Consider the following statements:
`S_(1)`: The antiderivative of every function is an odd function.
True or False?

Answer

To determine whether the statement "The antiderivative of every function is an odd function" is true or false, we can follow these steps:

### Step 1: Understand the Definition of an Odd Function
An odd function \( f(x) \) satisfies the condition:
\[ f(-x) = -f(x) \]
for all \( x \) in its domain.

### Step 2: Consider the Antiderivative
The antiderivative of a function \( f(x) \) is defined as:
\[ F(x) = \int f(x) \, dx \]
where \( F'(x) = f(x) \).

### Step 3: Choose a Simple Function
Let's take a simple function to analyze. For example, let:
\[ f(x) = x \]

### Step 4: Find the Antiderivative
Now, we find the antiderivative of \( f(x) \):
\[ F(x) = \int x \, dx = \frac{x^2}{2} + C \]
where \( C \) is the constant of integration.

### Step 5: Check if the Antiderivative is Odd
Now we need to check if \( F(x) \) is an odd function:
\[ F(-x) = \frac{(-x)^2}{2} + C = \frac{x^2}{2} + C \]
Now, we compare \( F(-x) \) with \( -F(x) \):
\[ -F(x) = -\left(\frac{x^2}{2} + C\right) = -\frac{x^2}{2} - C \]
Clearly, \( F(-x) \neq -F(x) \) unless \( C = 0 \).

### Step 6: Conclusion
Since we found that the antiderivative \( F(x) = \frac{x^2}{2} + C \) is not an odd function for all values of \( C \), we conclude that the statement "The antiderivative of every function is an odd function" is **False**.

### Summary
The statement is false because not all antiderivatives are odd functions. For example, the antiderivative of \( f(x) = x \) is \( F(x) = \frac{x^2}{2} + C \), which is not odd unless \( C = 0 \).