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Suma Superior de Darboux, por WikiMatematica.org
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Suma Superior de Darboux

De por WikiMatematica.org

Dada f(x) en el intervalo [a,b] para encontrar el área bajo la curva: Dividimos la región "S" en franjas de anchos iguales. El ancho de cada franja es: \(\Delta x =\frac{b-a}{n}


Teniendo los intervalos: \[[x_0,x_1],[x_1,x_2],[x_2,x_3],...,[x_n-1,x_n]

La ecuación para la suma superior de Darboux es la siguiente:

\ R_n=\sum_{i=1}^{n}f(x_{i})\Delta x

donde f(x_{i}) es la función valuada en el extremo derecho y \ x_i=\ a+\Delta x*(i).

Para esta suma es importante saber las siguientes identidades:

Sabiendo que: \sum_{i=1}^{n}a_{i}=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+...+a_{n}

Podemos obtener las siguientes igualdades:

\sum_{i=1}^{n}i=1+2+...+n= \frac{n(n+1)}{2}
\sum_{i=1}^{n}i^2=1^2+2^2+...+n^2= \frac{n(n+1)(2n+1)}{6}
\sum_{i=1}^{n}i^3=1^3+2^3+...+n^3= \frac{n^2(n+1)^2}{4}
\sum_{i=1}^{n}C=C*n (donde C es constante)


Contenido

Ejemplo 1

Evalúe la suma superior de Darboux para la función \ f(x)=x^3-6x con a=0, b=3, n=6

\int_{0}^{3}\ (x^3-6x) dx

\ \Delta x=\frac{3-0}{6}=\frac{1}{2}

\ x_i=0+\frac{i}{2}=\frac{i}{2}


R_6=\sum_{i=1}^{6}[(\frac{i}{2})^3-6(\frac{i}{2})]*\frac{1}{2}


\frac{1}{2}\sum_{i=1}^{6}\frac{i^3}{8}-3i


\frac{1}{2} [\frac{1}{8}(\frac{6^2(6+1)^2}{4})-3(\frac{6(6+1)}{2})]


\frac{1}{2}(55.125-63)


=-3.9375

Ejemplo 2

\int_{-1}^{5}(1+3x)dx

\lim_{\Delta x\rightarrow 0}\sum_{i=1}^{n}[1+3x_{i}]\Delta x

\Delta x = \frac{5-(-1)}{n}

x_{i}= -1+\frac{6}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[1+3(-1+\frac{6i}{n})]\frac{6}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[\frac{6}{n}-\frac{18}{n}+\frac{108i}{n^2}]

\lim_{ n\rightarrow \infty}[\sum_{i=1}^n}\frac{6}{n}-\sum_{i=1}^n}\frac{18}{n}+\sum_{i=1}^n}\frac{108i}{n^2}]

\lim_{ n\rightarrow \infty}[6-18 + \frac{180}{n^2}\sum_{i=1}^{n}i]

\lim_{ n\rightarrow \infty}[-12 + \frac{180}{n^2}*\frac{n(n+1)}{2}]

\lim_{ n\rightarrow \infty}[-12 + 54*\frac{n(n+1)}{n}]

\lim_{ n\rightarrow \infty}-12 +54\lim_{n\rightarrow \infty}\frac{n(1+\frac{1}{n})}{n}

-12*54*1

 42

--Antonio Moran 14:54 29 ago 2009 (CST)tonymoran

Ejemplo

\int_{0}^{2}(2-x^2)dx

\lim_{\Delta x\rightarrow 0} \sum_{i=1}^{n}[2-x^2_{i}]\Delta x

 \Delta x= \frac{2-0}{n}

 \Delta x= \frac{2}{n}

x_{i}= 0+\frac{2}{n}

\lim_{n\rigtharrow \infty}\sum_{i=1}^{n}[2 -(\frac{2i}{n})^2]\frac{2}{n}

\lim_{n\rigtharrow \infty}\sum_{i=1}^{n}[2-\frac{4i^2}{n^2}]\frac{2}{n}

\lim_{n\rigtharrow \infty}\sum_{i=1}^{n}[\frac{4}{n}-\frac{8i^2}{n^3}]

\lim_{n\rigtharrow \infty}[\sum_{i=1}^{n}\frac{4}{n}-\sum_{i=1}^{n}\frac{8i^2}{n^3}]

\lim_{n\rigtharrow \infty}[4- \frac{8}{n^3}\sum_{i=1}^{n}i^2

\lim_{n\rigtharrow \infty}[4-\frac{8}{n^3}\frac{n(n+1)(2n+1)}{6}

\lim_{n\rigtharrow \infty}[4-\frac{8}{6}\frac{n^3(2+\frac{3}{n}+\frac{1}{n^2})}{n^3}]

= 4-\frac{8}{6}*2

 =\frac{4}{3}

--Antonio Moran 20:12 31 ago 2009 (CST)tonymoran


Ejemplo 4

\int_{1}^{2}x^3dx

\lim_{\Delta x\rightarrow 0} \sum_{i=1}^{n}x^3\Delta x

\Delta x= \frac{2-1}{n}

\Delta x= \frac{1}{n}

 x_{i} = 1 + \frac{1i}{n}

\lim_{n\rightarrow \infty}\sum_{i=1}^{n}[1+\frac{i}{n}]^3*\frac{1}{n}

\lim_{n\rightarrow \infty}\sum_{i=1}^{n}[1+\frac{3i}{n}+\frac{3i^2}{n^2}+\frac{i^3}{n^3}]\frac{1}{n}
\lim_{n\rightarrow \infty}\sum_{i=1}^{n}[\frac{1}{n}+\frac{3i}{n^2}+\frac{3i^2}{n^3}+\frac{i^3}{n^4}]

\lim_{n\rightarrow \infty}[\sum_{i=1}^{n}\frac{1}{n}+\sum_{i=1}^{n}\frac{3i}{n^2}+\sum_{i=1}^{n}\frac{3i^2}{n^3}+\sum_{i=1}^{n}\frac{i^3}{n^4}]

\lim_{n\rightarrow \infty}[1 +\frac{3}{n^2}\sum_{i=1}^{n}i+\frac{3}{n^3}\sum_{i=1}^{n}i^2+\frac{1}{n^4}\sum_{i=1}^{n}i^3]

\lim_{n\rightarrow \infty}[1 +\frac{3}{n^2}\frac{n(n+1)}{2}+\frac{3}{n^3}\frac{n(n+1)(2n+1)}{6}+\frac{1}{n^4}\frac{n^2(n+1)^2}{4}]

\lim_{n\rightarrow \infty}[1 +\frac{3}{n^2}\frac{n^2+n}{2}+\frac{3}{n^3}\frac{2n^3+3n^2+n}{6}+\frac{1}{n^4}\frac{n^4+n^3+n^2}{4}]

\lim_{n\rightarrow \infty}[1 +\frac{3}{n^2}\frac{n^2(1+\frac{1}{n})}{2}+\frac{3}{n^3}\frac{n^3(2+\frac{3}{n}+\frac{1}{n^2}}{6}+\frac{1}{n^4}\frac{n^4(1+\frac{1}{n}+\frac{1}{n^2})}{4}

 =1+\frac{3}{2}+1+\frac{1}{4}

 = 3.75

--Antonio Moran 21:33 31 ago 2009 (CST)tonymoran



Ejemplo 5

\int_{1}^{4}(x^2+2x-5)dx

\Delta x = \frac{4-1}{n}= \frac{3}{n}

\lim_{\Delta x\rightarrow 0}\sum_{i=1}^{n}[x_{i}^2+2x_{i}-5]\Delta x

x_{i}= 1+\frac{3}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[(1+\frac{3i}{n})^2   +2(1+\frac{3i}{n})-5]\frac{3}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[\frac{9i^2 }{n^2 }+\frac{6i}{n}+1+\frac{6i}{n}+2-5]\frac{3}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[\frac{9i^2 }{n^2 }+\frac{12i}{n}-2]\frac{3}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[\frac{27i^2 }{n^3 }+\frac{36i}{n^2}-\frac{6}{n}]

\ \frac{27}{n^3}\sum_{i=1}^{n}[i^2]+\frac{36}{n^2}\sum_{i=1}^{n}[i] -\frac{6}{n}\sum_{i=1}^{n}[1 ]


\ \frac{27}{n^3}\frac{(n(n+1)(2n+1)}{6}+\frac{36}{n^2}\frac{(n(n+1)}{2} -6

\ \frac{27}{6}\frac{(n(n+1)(2n+1)}{n^3}+\frac{36}{2}\frac{(n(n+1)}{n^2} -6

\ \frac{27}{6}lim_{ n\rightarrow \infty} \frac{(n(n+1)(2n+1)}{n^3}+18 lim_{ n\rightarrow \infty}\frac{(n(n+1)}{n^2} -6

\ \frac{27}{6}lim_{ n\rightarrow \infty}(2)+18 lim_{ n\rightarrow \infty}(1) -6

\ \frac{27}{6}(2)+18(1) -6 = 21


\ R// 21

--Hersonjmc 23:26 31 ago 2009 (CST)Herson Marroquin


Ejemplo 6

\int_{0}^{3}(\frac{1}{2}x-1)dx

\Delta x = \frac{3-0}{n}= \frac{3}{n}

\lim_{\Delta x\rightarrow 0}\sum_{i=1}^{n}[\frac{1}{2}xi-1]\Delta x


x_{i}= \frac{3}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[\frac{1}{2}\frac{3}{n}i-1)]\frac{3}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[\frac{3/2 }{n}i-1]\frac{3}{n}

\lim_{ n\rightarrow \infty}\sum_{i=1}^{n}[\frac{9/2}{n^2}i-\frac{3}{n}]



\sum_{i=1}^{n}\frac{9/2}{n^2}i-\sum_{i=1}^{n}\frac{3}{n}


\ \frac{9/2}{n^2}\sum_{i=1}^{n}[i]-\frac{3}{n}\sum_{i=1}^{n}[1]


\ \frac{9/2}{n^2}\frac{(n(n+1))}{2}-3


\ \frac{9/2}{2}\frac{(n(n+1))}{n^2}-3

\ \frac{9}{4}\frac{(n(n+1))}{n^2}-3

\ \frac{9}{4}lim_{ n\rightarrow \infty} \frac{(n(n+1)}{n^2}-3

\ \frac{9}{4}lim_{ n\rightarrow \infty}(1)-3


\ \frac{9}{4}(1)-3 = - \frac{3}{4}

\ R//- \frac{3}{4}

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