If `f(-x)+f(x)=0` then `int_a^x f(t) dt` is

Options

A. An odd function
B. An even function
C. A periodic function
D. None of these

Answer

To solve the problem, we need to analyze the given condition and the integral.

### Step-by-Step Solution:

1. **Understanding the Condition**:
We are given that \( f(-x) + f(x) = 0 \). This implies that:
\[ f(-x) = -f(x) \]
This means that \( f(x) \) is an odd function.

**Hint**: Recognize that the property \( f(-x) = -f(x) \) defines odd functions.

2. **Defining the Integral**:
Let:
\[ \pi(x) = \int_a^x f(t) \, dt \]
We need to analyze \( \pi(-x) \).

**Hint**: Define the integral clearly and denote it with a simple function name.

3. **Calculating \( \pi(-x) \)**:
We can express \( \pi(-x) \) as follows:
\[ \pi(-x) = \int_a^{-x} f(t) \, dt \]
To evaluate this integral, we perform a change of variable. Let \( t = -y \), then \( dt = -dy \). The limits change as follows: when \( t = a \), \( y = -a \) and when \( t = -x \), \( y = x \). Thus:
\[ \pi(-x) = \int_{-a}^{x} f(-y)(-dy) = \int_{-a}^{x} -f(y) \, dy = -\int_{-a}^{x} f(y) \, dy \]

**Hint**: Use substitution to change the variable in the integral.

4. **Rearranging the Integral**:
Now we can express \( \pi(-x) \):
\[ \pi(-x) = -\left( \int_{-a}^{a} f(y) \, dy + \int_a^x f(y) \, dy \right) \]
This can be simplified to:
\[ \pi(-x) = -\int_{-a}^{a} f(y) \, dy - \int_a^x f(y) \, dy \]

**Hint**: Break down the integral into parts that can be analyzed separately.

5. **Using the Property of Odd Functions**:
Since \( f(y) \) is an odd function, the integral from \(-a\) to \(a\) will be zero:
\[ \int_{-a}^{a} f(y) \, dy = 0 \]
Therefore:
\[ \pi(-x) = -\int_a^x f(y) \, dy \]

**Hint**: Recall that the integral of an odd function over symmetric limits is zero.

6. **Conclusion**:
Since \( \pi(-x) = -\pi(x) \), we conclude that \( \pi(x) \) is an odd function.

**Final Answer**: The integral \( \int_a^x f(t) \, dt \) is an odd function.