Mathematics - Area under the curve Question with Solution | TestHub

$\begin{array}{rl}& x\left(1+y^2\right)=1\\ & y^2=2x\end{array}$

From equation (1) \& (2)
$\begin{array}{rl}x(1+2x)=1& \Rightarrow 2x^2+x-1=0\\ & \Rightarrow x=\frac{1}{2},x=-1\text{ (Reject) }\\ & \Rightarrow y^2=2\left(\frac{1}{2}\right)\\ & \Rightarrow y=\pm 1\end{array}$

$\begin{array}{rl}\text{ Area bounded }& ={\int }_{-1}^1\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy\\ & ={\left({\tan }^{-1}y-\frac{y^3}{6}\right)|}_{-1}^1\\ & =\frac{\pi }{2}-\frac{1}{3}\end{array}$