${\int }_{-3}^2\left\{\right|x+1|+|x+2|+|x-1\left|\right\}dx\text{ is }$

$\begin{array}{l}\text{ Let I}={\int }_{-3}^2\left\{\right|\text{x}+1|+|\text{x}+2|+|\text{x}-1\left|\right\}\text{dx}\\ \text{ Breaking points are }\\ \text{x}+1=0\Rightarrow \text{x}=-1\\ x+2=0\Rightarrow x=-2\\ x-1=0\Rightarrow x=1\\ \therefore \text{I}={\int }_{-3}^{-2}\text{f}\left(\text{x}\right)\text{dx}+{\int }_{-2}^{-1}\text{f}\left(\text{x}\right)\text{dx}+{\int }_{-1}^1\text{f}\left(\text{x}\right)\text{dx}+{\int }_1^2\text{f}\left(\text{x}\right)\text{dx}\\ \text{ where f}\left(\text{x}\right)=|\text{x}+1|+|\text{x}+2|+|\text{x}-1|\end{array}$

$\begin{array}{l}{\text{I}}_1={\int }_{-3}^{-2}\text{[-(x+1)-(x+2)-(x-1)]dx}\\ =-{\left[\frac{x^2}{2}+x+\frac{x^2}{2}+2x+\frac{x^2}{2}+x\right]}_{-3}^{-2}=\frac{7}{2}\\ {\text{I}}_2={\int }_{-2}^{-1}\text{[-(x+1)+(x+2)-(x-1)] d x}\\ =\frac{-x^2}{2}-x+\frac{x^2}{2}+2x-\frac{x^2}{2}+{\left.x\right|}_{-2}^{-1}\\ =\frac{-x^2}{2}+{\left.2x\right|}_{-2}^{-1}=\left(\frac{-1}{2}-2\right)-(-2-4)\\ =-\frac{5}{2}+6=\frac{7}{2}\\ {\text{I}}_3={\int }_{-1}^1\left[\right(x+1)+(x+2)-(x-1\left)\right]\text{d}x\\ ={\int }_{-1}^1(x+4)\text{d}x=\frac{x^2}{2}+{\left.4x\right|}_{-1}^1=8\\ {\text{I}}_4={\int }_1^2\text{[(x+1)+(x+2)+(x-1) d x]}\\ ={\int }_1^2(3x+2)\text{d}x=\frac{3x^2}{2}+{\left.2x\right|}_1^2=\frac{13}{2}\\ \therefore \text{I}={\text{I}}_1+{\text{I}}_2+{\text{I}}_3+{\text{I}}_4\\ =\frac{7}{2}+\frac{7}{2}+8+\frac{13}{2}=\frac{43}{2}\\ =21.5\end{array}$