The area of the region given by $$A = \{(x, y) : x^2 \le y \le \min\{x + 2, 4 - 3x\}\}$$ is

We need to find the area of the region $$A = \{(x, y) : x^2 \le y \le \min\{x + 2, 4 - 3x\}\}$$.

Set $$x + 2 = 4 - 3x$$:

$$4x = 2 \implies x = \dfrac{1}{2}$$

At $$x = \dfrac{1}{2}$$: $$y = \dfrac{1}{2} + 2 = \dfrac{5}{2}$$.

So $$\min\{x + 2, 4 - 3x\} = \begin{cases} x + 2, & x \le \dfrac{1}{2} \\ 4 - 3x, & x \ge \dfrac{1}{2}\end{cases}$$.

$$x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x = -1 \text{ or } x = 2$$

Since we need $$x^2 \le x + 2$$, this holds for $$-1 \le x \le 2$$.

$$x^2 + 3x - 4 = 0 \implies (x+4)(x-1) = 0 \implies x = -4 \text{ or } x = 1$$

Since we need $$x^2 \le 4 - 3x$$, this holds for $$-4 \le x \le 1$$.

For $$x \le \dfrac{1}{2}$$: we need $$x^2 \le x + 2$$, so $$x \in [-1, \dfrac{1}{2}]$$.

For $$x \ge \dfrac{1}{2}$$: we need $$x^2 \le 4 - 3x$$, so $$x \in [\dfrac{1}{2}, 1]$$.

$$\text{Area} = \int_{-1}^{1/2} [(x + 2) - x^2]\, dx + \int_{1/2}^{1} [(4 - 3x) - x^2]\, dx$$

First integral:

$$\int_{-1}^{1/2} (x + 2 - x^2)\, dx = \left[\dfrac{x^2}{2} + 2x - \dfrac{x^3}{3}\right]_{-1}^{1/2}$$

$$= \left(\dfrac{1}{8} + 1 - \dfrac{1}{24}\right) - \left(\dfrac{1}{2} - 2 + \dfrac{1}{3}\right)$$

$$= \left(\dfrac{3 + 24 - 1}{24}\right) - \left(\dfrac{3 - 12 + 2}{6}\right) = \dfrac{26}{24} - \left(-\dfrac{7}{6}\right) = \dfrac{13}{12} + \dfrac{7}{6} = \dfrac{13 + 14}{12} = \dfrac{27}{12} = \dfrac{9}{4}$$

Second integral:

$$\int_{1/2}^{1} (4 - 3x - x^2)\, dx = \left[4x - \dfrac{3x^2}{2} - \dfrac{x^3}{3}\right]_{1/2}^{1}$$

$$= \left(4 - \dfrac{3}{2} - \dfrac{1}{3}\right) - \left(2 - \dfrac{3}{8} - \dfrac{1}{24}\right)$$

$$= \dfrac{24 - 9 - 2}{6} - \dfrac{48 - 9 - 1}{24} = \dfrac{13}{6} - \dfrac{38}{24} = \dfrac{13}{6} - \dfrac{19}{12} = \dfrac{26 - 19}{12} = \dfrac{7}{12}$$

$$\text{Area} = \dfrac{9}{4} + \dfrac{7}{12} = \dfrac{27 + 7}{12} = \dfrac{34}{12} = \dfrac{17}{6}$$

The answer is Option B: $$\dfrac{17}{6}$$.