The area of the region bounded by the parabola $y^2=8 x$ and the line $x=2$ is :

$\frac{32}{3}$ sq. units

$y^2 = 8x~~~~~x = 2$

for x = 2

$y^2 = 8×2$

$\Rightarrow y^2=16 \Rightarrow y= \pm 4$

points of intersection $(2, \pm 4)$ as parabola is symmetric about x-axis about x-axis

area region I + area region II

= 2 × area region I

So area of parabola = A = $2 \int\limits_0^2 y d x$

So  A = $2 \int\limits_0^2 x 2 \sqrt{2} \sqrt{x} d x$

$y^2 = 8x$

$y=2\sqrt{2} \sqrt{x}$

$=4 \sqrt{2}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_0^2 \Rightarrow \frac{8 \sqrt{2}}{3}\left[2^{\frac{3}{2}}\right]$

$\frac{8}{3} \sqrt{2} \times 2 \sqrt{2}=\frac{32}{3}$ sq. units