so $|1-x^2|[cit{e}_{s}tart]=x^2-1$.Break the integral:$I={\int }_{-2}^{-1}(x^2-1)dx+{\int }_{-1}^1(1-x^2)dx+{\int }_1^3(x^2-1)dx$.
$I=[\frac{x^3}{3}-x{]}_{-2}^{-1}+[x-\frac{x^3}{3}{]}_{-1}^1+[\frac{x^3}{3}-x{]}_1^3$.
$I=[(-\frac{1}{3}+1)-(-\frac{8}{3}+2)]+[(1-\frac{1}{3})-(-1+\frac{1}{3})]+[(9-3)-(\frac{1}{3}-1)]$
$I=[\frac{2}{3}+\frac{2}{3}]+[\frac{2}{3}+\frac{2}{3}]+[6+\frac{2}{3}]$
$I=\frac{4}{3}+\frac{4}{3}+\frac{20}{3}=\frac{28}{3}$.