Question

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

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

$+\underset{3/4}{\overset{\frac{3+\sqrt{17}}{8}}{\int }}-2dx+\underset{\frac{3+\sqrt{17}}{8}}{\overset{1}{\int }}-3dx$
$=-[x{]}_0^{1/4}+0-[x{]}_{1/2}^{3/4}+-2[x{]}_{3/4}^{\frac{3+\sqrt{17}}{8}}-3[x{]}_{\frac{3+\sqrt{17}}{8}}^1$
$=-\left(\frac{1}{4}-0\right)-\left(\frac{3}{4}-\frac{1}{2}\right)-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$
$=-\frac{1}{4}-\frac{1}{4}+\frac{-6-2\sqrt{17}}{8}+\frac{3}{2}-3+\frac{9+3\sqrt{17}}{8}$
$=\frac{\sqrt{17}-13}{8}$