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(x) dx`
C. `(b-a)/(2) int_(a)^(b) f(x) dx`
D. None of these

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:

**Step 1: Define the integral.**
Let:
\[ I = \int_a^b x f(x) \, dx \]

**Step 2: Use the property of definite integrals.**
We can use the substitution \( u = a + b - x \). Then, \( du = -dx \). When \( x = a \), \( u = a + b - a = b \), and when \( x = b \), \( u = a + b - b = a \). Thus, we can rewrite the integral as:
\[ I = \int_b^a (a + b - u) f(a + b - u) (-du) \]
Reversing the limits of integration gives:
\[ I = \int_a^b (a + b - u) f(a + b - u) \, du \]

**Step 3: Substitute the condition.**
Using the condition \( f(a+b-u) = f(u) \), we can replace \( f(a+b-u) \) with \( f(u) \):
\[ I = \int_a^b (a + b - u) f(u) \, du \]

**Step 4: Expand the integral.**
Now, we can expand the integral:
\[ I = \int_a^b (a + b) f(u) \, du - \int_a^b u f(u) \, du \]
This simplifies to:
\[ I = (a + b) \int_a^b f(u) \, du - I \]

**Step 5: Solve for \( I \).**
Adding \( I \) to both sides gives:
\[ 2I = (a + b) \int_a^b f(u) \, du \]
Thus, we have:
\[ I = \frac{(a + b)}{2} \int_a^b f(x) \, dx \]

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