This document contains the solutions to 3 integral calculus problems using the Simpson's 3/8 rule. For each problem, the document calculates the interval width h, determines the interval points, calculates the function values f(xi) at those points, and applies the Simpson's 3/8 formula to approximate the integral. The three problems integrate the functions 1) 1 − e−x from 0 to 3, 2) 1 − x − 4x3 + x5 from -2 to 4, and 3) 8 + 4senx from 0 to π/2. The Simpson's rule is used to find approximate values of 2.0402115, 576, and 0.1842375 for

1. 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

2. 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 𝒙 𝟎 + 𝟑𝒇 𝒙 𝟏 + 𝟑𝒇 𝒙 𝟐 + 𝒇(𝒙 𝟑)]
Donde:
h =
𝑏−𝑎
𝑛

3. 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]

4. I = 𝑓 𝑥 𝑑𝑥 =
3
8
𝑏
𝑎
[5,440564]
I = 𝑓 𝑥 𝑑𝑥 =
𝑏
𝑎
2,0402115
I = 𝟏 − 𝒆−𝒙
𝒅𝒙
𝟑
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)

5. 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
𝑎
576
𝑰 = 𝟏 − 𝒙 − 𝟒𝒙 𝟑
+ 𝒙 𝟓
𝒅𝒙
𝟒
−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

6. 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
𝑎
0,06375 [0 + 0,51 + 1,53 + 0,85]
I = 𝑓 𝑥 𝑑𝑥 =
𝑏
𝑎
0,06375 [2,89]
I = 𝑓 𝑥 𝑑𝑥 =
𝑏
𝑎
0,1842375
𝟖 + 𝟒𝒔𝒆𝒏𝒙 𝒅𝒙 = 𝟎, 𝟏𝟖𝟒𝟐𝟑𝟕𝟓 𝒗𝒂𝒍𝒐𝒓 𝒂𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒅𝒐
𝒙/𝟐
𝟎