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The problem provided involves finding the coefficients \(b_n\) in the Fourier sine series expansion of the function \(f(x) = x\) over the interval \( (0, 2\pi) \). The general form of the Fourier sine series for \(f(x)\) is given by:

\[
f(x) = \sum_{n=1}^\infty b_n \sin(nx)
\]

where the coefficient \(b_n\) is determined by the integral:

\[
b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx
\]

For the function \(f(x) = x\) on the interval \((0, 2\pi)\), we take \(L = 2\pi\) and compute \(b_n\) as follows:

\[
b_n = \frac{2}{2\pi} \int_0^{2\pi} x \sin\left(\frac{n\pi x}{2\pi}\right) dx = \frac{1}{\pi} \int_0^{2\pi} x \sin(nx) \, dx
\]

This integral can be evaluated using integration by parts. Let:
- \( u = x \) and \( dv = \sin(nx) \, dx \)
- \( du = dx \) and \( v = -\frac{\cos(nx)}{n} \)

Using integration by parts, we have:

\[
\int x \sin(nx) \, dx = x \left(-\frac{\cos(nx)}{n}\right) - \int \left(-\frac{\cos(nx)}{n}\right) dx
\]

This simplifies to:

\[
- \frac{x \cos(nx)}{n} + \frac{\sin(nx)}{n^2}
\]

Evaluating this from \(0\) to \(2\pi\):

\[
\left[- \frac{2\pi \cos(2\pi n)}{n} + \frac{\sin(2\pi n)}{n^2}\right] - \left[- \frac{0 \cos(0)}{n} + \frac{\sin(0)}{n^2}\right]
\]

Since \(\sin(2\pi n) = 0\) and \(\cos(2\pi n) = 1\), we get:

\[
- \frac{2\pi}{n}
\]

Thus, 

\[
b_n = \frac{1}{\pi} \left(- \frac{2\pi}{n}\right) = -\frac{2}{n}
\]

Therefore, the correct choice is \((b)\ b_n = -\frac{2}{n}\).

Would you like any more detail on this computation or do you have any other questions on the topic?

1. What is the integral formula for Fourier cosine coefficients?
2. How would you determine the Fourier coefficients if the function was defined differently on the same interval?
3. What if \(f(x)\) was a piecewise function, how would the integration change?
4. Can you extend this analysis to find the Fourier cosine series for \(f(x) = x\)?
5. What are the implications of convergence for the Fourier series representing \(f(x) = x\)?

**Tip:** When solving integrals involving trigonometric functions for Fourier series, always check whether integration by parts can simplify the integral.

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

Find the Fourier series sine coefficient b_n for the function f(x) = x on the interval (0, 2π). This problem involves applying the Fourier sine series expansion formula, using integration by parts to solve the integral, and understanding trigonometric functions. Suitable for advanced calculus or mathematical methods courses.

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