If f(x) and g(x) are continuous functions satisfying `f(x)= f(a-x) and g(x) +g(a-x)=2`, then `int_(0)^(a) f(x) g(x) dx` is equal to

Options

A. `int_(0)^(a) g(x) dx`
B. `int_(0)^(a) f(x) dx`
C. 0
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Understanding the Properties**:
- We know that \( f(x) = f(a - x) \). This means that \( f(x) \) is symmetric about \( x = \frac{a}{2} \).
- We also know that \( g(x) + g(a - x) = 2 \). This implies that \( g(x) \) is symmetric around \( x = \frac{a}{2} \) as well, leading to \( g(a - x) = 2 - g(x) \).

2. **Substituting in the Integral**:
- We can rewrite the integral \( I \):
\[ I = \int_0^a f(x) g(x) \, dx \]
- Now, we will use the substitution \( x = a - u \). Thus, \( dx = -du \) and the limits change from \( x = 0 \) to \( x = a \) becoming \( u = a \) to \( u = 0 \):
\[ I = \int_a^0 f(a - u) g(a - u) (-du) = \int_0^a f(a - u) g(a - u) \, du \]

3. **Applying the Properties**:
- Using the properties of \( f \) and \( g \):
\[ I = \int_0^a f(a - u) g(a - u) \, du = \int_0^a f(u) g(a - u) \, du \]
- Since \( g(a - u) = 2 - g(u) \):
\[ I = \int_0^a f(u) (2 - g(u)) \, du \]

4. **Expanding the Integral**:
- We can expand this integral:
\[ I = \int_0^a 2f(u) \, du - \int_0^a f(u) g(u) \, du \]
- Let \( \int_0^a f(u) g(u) \, du = I \):
\[ I = 2 \int_0^a f(u) \, du - I \]

5. **Solving for \( I \)**:
- Rearranging gives:
\[ 2I = 2 \int_0^a f(u) \, du \]
- Dividing both sides by 2:
\[ I = \int_0^a f(u) \, du \]

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