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

Options

A. `(a-b)/(2)int_(a)^(b)f(x)dx`
B. `(a+b)/(2)int_(a)^(b)f(x)dx`
C. 0
D. `b-a`

Answer

To solve the integral \( I = \int_a^b x f(x) \, dx \) given the property \( f(a+b-x) = f(x) \), we can follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_a^b x f(x) \, dx \]

### Step 2: Use the property of the function
Using the property \( f(a+b-x) = f(x) \), we can change the variable in the integral. Let \( u = a + b - x \). Then, we have:
\[ du = -dx \quad \text{and when } x = a, \, u = a + b - a = b; \quad \text{when } x = b, \, u = a + b - b = a \]
Thus, the limits of integration change from \( a \) to \( b \) to \( b \) to \( a \) (which we can flip back to \( a \) to \( b \) with a negative sign):
\[ I = \int_b^a (a+b-u) f(a+b-u) (-du) = \int_a^b (a+b-u) f(u) \, du \]

### Step 3: Rewrite the integral
Now we can rewrite the integral:
\[ I = \int_a^b (a+b-u) f(u) \, du \]
This can be split into two integrals:
\[ I = \int_a^b (a+b) f(u) \, du - \int_a^b u f(u) \, du \]
Since \( \int_a^b u f(u) \, du \) is just \( I \), we have:
\[ I = (a+b) \int_a^b f(u) \, du - I \]

### Step 4: Solve for \( I \)
Now, we can add \( I \) to both sides:
\[ 2I = (a+b) \int_a^b f(u) \, du \]
Dividing both sides by 2 gives:
\[ I = \frac{a+b}{2} \int_a^b f(x) \, dx \]

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