`f: R rarr R, g: R rarr R` are continuous functions. The value of integral `int_(-pi//2)^(pi//2) [f(x) + f(-x)] [g(x) -g(-x)] dx` is

Options

A. `pi`
B. 1
C. `-1`
D. 0

Answer

To solve the integral \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x) + f(-x)] [g(x) - g(-x)] \, dx \), we will use the properties of definite integrals and the symmetry of the functions involved.

### Step-by-Step Solution:

1. **Define the Integral:**
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x) + f(-x)] [g(x) - g(-x)] \, dx \]

2. **Use the Property of Definite Integrals:**
We can use the property of definite integrals that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Here, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \), so \( a + b = 0 \).

3. **Change of Variables:**
Substitute \( x \) with \( -x \):
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(-x) + f(x)] [g(-x) - g(x)] \, (-dx) \]
This simplifies to:
\[ I = -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(-x) + f(x)] [g(-x) - g(x)] \, dx \]

4. **Rearranging the Integral:**
We can rewrite the integral:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x) + f(-x)] [g(x) - g(-x)] \, dx \]
This gives us:
\[ I = -I \]

5. **Solving for \( I \):**
From the equation \( I = -I \), we can add \( I \) to both sides:
\[ 2I = 0 \implies I = 0 \]

### Final Result:
The value of the integral is:
\[ I = 0 \]