Find the Fourier series for the function f(x) = 2x+2 for −1 < x < 0, and f(x) = 0 for 0 < x < 1.

Continuing from the previous step:

$$a_0 = (0^2 + 2 \cdot 0) - ((-1)^2 + 2(-1))$$ $$a_0 = 0 - (1 - 2)$$ $$a_0 = 0 - (-1)$$ $$a_0 = 1$$

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

Since $f(x)=0$ for $0 < x < 1$:

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

We use integration by parts, $\int u\,dv = uv - \int v\,du$.
Let $u = 2x+2$ and $dv = \cos(n\pi x)\,dx$.
Then $du = 2\,dx$ and $v = \frac{1}{n\pi} \sin(n\pi x)$.

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

Evaluate the first term:
At $x=0$: $(2(0)+2) \frac{1}{n\pi} \sin(0) = 2 \cdot \frac{1}{n\pi} \cdot 0 = 0$.
At $x=-1$: $(2(-1)+2) \frac{1}{n\pi} \sin(-n\pi) = 0 \cdot \frac{1}{n\pi} \cdot 0 = 0$.
So the first term is $0 - 0 = 0$.

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

Since $\cos(0)=1$ and $\cos(n\pi)=(-1)^n$:

$$a_n = \frac{2}{(n\pi)^2} (1 - (-1)^n)$$

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

Since $f(x)=0$ for $0 < x < 1$:

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

We use integration by parts, $\int u\,dv = uv - \int v\,du$.
Let $u = 2x+2$ and $dv = \sin(n\pi x)\,dx$.
Then $du = 2\,dx$ and $v = -\frac{1}{n\pi} \cos(n\pi x)$.

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

Evaluate the first term:
At $x=0$: $(2(0)+2) \left(-\frac{1}{n\pi} \cos(0)\right) = 2 \left(-\frac{1}{n\pi} \cdot 1\right) = -\frac{2}{n\pi}$.
At $x=-1$: $(2(-1)+2) \left(-\frac{1}{n\pi} \cos(-n\pi)\right) = 0 \cdot \left(-\frac{1}{n\pi} \cos(n\pi)\right) = 0$.
So the first term is $-\frac{2}{n\pi} - 0 = -\frac{2}{n\pi}$.

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

Since $\sin(0)=0$ and $\sin(-n\pi)=0$:

$$b_n = -\frac{2}{n\pi} + \frac{2}{(n\pi)^2} (0 - 0)$$ $$b_n = -\frac{2}{n\pi}$$

Substitute the coefficients $a_0=1$, $a_n = \frac{2}{(n\pi)^2} (1 - (-1)^n)$, and $b_n = -\frac{2}{n\pi}$ into the Fourier series formula:

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

The Fourier series for $f(x)$ is:

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