Mathematics - Definite Integration Question with Solution | TestHub

$I={\int }_0^1\left[2x-\left|3x^2-3x-2x+2\right|+1\right]dx$

$I={\int }_0^1\left[2x-\left|\left(3x-2\right)\left(x-1\right)\right|\right]dx+{\int }_0^{1}1dx$

$I={\int }_0^{\frac{2}{3}}\left[\left(2x-\left(3x^2-5x+2\right)\right)\right]dx+{\int }_{\frac{2}{3}}^1\left(2x+\left(3x^2-5x+2\right)\right)dx+1$

$I={\int }_0^{\frac{2}{3}}\left[-3x^2+7x-2\right]dx+{\int }_{\frac{2}{3}}^1\left(3x^2-3x+2\right)dx+1$

Let $I_1={\int }_0^{\frac{2}{3}}\left[-3x^2+7x-2\right]dx$

The graph of $y=-3x^2+7x-2$ and 

$I_2={\int }_{\frac{2}{3}}^1\left(3x^2-3x+2\right)dx$

So $I_1={\int }_0^{\alpha }\left(-2\right)dx+{\int }_{\alpha }^{\frac{1}{3}}\left(-1\right)dx+{\int }_{1/3}^{\beta }0dx+{\int }_{\beta }^{\frac{2}{3}}1·dx$

$=-2\alpha -\left(\frac{1}{3}-\alpha \right)+\frac{2}{3}-\beta =-\alpha -\beta +\frac{1}{3}$

$y=3x^2-3x+2$

When $x\in \left(\frac{2}{3},1\right)$

$3x^2-3x+2\in \left(\frac{4}{3},2\right)$

$\left[3x^2-3x+2\right]=1$

$\therefore I_2={\int }_{\frac{2}{3}}^1\left[3x^2-3x+2\right]dx=1\left(1-\frac{2}{3}\right)=\frac{1}{3}$

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{-2}{3}+\frac{\sqrt{37}+\sqrt{13}}{6}$

$=\frac{\sqrt{37}+\sqrt{13}-4}{6}$