Question

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

• A. 0.486
• B. 0.243
• C. 1.8
• D. 0

Answer 1

Answer: (D)

$\int\limits^{0.9}_{-0.9}\left\{\left[x^{2}\right]+\left(\frac{2-x}{2+x}\right)\right\}dx$
$= \int \limits^{0.9}_{-0.9} \left[x^{2}\right]dx+\int \limits^{0.9}_{-0.9} log \left(\frac{2-x}{2+x}\right)dx$
$= 0 +\int \limits^{0.9}_{-0.9} log \left(\frac{2-x}{2+x}\right)dx$
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.