Area (in sq. units) of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is

$\frac{9}{4}$

The correct answer is Option (2) → $\frac{9}{4}$

Given: $y^{2}=4x \Rightarrow x=\frac{y^{2}}{4}$

Region bounded by the parabola $x=\frac{y^{2}}{4}$, y-axis ($x=0$), and $y=3$.

Area = $\displaystyle\int_{0}^{3}\left[\frac{y^{2}}{4}-0\right]dy$

$=\frac{1}{4}\displaystyle\int_{0}^{3}y^{2}dy=\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{0}^{3}=\frac{1}{4}\times 9=\frac{9}{4}$

Required area = $\frac{9}{4}$ sq. units