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

Options

A. `(a+b)overset(b)underset(a)int f(x)dx`
B. `(1)/(2)(a+b)overset(b)underset(a)intf(x)dx`
C. `(b-a)overset(b)underset(a)int f(x)dx`
D. `(1)/(2)(b-a)overset(b-a)overset(b)underset(a)int f(x)dx`

Explanation

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

### Step-by-Step Solution:

1. **Define the Integral**:
Let \( I = \int_a^b x f(x) \, dx \).

2. **Use the Property of the Function**:
Since \( f(x) = f(a + b - x) \), we can substitute \( x \) with \( a + b - x \) in the integral. Thus, we can write:
\[ I = \int_a^b (a + b - x) f(a + b - x) \, dx \]

3. **Change of Variables**:
Now, we perform a change of variable. Let \( u = a + b - x \). Then, \( du = -dx \). When \( x = a \), \( u = b \) and when \( x = b \), \( u = a \). Therefore, we can rewrite the integral as:
\[ I = \int_b^a (a + b - u) f(u) (-du) = \int_a^b (a + b - u) f(u) \, du \]

4. **Rewrite the Integral**:
Thus, we have:
\[ I = \int_a^b (a + b - x) f(x) \, dx \]

5. **Combine the Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_a^b x f(x) \, dx \]
\[ I = \int_a^b (a + b - x) f(x) \, dx \]
Adding these two equations gives:
\[ 2I = \int_a^b \left( x f(x) + (a + b - x) f(x) \right) dx \]

6. **Simplify the Expression**:
This simplifies to:
\[ 2I = \int_a^b (a + b) f(x) \, dx \]

7. **Solve for \( I \)**:
Dividing both sides by 2, we obtain:
\[ I = \frac{1}{2} (a + b) \int_a^b f(x) \, dx \]

### Final Result:
Thus, the value of the integral \( \int_a^b x f(x) \, dx \) is:
\[ \int_a^b x f(x) \, dx = \frac{1}{2} (a + b) \int_a^b f(x) \, dx \]