Let f and g be continuous fuctions on [0, a] such that `f(x)=f(a-x)" and "g(x)+g(a-x)=4 " then " int_(0)^(a)f(x)g(x)dx` is equal to

Options

A. `4underset0oversetaintf(x)dx`
B. `2underset0oversetaintf(x)dx`
C. `-3underset0oversetaintf(x)dx`
D. `underset0oversetaintf(x)dx`

Answer

To solve the problem, we need to evaluate the integral \( I = \int_0^a f(x) g(x) \, dx \) given the conditions on the functions \( f \) and \( g \).

### Step-by-Step Solution:

1. **Understand the properties of the functions:**
We are given two properties:
- \( f(x) = f(a - x) \) (which means \( f \) is symmetric about \( x = \frac{a}{2} \))
- \( g(x) + g(a - x) = 4 \) (which means the sum of \( g(x) \) and its reflection is constant)

2. **Express \( g(a - x) \):**
From the second property, we can express \( g(a - x) \) as:
\[ g(a - x) = 4 - g(x) \]

3. **Change of variable in the integral:**
We can change the variable in the integral \( I \):
\[ I = \int_0^a f(x) g(x) \, dx \]
By substituting \( x \) with \( a - x \), we have:
\[ I = \int_0^a f(a - x) g(a - x) \, dx \]
Using the properties of \( f \) and \( g \):
\[ I = \int_0^a f(x) g(a - x) \, dx \]
Now substituting for \( g(a - x) \):
\[ I = \int_0^a f(x) (4 - g(x)) \, dx \]

4. **Distributing the integral:**
We can distribute the integral:
\[ I = \int_0^a f(x) \cdot 4 \, dx - \int_0^a f(x) g(x) \, dx \]
This can be rewritten as:
\[ I = 4 \int_0^a f(x) \, dx - I \]

5. **Combine like terms:**
Adding \( I \) to both sides gives:
\[ 2I = 4 \int_0^a f(x) \, dx \]

6. **Solve for \( I \):**
Dividing both sides by 2, we find:
\[ I = 2 \int_0^a f(x) \, dx \]

### Final Result:
Thus, the value of the integral \( \int_0^a f(x) g(x) \, dx \) is:
\[ I = 2 \int_0^a f(x) \, dx \]