Question

The area bounded by the curve $y=x^2$ and $y=\frac{2}{1+x^2}$ is $\lambda$ sq. units, then the value of $[\lambda ]$ is
[$[k]$ denotes greatest integer less than or equal to $k$.]

Answer 1

Answer: 2

Required area $=\underset{-1}{\overset{1}{\int }}\left(\frac{2}{1+x^2}-x^2\right)dx$
$=2\underset{0}{\overset{1}{\int }}\left(\frac{2}{1+x^2}-x^2\right)dx$
$=2{\left(2{\tan }^{-1}x-\frac{x^3}{3}\right)}_0^1=2\left(\frac{\pi }{2}-\frac{1}{3}\right)=\pi -\frac{2}{3}$ sq. units

$\therefore \left[\pi -\frac{2}{3}\right]=2$