Area lying between the curves $y^2=9x $ and $y =3 x$ is :

$\frac{1}{4}$ sq. units $\frac{1}{2}$ sq. units $\frac{2}{3}$ sq. units $\frac{1}{3}$ sq. units

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

The correct answer is Option (2) → $\frac{1}{2}$ sq. units $y^2=9x$ and $y =3x$ as $y=3x$ $⇒y^2=9x$ so $(3x)^2=9x$ $x^2=x$ $x=0,1$, $y=0,3$ points of intersection area = $\int\limits_0^13\sqrt{x}-3xdx$ $=3\left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_0^1=\frac{1}{2}$ sq. units