If [t] denotes the greatest integer ≤ t , then the value of ∫ _ 0 ^...

$I = {∫}_{0}^{1}\left[2x - \left|3{x}^{2} - 3x - 2x + 2\right| + 1\right]dx$

$I = {∫}_{0}^{1}\left[2x - \left|\left(3x - 2\right)\left(x - 1\right)\right|\right]dx + {∫}_{0}^{1}1dx$

$I = {∫}_{0}^{\dfrac{2}{3}}\left[\left(2x - \left(3{x}^{2} - 5x + 2\right)\right)\right]dx + {∫}_{\dfrac{2}{3}}^{1}\left(2x + \left(3{x}^{2} - 5x + 2\right)\right)dx + 1$

$I = {∫}_{0}^{\dfrac{2}{3}}\left[- 3{x}^{2} + 7x - 2\right]dx + {∫}_{\dfrac{2}{3}}^{1}\left(3{x}^{2} - 3x + 2\right)dx + 1$

Let ${I}_{1} = {∫}_{0}^{\dfrac{2}{3}}\left[- 3{x}^{2} + 7x - 2\right]dx$

The graph of $y = - 3{x}^{2} + 7x - 2$ and

${I}_{2} = {∫}_{\dfrac{2}{3}}^{1}\left(3{x}^{2} - 3x + 2\right)dx$

So ${I}_{1} = {∫}_{0}^{\alpha}\left(- 2\right)dx + {∫}_{\alpha}^{\dfrac{1}{3}}\left(- 1\right)dx + {∫}_{1/3}^{\beta}0dx + {∫}_{\beta}^{\dfrac{2}{3}}1 \cdot dx$

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

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

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

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

\left[3{x}^{2} - 3x + 2\right] = 1$</p><p>$∴ {I}_{2} = {∫}_{\dfrac{2}{3}}^{1}\left[3{x}^{2} - 3x + 2\right]dx = 1\left(1 - \dfrac{2}{3}\right) = \dfrac{1}{3}$</p><p>Hence$I = \left(\dfrac{1}{3} - ( \alpha + \beta )\right) + \left(\dfrac{1}{3}\right) + 1$</p><p>$= \dfrac{5}{3} - \left(\dfrac{7 - \sqrt{37}}{6} + \dfrac{7 - \sqrt{13}}{6}\right)$</p><p>$= \dfrac{- 2}{3} + \dfrac{\sqrt{37} + \sqrt{13}}{6}$$= \dfrac{- 2}{3} + \dfrac{\sqrt{37} + \sqrt{13}}{6}$

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