If $\left[\mathrm{t}\right]$ denotes the greatest integer $⩽\mathrm{t}$, then the value of ${\int }_0^1 \left[2\mathrm{x}-\left|3{\mathrm{x}}^2-5\mathrm{x}+2\right|+1\right]$dx is:

$\begin{array}{l}\mathrm{I}={\int }_0^1 \left[2\mathrm{x}-\left|3{\mathrm{x}}^2-3\mathrm{x}-2\mathrm{x}+2\right|+1\right]\mathrm{dx}\\ \mathrm{I}={\int }_0^1 [2\mathrm{x}-|(3\mathrm{x}-2)(\mathrm{x}-1)\left|\right]\mathrm{dx}+{\int }_0^1 1\mathrm{dx}\\ \mathrm{I}={\int }_0^{2/3} \left[\left(2\mathrm{x}-\left(3{\mathrm{x}}^2-5\mathrm{x}+2\right)\right)\right]\mathrm{dx}+{\int }_{2/3}^1 \left(2\mathrm{x}+\left(3{\mathrm{x}}^2-5\mathrm{x}+2\right)\right)\mathrm{dx}+1\\ \mathrm{I}={\int }_0^{2/3} \left[-3{\mathrm{x}}^2+7\mathrm{x}-2\right]\mathrm{dx}+{\int }_{2/3}^1 \left(3{\mathrm{x}}^2-3\mathrm{x}+2\right)\mathrm{dx}+1\\ \mathrm{y}=-3{\mathrm{x}}^2+7\mathrm{x}-2\end{array}$

$\begin{array}{l}{\int }_0^{\mathrm{\alpha }} (-2)\mathrm{dx}+{\int }_{\mathrm{\alpha }}^{1/3} (-1)\mathrm{dx}+{\int }_{1/3}^{\mathrm{\beta }} 0\mathrm{dx}+{\int }_{\mathrm{\beta }}^{2/3} 1.\mathrm{dx}\\ =-2\mathrm{\alpha }-\left(\frac{1}{3}-\mathrm{\alpha }\right)+\frac{2}{3}-\mathrm{\beta }=-\mathrm{\alpha }-\mathrm{\beta }+\frac{1}{3}\\ \mathrm{y}=3{\mathrm{x}}^2-3\mathrm{x}+2\end{array}$

When $\mathrm{x}\in \left(\frac{2}{3},1\right)$
$\begin{array}{l}3{\mathrm{x}}^2-3\mathrm{x}+2\in \left(\frac{4}{3},2\right)\\ \left[3{\mathrm{x}}^2-3\mathrm{x}+2\right]=1\\ \therefore {\int }_{2/3}^1 \left[3{\mathrm{x}}^2-3\mathrm{x}+2\right]\mathrm{dx}=1\left(1-\frac{2}{3}\right)=\frac{1}{3}\end{array}$
Hence $\mathrm{I}=\left(\frac{1}{3}-(\mathrm{\alpha }+\mathrm{\beta })\right)+\left(\frac{1}{3}\right)+1$
$\begin{array}{l}=\frac{5}{3}-\left(\frac{7-\sqrt{37}}{6}+\frac{7-\sqrt{13}}{6}\right)\\ =\frac{-2}{3}+\frac{\sqrt{37}+\sqrt{13}}{6}\\ =\frac{\sqrt{37}+\sqrt{13}-4}{6}\end{array}$