${\int }_0^2 \left(\left|2{\mathrm{x}}^2-3\mathrm{x}\right|+\left[\mathrm{x}-\frac{1}{2}\right]\right)\mathrm{dx}$, where [t] is the greatest integer function, is equal to :

$\begin{array}{l}{\int }_0^2 \left|2{\mathrm{x}}^2-3\mathrm{x}\right|\mathrm{dx}\\ ={\int }_0^{\frac{3}{2}} \left(3\mathrm{x}-2{\mathrm{x}}^2\right)\mathrm{dx}+{\int }_{\frac{3}{2}}^2 \left(2{\mathrm{x}}^2-3\mathrm{x}\right)\mathrm{dx}=\frac{19}{12}\\ {\int }_0^2 \left[\mathrm{x}-\frac{1}{2}\right]\mathrm{dx}={\int }_{\frac{-1}{2}}^{\frac{3}{2}} \left[\mathrm{t}\right] \mathrm{dt}\\ ={\int }_{-\frac{1}{2}}^0 (-1)\mathrm{dt}+{\int }_0^1 0\cdot \mathrm{dt}+{\int }_1^{\frac{3}{2}} 1\cdot \mathrm{dt}=0.\end{array}$