${\int }_{-1}^1 max\{\mathrm{x}-[\mathrm{x}],-\mathrm{x}-[-\mathrm{x}\left]\right\}$ dx is equal to

$\begin{array}{l}{\int }_{-1}^1 max\left(\right\{\mathrm{x}\},\{-\mathrm{x}\left\}\right)\mathrm{dx}\\ ={\int }_{-1}^{-0.5} \{-\mathrm{x}\}\mathrm{dx}+{\int }_{-0.5}^0 \left\{\mathrm{x}\right\}\mathrm{dx}+{\int }_0^{0.5} \{-\mathrm{x}\}\mathrm{dx}+{\int }_{0.5}^1 \left\{\mathrm{x}\right\}\mathrm{dx}\\ ={\int }_{-1}^{-0.5} (-\mathrm{x})\mathrm{dx}+{\int }_{-0.5}^0 \{\mathrm{x}-[\mathrm{x}\left]\right\}\mathrm{dx}+{\int }_0^{0.5} (-\mathrm{x}-[-\mathrm{x}\left]\right)\mathrm{dx}\\ +{\int }_{0.5}^1 (\mathrm{x}-[\mathrm{x}\left]\right)\mathrm{dx}\end{array}$

$\begin{array}{l}={\left[-\frac{{\mathrm{x}}^2}{2}\right]}_{-1}^{-0.5}+{\left[\frac{{\mathrm{x}}^2}{2}+\mathrm{x}\right]}_{-0.5}^0+{\left[\mathrm{x}-\frac{{\mathrm{x}}^2}{2}\right]}_0^{0.5}+{\left[\frac{{\mathrm{x}}^2}{2}\right]}_{0.5}^1\\ =-\frac{0.25}{2}+\frac{1}{2}+\left(-\frac{0.25}{2}+0.5\right)+0.5-\frac{0.25}{2}+\frac{1}{2}-\frac{0.25}{2}\\ =2-\frac{1}{2}=\frac{3}{2}\end{array}$