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 (1) → $\frac{9}{4}$

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

Boundaries: the curve, the y-axis $(x=0)$, and the line $y = 3$.

The curve intersects the y-axis at $y = 0$.

Required area = $\displaystyle \int_{0}^{3} x\,dy = \int_{0}^{3} \frac{y^{2}}{4}\,dy$

$= \frac{1}{4}\left[\frac{y^{3}}{3}\right]_{0}^{3} = \frac{1}{4}\times\frac{27}{3} = \frac{9}{4}$

Area = $\frac{9}{4}$ square units.