Area Under a Parabola (Standard Form)

A parabola in standard form, such as $y^2 = 4ax$ (opening to the right, vertex at the origin, with $a>0$), is not a function of $x$ by itself -- solving for $y$ gives two branches, $y = \pm\sqrt{4ax} = \pm 2\sqrt{a}\sqrt{x}$, symmetric about the $x$-axis. The regions this syllabus asks for are bounded by both branches together with a vertical line $x=h$ (for some $h>0$), so the region is symmetric about the $x$-axis and its area is found by doubling the area under just the upper branch.

Setting up the integral. The upper branch is $y = 2\sqrt{a}\,\sqrt{x} = 2\sqrt{ax}$. The area under this branch alone, from the vertex $x=0$ to $x=h$, is $\int_0^h 2\sqrt{ax}\,dx$; doubling it (to include the lower branch, the mirror image below the $x$-axis) gives the total enclosed area:

Evaluating the integral. Since $\int \sqrt{x}\,dx = \int x^{1/2}\,dx = \dfrac{2}{3}x^{3/2} + C$,

The special case of the latus rectum. The chord through the focus $(a,0)$, perpendicular to the axis, is called the latus rectum; it is exactly the vertical line $x=a$. Setting $h=a$ in the formula above,

using $a^{3/2} = a\sqrt{a}$ and $\sqrt{a}\cdot\sqrt{a}=a$. This gives the well-known result: the area enclosed between a standard parabola and its own latus rectum is $\dfrac{8}{3}a^2$, a useful benchmark to sanity-check any latus-rectum question against.

Alternative technique: integrating with respect to $y$. When a region is bounded more naturally by the $y$-axis and a horizontal line $y=k$ -- for instance, the parabola $y^2=4ax$ together with the $y$-axis and the line $y=k$, in the first quadrant -- it is more direct to solve the parabola's equation for $x$ instead of $y$:

Now a thin horizontal strip at height $y$, of thickness $dy$, has length $x = \dfrac{y^2}{4a}$ (measured from the $y$-axis out to the curve), so the area swept out as $y$ runs from $0$ to $k$ is …