Let `f:[-1,2]->[0,oo)` be a continuous function such that `f(x)=f(1-x)fora l lx in [-1,2]dot` Let `R_1=int_(-1)^2xf(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 `R_1=2R_2` (b) `R_1=3R_2` (c) `2R_1=R_2` (d) `3R_1=R_2`

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Given Function**:
We have a continuous function \( f: [-1, 2] \to [0, \infty) \) such that \( f(x) = f(1 - x) \) for all \( x \in [-1, 2] \). This property indicates that the function is symmetric about \( x = 1 \).

2. **Setting Up the Integrals**:
We define:
\[ R_1 = \int_{-1}^{2} x f(x) \, dx \]
and
\[ R_2 = \text{Area under the curve } y = f(x) \text{ from } x = -1 \text{ to } x = 2 = \int_{-1}^{2} f(x) \, dx. \]

3. **Using the Symmetry Property**:
We can use the symmetry property of \( f(x) \) to rewrite \( R_1 \):
\[ R_1 = \int_{-1}^{2} x f(x) \, dx. \]
We can split this integral into two parts:
\[ R_1 = \int_{-1}^{1} x f(x) \, dx + \int_{1}^{2} x f(x) \, dx. \]
For the second integral, we perform a substitution \( x = 1 + t \) where \( t \) goes from \( 0 \) to \( 1 \):
\[ \int_{1}^{2} x f(x) \, dx = \int_{0}^{1} (1 + t) f(1 - t) \, dt = \int_{0}^{1} (1 + t) f(t) \, dt. \]
Thus,
\[ R_1 = \int_{-1}^{1} x f(x) \, dx + \int_{0}^{1} (1 + t) f(t) \, dt. \]

4. **Combining the Integrals**:
Now, we can combine the integrals:
\[ R_1 = \int_{-1}^{1} x f(x) \, dx + \int_{0}^{1} f(t) \, dt + \int_{0}^{1} t f(t) \, dt. \]
Using the symmetry \( f(1 - t) = f(t) \), we can express \( R_1 \) in terms of \( R_2 \).

5. **Finding the Relationship**:
By adding the two integrals \( R_1 \) and its symmetric counterpart, we find:
\[ 2R_1 = \int_{-1}^{2} f(x) \, dx. \]
Since \( R_2 = \int_{-1}^{2} f(x) \, dx \), we have:
\[ 2R_1 = R_2. \]

### Conclusion:
Thus, the relationship between \( R_1 \) and \( R_2 \) is:
\[ 2R_1 = R_2. \]
The correct option is (c) \( 2R_1 = R_2 \).