Question

If $[x]$ is the greatest integer $\le x,$ then the value of the integral $\underset{-0.9}{\overset{0.9}{\int }}\left(\left[x^2\right]+\left(\frac{2-x}{2+x}\right)\right)dx$ is

• A. 0.486
• B. 0.243
• C. 1.8
• D. 0

Answer 1

Answer: (D)

$\underset{-0.9}{\overset{0.9}{\int }}\left\{\left[x^2\right]+\left(\frac{2-x}{2+x}\right)\right\}dx$
$=\underset{-0.9}{\overset{0.9}{\int }}\left[x^2\right]dx+\underset{-0.9}{\overset{0.9}{\int }}log\left(\frac{2-x}{2+x}\right)dx$
$=0+\underset{-0.9}{\overset{0.9}{\int }}log\left(\frac{2-x}{2+x}\right)dx$
Put $x=-x\Rightarrow f\left(x\right)=log\frac{2-x}{2+x}$
and $f\left(-x\right)=log\frac{2-x}{2+x}$
$=-log\frac{\left(2-x\right)}{2+x}=-f\left(x\right)$
So, it is an odd function, hence Required integral = 0.