Application of Integrals: Area Under Parabola and Latus Rectum

The parabola $y^2 = 4ax$ opens to the right along the $x$-axis with vertex $(0,0)$. The latus rectum is the vertical line $x = a$. Due to symmetry about the $x$-axis, total area $A = 2 \int_{0}^{a} y \, dx$.

For the upper half, $y = \sqrt{4ax} = 2\sqrt{a} \cdot x^{1/2}$. $A = 2 \int_{0}^{a} 2\sqrt{a} x^{1/2} \, dx = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}$ $A = 4\sqrt{a} \cdot \frac{2}{3} [a^{3/2} - 0] = \frac{8\sqrt{a}}{3} \cdot a\sqrt{a} = \frac{8}{3} a^2\text{ sq. units}$