Area bounded by the curves $y^2=9x $ and $y=3x$ is :

$\frac{2}{3}sq\, units$ $\frac{1}{3}sq\, units$ $\frac{1}{2}sq\, units$ $2sq\, units $

$\frac{1}{2}sq\, units$

The correct answer is Option (3) → $\frac{1}{2}sq\, units$ $y^2=9x$, $y=3x$ they interact at $(3x)^2=9x$ so $x^2=x, x=0,1$ $y=0,3$ Area required = $\int\limits_0^13\sqrt{x}-3xdx$ $=\left[\frac{2}{3}x^{\frac{3}{2}}-\frac{x^2}{2}\right]_0^1=\frac{1}{2}$ sq. units