Question

The area bounded by the curves $y=x^{2}$ and $y=\frac{2}{\left(1 \, + \, x^{2}\right)}$ is

• A. $\left(\right. \pi - \frac{1}{3} \left.\right)$ sq. units
• B. $\left(\right. \pi - \frac{2}{3} \left.\right)$ sq. units
• C. $\frac{\left(\right. 2 \pi - 1 \left.\right)}{3}$ sq. units
• D. None of these

Answer 1

Answer: (B)

For point of intersection $x^{2}=\frac{2}{1 + x^{2}}$
$\Rightarrow \, x^{4}+ \, x^{2}- \, 2=0$
$\Rightarrow \, \left(x^{2} - \, 1\right)\left(x^{2 \, } + \, 2\right)=0$
$\Rightarrow \, x=1, \, -1, \, \Rightarrow y=1, \, -1$
$\therefore \, area=2\displaystyle \int _{0}^{1} \left(\frac{2}{1 + x^{2}} - x^{2}\right) \, dx$
$=4\left[tan^{- 1} x\right]_{ \, 0}^{ \, 1}- \, \frac{2}{3}\left[x^{3}\right]_{ \, 0}^{ \, 1}$
$=4 \, \left(\pi / 4\right)-2 / 3=\pi \, -2 / 3$