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The problem asks for the value of \( b_n \) in a Fourier series expansion of \( f(x) = x \) over the interval \( (0, 2\pi) \). Let's solve it step by step:

### 1. Fourier Coefficients for \( f(x) = x \)
For a function \( f(x) \), the Fourier coefficients \( b_n \) for the sine terms are given by:
\[
b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) \, dx
\]
where \( f(x) = x \).

### 2. Calculating \( b_n \)
Substitute \( f(x) = x \) into the formula:
\[
b_n = \frac{1}{\pi} \int_0^{2\pi} x \sin(nx) \, dx
\]
This is an integration by parts problem. Let \( u = x \) and \( dv = \sin(nx) \, dx \), so \( du = dx \) and \( v = -\frac{1}{n} \cos(nx) \).

Apply integration by parts:
\[
b_n = \frac{1}{\pi} \left[ -\frac{x \cos(nx)}{n} \Big|_0^{2\pi} + \frac{1}{n} \int_0^{2\pi} \cos(nx) \, dx \right]
\]

### 3. Evaluate the Boundary Terms
At \( x = 2\pi \) and \( x = 0 \):
\[
\left[ -\frac{x \cos(nx)}{n} \right] \Big|_0^{2\pi} = -\frac{2\pi \cos(2n\pi)}{n} + \frac{0 \cdot \cos(0)}{n} = -\frac{2\pi}{n}
\]
since \( \cos(2n\pi) = 1 \).

### 4. Solve the Remaining Integral
\[
\int_0^{2\pi} \cos(nx) \, dx = 0
\]
because the integral of \( \cos(nx) \) over a full period \( [0, 2\pi] \) is zero.

### 5. Final Result
Thus, the value of \( b_n \) is:
\[
b_n = \frac{1}{\pi} \left( -\frac{2\pi}{n} \right) = -\frac{2}{n}
\]

### Conclusion:
The correct value for \( b_n \) is \( -\frac{2}{n} \), which corresponds to option **I. (b)**.

Would you like more details or explanations on this problem? 

Here are some related questions for further exploration:
1. What are the general formulas for Fourier sine and cosine coefficients?
2. How would the Fourier series change if the interval was \( (-\pi, \pi) \)?
3. What is the role of the constant term \( a_0 \) in the Fourier series?
4. How do boundary conditions affect the Fourier series expansion?
5. Can you perform the same Fourier expansion for other simple functions like \( f(x) = x^2 \)?

**Tip:** Remember that integrating by parts is a key technique when dealing with products of functions like \( x \sin(nx) \).

If f(x) = x in (0, 2π), then what is the value of the Fourier sine coefficient b_n?

Learn how to calculate the Fourier sine coefficient b_n for the function f(x) = x on the interval (0, 2π). This problem covers key concepts in Fourier series, integration by parts, and the orthogonality of sine and cosine. Suitable for advanced math students.

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