Let $$a$$ and $$b$$ be real constants such that the function $$f$$ defined by $$f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 \\ bx + 2, & x > 1 \end{cases}$$ be differentiable on $$R$$. Then, the value of $$\int_{-2}^{2} f(x) \, dx$$ equals

For differentiability at $$x = 1$$: continuity and equal derivatives.

Continuity: $$1 + 3 + a = b + 2 \Rightarrow a + 4 = b + 2 \Rightarrow b = a + 2$$ ... (1)

Derivatives: $$f'(x) = 2x + 3$$ for $$x \leq 1$$, $$f'(x) = b$$ for $$x > 1$$.

At $$x = 1$$: $$2 + 3 = b \Rightarrow b = 5$$. From (1): $$a = 3$$.

$$f(x) = \begin{cases} x^2 + 3x + 3, & x \leq 1 \\ 5x + 2, & x > 1 \end{cases}$$

$$\int_{-2}^{2} f(x)dx = \int_{-2}^{1}(x^2+3x+3)dx + \int_{1}^{2}(5x+2)dx$$

$$= \left[\frac{x^3}{3}+\frac{3x^2}{2}+3x\right]_{-2}^{1} + \left[\frac{5x^2}{2}+2x\right]_{1}^{2}$$

At $$x=1$$: $$\frac{1}{3}+\frac{3}{2}+3 = \frac{2+9+18}{6} = \frac{29}{6}$$

At $$x=-2$$: $$-\frac{8}{3}+6-6 = -\frac{8}{3}$$

First integral: $$\frac{29}{6}+\frac{8}{3} = \frac{29+16}{6} = \frac{45}{6}$$

At $$x=2$$: $$10+4 = 14$$. At $$x=1$$: $$\frac{5}{2}+2 = \frac{9}{2}$$.

Second integral: $$14 - \frac{9}{2} = \frac{19}{2}$$

Total: $$\frac{45}{6} + \frac{19}{2} = \frac{45}{6} + \frac{57}{6} = \frac{102}{6} = 17$$.

The answer is Option (4): $$\boxed{17}$$.