$\underset{-2}{\overset{4}{\int }}\left(\left|x-2\right|+\left[x\right]\right)dx=\left(where\left[x\right]isthegreater\mathrm{int}egernotmorethanx\right)$

$$\begin{gathered} {\int }_{-2}^4 \left(\left|\mathrm{x}-2\right|+\left[\mathrm{x}\right]\right)\mathrm{dx} \\ ={\int }_{-2}^4(\mathrm{x}-2)+{\int }_{-2}^4\left[\mathrm{x}\right]\mathrm{dx} \\ ={\int }_{-2}^2\left|\mathrm{x}-2\right|+{\int }_2^4\left|\mathrm{x}-2\right|+{\int }_{-2}^{-1}\left[\mathrm{x}\right]+{\int }_{-1}^0\left[\mathrm{x}\right]+{\int }_0^1\left[\mathrm{x}\right]+{\int }_1^2\left[\mathrm{x}\right]+{\int }_2^3\left[\mathrm{x}\right]+{\int }_3^4\left[\mathrm{x}\right]\mathrm{dx} \\ ={\int }_{-2}^2-(\mathrm{x}-2)+{\int }_2^4(\mathrm{x}-2)+(-2)+(-1)+0+1+2+3 \\ ={\left(2\mathrm{x}-\frac{{\mathrm{x}}^2}{2}\right)}_{-2}^2+{\left(\frac{{\mathrm{x}}^2}{2}-2\mathrm{x}\right)}_2^4+3 \\ =2-(-6)+\left((8-8)-(2-4)\right)-3 \\ =8+2+3=13 \end{gathered}$$