The area enclosed between the curves $$y^2 + 4x = 4$$ and $$y - 2x = 2$$ is

We need to find the area enclosed between $$y^2 + 4x = 4$$ (i.e., $$y^2 = 4 - 4x = -4(x - 1)$$, a left-opening parabola with vertex $$(1, 0)$$) and $$y - 2x = 2$$ (i.e., $$y = 2x + 2$$).

We start by finding the intersection points. Substituting $$y = 2x + 2$$ into $$y^2 + 4x = 4$$ gives:

$$ (2x + 2)^2 + 4x = 4 $$

which simplifies to

$$ 4x^2 + 8x + 4 + 4x = 4 $$

and then to

$$ 4x^2 + 12x = 0, \quad 4x(x + 3) = 0 $$

giving $$x = 0$$ or $$x = -3$$. When $$x = 0$$, $$y = 2$$, and when $$x = -3$$, $$y = -4$$. Therefore, the intersection points are $$(0, 2)$$ and $$(-3, -4)$$.

Next, we express $$x$$ from each curve in terms of $$y$$. From the parabola, $$x = \frac{4 - y^2}{4} = 1 - \frac{y^2}{4}$$, and from the line, $$x = \frac{y - 2}{2}$$. For $$y \in [-4, 2]$$, the parabola lies to the right of the line. Therefore, the area is given by:

$$ \text{Area} = \int_{-4}^{2} \Bigl[\Bigl(1 - \frac{y^2}{4}\Bigr) - \frac{y - 2}{2}\Bigr] \,dy. $$

This simplifies to:

$$ \int_{-4}^{2} \Bigl(1 - \frac{y^2}{4} - \frac{y}{2} + 1\Bigr)\,dy = \int_{-4}^{2} \Bigl(2 - \frac{y}{2} - \frac{y^2}{4}\Bigr)\,dy. $$

We then evaluate the integral:

$$ \Bigl[2y - \frac{y^2}{4} - \frac{y^3}{12}\Bigr]_{-4}^{2}. $$

At $$y = 2$$ this equals $$4 - 1 - \frac{8}{12} = 4 - 1 - \frac{2}{3} = \frac{7}{3}$$, and at $$y = -4$$ it equals $$-8 - 4 + \frac{16}{3} = -12 + \frac{16}{3} = -\frac{20}{3}$$. Thus, the area is:

$$ \frac{7}{3} - \bigl(-\frac{20}{3}\bigr) = \frac{27}{3} = 9. $$

Hence, the enclosed area is 9.