If `f(a+b-x) =f(x)` , then `int_(a)^(b) x f(x) dx ` is equal to

Options

A. `(a+b)/( 2) int_(a)^(b) f(b-x) dx`
B. `(a+b)/( 2) int_(a)^(b) f(a-x) dx`
C. `(b-a)/( 2) int_(a)^(b) f(x) dx`
D. `(a+b)/( 2) int_(a)^(b) f(x) dx`

Answer

To solve the problem, we need to evaluate the integral \( I = \int_{a}^{b} x f(x) \, dx \) given the condition \( f(a + b - x) = f(x) \).

### Step-by-Step Solution:

1. **Define the Integral:**
Let \( I = \int_{a}^{b} x f(x) \, dx \).

2. **Use the Property of Integration:**
We know that \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \). This property allows us to transform the variable in the integral.

3. **Change of Variable:**
We can express the integral \( I \) in terms of \( a + b - x \):
\[ I = \int_{a}^{b} (a + b - x) f(a + b - x) \, dx \]
Since \( f(a + b - x) = f(x) \), we can substitute:
\[ I = \int_{a}^{b} (a + b - x) f(x) \, dx \]

4. **Rewrite the Integral:**
Now we can rewrite this integral:
\[ I = \int_{a}^{b} (a + b) f(x) \, dx - \int_{a}^{b} x f(x) \, dx \]
The second term is just \( I \):
\[ I = (a + b) \int_{a}^{b} f(x) \, dx - I \]

5. **Combine Like Terms:**
Adding \( I \) to both sides gives:
\[ 2I = (a + b) \int_{a}^{b} f(x) \, dx \]

6. **Solve for \( I \):**
Dividing both sides by 2, we find:
\[ I = \frac{(a + b)}{2} \int_{a}^{b} f(x) \, dx \]

### Final Answer:
Thus, the value of the integral \( \int_{a}^{b} x f(x) \, dx \) is:
\[ \int_{a}^{b} x f(x) \, dx = \frac{(a + b)}{2} \int_{a}^{b} f(x) \, dx \]