Question

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

• A. $-1$
• B. $-\frac{5}{4}$
• C. $\frac{\sqrt{17}-13}{8}$
• D. $\frac{\sqrt{17}-16}{8}$

Answer 1

Answer: (C)

$\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$
$=\int\limits_{0}^{1 / 4}-1 d x+\int\limits_{1 / 4}^{1 / 2} 0 d x+\int\limits_{1 / 2}^{3 / 4}-1 d x$

$+\int\limits_{3 / 4}^{\frac{3+\sqrt{17}}{8}}-2 dx +\int\limits_{\frac{3+\sqrt{17}}{8}}^{1}-3 dx$
$=-[x]_{0}^{1 / 4}+0-[x]_{1 / 2}^{3 / 4}+-2[x]_{3 / 4}^{\frac{3+\sqrt{17}}{8}}-3[x]_{\frac{3+\sqrt{17}}{8}}^{1}$
$=-\left(\frac{1}{4}-0\right)-\left(\frac{3}{4}-\frac{1}{2}\right)-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$
$=-\frac{1}{4}-\frac{1}{4}+\frac{-6-2 \sqrt{17}}{8}+\frac{3}{2}-3+\frac{9+3 \sqrt{17}}{8}$
$=\frac{\sqrt{17}-13}{8}$