Question

The area bounded by the curves $x=y^2$ and $x=3-2y^2$ is:

Answer 1

From figure, the two curves represent parabolas with vertices at $(0,0)$ and $(3,0)$. They intersect at $(1,1)$ and $(1,-1),$ so the required area is
here,

$x=y^2$ $\Rightarrow$ $y=\sqrt{x}$

also

$x=3-2y^2$ $\Rightarrow$ $-2y^2=x-3$$\Rightarrow$$y=\sqrt{\frac{3-x}{2}}$

$\Rightarrow$ area of $OPMQO=2$ (area of $OPMO)$

$=2({\int }_0^1\sqrt{x}dx+{\int }_1^3\sqrt{\frac{3-x}{2}}dx)$

$={\left(\frac{2}{3}{x}^{3/2}\right|}_0^1{-\frac{1}{\sqrt{2}}\cdot \frac{2}{3}(3-x{)}^{3/2}|}_1^3)$

$=2[\frac{2}{3}-(0-\frac{1}{\sqrt{2}}\cdot \frac{2}{3}{2}^{3/2})]=2(\frac{2}{3}+\frac{4}{3})=4$