The area (in sq. units) of the region bounded by the parabola $y^2 = 4x$ and the line $x = 1$ is

$\frac{8}{3}$

The correct answer is Option (4) → $\frac{8}{3}$

The curve is

$y^{2}=4x$

and the vertical line is

$x=1$

For $y^{2}=4x$ we have

$x=\frac{y^{2}}{4}$

The region lies between $x=\frac{y^{2}}{4}$ and $x=1$. Find corresponding limits of $y$:

$y^{2}=4(1)\Rightarrow y=\pm 2$

Area:

$A=\int_{-2}^{2}\left(1-\frac{y^{2}}{4}\right)\,dy$

Compute the integral:

$A=\int_{-2}^{2}1\,dy-\int_{-2}^{2}\frac{y^{2}}{4}\,dy$

$A=\left[y\right]_{-2}^{2}-\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{-2}^{2}$

$A=(2-(-2))-\frac{1}{4}\left(\frac{8}{3}-\frac{-8}{3}\right)$

$A=4-\frac{1}{4}\left(\frac{16}{3}\right)$

$A=4-\frac{4}{3}$

$A=\frac{12}{3}-\frac{4}{3}$

$A=\frac{8}{3}$