Let `f:[1,2] to [0,oo)` be a continuous function such that `f(x)=f(1-x)` for all `x in [-1,2]. ` Let `R_(1)=int_(-1)^(2) xf(x) dx,` and `R_(2)` be the area of the region bounded by `y=f(x),x=-1,x=2` and the x-axis . Then,

Options

A. `R_(1)=2R_(2)`
B. `R_(1)=3R_(2)`
C. `2R_(1)=3R_(2)`
D. `3R_(1)=R_(2)`

Explanation

To solve the problem, we need to evaluate the integrals \( R_1 \) and \( R_2 \) based on the properties of the function \( f \) and the given conditions.

### Step-by-step Solution:

1. **Understanding the Function**:
We have a continuous function \( f: [1, 2] \to [0, \infty) \) such that \( f(x) = f(1 - x) \) for all \( x \) in the interval \([-1, 2]\). This symmetry will be useful in our calculations.

2. **Defining \( R_1 \)**:
The first integral \( R_1 \) is defined as:
\[ R_1 = \int_{-1}^{2} x f(x) \, dx \]

3. **Using the Symmetry Property**:
We can use the property of the function to rewrite \( R_1 \). According to the property, we can substitute \( x \) with \( 1 - x \):
\[ R_1 = \int_{-1}^{2} x f(x) \, dx = \int_{-1}^{2} (1 - (1 - x)) f(1 - x) \, dx \]
Simplifying this gives:
\[ R_1 = \int_{-1}^{2} (1 - x) f(1 - x) \, dx \]

4. **Changing the Variable**:
To evaluate this integral, we change the variable. Let \( u = 1 - x \), then \( du = -dx \). The limits change as follows:
- When \( x = -1 \), \( u = 2 \)
- When \( x = 2 \), \( u = -1 \)

Thus, we have:
\[ R_1 = \int_{2}^{-1} (1 - (1 - u)) f(u) (-du) = \int_{-1}^{2} u f(u) \, du \]

5. **Combining the Integrals**:
Now we can combine the two expressions for \( R_1 \):
\[ R_1 = \int_{-1}^{2} x f(x) \, dx = \int_{-1}^{2} (1 - x) f(x) \, dx \]
This gives us:
\[ 2R_1 = \int_{-1}^{2} f(x) \, dx \]

6. **Defining \( R_2 \)**:
The second integral \( R_2 \) is defined as the area under the curve \( y = f(x) \) from \( x = -1 \) to \( x = 2 \):
\[ R_2 = \int_{-1}^{2} f(x) \, dx \]

7. **Relating \( R_1 \) and \( R_2 \)**:
From the previous steps, we have:
\[ 2R_1 = R_2 \]

### Conclusion:
Thus, we conclude that:
\[ R_2 = 2R_1 \]