We are asked to find the Fourier series of the function $f(x)$ defined as: $f(x) = \begin{cases} 2, & -2 < x < 0 \\ x, & 0 < x < 2 \end{cases}$

1. Problem Description

We are asked to find the Fourier series of the function $f(x)$ defined as:

$f(x) = \begin{cases} 2, & -2 < x < 0 \\ x, & 0 < x < 2 \end{cases}$

2. Solution Steps

The Fourier series of a function $f(x)$ defined on the interval $(-L, L)$ is given by:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right)$

where the Fourier coefficients are given by:

$a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx$

$a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx$

$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$

In our case, $L = 2$. Let's calculate the coefficients:

$a_0 = \frac{1}{2} \int_{-2}^{2} f(x) dx = \frac{1}{2} \left( \int_{-2}^{0} 2 dx + \int_{0}^{2} x dx \right) = \frac{1}{2} \left( [2x]_{-2}^{0} + [\frac{x^2}{2}]_{0}^{2} \right) = \frac{1}{2} \left( (0 - (-4)) + (\frac{4}{2} - 0) \right) = \frac{1}{2} (4 + 2) = \frac{6}{2} = 3$

$a_n = \frac{1}{2} \int_{-2}^{2} f(x) \cos\left(\frac{n\pi x}{2}\right) dx = \frac{1}{2} \left( \int_{-2}^{0} 2 \cos\left(\frac{n\pi x}{2}\right) dx + \int_{0}^{2} x \cos\left(\frac{n\pi x}{2}\right) dx \right)$

$\int_{-2}^{0} 2 \cos\left(\frac{n\pi x}{2}\right) dx = 2 \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_{-2}^{0} = \frac{4}{n\pi} \left( \sin(0) - \sin(-n\pi) \right) = \frac{4}{n\pi} (0 - 0) = 0$

$\int_{0}^{2} x \cos\left(\frac{n\pi x}{2}\right) dx = \left[ \frac{2x}{n\pi} \sin\left(\frac{n\pi x}{2}\right) + \frac{4}{n^2 \pi^2} \cos\left(\frac{n\pi x}{2}\right) \right]_{0}^{2} = \left( \frac{4}{n\pi} \sin(n\pi) + \frac{4}{n^2 \pi^2} \cos(n\pi) \right) - \left( 0 + \frac{4}{n^2 \pi^2} \cos(0) \right) = 0 + \frac{4}{n^2 \pi^2} (-1)^n - \frac{4}{n^2 \pi^2} = \frac{4}{n^2 \pi^2} ((-1)^n - 1)$

$a_n = \frac{1}{2} \left( 0 + \frac{4}{n^2 \pi^2} ((-1)^n - 1) \right) = \frac{2}{n^2 \pi^2} ((-1)^n - 1)$

$b_n = \frac{1}{2} \int_{-2}^{2} f(x) \sin\left(\frac{n\pi x}{2}\right) dx = \frac{1}{2} \left( \int_{-2}^{0} 2 \sin\left(\frac{n\pi x}{2}\right) dx + \int_{0}^{2} x \sin\left(\frac{n\pi x}{2}\right) dx \right)$

$\int_{-2}^{0} 2 \sin\left(\frac{n\pi x}{2}\right) dx = 2 \left[ -\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right) \right]_{-2}^{0} = -\frac{4}{n\pi} \left( \cos(0) - \cos(-n\pi) \right) = -\frac{4}{n\pi} (1 - (-1)^n)$

$\int_{0}^{2} x \sin\left(\frac{n\pi x}{2}\right) dx = \left[ -\frac{2x}{n\pi} \cos\left(\frac{n\pi x}{2}\right) + \frac{4}{n^2 \pi^2} \sin\left(\frac{n\pi x}{2}\right) \right]_{0}^{2} = \left( -\frac{4}{n\pi} \cos(n\pi) + \frac{4}{n^2 \pi^2} \sin(n\pi) \right) - (0 + 0) = -\frac{4}{n\pi} (-1)^n$

$b_n = \frac{1}{2} \left( -\frac{4}{n\pi} (1 - (-1)^n) - \frac{4}{n\pi} (-1)^n \right) = \frac{1}{2} \left( -\frac{4}{n\pi} + \frac{4}{n\pi} (-1)^n - \frac{4}{n\pi} (-1)^n \right) = -\frac{2}{n\pi}$

Therefore, the Fourier series is:

$f(x) = \frac{3}{2} + \sum_{n=1}^{\infty} \left( \frac{2}{n^2 \pi^2} ((-1)^n - 1) \cos\left(\frac{n\pi x}{2}\right) - \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right)$

3. Final Answer

$f(x) = \frac{3}{2} + \sum_{n=1}^{\infty} \left( \frac{2}{n^2 \pi^2} ((-1)^n - 1) \cos\left(\frac{n\pi x}{2}\right) - \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right)$