To determine the integral of the piecewise function $$f$$ over the interval $$[-2, 2]$$, we first ensure that $$f$$ is differentiable on $$\mathbb{R}$$, as given in the problem statement. Differentiability implies continuity, so $$f$$ must also be continuous at $$x=1$$.

The condition for continuity at $$x=1$$ is:

$$x^2 + 3x + a = bx + 2$$ at $$x=1$$.

This simplifies to:

$$1 + 3 + a = b(1) + 2$$

$$\Rightarrow a = b - 2$$

The first derivative of $$f$$ gives us two different expressions depending on the value of $$x$$:

For $$x \leq 1$$, $$f'(x) = 2x + 3$$; and for $$x > 1$$, $$f'(x) = b$$.

For $$f$$ to be differentiable at $$x=1$$, these derivatives must be equal at that point. Setting $$f'(1)$$ from both expressions equal to each other gives $$2(1) + 3 = b$$, thus $$b = 5$$. And from $$a = b - 2$$, we have $$a = 3$$.

Now, we can calculate the integral of $$f$$ over the specified interval:

$$\int\limits_{-2}^1 (x^2 + 3x + 3) \, dx + \int\limits_{1}^2 (5x + 2) \, dx$$

$$\begin{aligned} & =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\\\ & =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\\\ & =6+\frac{3}{2}+12-\frac{5}{2}=17 \end{aligned}$$