If `int_(-1)^(1)f(x)dx=0` then

Options

A. `f(-x)=f(x)`
B. `f(-x)=-f(x)`
C. `f(x)=2f(x)`
D. `f(-x)=2f(x)`

Answer

To solve the problem, we start with the given integral condition:

\[ \int_{-1}^{1} f(x) \, dx = 0 \]

### Step 1: Use the property of definite integrals
We can apply the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx \]

This property is particularly useful because it allows us to relate the function \( f(x) \) with \( f(-x) \).

### Step 2: Express the integral in terms of \( f(-x) \)
Using the property mentioned above, we can write:

\[ \int_{-1}^{1} f(x) \, dx = \int_{-1}^{1} f(-x) \, dx \]

Since we know that:

\[ \int_{-1}^{1} f(x) \, dx = 0 \]

It follows that:

\[ \int_{-1}^{1} f(-x) \, dx = 0 \]

### Step 3: Combine the two integrals
Now, we can combine the two integrals:

\[ \int_{-1}^{1} f(x) \, dx + \int_{-1}^{1} f(-x) \, dx = 0 + 0 = 0 \]

This can be rewritten as:

\[ \int_{-1}^{1} \left( f(x) + f(-x) \right) \, dx = 0 \]

### Step 4: Analyze the integrand
For the integral of \( f(x) + f(-x) \) to equal zero over the symmetric interval \([-1, 1]\), it must be that:

\[ f(x) + f(-x) = 0 \]

This implies:

\[ f(-x) = -f(x) \]

### Conclusion
The conclusion we reach is that the function \( f(x) \) is an odd function.