$ I =\int\limits_0^1\left[2 x-\left|3 x^2-3 x-2 x+2\right|+1\right] d x $
$ I=\int\limits_0^1[2 x-|(3 x-2)(x-1)|] d x+\int\limits_0^1 1 d x$
$ I =\int\limits_0^{2 / 3}\left[\left(2 x -\left(3 x ^2-5 x +2\right)\right)\right] dx +\int_{2 / 3}^1\left(2 x +\left(3 x ^2-5 x +2\right)\right) dx +1 $
$ I =\int\limits_0^{2 / 3}\left[-3 x ^2+7 x -2\right] dx +\int\limits_{2 / 3}^1\left(3 x ^2-3 x +2\right) dx +1$
$\int\limits_0^\alpha(-2) d x+\int\limits_\alpha^{1 / 3}(-1) d x+\int\limits_{1 / 3}^\beta 0 d x+\int\limits_\beta^{2 / 3} 1 . d x $
$=-2 \alpha-\left(\frac{1}{3}-\alpha\right)+\frac{2}{3}-\beta=-\alpha-\beta+\frac{1}{3}$
When $x \in\left(\frac{2}{3}, 1\right)$
$ 3 x^2-3 x+2 \in\left(\frac{4}{3}, 2\right)$
$ {\left[3 x^2-3 x+2\right]=1} $
$ \therefore \int\limits_{2 / 3}^1\left[3 x^2-3 x+2\right] d x=1\left(1-\frac{2}{3}\right)=\frac{1}{3} $
$\text { Hence } I=\left(\frac{1}{3}-(\alpha+\beta)\right)+\left(\frac{1}{3}\right)+1 $
$=\frac{5}{3}-\left(\frac{7-\sqrt{37}}{6}+\frac{7-\sqrt{13}}{6}\right)$
$ =\frac{-2}{3}+\frac{\sqrt{37}+\sqrt{13}}{6} $
$ =\frac{\sqrt{37}+\sqrt{13}-4}{6}$
