Q.Find the area of the region bounded by the parabola $y^2 = 8x$ and its latus rectum (the line $x = 2$).

Comparing $y^2=8x$ with the standard form $y^2=4ax$ gives $4a=8$, so $a=2$ -- and the line $x=2$ is exactly $x=a$, the latus rectum. The region is symmetric about the $x$-axis, so the area is twice the area under the upper branch $y=\sqrt{8x}$ from $x=0$ to $x=2$:

Now $\sqrt{8}=2\sqrt{2}$ and $2^{3/2}=2\sqrt{2}$, so …

Identify $a$ by comparing with $y^2=4ax$, recognise $x=a$ as the latus rectum, and either integrate $2\int_0^a\sqrt{4ax} …

Forgetting to double the upper-branch integral for the lower (mirror-image) …