If a is a fixed real number such that f(a-x)+f(a+x)=0, then `int_(0)^(2a) f(x)` dx=

Options

A. `(a)/(2)`
B. 0
C. `-(a)/(2)`
D. 2a

Answer

To solve the problem, we need to evaluate the integral \( \int_{0}^{2a} f(x) \, dx \) given the condition \( f(a - x) + f(a + x) = 0 \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We start with the condition:
\[ f(a - x) + f(a + x) = 0 \]
This implies that:
\[ f(a - x) = -f(a + x) \]

2. **Substituting \( x \) with \( a - x \)**:
Let's replace \( x \) with \( a - x \) in the original equation:
\[ f(a - (a - x)) + f(a + (a - x)) = 0 \]
Simplifying this gives:
\[ f(x) + f(2a - x) = 0 \]
Thus, we can conclude:
\[ f(2a - x) = -f(x) \]

3. **Setting Up the Integral**:
Now we can evaluate the integral:
\[ \int_{0}^{2a} f(x) \, dx \]
We can split this integral into two parts:
\[ \int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{a}^{2a} f(x) \, dx \]

4. **Changing the Variable in the Second Integral**:
In the second integral, we will use the substitution \( x = 2a - u \), where \( dx = -du \). When \( x = a \), \( u = a \) and when \( x = 2a \), \( u = 0 \). Thus:
\[ \int_{a}^{2a} f(x) \, dx = \int_{a}^{0} f(2a - u) (-du) = \int_{0}^{a} f(2a - u) \, du \]
Using our earlier result \( f(2a - u) = -f(u) \), we get:
\[ \int_{a}^{2a} f(x) \, dx = \int_{0}^{a} -f(u) \, du = -\int_{0}^{a} f(x) \, dx \]

5. **Combining the Results**:
Now substituting back into our integral:
\[ \int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + (-\int_{0}^{a} f(x) \, dx) = 0 \]

### Final Result:
Thus, we conclude:
\[ \int_{0}^{2a} f(x) \, dx = 0 \]