If the function `f(x)` is symmetric about the line `x=3`, then the value of the integral `I=int_(-2)^(8)(f(x))/(f(x)+f(6-x))dx` is

Options

A. 0
B. 5
C. 10
D. 16

Answer

To solve the integral \( I = \int_{-2}^{8} \frac{f(x)}{f(x) + f(6-x)} \, dx \), given that the function \( f(x) \) is symmetric about the line \( x = 3 \), we can follow these steps:

### Step 1: Understanding Symmetry
Since \( f(x) \) is symmetric about \( x = 3 \), we have:
\[ f(6 - x) = f(x) \]
This means that for any \( x \), the value of the function at \( 6 - x \) is the same as the value at \( x \).

### Step 2: Rewrite the Integral
We can rewrite the integral using the property of definite integrals:
\[ I = \int_{-2}^{8} \frac{f(x)}{f(x) + f(6-x)} \, dx \]
Substituting \( x \) with \( 6 - x \):
\[ I = \int_{-2}^{8} \frac{f(6-x)}{f(6-x) + f(x)} \, dx \]
Using the symmetry property, we can replace \( f(6-x) \) with \( f(x) \):
\[ I = \int_{-2}^{8} \frac{f(x)}{f(x) + f(x)} \, dx = \int_{-2}^{8} \frac{f(x)}{2f(x)} \, dx = \int_{-2}^{8} \frac{1}{2} \, dx \]

### Step 3: Calculate the Integral
Now, we can calculate the integral:
\[ I = \frac{1}{2} \int_{-2}^{8} 1 \, dx \]
The integral of \( 1 \) from \( -2 \) to \( 8 \) is simply the length of the interval:
\[ \int_{-2}^{8} 1 \, dx = 8 - (-2) = 10 \]
Thus,
\[ I = \frac{1}{2} \cdot 10 = 5 \]

### Conclusion
The value of the integral \( I \) is:
\[ \boxed{5} \]