Question

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

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

Answer 1

The correct option is D $\frac{\sqrt{17}-13}{8}$

$\underset{0}{\overset{1}{\int }}[-8x^2+6x-1]dx$


$\underset{0}{\overset{\frac{1}{4}}{\int }}(-1)dx+\underset{\frac{1}{4}}{\overset{\frac{1}{2}}{\int }}0dx+\underset{\frac{1}{2}}{\overset{\frac{3}{4}}{\int }}-1dx+\underset{\frac{3}{4}}{\overset{\frac{3+\sqrt{17}}{8}}{\int }}-2dx+\underset{\frac{3+\sqrt{17}}{8}}{\overset{1}{\int }}-3dx$
$=-\frac{2}{4}-\frac{1}{4}-2(\frac{3+\sqrt{17}}{8}-\frac{3}{4})-3(1-\frac{3+\sqrt{17}}{8})$
$=\frac{\sqrt{17}-13}{8}$