Let t denote the greatest integer less than or equal to t . Then, the value of the integral ∫ 0 1 - 8…

The graph of $y=-8x^2+6x-1$ will be as shown 

Now ${\int }_0^1\left[-8x^2+6x-1\right]dx$$={\int }_0^{\frac{1}{4}}\left(-1\right)dx+{\int }_{\frac{1}{4}}^{\frac{1}{2}}\left(0\right)dx+{\int }_{\frac{1}{2}}^{\frac{3}{4}}\left(-1\right)dx$

$+{\int }_{\frac{3}{4}}^{\frac{3+\sqrt{17}}{8}}\left(-2\right)dx+{\int }_{\frac{3+\sqrt{17}}{8}}^1\left(-3\right)dx$

$=-{\left[x\right]}_0^{\frac{1}{4}}+0-{\left[x\right]}_{\frac{1}{2}}^{\frac{3}{4}}+-2{\left[x\right]}_{\frac{3}{4}}^{\frac{3+\sqrt{17}}{8}}-3{\left[x\right]}_{\frac{3+\sqrt{17}}{8}}^1$

$=-\left(\frac{1}{4}-0\right)-\left(\frac{3}{4}-\frac{1}{2}\right)-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$

$=-\frac{1}{4}-\frac{1}{4}+\frac{-6-2\sqrt{17}}{8}+\frac{3}{2}-3+\frac{9+3\sqrt{17}}{8}$

$=\frac{\sqrt{17}-13}{8}$