The other integral gives $$\begin{equation*} \int_{x_0-h}^{x_0}f(x)dx=\frac{h}{2}\left(f(x_0) + f(x_0-h)\right)+O(h^3), \end{equation*}$$ and adding up we obtain $$\begin{equation} \int_{x_0-h}^{x_0+h}f(x)dx=\frac{h}{2}\left(f(x_0+h) + 2f(x_0) + f(x_0-h)\right)+O(h^3), \tag{3} \end{equation}$$ which is the well-known trapezoidal rule. Concerning the error in the approximation made, $O(h^3)=O((b-a)^3/N^3)$, you should note that this is the local error. Since we are splitting the integral from $a$ to $b$ in $N$ pieces, we will have to perform approximately $N$ such operations.