Question

If $[t]$ denotes the greatest integer $\le t$, then the value of ${\int }_0^1\left[2x-\left|3x^2-5x+2\right|+1\right]dx$ is:

• A. $\frac{\sqrt{37}+\sqrt{13}-4}{6}$
• B. $\frac{\sqrt{37}-\sqrt{13}-4}{6}$
• C. $\frac{-\sqrt{37}-\sqrt{13}+4}{6}$
• D. $\frac{-\sqrt{37}+\sqrt{13}+4}{6}$

Answer 1

Answer: (A)

$I=\underset{0}{\overset{1}{\int }}\left[2x-\left|3x^2-3x-2x+2\right|+1\right]dx$
$I=\underset{0}{\overset{1}{\int }}[2x-|(3x-2)(x-1)|]dx+\underset{0}{\overset{1}{\int }}1dx$
$I=\underset{0}{\overset{2/3}{\int }}\left[\left(2x-\left(3x^2-5x+2\right)\right)\right]dx+{\int }_{2/3}^1\left(2x+\left(3x^2-5x+2\right)\right)dx+1$
$I=\underset{0}{\overset{2/3}{\int }}\left[-3x^2+7x-2\right]dx+\underset{2/3}{\overset{1}{\int }}\left(3x^2-3x+2\right)dx+1$

$\underset{0}{\overset{\alpha }{\int }}(-2)dx+\underset{\alpha }{\overset{1/3}{\int }}(-1)dx+\underset{1/3}{\overset{\beta }{\int }}0dx+\underset{\beta }{\overset{2/3}{\int }}1.dx$
$=-2\alpha -\left(\frac{1}{3}-\alpha \right)+\frac{2}{3}-\beta =-\alpha -\beta +\frac{1}{3}$

When $x\in \left(\frac{2}{3},1\right)$
$3x^2-3x+2\in \left(\frac{4}{3},2\right)$
$\left[3x^2-3x+2\right]=1$
$\therefore \underset{2/3}{\overset{1}{\int }}\left[3x^2-3x+2\right]dx=1\left(1-\frac{2}{3}\right)=\frac{1}{3}$
$\text{ Hence }I=\left(\frac{1}{3}-(\alpha +\beta )\right)+\left(\frac{1}{3}\right)+1$
$=\frac{5}{3}-\left(\frac{7-\sqrt{37}}{6}+\frac{7-\sqrt{13}}{6}\right)$
$=\frac{-2}{3}+\frac{\sqrt{37}+\sqrt{13}}{6}$
$=\frac{\sqrt{37}+\sqrt{13}-4}{6}$