Regla de Boole

$$ \begin{equation} \int_{x_{0}}^{x_{0}+4h} f(x) \ dx = a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) + a_{4} f(x_{0}+4h) \tag{10} \end{equation} $$

Usando \( f(x) = 1 \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) + a_{4} f(x_{0}+4h) = \int_{x_{0}}^{x_{0}+4h} 1 \ dx \end{equation*} $$

Reemplazando valores e integrando $$ \begin{equation*} a_{0} + a_{1} + a_{2} + a_{3} + a_{4} = 4h \end{equation*} $$

Usando \( f(x) = x \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) + a_{4} f(x_{0}+4h) = \int_{x_{0}}^{x_{0}+4h} x \ dx \end{equation*} $$

Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0} + a_{1} (x_{0}+h) + a_{2} (x_{0}+2h) + a_{3} (x_{0}+3h) + a_{4} (x_{0}+4h) = \frac{1}{2} (x_{0}+4h)^{2} - \frac{1}{2} x_{0}^{2} \end{equation*} $$

Usando \( f(x) = x^{2} \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) + a_{4} f(x_{0}+4h) = \int_{x_{0}}^{x_{0}+4h} x^{2} \ dx \end{equation*} $$

Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0}^{2} + a_{1} (x_{0}+h)^{2} + a_{2} (x_{0}+2h)^{2} + a_{3} (x_{0}+3h)^{2} + a_{4} (x_{0}+4h)^{2} = \frac{1}{3} (x_{0}+4h)^{3} - \frac{1}{3} x_{0}^{3} \end{equation*} $$

Usando \( f(x) = x^{3} \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) + a_{4} f(x_{0}+4h) = \int_{x_{0}}^{x_{0}+4h} x^{3} \ dx \end{equation*} $$

Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0}^{3} + a_{1} (x_{0}+h)^{3} + a_{2} (x_{0}+2h)^{3} + a_{3} (x_{0}+3h)^{3} + a_{4} (x_{0}+4h)^{3} = \frac{1}{4} (x_{0}+4h)^{4} - \frac{1}{4} x_{0}^{4} \end{equation*} $$

Usando \( f(x) = x^{4} \) $$ \begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) + a_{4} f(x_{0}+4h) = \int_{x_{0}}^{x_{0}+4h} x^{4} \ dx \end{equation*} $$

Reemplazando valores e integrando $$ \begin{equation*} a_{0} x_{0}^{4} + a_{1} (x_{0}+h)^{4} + a_{2} (x_{0}+2h)^{4} + a_{3} (x_{0}+3h)^{4} + a_{4} (x_{0}+4h)^{4} = \frac{1}{5} (x_{0}+4h)^{5} - \frac{1}{5} x_{0}^{5} \end{equation*} $$

Formando un sistema de ecuaciones $$ \begin{equation*} \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ x_{0} & x_{0}+h & x_{0}+2h & x_{0}+3h & x_{0}+4h \\ x_{0}^{2} & (x_{0}+h)^{2} & (x_{0}+2h)^{2} & (x_{0}+3h)^{2} & (x_{0}+4h)^{2} \\ x_{0}^{3} & (x_{0}+h)^{3} & (x_{0}+2h)^{3} & (x_{0}+3h)^{3} & (x_{0}+4h)^{3} \\ x_{0}^{4} & (x_{0}+h)^{4} & (x_{0}+2h)^{4} & (x_{0}+3h)^{4} & (x_{0}+4h)^{4} \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{bmatrix} = \begin{bmatrix} 4h \\ \frac{1}{2} (x_{0}+4h)^{2} - \frac{1}{2} x_{0}^{2} \\ \frac{1}{3} (x_{0}+4h)^{3} - \frac{1}{3} x_{0}^{3} \\ \frac{1}{4} (x_{0}+4h)^{4} - \frac{1}{4} x_{0}^{4} \\ \frac{1}{5} (x_{0}+4h)^{5} - \frac{1}{5} x_{0}^{5} \end{bmatrix} \end{equation*} $$

Resolviendo $$ \begin{align*} a_{0} &= \frac{14}{45} h \\ a_{1} &= \frac{64}{45} h \\ a_{2} &= \frac{8}{45} h \\ a_{3} &= \frac{64}{45} h \\ a_{4} &= \frac{14}{45} h \end{align*} $$

Reemplazando en (10) $$ \begin{equation*} \int_{x_{0}}^{x_{0}+4h} f(x) \ dx = \frac{2}{45} h \ [7 f(x_{0}) + 32 f(x_{0}+h) + 12 f(x_{0}+2h) + 32 f(x_{0}+3h) + 7 f(x_{0}+4h)] \end{equation*} $$