Let `f : [-1, 2]to [0, oo)` be a continuous function such that `f(x)=f(1-x), AA x in [-1, 2]`. If `R_(1)=int_(-1)^(2)xf(x)dx` and `R_(2)` are 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)=R_(2)`
D. `3R_(1)=R_(2)`

Answer

To solve the problem, we need to establish the relationship between the integrals \( R_1 \) and \( R_2 \) given the properties of the function \( f \).

### Step-by-Step Solution:

1. **Define the integrals**:
- We have \( R_1 = \int_{-1}^{2} x f(x) \, dx \)
- We also need to find \( R_2 \), which represents the area under the curve \( y = f(x) \) from \( x = -1 \) to \( x = 2 \):
\[ R_2 = \int_{-1}^{2} f(x) \, dx \]

2. **Use the property of the function**:
- We know that \( f(x) = f(1 - x) \) for \( x \in [-1, 2] \). This symmetry will help us evaluate \( R_1 \).

3. **Transform \( R_1 \)**:
- We can split the integral \( R_1 \) using the property of \( f \):
\[ R_1 = \int_{-1}^{2} x f(x) \, dx \]
- Now, we can change the variable in the integral by letting \( u = 1 - x \). Then, \( du = -dx \) and the limits change as follows:
- When \( x = -1 \), \( u = 2 \)
- When \( x = 2 \), \( u = -1 \)
- Thus, we have:
\[ R_1 = \int_{2}^{-1} (1 - u) f(1 - u) (-du) = \int_{-1}^{2} (1 - u) f(u) \, du \]
- Since \( f(1 - u) = f(u) \), we can rewrite this as:
\[ R_1 = \int_{-1}^{2} (1 - x) f(x) \, dx \]

4. **Combine the two expressions for \( R_1 \)**:
- Now we have two expressions for \( R_1 \):
\[ R_1 = \int_{-1}^{2} x f(x) \, dx \quad \text{and} \quad R_1 = \int_{-1}^{2} (1 - x) f(x) \, dx \]
- Adding these two equations gives:
\[ 2R_1 = \int_{-1}^{2} (x + (1 - x)) f(x) \, dx = \int_{-1}^{2} f(x) \, dx = R_2 \]

5. **Final relationship**:
- From the above, we can conclude:
\[ 2R_1 = R_2 \quad \Rightarrow \quad R_1 = \frac{R_2}{2} \]

### Conclusion:
The relationship between \( R_1 \) and \( R_2 \) is:
\[ R_1 = \frac{R_2}{2} \]