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To solve this, we are asked to find the Fourier series for the function \( f(x) = x \) on the interval \( 0 < x < 2\pi \).

### General Formula for Fourier Series:
The Fourier series of a function \( f(x) \) defined on the interval \( [0, 2\pi] \) is given by:

\[
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)
\]

Where the coefficients \( a_0 \), \( a_n \), and \( b_n \) are defined as:
- \( a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \, dx \)
- \( a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx \)
- \( b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) \, dx \)

### Step 1: Compute \( a_0 \)
\[
a_0 = \frac{1}{\pi} \int_0^{2\pi} x \, dx
\]
\[
a_0 = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_0^{2\pi} = \frac{1}{\pi} \times \frac{(2\pi)^2}{2} = 2\pi
\]
Thus, \( a_0 = 2\pi \).

### Step 2: Compute \( a_n \)
\[
a_n = \frac{1}{\pi} \int_0^{2\pi} x \cos(nx) \, dx
\]
This requires integration by parts. Let \( u = x \) and \( dv = \cos(nx) \, dx \).

- \( du = dx \)
- \( v = \frac{\sin(nx)}{n} \)

Thus, applying integration by parts:
\[
a_n = \frac{1}{\pi} \left[ \frac{x \sin(nx)}{n} \right]_0^{2\pi} - \frac{1}{\pi} \int_0^{2\pi} \frac{\sin(nx)}{n} \, dx
\]
Evaluating the boundary terms at 0 and \( 2\pi \), they vanish because \( \sin(n \cdot 0) = \sin(n \cdot 2\pi) = 0 \). The remaining integral is:
\[
a_n = -\frac{1}{\pi n} \int_0^{2\pi} \sin(nx) \, dx = 0
\]
Thus, \( a_n = 0 \).

### Step 3: Compute \( b_n \)
\[
b_n = \frac{1}{\pi} \int_0^{2\pi} x \sin(nx) \, dx
\]
Again, use integration by parts with \( u = x \) and \( dv = \sin(nx) \, dx \).

- \( du = dx \)
- \( v = -\frac{\cos(nx)}{n} \)

Applying integration by parts:
\[
b_n = \frac{1}{\pi} \left[ -\frac{x \cos(nx)}{n} \right]_0^{2\pi} + \frac{1}{\pi} \int_0^{2\pi} \frac{\cos(nx)}{n} \, dx
\]
Evaluating the boundary terms at 0 and \( 2\pi \) again gives 0 because \( \cos(n \cdot 0) = \cos(n \cdot 2\pi) = 1 \). The remaining integral is:
\[
b_n = \frac{1}{n} \cdot 2 \quad \text{(since the integral evaluates to \( 2 \) for odd \( n \))}.
\]

### Final Fourier Series:
The Fourier series for \( f(x) = x \) over \( 0 < x < 2\pi \) is:
\[
f(x) = \pi - 2\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)
\]

Would you like further clarification on any step or additional details? Here are five related questions to deepen your understanding:

1. How does the Fourier series differ when the function is defined on a symmetric interval, such as \( [-\pi, \pi] \)?
2. What happens to the Fourier series if we extend \( f(x) = x \) periodically beyond \( [0, 2\pi] \)?
3. Could we represent non-periodic functions using Fourier series over a finite interval?
4. Why do the cosine terms vanish for this function in the Fourier series?
5. What are the practical applications of Fourier series in engineering and physics?

**Tip**: For odd functions, the Fourier series will only contain sine terms, while for even functions, it will only contain cosine terms.

Find the Fourier series to represent f(x) = x for 0 < x < 2π.

Learn how to compute the Fourier series for the function f(x) = x over the interval 0 < x < 2π. This solution includes detailed steps with integration by parts, leading to the final Fourier series representation. Perfect for advanced calculus students.

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