${\int }_{-2}^3 \left|{\mathrm{x}}^2-1\right|\mathrm{dx}$ is equal to

${\int }_{-2}^3 \left|{\mathrm{x}}^2-1\right|\mathrm{dx}={\int }_{-2}^{-1} \left|{\mathrm{x}}^2-1\right|\mathrm{dx}+{\int }_{-1}^1 \left|{\mathrm{x}}^2-1\right|\mathrm{dx}+{\int }_1^3 \left|{\mathrm{x}}^2-1\right|\mathrm{dx}$

[here, modulus function will change at the points, ,when ${\mathrm{x}}^2-1=0\mathrm{i}\text{. . , at }\mathrm{x}=\pm 1$] 

so,

$\begin{array}{l}\mathrm{I}={\int }_{-2}^{-1} \left({\mathrm{x}}^2-1\right)\mathrm{dx}+{\int }_{-1}^1 \left(1-{\mathrm{x}}^2\right)\mathrm{dx}+{\int }_1^3 \left({\mathrm{x}}^2-1\right)\mathrm{dx}\\ ={\left[\frac{{\mathrm{x}}^3}{3}-\mathrm{x}\right]}_{-2}^{-1}+{\left[\mathrm{x}-\frac{{\mathrm{x}}^3}{3}\right]}_{-1}^1+{\left[\frac{{\mathrm{x}}^3}{3}-\mathrm{x}\right]}_1^3\\ =\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+6+\frac{2}{3}=\frac{28}{3}\end{array}$