Question

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

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

Answer 1

Answer: (B)

For point of intersection $x^2=\frac{2}{1+x^2}$
$\Rightarrow x^4+x^2-2=0$
$\Rightarrow \left(x^2-1\right)\left(x^2+2\right)=0$
$\Rightarrow x=1,-1,\Rightarrow y=1,-1$
$\therefore area=2{\int }_0^1\left(\frac{2}{1+x^2}-x^2\right)dx$
$=4{\left[ta{n}^{-1}x\right]}_0^1-\frac{2}{3}{\left[x^3\right]}_0^1$
$=4\left(\pi /4\right)-2/3=\pi -2/3$