If f(6-x)=f(x), for all x, then `1/5 int_2^3 x[f(x)+f(x+1)]dx` is equal to :

Options

A. `int_3^4 f(x+2)dx`
B. `int_3^4 f(x+1)dx`
C. `int_1^2 f(x+1)dx`
D. `int_1^3 f(x)dx`

Explanation

To solve the given problem, we need to evaluate the integral

\[ I = \frac{1}{5} \int_2^3 x [f(x) + f(x+1)] \, dx. \]

Given the property \( f(6-x) = f(x) \), we can utilize this symmetry in our calculations.

### Step-by-Step Solution:

1. **Substituting the Integral**:
We can use the property of definite integrals that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx. \]
Here, \( a = 2 \) and \( b = 3 \), so \( a + b = 5 \). Thus, we can write:
\[ I = \frac{1}{5} \int_2^3 x [f(x) + f(x+1)] \, dx = \frac{1}{5} \int_2^3 x [f(5-x) + f(6-x)] \, dx. \]

2. **Using the Property of \( f \)**:
From the property \( f(6-x) = f(x) \), we can replace \( f(6-x) \) with \( f(x) \):
\[ I = \frac{1}{5} \int_2^3 x [f(5-x) + f(x)] \, dx. \]

3. **Splitting the Integral**:
Now, we can split the integral into two parts:
\[ I = \frac{1}{5} \left( \int_2^3 x f(5-x) \, dx + \int_2^3 x f(x) \, dx \right). \]

4. **Changing Variables in the First Integral**:
For the first integral, we can change the variable by letting \( u = 5 - x \), which gives \( du = -dx \). When \( x = 2 \), \( u = 3 \) and when \( x = 3 \), \( u = 2 \). Thus:
\[ \int_2^3 x f(5-x) \, dx = \int_3^2 (5-u) f(u) (-du) = \int_2^3 (5-u) f(u) \, du. \]
This simplifies to:
\[ \int_2^3 (5-u) f(u) \, du = \int_2^3 5f(u) \, du - \int_2^3 u f(u) \, du. \]

5. **Combining the Integrals**:
Now substituting back, we have:
\[ I = \frac{1}{5} \left( \int_2^3 (5f(u) - u f(u)) \, du + \int_2^3 x f(x) \, dx \right). \]
Notice that \( \int_2^3 u f(u) \, du \) cancels out:
\[ I = \frac{1}{5} \left( 5 \int_2^3 f(x) \, dx \right) = \int_2^3 f(x) \, dx. \]

6. **Final Result**:
Therefore, the final result is:
\[ I = \int_2^3 f(x) \, dx. \]