The area of the region bounded by $$y - x = 2$$ and $$x^2 = y$$ is equal to:

We have two bounding curves, namely the straight line $$y - x = 2$$ and the parabola $$x^{2} = y$$. First we write the line in the explicit form $$y = x + 2$$ so that both relations give $$y$$ in terms of $$x$$.

To locate the intersection points we equate the two expressions for $$y$$:

$$x + 2 = x^{2}.$$

Bringing all terms to one side,

$$x^{2} - x - 2 = 0.$$

Now we factor the quadratic:

$$x^{2} - x - 2 = (x - 2)(x + 1) = 0.$$

Setting each factor equal to zero gives the solutions

$$x - 2 = 0 \;\Rightarrow\; x = 2,$$

$$x + 1 = 0 \;\Rightarrow\; x = -1.$$

For these $$x$$-values, the corresponding $$y$$-values can be found from either curve; using the line $$y = x + 2$$ we get

$$y(2) = 2 + 2 = 4,$$

$$y(-1) = -1 + 2 = 1.$$

Thus the curves meet at the points $$(-1,\,1)$$ and $$(2,\,4)$$, so the horizontal span of the enclosed region is from $$x = -1$$ to $$x = 2$$.

Next we decide which curve lies above the other within this interval. Choosing an easy test point, say $$x = 0,$$ we find

For the line: $$y = 0 + 2 = 2,$$

For the parabola: $$y = 0^{2} = 0.$$

Because $$2 > 0,$$ the line $$y = x + 2$$ is above the parabola $$y = x^{2}$$ throughout the interval from $$x = -1$$ to $$x = 2$$.

The standard formula for the area between two curves $$y = f(x)$$ (upper) and $$y = g(x)$$ (lower) from $$x = a$$ to $$x = b$$ is

$$A = \int_{a}^{b} \bigl[\,f(x) - g(x)\,\bigr] \, dx.$$

Here $$f(x) = x + 2$$ and $$g(x) = x^{2},$$ with $$a = -1$$ and $$b = 2.$$ Substituting we obtain

$$A = \int_{-1}^{2} \bigl[(x + 2) - x^{2}\bigr] \, dx.$$

Simplifying the integrand,

$$A = \int_{-1}^{2} \bigl(-x^{2} + x + 2\bigr) \, dx.$$

We now integrate term by term:

$$\int \left(-x^{2}\right) dx = -\dfrac{x^{3}}{3},$$

$$\int x \, dx = \dfrac{x^{2}}{2},$$

$$\int 2 \, dx = 2x.$$

Putting these together, the antiderivative is

$$-\dfrac{x^{3}}{3} + \dfrac{x^{2}}{2} + 2x.$$

We evaluate this expression from $$x = -1$$ to $$x = 2$$.

At the upper limit $$x = 2$$:

$$-\dfrac{(2)^{3}}{3} + \dfrac{(2)^{2}}{2} + 2(2) = -\dfrac{8}{3} + \dfrac{4}{2} + 4 = -\dfrac{8}{3} + 2 + 4 = -\dfrac{8}{3} + 6 = \dfrac{-8 + 18}{3} = \dfrac{10}{3}.$$

At the lower limit $$x = -1$$:

$$-\dfrac{(-1)^{3}}{3} + \dfrac{(-1)^{2}}{2} + 2(-1) = -\left(-\dfrac{1}{3}\right) + \dfrac{1}{2} - 2 = \dfrac{1}{3} + \dfrac{1}{2} - 2.$$

Combining the fractions,

$$\dfrac{1}{3} + \dfrac{1}{2} = \dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6},$$

so the whole value at $$x = -1$$ equals

$$\dfrac{5}{6} - 2 = \dfrac{5}{6} - \dfrac{12}{6} = -\dfrac{7}{6}.$$

The required area is the difference of these two numerical values:

$$A = \left[\dfrac{10}{3}\right] - \left[-\dfrac{7}{6}\right] = \dfrac{10}{3} + \dfrac{7}{6}.$$

We take a common denominator of $$6$$:

$$\dfrac{10}{3} = \dfrac{20}{6},$$

so

$$A = \dfrac{20}{6} + \dfrac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2}.$$

Thus the area of the region bounded by the given curves is $$\dfrac{9}{2}$$ square units.

Hence, the correct answer is Option C.