If f and g are two continuous functions being even and and odd, respectively, then `int_(-a)^(a)(f(x))/(b^(g(x)+1))dx` is equal to (a being any non-zero number and b is positive real number, `b ne 1` )

Options

A. independent of f
B. independent of g
C. independent of both f and g
D. none of these

Explanation

To solve the problem, we need to evaluate the definite integral

\[ I = \int_{-a}^{a} \frac{f(x)}{b^{g(x) + 1}} \, dx \]

where \( f \) is an even function and \( g \) is an odd function.

### Step-by-Step Solution:

1. **Understanding the properties of functions**:
- Since \( f(x) \) is an even function, we have \( f(-x) = f(x) \).
- Since \( g(x) \) is an odd function, we have \( g(-x) = -g(x) \).

2. **Substituting \(-x\) into the integral**:
We will evaluate the integral by substituting \(-x\) for \(x\):
\[ I = \int_{-a}^{a} \frac{f(-x)}{b^{g(-x) + 1}} \, dx \]
Using the properties of \(f\) and \(g\):
\[ I = \int_{-a}^{a} \frac{f(x)}{b^{-g(x) + 1}} \, dx \]

3. **Rewriting the integral**:
The expression \(b^{-g(x) + 1}\) can be rewritten as:
\[ b^{-g(x) + 1} = \frac{b}{b^{g(x)}} \]
Thus, we have:
\[ I = \int_{-a}^{a} \frac{f(x)}{\frac{b}{b^{g(x)}}} \, dx = \int_{-a}^{a} \frac{f(x) b^{g(x}}{b} \, dx \]
This simplifies to:
\[ I = \frac{1}{b} \int_{-a}^{a} f(x) b^{g(x)} \, dx \]

4. **Combining both integrals**:
Now, we can add the two expressions for \(I\):
\[ 2I = \int_{-a}^{a} \frac{f(x)}{b^{g(x) + 1}} \, dx + \int_{-a}^{a} \frac{f(x) b^{g(x)}}{b} \, dx \]
This gives:
\[ 2I = \int_{-a}^{a} f(x) \left( \frac{1}{b^{g(x) + 1}} + \frac{b^{g(x)}}{b} \right) \, dx \]

5. **Simplifying the integral**:
The terms inside the integral can be combined:
\[ 2I = \int_{-a}^{a} f(x) \left( \frac{1 + b^{g(x)}}{b^{g(x) + 1}} \right) \, dx \]

6. **Using the property of even functions**:
Since \(f(x)\) is even, we can simplify the integral:
\[ I = \int_{0}^{a} f(x) \, dx \]
This integral does not depend on \(g\).

### Conclusion:
The final result shows that the integral \(I\) is independent of the function \(g\). Therefore, the answer to the question is:

\[ \text{The integral } I \text{ is independent of } g. \]