`int_(2-a)^(2+a)f(x)dx` is equal to [where `f(2-alpha)=f(2+alpha) AAalpha in R`
(a) 2`int_2^(2+a)f(x)dx` (b) `2int_0^af(x)dx` (c) `2int_2^2f(x)dx` (d) none of these

Options

A. `2int_(2)^(2+a)f(x)dx`
B. `2int_(0)^(a)f(x)dx`
C. `2int_(2)^(2)f(x)dx`
D. none of these

Explanation

To solve the integral \( \int_{2-a}^{2+a} f(x) \, dx \) given the condition \( f(2 - \alpha) = f(2 + \alpha) \) for all \( \alpha \in \mathbb{R} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Symmetry**:
The condition \( f(2 - \alpha) = f(2 + \alpha) \) indicates that the function \( f(x) \) is symmetric about the line \( x = 2 \). This means that for any \( x \) value that is a distance \( a \) from 2, the function takes the same value.

2. **Setting Up the Integral**:
We need to evaluate the integral:
\[ I = \int_{2-a}^{2+a} f(x) \, dx \]

3. **Changing Variables**:
To exploit the symmetry, we can make a substitution. Let \( x = 2 + u \), then \( dx = du \). The limits change as follows:
- When \( x = 2 - a \), \( u = -a \)
- When \( x = 2 + a \), \( u = a \)

Therefore, the integral becomes:
\[ I = \int_{-a}^{a} f(2 + u) \, du \]

4. **Using the Symmetry**:
Using the symmetry property of \( f \):
\[ f(2 + u) = f(2 - u) \]
Thus, we can rewrite the integral:
\[ I = \int_{-a}^{a} f(2 - u) \, du \]

5. **Combining the Integrals**:
We can combine both forms of the integral:
\[ I = \int_{-a}^{a} f(2 + u) \, du + \int_{-a}^{a} f(2 - u) \, du \]
Since both integrals are equal due to symmetry, we can write:
\[ I = 2 \int_{0}^{a} f(2 + u) \, du \]

6. **Final Result**:
Since \( u \) is just a dummy variable, we can replace \( u \) back to \( x \):
\[ I = 2 \int_{2}^{2+a} f(x) \, dx \]

### Conclusion:
Thus, we conclude that:
\[ \int_{2-a}^{2+a} f(x) \, dx = 2 \int_{2}^{2+a} f(x) \, dx \]
This corresponds to option (a).