If f(a+x)=f(x), then `int_(0)^(na) f(x)dx` is equal to `(n in N)`

Options

A. `(n-1)overset(a)underset(0)int f(x)dx`
B. `noverset(a)underset(0)int f(x)dx`
C. `overset((n-1a))underset(0)int f(x)dx`
D. none of these

Answer

To solve the problem, we need to analyze the given condition and the integral we want to evaluate.

### Step-by-Step Solution:

1. **Understanding the Given Condition:**
We are given that \( f(a + x) = f(x) \). This implies that the function \( f(x) \) is periodic with a period of \( a \). This means that the function repeats its values every \( a \) units.

**Hint:** Recognize that the condition indicates periodicity, which can be used to simplify the integral.

2. **Setting Up the Integral:**
We want to evaluate the integral \( \int_{0}^{na} f(x) \, dx \) for \( n \in \mathbb{N} \).

**Hint:** Identify the limits of integration and the periodic nature of the function to simplify the integral.

3. **Using the Periodicity:**
Since \( f(x) \) is periodic with period \( a \), we can express the integral over the interval \( [0, na] \) as a sum of integrals over the intervals of length \( a \):
\[ \int_{0}^{na} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{a}^{2a} f(x) \, dx + \int_{2a}^{3a} f(x) \, dx + \ldots + \int_{(n-1)a}^{na} f(x) \, dx \]

**Hint:** Break the integral into segments that match the period of the function.

4. **Evaluating Each Integral:**
Each integral over one period is the same due to the periodicity:
\[ \int_{ka}^{(k+1)a} f(x) \, dx = \int_{0}^{a} f(x) \, dx \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
Therefore, we can write:
\[ \int_{0}^{na} f(x) \, dx = n \int_{0}^{a} f(x) \, dx \]

**Hint:** Use the fact that the integral over each period is equal to the integral over the first period.

5. **Final Result:**
Thus, we conclude that:
\[ \int_{0}^{na} f(x) \, dx = n \int_{0}^{a} f(x) \, dx \]

**Hint:** The final expression shows how the integral over multiple periods relates to the integral over a single period.

### Conclusion:
The integral \( \int_{0}^{na} f(x) \, dx \) equals \( n \int_{0}^{a} f(x) \, dx \), confirming that the periodic nature of \( f(x) \) allows us to simplify the evaluation of the integral over multiple periods.