The value of  ${\int }_{-1}^3 \left\{\right|x-2|+[x\left]\right\}dx$, where $\left[x\right]$ denotes the greatest integer less than or equal to x is

$$\begin{gathered} {\int }_{-1}^3 \left\{\right|x-2|+[x\left]\right\}\mathrm{d}x={\int }_{-1}^0 \left\{\right|x-2|+[x\left]\right\}\mathrm{d}x+ \\ \begin{array}{r}{\int }_0^1 \left\{\right|x-2|+[x\left]\right\}\mathrm{d}x+{\int }_1^2 \{\mid x-2\mid +\left[x\right]\}\mathrm{d}x\\ +{\int }_2^3 \left\{\right|x-2|+[x\left]\right\}\mathrm{d}x\end{array} \end{gathered}$$

$\begin{array}{r}={\int }_{-1}^0 (2-x-1)\mathrm{d}x+{\int }_0^1 (2-x+0)\mathrm{d}x+\\ {\int }_1^2 (2-x+1)\mathrm{d}x+{\int }_2^3 (x-2+2)\mathrm{d}x\end{array}$

$\begin{array}{l}=x-{\left.\frac{x^2}{2}\right|}_{-1}^0+2x-{\left.\frac{x^2}{2}\right|}_0^1+3x-{\left.\frac{x^2}{2}\right|}_1^2+{\left.\frac{x^2}{2}\right|}_2^3\\ =-\left(-1-\frac{1}{2}\right)+\left(2-\frac{1}{2}\right)+(6-2)-\left(3-\frac{1}{2}\right)+\frac{9}{2}-2=7\end{array}$