Let $$f : [-2, 3] \rightarrow [0, \infty)$$ be a continuous function such that $$f(1-x) = f(x)$$ for all $$x \in [-2, 3]$$. If $$R_1$$ is the numerical value of the area of the region bounded by $$y = f(x)$$, $$x = -2$$, $$x = 3$$ and the axis of x and $$R_2 = \int_{-2}^{3} xf(x)dx$$, then :

We are given a continuous function $$f : [-2, 3] \rightarrow [0, \infty)$$ such that $$f(1 - x) = f(x)$$ for all $$x \in [-2, 3]$$. This symmetry condition means the graph of $$f(x)$$ is symmetric about the line $$x = \frac{1}{2}$$. The interval $$[-2, 3]$$ is symmetric about $$x = \frac{1}{2}$$ because the midpoint is $$\frac{-2 + 3}{2} = \frac{1}{2}$$.

We define $$R_1$$ as the area of the region bounded by $$y = f(x)$$, the lines $$x = -2$$, $$x = 3$$, and the x-axis. Since $$f(x) \geq 0$$, this area is given by the integral: $$R_1 = \int_{-2}^{3} f(x) dx.$$

We also define $$R_2$$ as: $$R_2 = \int_{-2}^{3} x f(x) dx.$$

To find a relation between $$R_1$$ and $$R_2$$, we use the symmetry $$f(1 - x) = f(x)$$. Consider the integral for $$R_2$$ and apply the substitution $$u = 1 - x$$. Then, when $$x = -2$$, $$u = 1 - (-2) = 3$$, and when $$x = 3$$, $$u = 1 - 3 = -2$$. Also, $$du = -dx$$, so $$dx = -du$$.

Substituting into $$R_2$$: $$R_2 = \int_{-2}^{3} x f(x) dx = \int_{3}^{-2} (1 - u) f(1 - u) (-du).$$

Reversing the limits of integration removes the negative sign: $$R_2 = \int_{-2}^{3} (1 - u) f(1 - u) du.$$

Using the symmetry $$f(1 - u) = f(u)$$, we get: $$R_2 = \int_{-2}^{3} (1 - u) f(u) du.$$

Since the variable of integration is a dummy variable, we can replace $$u$$ with $$x$$: $$R_2 = \int_{-2}^{3} (1 - x) f(x) dx.$$

We now have two expressions for $$R_2$$: $$R_2 = \int_{-2}^{3} x f(x) dx \quad \text{(original definition)},$$ $$R_2 = \int_{-2}^{3} (1 - x) f(x) dx \quad \text{(from symmetry)}.$$

Adding these two equations together: $$2R_2 = \int_{-2}^{3} x f(x) dx + \int_{-2}^{3} (1 - x) f(x) dx = \int_{-2}^{3} \left[ x + (1 - x) \right] f(x) dx = \int_{-2}^{3} 1 \cdot f(x) dx.$$

The integral simplifies to: $$2R_2 = \int_{-2}^{3} f(x) dx = R_1.$$

Therefore, we have: $$R_1 = 2R_2.$$

Comparing with the options: - Option A: $$3R_1 = 2R_2$$ implies $$R_1 = \frac{2}{3}R_2$$, which does not match. - Option B: $$2R_1 = 3R_2$$ implies $$R_1 = \frac{3}{2}R_2$$, which does not match. - Option C: $$R_1 = R_2$$, which does not match. - Option D: $$R_1 = 2R_2$$, which matches.

Hence, the correct answer is Option D.