Using integration, find the area of the region bounded by the parabola $\displaystyle y^{2}=4 a x$ and its latus rectum.

(figure labels as printed: $\displaystyle y^2 = 4ax$, $\displaystyle x = a$, "Latus Rectum", axes $\displaystyle X$, $\displaystyle Y$, origin $\displaystyle O$)Required area $\displaystyle = 2\int_0^a 2\sqrt{ax}\ dx$$\displaystyle = 4\sqrt{a} \int_0^a \sqrt{x}\ dx$$\displaystyle = 4\sqrt{a}\ \left| \dfrac{2x^{3/2}}{3} \right|_0^a$$\displaystyle = \dfrac{8}{3} a \sqrt{a}\ \sqrt{a}$$\displaystyle = \dfrac{8}{3} a^2$