Question

${\int }_{-1}^2\left(\left|x+1\right|+\left|x\right|+\left|x-1\right|\right)dx$

Answer 1

We know that

$\left|x+1\right|=\left\{\begin{array}{ll}x+1,& \mathrm{if}x+1\ge 0\\ -\left(x+1\right),& \mathrm{if}x+1<0\end{array}\right.=\left\{\begin{array}{ll}x+1,& \mathrm{if}x\ge -1\\ -\left(x+1\right),& \mathrm{if}x<-1\end{array}\right.$

$\left|x\right|=\left\{\begin{array}{ll}x,& \mathrm{if}x\ge 0\\ -x,& \mathrm{if}x<0\end{array}\right.$

$\left|x-1\right|=\left\{\begin{array}{ll}x-1,& \mathrm{if}x-1\ge 0\\ -\left(x-1\right),& \mathrm{if}x-1<0\end{array}\right.=\left\{\begin{array}{ll}x-1,& \mathrm{if}x\ge 1\\ -\left(x-1\right),& \mathrm{if}x<1\end{array}\right.$

When $-1\le x\le 0,$

$\left|x+1\right|+\left|x\right|+\left|x-1\right|=x+1+\left(-x\right)+\left[-\left(x-1\right)\right]=2-x$

When $0\le x\le 1,$

$\left|x+1\right|+\left|x\right|+\left|x-1\right|=x+1+x+\left[-\left(x-1\right)\right]=x+2$

When $1\le x\le 2,$

$\left|x+1\right|+\left|x\right|+\left|x-1\right|=x+1+x+x-1=3x$

$$\begin{gathered} \therefore {\int }_{-1}^2\left(\left|x+1\right|+\left|x\right|+\left|x-1\right|\right)dx \\ ={\int }_{-1}^0\left(2-x\right)dx+{\int }_0^1\left(x+2\right)dx+{\int }_1^{2}3xdx \\ ={\overline{)\frac{{\left(2-x\right)}^2}{2\times \left(-1\right)}}}_{-1}^0+{\overline{)\frac{{\left(x+2\right)}^2}{2}}}_0^1+3\times {\overline{)\frac{x^2}{2}}}_1^2 \\ =-\frac{1}{2}\left(4-9\right)+\frac{1}{2}\left(9-4\right)+\frac{3}{2}\left(4-1\right) \\ =\frac{5}{2}+\frac{5}{2}+\frac{9}{2} \\ =\frac{19}{2} \end{gathered}$$