The area (in sq. units) of the region enclosed between the parabola $$y^2 = 2x$$ and the line $$x + y = 4$$ is ______.

We need to find the area enclosed between the parabola $$y^2 = 2x$$ and the line $$x + y = 4$$, i.e., $$x = 4 - y$$.

Finding the intersection points: Substituting $$x = y^2/2$$ into $$x + y = 4$$:

$$\frac{y^2}{2} + y = 4$$

$$y^2 + 2y - 8 = 0$$

$$(y + 4)(y - 2) = 0$$

So $$y = -4$$ and $$y = 2$$.

The corresponding $$x$$ values: when $$y = 2$$, $$x = 2$$; when $$y = -4$$, $$x = 8$$.

Integrating with respect to $$y$$ (the line is to the right of the parabola for $$y \in [-4, 2]$$):

$$A = \int_{-4}^{2} \left[(4-y) - \frac{y^2}{2}\right] dy$$

$$= \int_{-4}^{2} \left(4 - y - \frac{y^2}{2}\right) dy$$

$$= \left[4y - \frac{y^2}{2} - \frac{y^3}{6}\right]_{-4}^{2}$$

At $$y = 2$$: $$8 - 2 - \frac{8}{6} = 6 - \frac{4}{3} = \frac{14}{3}$$

At $$y = -4$$: $$-16 - 8 + \frac{64}{6} = -24 + \frac{32}{3} = \frac{-72 + 32}{3} = \frac{-40}{3}$$

$$A = \frac{14}{3} - \frac{-40}{3} = \frac{14 + 40}{3} = \frac{54}{3} = 18$$

The correct answer is $$18$$.