Find the Fourier series of the function f(x) = 2x+2, & -1 < x < 0 0, & 0 < x < 1

Step 1: Determine the period and Fourier series formulas.
The function is given as $f(x) = \begin{cases} 2x+2, & -1 < x < 0 \\ 0, & 0 < x < 1 \end{cases}$.
The condition $f(x+2) = f(x)$ indicates that the function is periodic with period $T=2$.
For a Fourier series, the period is $2L$, so $2L=2$, which means $L=1$.
The Fourier series is given by:

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

With $L=1$, this becomes:

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

The coefficients are calculated using the following integrals over one period, for example, from $-L$ to $L$:

$$a_0 = \frac{1}{L} \int_{-L}^{L} f(x)\,dx = \int_{-1}^{1} f(x)\,dx$$ $$a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right)\,dx = \int_{-1}^{1} f(x) \cos(n\pi x)\,dx$$ $$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right)\,dx = \int_{-1}^{1} f(x) \sin(n\pi x)\,dx$$

Step 2: Calculate $a_0$.

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

Since $f(x)$ is defined piecewise:

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