If f(x) and g(x) are two continuous functions defined on [-a,a] then the the value of `int_(-a)^(a) {f(x)f+(-x) } {g(x)-g(-x)}dx` is,

Options

A. 2a
B. f(a) g(a)
C. a
D. none of these

Explanation

To solve the integral \( I = \int_{-a}^{a} (f(x) + f(-x))(g(x) - g(-x)) \, dx \), we will follow these steps:

### Step 1: Define the Integral
Let:
\[ I = \int_{-a}^{a} (f(x) + f(-x))(g(x) - g(-x)) \, dx \]

### Step 2: Apply the Property of Integrals
We can use the property of integrals that states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, we will apply this property to the integral \( I \).

### Step 3: Substitute in the Integral
Using the property, we can rewrite the integral:
\[ I = \int_{-a}^{a} (f(-x) + f(x))(g(-x) - g(x)) \, dx \]

### Step 4: Change of Variables
Now, let’s change the variable \( x \) to \( -x \):
\[ I = \int_{-a}^{a} (f(-x) + f(x))(g(-x) - g(x)) \, dx \]
This gives us a new expression for \( I \):
\[ I = \int_{-a}^{a} (f(-x) + f(x))(g(-x) - g(x)) \, dx \]

### Step 5: Combine the Two Expressions
Now we have two expressions for \( I \):
1. \( I = \int_{-a}^{a} (f(x) + f(-x))(g(x) - g(-x)) \, dx \)
2. \( I = \int_{-a}^{a} (f(-x) + f(x))(g(-x) - g(x)) \, dx \)

Adding these two equations:
\[ 2I = \int_{-a}^{a} (f(x) + f(-x)) \left( (g(x) - g(-x)) + (g(-x) - g(x)) \right) \, dx \]

### Step 6: Simplify the Expression
Notice that the terms \( (g(x) - g(-x)) + (g(-x) - g(x)) \) cancel each other out:
\[ (g(x) - g(-x)) + (g(-x) - g(x)) = 0 \]
Thus, we have:
\[ 2I = \int_{-a}^{a} (f(x) + f(-x)) \cdot 0 \, dx = 0 \]

### Step 7: Solve for \( I \)
Since \( 2I = 0 \), we conclude:
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \int_{-a}^{a} (f(x) + f(-x))(g(x) - g(-x)) \, dx = 0 \]