The problem asks us to find the Fourier series representation of the function $f(x)$ defined as: $f(x) = \begin{cases} 2, & -2 < x < 0 \\ x, & 0 < x < 2 \end{cases}$

1. Problem Description

The problem asks us to find the Fourier series representation 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 representation of a function $f(x)$ defined on the interval $(-L, L)$ is given by:

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

Here, $L=2$. Thus, the Fourier series is

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

The coefficients are given by:

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

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

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

First, we calculate $a_0$:

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

$a_0 = \frac{1}{2} \left[ 2x |_{-2}^{0} + \frac{x^2}{2} |_{0}^{2} \right] = \frac{1}{2} \left[ 2(0 - (-2)) + \frac{2^2}{2} - \frac{0^2}{2} \right]$

$a_0 = \frac{1}{2} [ 4 + 2 ] = \frac{6}{2} = 3$

Next, we calculate $a_n$:

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

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

$\int_{0}^{2} x \cos(\frac{n\pi x}{2}) dx$ Using integration by parts, let $u=x$, $dv = \cos(\frac{n\pi x}{2}) dx$. Then $du = dx$, $v = \frac{2}{n\pi} \sin(\frac{n\pi x}{2})$.

$\int_{0}^{2} x \cos(\frac{n\pi x}{2}) dx = x \cdot \frac{2}{n\pi} \sin(\frac{n\pi x}{2}) |_{0}^{2} - \int_{0}^{2} \frac{2}{n\pi} \sin(\frac{n\pi x}{2}) dx$

$= [2 \cdot \frac{2}{n\pi} \sin(n\pi) - 0] - \frac{2}{n\pi} \cdot (-\frac{2}{n\pi} \cos(\frac{n\pi x}{2})) |_{0}^{2}$

$= 0 + \frac{4}{n^2 \pi^2} [\cos(n\pi) - \cos(0)] = \frac{4}{n^2 \pi^2} [\cos(n\pi) - 1]$

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

Since $\cos(n\pi) = (-1)^n$, $a_n = \frac{2}{n^2 \pi^2} [(-1)^n - 1]$

If $n$ is even, $a_n = 0$. If $n$ is odd, $a_n = \frac{2}{n^2 \pi^2} (-1 - 1) = -\frac{4}{n^2 \pi^2}$

Now, we calculate $b_n$:

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

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

$\int_{0}^{2} x \sin(\frac{n\pi x}{2}) dx$ Using integration by parts, let $u=x$, $dv = \sin(\frac{n\pi x}{2}) dx$. Then $du = dx$, $v = -\frac{2}{n\pi} \cos(\frac{n\pi x}{2})$.

$\int_{0}^{2} x \sin(\frac{n\pi x}{2}) dx = x \cdot (-\frac{2}{n\pi} \cos(\frac{n\pi x}{2})) |_{0}^{2} - \int_{0}^{2} (-\frac{2}{n\pi} \cos(\frac{n\pi x}{2})) dx$

$= [2 \cdot (-\frac{2}{n\pi} \cos(n\pi)) - 0] + \frac{2}{n\pi} \cdot (\frac{2}{n\pi} \sin(\frac{n\pi x}{2})) |_{0}^{2}$

$= -\frac{4}{n\pi} \cos(n\pi) + \frac{4}{n^2 \pi^2} [\sin(n\pi) - \sin(0)] = -\frac{4}{n\pi} \cos(n\pi) + 0$

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

$b_n = \frac{1}{2} \left[ -\frac{4}{n\pi} \right] = -\frac{2}{n\pi}$

Therefore,

$f(x) = \frac{3}{2} + \sum_{n=1}^{\infty} [a_n \cos(\frac{n\pi x}{2}) + b_n \sin(\frac{n\pi x}{2})]$

$f(x) = \frac{3}{2} + \sum_{n=1,3,5,...}^{\infty} [-\frac{4}{n^2 \pi^2} \cos(\frac{n\pi x}{2})] + \sum_{n=1}^{\infty} [-\frac{2}{n\pi} \sin(\frac{n\pi x}{2})]$

$f(x) = \frac{3}{2} + \sum_{k=0}^{\infty} [-\frac{4}{(2k+1)^2 \pi^2} \cos(\frac{(2k+1)\pi x}{2})] + \sum_{n=1}^{\infty} [-\frac{2}{n\pi} \sin(\frac{n\pi x}{2})]$

$f(x) = \frac{3}{2} - \sum_{k=0}^{\infty} \frac{4}{(2k+1)^2 \pi^2} \cos(\frac{(2k+1)\pi x}{2}) - \sum_{n=1}^{\infty} \frac{2}{n\pi} \sin(\frac{n\pi x}{2})$

3. Final Answer

$f(x) = \frac{3}{2} - \sum_{k=0}^{\infty} \frac{4}{(2k+1)^2 \pi^2} \cos(\frac{(2k+1)\pi x}{2}) - \sum_{n=1}^{\infty} \frac{2}{n\pi} \sin(\frac{n\pi x}{2})$