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

$\frac{1}{2}$ 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