# Método de los coeficientes indeterminados

$$\int_{x_{0}}^{x_{0}+ih} f(x) \ dx \approx \sum_{i=0}^{n} a_{i} f(x_{0}+ih)$$

# Regla del trapecio

$$\int_{x_{0}}^{x_{0}+h} f(x) \ dx = a_{0} f(x_{0}) + a_{1} f(x_{0}+h) \label{ecuacion2}$$

Usando $f(x) = 1$

\begin{equation*} a_{0} f(x_{0}) + a_{1} f(x_{0}+h) = \int_{x_{0}}^{x_{0}+h} 1 \ dx \end{equation*}

Reemplazando valores e integrando

\begin{equation*} a_{0} + a_{1} = h \end{equation*}

Usando $f(x) = x$

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

Reemplazando valores e integrando

\begin{equation*} a_{0} x_{0} + a_{1} (x_{0}+h) = \frac{1}{2} (x_{0}+h)^{2} - \frac{1}{2} x_{0}^{2} \end{equation*}

Formando un sistema de ecuaciones

\begin{equation*} \begin{bmatrix} 1 & 1 \\ x_{0} & x_{0}+h \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \end{bmatrix} = \begin{bmatrix} h \\ \frac{1}{2} (x_{0}+h)^{2} - \frac{1}{2} x_{0}^{2} \end{bmatrix} \end{equation*}

Resolviendo

\begin{align*} a_{0} &= \frac{1}{2} h \\ a_{1} &= \frac{1}{2} h \end{align*}

Reemplazando en \eqref{ecuacion2}

\begin{equation*} \int_{x_{0}}^{x_{0}+h} f(x) \ dx = \frac{1}{2} h \ [f(x_{0}) + f(x_{0}+h)] \end{equation*}

# Regla de Simpson 1/3

$$\int_{x_{0}}^{x_{0}+2h} f(x) \ dx = a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) \label{ecuacion3}$$

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 \eqref{ecuacion3}

\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*}

# Regla de Simpson 3/8

$$\int_{x_{0}}^{x_{0}+3h} f(x) \ dx = a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) \label{ecuacion4}$$

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) = \int_{x_{0}}^{x_{0}+3h} 1 \ dx \end{equation*}

Reemplazando valores e integrando

\begin{equation*} a_{0} + a_{1} + a_{2} + a_{3} = 3h \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) = \int_{x_{0}}^{x_{0}+3h} 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) = \frac{1}{2} (x_{0}+3h)^{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) = \int_{x_{0}}^{x_{0}+3h} 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} = \frac{1}{3} (x_{0}+3h)^{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) = \int_{x_{0}}^{x_{0}+3h} 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} = \frac{1}{4} (x_{0}+3h)^{4} - \frac{1}{4} x_{0}^{4} \end{equation*}

Formando un sistema de ecuaciones

\begin{equation*} \begin{bmatrix} 1 & 1 & 1 & 1 \\ x_{0} & x_{0}+h & x_{0}+2h & x_{0}+3h \\ x_{0}^{2} & (x_{0}+h)^{2} & (x_{0}+2h)^{2} & (x_{0}+3h)^{2} \\ x_{0}^{3} & (x_{0}+h)^{3} & (x_{0}+2h)^{3} & (x_{0}+3h)^{3} \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{bmatrix} = \begin{bmatrix} 3h \\ \frac{1}{2} (x_{0}+3h)^{2} - \frac{1}{2} x_{0}^{2} \\ \frac{1}{3} (x_{0}+3h)^{3} - \frac{1}{3} x_{0}^{3} \\ \frac{1}{4} (x_{0}+3h)^{4} - \frac{1}{4} x_{0}^{4} \end{bmatrix} \end{equation*}

Resolviendo

\begin{align*} a_{0} &= \frac{3}{8} h \\ a_{1} &= \frac{9}{8} h \\ a_{2} &= \frac{9}{8} h \\ a_{3} &= \frac{3}{8} h \end{align*}

Reemplazando en \eqref{ecuacion4}

\begin{equation*} \int_{x_{0}}^{x_{0}+3h} f(x) \ dx = \frac{3}{8} h \ [f(x_{0}) + 3 f(x_{0}+h) + 3 f(x_{0}+2h) + f(x_{0}+3h)] \end{equation*}

# Regla de Boole

$$\int_{x_{0}}^{x_{0}+4h} f(x) \ dx = a_{0} f(x_{0}) + a_{1} f(x_{0}+h) + a_{2} f(x_{0}+2h) + a_{3} f(x_{0}+3h) + a_{4} f(x_{0}+4h) \label{ecuacion5}$$

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 \eqref{ecuacion5}

\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*}


In [ ]: