Mathematics - Area Under Curves Question with Solution | TestHub

We have, $x+2y^2=0$ $\Rightarrow y^2=-\frac{x}{2}\dots \text{ (i) }$ a parabola with vertex $(0,0)$ and $x+3y^2=1$ $\Rightarrow y^2=\frac{1-x}{3}=-\left(\frac{x-1}{3}\right)\dots$ (ii), a parabola with vertex $(1,0)$ Solving (i) and (ii), we get $y=\pm 1$
$\mathrm{A}={\int }_{-1}^1\left(\left(1-3y^2\right)-\left(-2y^2\right)\right)dy$ $=2{\int }_0^1\left(1-y^2\right)dy=2{\left[y-\frac{y^3}{3}\right]}_0^1$ $=\frac{4}{3}$ sq.units