The area (in sq. units) of the region $$A = \{(x, y) : x^2 \le y \le x + 2\}$$ is:

We have to find the area enclosed by the set $$A=\{(x,y):x^{2}\le y\le x+2\}.$$

The lower boundary of the region is the parabola $$y=x^{2}$$ and the upper boundary is the straight line $$y=x+2.$$

To use integration we first locate the points where the two curves meet. These points satisfy $$x^{2}=x+2.$$

Rearranging we get $$x^{2}-x-2=0.$$

Factoring, $$\bigl(x-2\bigr)\bigl(x+1\bigr)=0,$$ hence $$x=2 \text{ or } x=-1.$$

So, throughout the interval $$-1\le x\le 2,$$ the line lies above the parabola (because $$x+2 - x^{2}\ge 0$$ there). Therefore the area is obtained by integrating the vertical “height” $$\bigl[(x+2)-(x^{2})\bigr]$$ with respect to $$x$$ between $$x=-1$$ and $$x=2.$$

The standard formula we are using is: “If a region is bounded by $$y=f(x)$$ on top and $$y=g(x)$$ below, the area is $$\displaystyle\int_{a}^{b}\bigl[f(x)-g(x)\bigr]\,dx$$, where $$a$$ and $$b$$ are the points of intersection.” Here $$f(x)=x+2$$ and $$g(x)=x^{2}.$$

So the desired area is

$$\text{Area}= \int_{-1}^{2}\Bigl[(x+2)-(x^{2})\Bigr]\,dx =\int_{-1}^{2}\Bigl(-x^{2}+x+2\Bigr)\,dx.$$

We now integrate term by term. The antiderivative of $$-x^{2}$$ is $$-\dfrac{x^{3}}{3},$$ of $$x$$ is $$\dfrac{x^{2}}{2},$$ and of the constant $$2$$ is $$2x.$$ Thus

$$\int\bigl(-x^{2}+x+2\bigr)\,dx = -\dfrac{x^{3}}{3}+\dfrac{x^{2}}{2}+2x.$$

Evaluating this expression from $$x=-1$$ to $$x=2$$ we proceed step by step.

First at $$x=2:$$

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

Next at $$x=-1:$$

$$-\dfrac{(-1)^{3}}{3}+\dfrac{(-1)^{2}}{2}+2(-1)= -\Bigl(-\dfrac{1}{3}\Bigr)+\dfrac{1}{2}-2 = \dfrac{1}{3}+\dfrac{1}{2}-2.$$

Combining the fractions, $$\dfrac{1}{3}+\dfrac{1}{2}= \dfrac{2}{6}+\dfrac{3}{6}= \dfrac{5}{6},$$ so

$$\dfrac{1}{3}+\dfrac{1}{2}-2 = \dfrac{5}{6}-2 = \dfrac{5}{6}-\dfrac{12}{6}= -\dfrac{7}{6}.$$

Now subtract the lower value from the upper value:

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

Writing both terms with denominator $$6$$, $$\dfrac{10}{3}= \dfrac{20}{6},$$ hence

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

This positive value indeed represents the required area in square units.

Hence, the correct answer is Option C.