Usando \( f(x) = 1 \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) = \int_{x_{0}}^{x_{0}+2h} 1 \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} + a_{1} + a_{2} = 2h \end{equation*} $$
Usando \( f(x) = x \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) = \int_{x_{0}}^{x_{0}+2h} x \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0} + a_{1} (x_{0}+h) + a_{2} (x_{0}+2h) = \frac{1}{2} (x_{0}+2h)^{2} - \frac{1}{2} x_{0}^{2} \end{equation*} $$
Usando \( f(x) = x^{2} \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) = \int_{x_{0}}^{x_{0}+2h} x^{2} \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0}^{2} + a_{1} (x_{0}+h)^{2} + a_{2} (x_{0}+2h)^{2} = \frac{1}{3} (x_{0}+2h)^{3} - \frac{1}{3} x_{0}^{3} \end{equation*} $$
Formando un sistema de ecuaciones $$ \begin{equation*} \begin{bmatrix} 1 & 1 & 1 \\ x_{0} & x_{0}+h & x_{0}+2h \\ x_{0}^{2} & (x_{0}+h)^{2} & (x_{0}+2h)^{2} \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \end{bmatrix} = \begin{bmatrix} 2h \\ \frac{1}{2} (x_{0}+2h)^{2} - \frac{1}{2} x_{0}^{2} \\ \frac{1}{3} (x_{0}+2h)^{3} - \frac{1}{3} x_{0}^{3} \end{bmatrix} \end{equation*} $$
Resolviendo $$ \begin{align*} a_{0} &= \frac{1}{3} h \\ a_{1} &= \frac{4}{3} h \\ a_{2} &= \frac{1}{3} h \end{align*} $$
Reemplazando en (8) $$ \begin{equation*} \int_{x_{0}}^{x_{0}+2h} f(x) \ dx = \frac{1}{3} h \ [f(x_{0}) + 4 f(x_{0}+h) + f(x_{0}+2h)] \end{equation*} $$