The antiderivative of every odd function is an

Options

A. even function
B. odd function
C. neither even nor odd
D. none

Answer

To solve the problem, we need to demonstrate that the antiderivative of every odd function is an even function. Let's proceed step by step.

### Step 1: Define an Odd Function
An odd function \( f(x) \) satisfies the property:
\[ f(-x) = -f(x) \]
For our example, let's take \( f(x) = x \).

### Step 2: Verify that \( f(x) \) is Odd
To check if \( f(x) = x \) is odd, we substitute \(-x\) into the function:
\[ f(-x) = -x \]
Since \( f(-x) = -f(x) \), we confirm that \( f(x) = x \) is indeed an odd function.

### Step 3: Find the Antiderivative
Now, we need to find the antiderivative of \( f(x) \):
\[ F(x) = \int f(x) \, dx = \int x \, dx \]
Calculating the integral:
\[ F(x) = \frac{x^2}{2} + C \]
where \( C \) is the constant of integration.

### Step 4: Analyze the Antiderivative
Now, we need to check if \( F(x) \) is an even function. An even function \( F(x) \) satisfies the property:
\[ F(-x) = F(x) \]
Let's calculate \( F(-x) \):
\[ F(-x) = \frac{(-x)^2}{2} + C = \frac{x^2}{2} + C = F(x) \]
Since \( F(-x) = F(x) \), we conclude that \( F(x) \) is an even function.

### Conclusion
Thus, we have shown that the antiderivative of the odd function \( f(x) = x \) is the even function \( F(x) = \frac{x^2}{2} + C \). Therefore, we can conclude that the antiderivative of every odd function is an even function.

### Final Answer
The antiderivative of every odd function is an **even function**.

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