If `f(3-x)= f(x)`, then `int_(1)^(2) xf(x) dx` is equal to

Options

A. `(3)/(2) int_(1)^(2) f(2-x) dx`
B. `(3)/(2) int_(1)^(2) f(x) dx`
C. `(1)/(2) int_(1)^(2) f(x) dx`
D. None of these

Answer

To solve the integral \( I = \int_{1}^{2} x f(x) \, dx \) given the condition \( f(3-x) = f(x) \), we can use a property of definite integrals. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_{1}^{2} x f(x) \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property of definite integrals which states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]
In our case, \( a = 1 \) and \( b = 2 \), so \( a + b = 3 \). Therefore, we can write:
\[ I = \int_{1}^{2} x f(x) \, dx = \int_{1}^{2} (3 - x) f(3 - x) \, dx \]

### Step 3: Substitute the Given Condition
Since we know \( f(3 - x) = f(x) \), we can replace \( f(3 - x) \) with \( f(x) \):
\[ I = \int_{1}^{2} (3 - x) f(x) \, dx \]

### Step 4: Expand the Integral
Now, we can expand the integral:
\[ I = \int_{1}^{2} 3f(x) \, dx - \int_{1}^{2} x f(x) \, dx \]
This gives us:
\[ I = 3 \int_{1}^{2} f(x) \, dx - I \]

### Step 5: Solve for \( I \)
Now we can add \( I \) to both sides:
\[ I + I = 3 \int_{1}^{2} f(x) \, dx \]
\[ 2I = 3 \int_{1}^{2} f(x) \, dx \]
Dividing both sides by 2:
\[ I = \frac{3}{2} \int_{1}^{2} f(x) \, dx \]

### Conclusion
Thus, the value of the integral \( I \) is:
\[ I = \frac{3}{2} \int_{1}^{2} f(x) \, dx \]