The area (in sq. units) bounded by the parabola $y^2 = 4ax$, its latus rectum and the $x$-axis in the first quadrant is:

The parabola $y^2 = 4ax$ opens towards the right with vertex at the origin $(0, 0)$ and focus at $(a, 0)$.

The latus rectum is a vertical line passing through the focus, perpendicular to the x-axis.

For this parabola, the latus rectum is the line $x = a$.

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To find where the latus rectum meets the parabola, substitute $x = a$ into $y^2 = 4ax$:

$y^2 = 4a(a)$

$y^2 = 4a^2$

$y = \pm 2a$

In the first quadrant, $y = 2a$.

The latus rectum goes from $(a, 0)$ to $(a, 2a)$ in the first quadrant.

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The bounded region has:

Left boundary: The parabola from $(0, 0)$ to $(a, 2a)$

Right boundary: The latus rectum from $(a, 0)$ to $(a, 2a)$

Bottom boundary: The x-axis from $(0, 0)$ to $(a, 0)$

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From $y^2 = 4ax$, solving for $y$:

$y = 2\sqrt{ax}$

The area is:

$\text{Area} = \int_0^a 2\sqrt{ax} \, dx$

$= 2\sqrt{a} \int_0^a \sqrt{x} \, dx$

$= 2\sqrt{a} \int_0^a x^{1/2} \, dx$

$= 2\sqrt{a} \left[\frac{x^{3/2}}{3/2}\right]_0^a$

$= 2\sqrt{a} \cdot \frac{2}{3} \left[x^{3/2}\right]_0^a$

$= \frac{4\sqrt{a}}{3} \left[a^{3/2} - 0\right]$

$= \frac{4\sqrt{a}}{3} \cdot a^{3/2}$

$= \frac{4}{3} \cdot a^{1/2} \cdot a^{3/2}$

$= \frac{4}{3}a^2$

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Therefore, the area bounded by the parabola, its latus rectum, and the x-axis in the first quadrant is $\dfrac{4}{3}a^2$ square units.