Usando \( f(x) = 1 \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) + a_{2} f(x_{2}) = \int_{x_{0}}^{x_{2}} 1 \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} + a_{1} + a_{2} = x_{2} - x_{0} \end{equation*} $$
Usando \( f(x) = x \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) + a_{2} f(x_{2}) = \int_{x_{0}}^{x_{2}} x \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0} + a_{1} x_{1} + a_{2} x_{2} = \frac{1}{2} x_{2}^{2} - \frac{1}{2} x_{0}^{2} \end{equation*} $$
Usando \( f(x) = x^{2} \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) + a_{2} f(x_{2}) = \int_{x_{0}}^{x_{2}} x^{2} \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0}^{2} + a_{1} x_{1}^{2} + a_{2} x_{2}^{2} = \frac{1}{3} x_{2}^{3} - \frac{1}{3} x_{0}^{3} \end{equation*} $$
Formando un sistema de ecuaciones $$ \begin{equation*} \begin{bmatrix} 1 & 1 & 1 \\ x_{0} & x_{1} & x_{2} \\ x_{0}^{2} & x_{1}^{2} & x_{2}^{2} \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \end{bmatrix} = \begin{bmatrix} x_{2} - x_{0} \\ \frac{1}{2} x_{2}^{2} - \frac{1}{2} x_{0}^{2} \\ \frac{1}{3} x_{2}^{3} - \frac{1}{3} x_{0}^{3} \end{bmatrix} \end{equation*} $$
Resolviendo $$ \begin{align*} a_{0} &= \frac{2 x_{0}^{2} - 3 x_{0} x_{1} - x_{0} x_{2} + 3 x_{1} x_{2} - x_{2}^{2}}{6 (x_{1} - x_{0})} \\ a_{1} &= -\frac{x_{0}^{3} - 3 x_{0}^{2} x_{2} + 3 x_{0} x_{2}^{2} - x_{2}^{3}}{6 (x_{2} - x_{1}) (x_{1} - x_{0})} \\ a_{2} &= -\frac{x_{0}^{2} - 3 x_{0} x_{1} + x_{0} x_{2} + 3 x_{1} x_{2} - 2 x_{2}^{2}}{6 (x_{2} - x_{1})} \end{align*} $$
Reemplazando en (3) $$ \begin{equation*} \int_{x_{0}}^{x_{2}} f(x) \ dx = \biggl[ \frac{2 x_{0}^{2} - 3 x_{0} x_{1} - x_{0} x_{2} + 3 x_{1} x_{2} - x_{2}^{2}}{6 (x_{1} - x_{0})} \biggr] f(x_{0}) - \biggl[ \frac{x_{0}^{3} - 3 x_{0}^{2} x_{2} + 3 x_{0} x_{2}^{2} - x_{2}^{3}}{6 (x_{2} - x_{1}) (x_{1} - x_{0})} \biggr] f(x_{1}) - \biggl[ \frac{x_{0}^{2} - 3 x_{0} x_{1} + x_{0} x_{2} + 3 x_{1} x_{2} - 2 x_{2}^{2}}{6 (x_{2} - x_{1})} \biggr] f(x_{2}) \end{equation*} $$