Question

The value of the integral $\underset{-2}{\overset{2}{\int }}\left(\right[x]+\log \left(\frac{1+x}{1-x}\right)+\sin \left(\log \left(\frac{2+x}{2-x}\right)\right)+x^{2011})dx$ is
(where [.] denotes greatest integer function)

• A. $4$
• B. $2$
• C. $-2$
• D. $-4$

Answer 1

The correct option is C $-2$

Let $I=\underset{-2}{\overset{2}{\int }}\left(\right[x]+\log \left(\frac{1+x}{1-x}\right)+\sin \left(\log \left(\frac{2+x}{2-x}\right)\right)+x^{2011})dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$=\underset{-2}{\overset{2}{\int }}\left[x\right]dx+0+0+0$
$=\underset{-2}{\overset{-1}{\int }}\left[x\right]dx+\underset{-1}{\overset{0}{\int }}\left[x\right]dx+\underset{0}{\overset{1}{\int }}\left[x\right]dx+\underset{1}{\overset{2}{\int }}\left[x\right]dx$
$=\underset{-2}{\overset{-1}{\int }}(-2)dx+\underset{-1}{\overset{0}{\int }}(-1)dx+\underset{0}{\overset{1}{\int }}0.dx+\underset{1}{\overset{2}{\int }}1dx$

$=-2(-1+2)+-1(0+1)+0+1(2-1)$
$=-2-1+1$
$=-2$