Regla de simpson
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![Page 1: Regla de simpson](https://reader038.fdocumento.com/reader038/viewer/2022100802/587ad0b21a28ab760f8b6b6f/html5/thumbnails/1.jpg)
REPÚBLICA BOLIVARIANA DE VENEZUELA MINISTERIO DEL PODER POPULAR PARA LA EDUCACIÓN SUPERIOR
INSTITUTO UNIVERSITARIO POLITÉCNICO “SANTIAGO MARIÑO”
SAIA BARINAS PROGRAMACIÓN NUMERICA
INTEGRACIÓN NUMERICA
Regla de Simpson
Autor: Nestor Moreno
C.I. 14.331.859
Guarenas, Agosto del 2016
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Dadas las siguientes integrales:
1 − 𝑒−𝑥 𝑑𝑥
3
0
1 − 𝑥 − 4𝑥3 + 𝑥5 𝑑𝑥
4
−2
8 + 4𝑠𝑒𝑛𝑥 𝑑𝑥
𝑥/2
0
Resuelva las integrales utilizando la Regla de Simpson 3/8
Formula a usar:
I = 𝒇 𝒙 𝒅𝒙 =𝟑𝒉
𝟖
𝒃
𝒂[𝒇 𝒙𝟎 + 𝟑𝒇 𝒙𝟏 + 𝟑𝒇 𝒙𝟐 + 𝒇(𝒙𝟑)]
Donde:
h = 𝑏−𝑎
𝑛
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Ejercicio N° 1
1 − 𝑒−𝑥 𝑑𝑥
3
0
Cálculo de h (ancho de los sub-intervalos)
h = 𝑏−𝑎
𝑛
Donde: a = 0 b = 3 n = 3
h = 3−0
3=
3
3 = 1
Ahora cálculo de los puntos de sub intervalos
X0 = a = 0
X1 = x0 + h = 0 + 1 = 1
X2 = x1 + h = 1 + 1 = 2
X3 = x2 + h = 2 + 1 = 3
Cálculo de f(xi)
x f(x) = 1 – e-x
X0 = 0 f(0) = (1 – e-0) = 0 = f(x0)
X1 = 1 f(1) = (1 – e-1) = 0,632120 = f(x1)
X2 = 2 f(2) = (1 – e-2) = 0,864664 = f(x2)
X3 = 3 f(3) = (1 – e-3) = 0,950212 = f(x3)
Formula de Simpson para n = 3
I = 𝑓 𝑥 𝑑𝑥 =3ℎ
8
𝑏
𝑎[𝑓 𝑥0 + 3𝑓 𝑥1 + 3𝑓 𝑥2 + 𝑓(𝑥3)]
I = 𝑓 𝑥 𝑑𝑥 =3(1)
8
𝑏
𝑎[0 + 3(0,632120) + 3(0,864664) + 0,950212]
I = 𝑓 𝑥 𝑑𝑥 =3
8
𝑏
𝑎[1,89636 + 2,593992 + 0,950212]
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I = 𝑓 𝑥 𝑑𝑥 =3
8
𝑏
𝑎[5,440564]
I = 𝑓 𝑥 𝑑𝑥 =𝑏
𝑎2,0402115
I = 𝟏 − 𝒆−𝒙 𝒅𝒙𝟑
𝟎 = 2,0402115 valor aproximado
Ejercicio N° 2
1 − 𝑥 − 4𝑥3 + 𝑥5 𝑑𝑥
4
−2
Cálculo de h (ancho de los sub-intervalos)
h = 𝑏−𝑎
𝑛
Donde: a = - 2 b = 4 n = 3
h = 4−(−2)
3=
6
3 = 2
Ahora cálculo de los puntos de sub intervalos
X0 = a = - 2
X1 = x0 + h = -2 + 2 = 0
X2 = x1 + h = 0 + 2 = 2
X3 = x2 + h = 2 + 2 = 4
Cálculo de f(xi)
x f(x) = 1 – x – 4x-3 + x5
X0 = -2 f(0) = 1 – (-2) - 4(-2)3 + (-2)5 = 3 = f(x0)
X1 = 0 f(1) = 1 – 0 – 4(0)3 + 05 = 1 = f(x1)
X2 = 2 f(2) = 1 – 2 – 4(2)3 + (2)5 = -1 = f(x2)
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X3 = 4 f(3) = 1 – 4 – 4(4)3 + 45 = 765 = f(x3)
Formula de Simpson para n = 3
I = 𝑓 𝑥 𝑑𝑥 =3ℎ
8
𝑏
𝑎[𝑓 𝑥0 + 3𝑓 𝑥1 + 3𝑓 𝑥2 + 𝑓(𝑥3)]
I = 𝑓 𝑥 𝑑𝑥 =3(2)
8
𝑏
𝑎[3 + 3 1 + 3 −1 + 765]
I = 𝑓 𝑥 𝑑𝑥 =6
8
𝑏
𝑎[768]
I = 𝑓 𝑥 𝑑𝑥 =𝑏
𝑎576
𝑰 = 𝟏 − 𝒙 − 𝟒𝒙𝟑 + 𝒙𝟓 𝒅𝒙𝟒
−𝟐 = 576 valor aproximado
Ejercicio N° 3
8 + 4𝑠𝑒𝑛𝑥 𝑑𝑥
𝑥/2
0
Cálculo de h (ancho de los sub-intervalos)
h = 𝑏−𝑎
𝑛
Donde: a = 0 b =𝜋
2 n = 3
h = 𝜋
2− 0
3= 0,17
Ahora cálculo de los puntos de sub intervalos
X0 = a = 0
X1 = x0 + h = 0 + 0,17 = 0,17
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X2 = x1 + h = 0,17 + 0,17 = 0,51
X3 = x2 + h = 0,51 + 0,17 = 0,85
Cálculo de f(xi)
x f(x) = 8 + 4 senx
X0 = 0 f(0) = 8 + 4sen(0) = 8 = f(x0)
X1 = 0,17 f(1) = 8 + 4sen(0,17) = 8,011868 = f(x1)
X2 = 0,51 f(2) = 8 + 4sen(0,51) = 8,035604 = f(x2)
X3 = 0,85 f(3) = 8 + 4sen(0,85) = 8,059339 = f(x3)
Formula de Simpson para n = 3
I = 𝑓 𝑥 𝑑𝑥 =3ℎ
8
𝑏
𝑎[𝑓 𝑥0 + 3𝑓 𝑥1 + 3𝑓 𝑥2 + 𝑓(𝑥3)]
I = 𝑓 𝑥 𝑑𝑥 =3(0,17)
8
𝑏
𝑎[0 + 3 0,17 + 3 0,51 + 0,85]
I = 𝑓 𝑥 𝑑𝑥 =𝑏
𝑎0,06375 [0 + 0,51 + 1,53 + 0,85]
I = 𝑓 𝑥 𝑑𝑥 =𝑏
𝑎0,06375 [2,89]
I = 𝑓 𝑥 𝑑𝑥 =𝑏
𝑎0,1842375
𝟖 + 𝟒𝒔𝒆𝒏𝒙 𝒅𝒙 = 𝟎, 𝟏𝟖𝟒𝟐𝟑𝟕𝟓 𝒗𝒂𝒍𝒐𝒓 𝒂𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒅𝒐
𝒙/𝟐
𝟎