The integral `int_(0)^(a) (g(x))/(f(x)+f(a-x))dx` vanishes, if

Options

A. g(x) is odd
B. f(x) =f(a-x)
C. g(x)=-g(a-x)
D. f(a-x)g=g(x)

Explanation

To solve the integral \( I = \int_{0}^{a} \frac{g(x)}{f(x) + f(a-x)} \, dx \) and determine the condition under which it vanishes, we can follow these steps:

### Step-by-Step Solution

1. **Define the Integral**:
Let \( I = \int_{0}^{a} \frac{g(x)}{f(x) + f(a-x)} \, dx \).

2. **Change of Variable**:
We will use the substitution \( x = a - t \). Then, \( dx = -dt \). When \( x = 0 \), \( t = a \) and when \( x = a \), \( t = 0 \). Thus, we can rewrite the integral:
\[ I = \int_{a}^{0} \frac{g(a - t)}{f(a - t) + f(t)} (-dt) = \int_{0}^{a} \frac{g(a - t)}{f(a - t) + f(t)} \, dt \]

3. **Rearranging the Integral**:
Now we have two expressions for \( I \):
\[ I = \int_{0}^{a} \frac{g(x)}{f(x) + f(a-x)} \, dx \]
and
\[ I = \int_{0}^{a} \frac{g(a - x)}{f(a - x) + f(x)} \, dx \]

4. **Adding the Two Integrals**:
Adding these two equations gives:
\[ 2I = \int_{0}^{a} \left( \frac{g(x)}{f(x) + f(a-x)} + \frac{g(a - x)}{f(a - x) + f(x)} \right) dx \]
Since the denominators are the same, we can combine the numerators:
\[ 2I = \int_{0}^{a} \frac{g(x) + g(a - x)}{f(x) + f(a-x)} \, dx \]

5. **Condition for Vanishing**:
For \( I \) to vanish, the integral must equal zero:
\[ I = 0 \implies \int_{0}^{a} \frac{g(x) + g(a - x)}{f(x) + f(a-x)} \, dx = 0 \]
This integral will be zero if the numerator \( g(x) + g(a - x) = 0 \) for all \( x \) in the interval \( [0, a] \).

6. **Conclusion**:
Therefore, the condition for the integral \( I \) to vanish is:
\[ g(x) = -g(a - x) \]

### Final Answer
The integral \( I = \int_{0}^{a} \frac{g(x)}{f(x) + f(a-x)} \, dx \) vanishes if \( g(x) = -g(a - x) \).