Usando \( f(x) = 1 \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) = \int_{a}^{b} 1 \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} + a_{1} = b - a \end{equation*} $$
Usando \( f(x) = x \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) = \int_{a}^{b} x \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0} + a_{1} x_{1} = \frac{b^{2}}{2} - \frac{a^{2}}{2} \end{equation*} $$
Usando \( f(x) = x^{2} \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) = \int_{a}^{b} x^{2} \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0}^{2} + a_{1} x_{1}^{2} = \frac{b^{3}}{3} - \frac{a^{3}}{3} \end{equation*} $$
Usando \( f(x) = x^{3} \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) = \int_{a}^{b} x^{3} \ dx \end{equation*} $$
Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0}^{3} + a_{1} x_{1}^{3} = \frac{b^{4}}{4} - \frac{a^{4}}{4} \end{equation*} $$
Formando un sistema de ecuaciones $$ \begin{align*} a_{0} + a_{1} &= b - a \\ a_{0} x_{0} + a_{1} x_{1} &= \frac{b^{2}}{2} - \frac{a^{2}}{2} \\ a_{0} x_{0}^{2} + a_{1} x_{1}^{2} &= \frac{b^{3}}{3} - \frac{a^{3}}{3} \\ a_{0} x_{0}^{3} + a_{1} x_{1}^{3} &= \frac{b^{4}}{4} - \frac{a^{4}}{4} \end{align*} $$
Resolviendo $$ \begin{align*} a_{0} &= - \frac{1}{2} a + \frac{1}{2} b \\ a_{1} &= - \frac{1}{2} a + \frac{1}{2} b \\ x_{0} &= \frac{\sqrt{3}}{6} \bigl[ \sqrt{3} (a+b) + a - b \bigr] \\ x_{1} &= \frac{1}{2} (a+b) - \frac{\sqrt{3}}{6}(a-b) \end{align*} $$