Let f (x) be a conitnuous function defined on [0,a] such that `f(a-x)=f(x)"for all" x in [ 0,a]`. If ` int_(0)^(a//2) f(x) dx=alpha,` then `int _(0)^(a) f(x) dx` is equal to

Options

A. `alpha`
B. `2 alpha`
C. 0
D. none of these

Explanation

To solve the problem, we need to evaluate the integral \( \int_0^a f(x) \, dx \) given the condition \( f(a - x) = f(x) \) for all \( x \) in the interval \([0, a]\) and the information that \( \int_0^{a/2} f(x) \, dx = \alpha \).

### Step-by-Step Solution:

1. **Define the Integral**:
Let \( I = \int_0^a f(x) \, dx \).

2. **Use the Property of the Function**:
Given that \( f(a - x) = f(x) \), we can change the variable in the integral:
\[ \int_0^a f(a - x) \, dx = \int_0^a f(x) \, dx = I \]

3. **Change of Variable**:
Let \( u = a - x \). Then, \( du = -dx \). When \( x = 0 \), \( u = a \) and when \( x = a \), \( u = 0 \). Thus, we have:
\[ \int_0^a f(a - x) \, dx = \int_a^0 f(u) (-du) = \int_0^a f(u) \, du = I \]

4. **Split the Integral**:
Now we can express \( I \) as:
\[ I = \int_0^{a/2} f(x) \, dx + \int_{a/2}^a f(x) \, dx \]

5. **Evaluate the Second Integral**:
For the second integral, we can use the property of the function again:
\[ \int_{a/2}^a f(x) \, dx = \int_{a/2}^a f(a - (a - x)) \, dx = \int_{a/2}^a f(a - x) \, dx \]
By changing the variable \( v = a - x \), we have \( dv = -dx \), and when \( x = a/2 \), \( v = a/2 \) and when \( x = a \), \( v = 0 \):
\[ \int_{a/2}^a f(a - x) \, dx = \int_{a/2}^0 f(v) (-dv) = \int_0^{a/2} f(v) \, dv = \int_0^{a/2} f(x) \, dx \]

6. **Combine the Results**:
Thus, we can write:
\[ I = \int_0^{a/2} f(x) \, dx + \int_0^{a/2} f(x) \, dx = 2 \int_0^{a/2} f(x) \, dx \]

7. **Substitute the Given Value**:
We know that \( \int_0^{a/2} f(x) \, dx = \alpha \), so:
\[ I = 2 \alpha \]

8. **Final Result**:
Therefore, the value of the integral \( \int_0^a f(x) \, dx \) is:
\[ \int_0^a f(x) \, dx = 2\alpha \]

### Conclusion:
The final answer is:
\[ \int_0^a f(x) \, dx = 2\alpha \]