Mathematics - Area Under Curves Question with Solution | TestHub

We have, $x=-2y^2$ $\Rightarrow y^2=-\frac{1}{2}x$ ...(i) and $x=1-3y^2$ $\Rightarrow y^2=-\frac{1}{3}(x-1)$ ...(ii) From Eqs. (i) and (ii),we get. $\begin{array}{rl}& -\frac{1}{2}x=-\frac{1}{3}(x-1)\\ \Rightarrow & 3x=2(x-1)\\ \Rightarrow & 3x=2x-2\\ \Rightarrow & x=-2\\ \Rightarrow & y^2=-\frac{1}{2}(-2)=1\\ \Rightarrow & y=\pm 1\end{array}$ $\therefore$ Point of intersection of two curves is $(-2\pm 1)$$\therefore$ Required Area $=2{\int }_0^1\sin xdx$ $=2{\left[y-\frac{y^3}{3}\right]}_0^1$ $=2\left[1-\frac{1}{3}\right]$ $=\frac{4}{3}\mathrm{sq}$ units