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*}
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Usando $f(x) = 1$
\begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) + a_{2} f(x_{2}) = \int_{a}^{b} 1 \ dx \end{equation*}Reemplazando valores e integrando
\begin{equation*} a_{0} + a_{1} + a_{2} = b - a \end{equation*}Usando $f(x) = x$
\begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) + a_{2} f(x_{2}) = \int_{a}^{b} x \ dx \end{equation*}Reemplazando valores e integrando
\begin{equation*} a_{0} x_{0} + a_{1} x_{1} + a_{2} x_{2} = \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}) + a_{2} f(x_{2}) = \int_{a}^{b} 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{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}) + a_{2} f(x_{2}) = \int_{a}^{b} x^{3} \ dx \end{equation*}Reemplazando valores e integrando
\begin{equation*} a_{0} x_{0}^{3} + a_{1} x_{1}^{3} + a_{2} x_{2}^{3} = \frac{b^{4}}{4} - \frac{a^{4}}{4} \end{equation*}Usando $f(x) = x^{4}$
\begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) + a_{2} f(x_{2}) = \int_{a}^{b} x^{4} \ dx \end{equation*}Reemplazando valores e integrando
\begin{equation*} a_{0} x_{0}^{4} + a_{1} x_{1}^{4} + a_{2} x_{2}^{4} = \frac{b^{5}}{5} - \frac{a^{5}}{5} \end{equation*}Usando $f(x) = x^{5}$
\begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{1}) + a_{2} f(x_{2}) = \int_{a}^{b} x^{5} \ dx \end{equation*}Reemplazando valores e integrando
\begin{equation*} a_{0} x_{0}^{5} + a_{1} x_{1}^{5} + a_{2} x_{2}^{5} = \frac{b^{6}}{6} - \frac{a^{6}}{6} \end{equation*}Formando un sistema de ecuaciones
\begin{align*} a_{0} + a_{1} + a_{2} &= b - a \\ a_{0} x_{0} + a_{1} x_{1} + a_{2} x_{2} &= \frac{b^{2}}{2} - \frac{a^{2}}{2} \\ a_{0} x_{0}^{2} + a_{1} x_{1}^{2} + a_{2} x_{2}^{2} &= \frac{b^{3}}{3} - \frac{a^{3}}{3} \\ a_{0} x_{0}^{3} + a_{1} x_{1}^{3} + a_{2} x_{2}^{3} &= \frac{b^{4}}{4} - \frac{a^{4}}{4} \\ a_{0} x_{0}^{4} + a_{1} x_{1}^{4} + a_{2} x_{2}^{4} &= \frac{b^{5}}{5} - \frac{a^{5}}{5} \\ a_{0} x_{0}^{5} + a_{1} x_{1}^{5} + a_{2} x_{2}^{5} &= \frac{b^{6}}{6} - \frac{a^{6}}{6} \end{align*}Resolviendo
\begin{align*} a_{0} &= - \frac{5}{18} a + \frac{5}{18} b \\ a_{1} &= - \frac{4}{9} a + \frac{4}{9} b \\ a_{2} &= - \frac{5}{18} a + \frac{5}{18} b \\ x_{0} &= \frac{1}{2} (a+b) + \frac{\sqrt{15}}{10} (a-b) \\ x_{1} &= \frac{1}{2} (a+b) \\ x_{2} &= \frac{1}{2} (a+b) - \frac{\sqrt{15}}{10} (a-b) \end{align*}
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