Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $\int_{0}^{1} [- 8 x^{2} + 6 x - 1] dx$ is equal to

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$\frac{\sqrt{17} - 16}{8}$

$\frac{- 5}{4}$

$- 1$

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

answer is C.

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Detailed Solution

$\begin{array}{l} \int_{0}^{1} [- 8 x^{2} + 6 x - 1] d x \\ = \int_{0}^{1 / 4} - 1 d x + \int_{1 / 4}^{1 / 2} 0 d x + \int_{1 / 2}^{3 / 4} - 1 d x + \int_{3 / 4}^{\frac{3 + \sqrt{17}}{8}} - 2 d x + \int_{\frac{3 + \sqrt{17}}{8}}^{1} - 3 d x \\ = - [x]_{0}^{1 / 4} + 0 - [x]_{1 / 2}^{3 / 4} + - 2 [x]_{3 / 4}^{\frac{3 + \sqrt{17}}{8}} - 3 [x]_{\frac{3 + \sqrt{17}}{1}}^{8} \\ \begin{array}{l} = - (\frac{1}{4} - 0) - (\frac{3}{4} - \frac{1}{2}) - 2 (\frac{3 + \sqrt{17}}{8} - \frac{3}{4}) - 3 (1 - \frac{3 + \sqrt{17}}{8}) \\ = - \frac{1}{4} - \frac{1}{4} + \frac{- 6 - 2 \sqrt{17}}{8} + \frac{3}{2} - 3 + \frac{9 + 3 \sqrt{17}}{8} \\ = \frac{\sqrt{17} - 13}{8} \end{array} \end{array}$