Relation Between Definite Integral and Area Bounded by Continuous Function

To find the relationship between $R_1$ and $R_2$, we begin by writing their integral definitions.

Since $f(x) \ge 0$ for all $x \in [-1, 2]$, the area $R_2$ of the region bounded by $y = f(x)$, the lines $x = -1$, $x = 2$, and the $x$-axis is given by:
$R_2 = \int_{-1}^{2} f(x) \, dx$

We are given:
$R_1 = \int_{-1}^{2} x f(x) \, dx \quad \text{--- (1)}$

Using the standard definite integral property $\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a + b - x) \, dx$, where $a = -1$ and $b = 2$, we have $a + b = -1 + 2 = 1$.

Applying this substitution to the integral for $R_1$:
$R_1 = \int_{-1}^{2} (1 - x) f(1 - x) \, dx$

We are given that $f(x) = f(1 - x)$ for all $x \in [-1, 2]$. Substituting $f(1 - x) = f(x)$ gives:
$R_1 = \int_{-1}^{2} (1 - x) f(x) \, dx$

Splitting the integral:
$R_1 = \int_{-1}^{2} f(x) \, dx - \int_{-1}^{2} x f(x) \, dx$

Using the definitions of $R_1$ and $R_2$:
$R_1 = R_2 - R_1$

Adding $R_1$ to both sides:
$2R_1 = R_2$

Thus, the correct option is (C).