# Practica 3 an

Páginas: 2 (424 palabras) Publicado: 17 de enero de 2016
﻿Problema #1

Problema #2

La solución de la integral definida es: 0.9524+0.4082475+0.195383333 = 1.5560
Problema #3

M0=0
V(x)=5+0.25x2
M=5(11)+0.25/3(11)3=165.916
∆x=1Xi=x0+∆x1
x0=0
x1=0+(1)(1)=1
x2=0+1(2)=2
x3 = 3
x4=4
x5=5
x6=6
x7=7
x8=8
x9=9
x10=10
x11=11
f(x0)=5+0.25(0)2=5
f(x1)= 5+0.25(1)2=5.25
f(x2)= 5+0.25(2)2=6
f(x3)= 5+0.25(3)2=7.25
f(x4)=5+0.25(4)2=9
f(x5)= 5+0.25(5)2=11.25
f(x6)= 5+0.25(6)2=14
f(x7)= 5+0.25(7)2=17.25
f(x8)= 5+0.25(8)2=21
f(x9) 5+0.25(9)2=25.25
f(x10)= 5+0.25(10)2=30
f(x11)= 5+0.25(11)2=35.25I=1/2[5+2(5.25+6+7.25+9+11.25+14+17.25+21+25.25+30)+35.25]=166.375
C)
Tramo(0-9) simpson 3/8
Tramo (9-11) simpson 1/3
I=3∆x/8[5+3(5.25)+3(6)+2(7.25)+3(9)+3(11.25)+2(14)+3(17.25)+3(21)+25.25]+∆x/3[25.25+4(30)+35.25]=165.75

Problema #4

Para h=4(Trapecio)
I=20[0+2(2+4+4+3.4)+0/10]=53.6
Para h=2(Trapecio)
I=20[0+2(1.8+2+4+4+6+4+3.6+3.4+2.8)+0/20]=63.2
Para h=2(simpson1/3)I=20[0+4(1.8+4+6+3.6+2.8)+2(2+4+4+3.4)+0/30]=66.4

Problema #5

Tramo 1 (0-2) Trapecio
∆x = 2
n=1
T=∆ x/2[f(x1)+f(x2)]=2/2[0.04+0.040685]=0.08068
Tramo 2 (2-4) Simpson 1/3
∆x =1I=∆x/3[f(x0)+4f(x1)+f(x2)]=(1)(1/3)= [(0.040685)+4(0.041234)+0.0418]=0.082473
Tramo 3(4-10) simpson 3/8
∆x=2
I=3∆x/8[f(x)+3f(x1)+2f(x2)+f(x2)]=(3/8)(2)[0.040685+4(0.041234)+0.05115]=0.26895Total = tramo 1+ Tramo2+ Tramo 3 = 0.432108 Gramos = 0.000432108 Kilogramos
Problema #6

a-)
X
t
7:30
7:45
8:00
8:15
8:45
9:15
Y

18
24
14
24
21
9--------0.25 ---------- 0.25 ----------0.25 ---------- 0.5 ----------- 0.5 ---------
Simpson 3/8Simpson 1/3

Simpson 3/8 = 0.25(3)/8 (18 + 3(24) + 3(14) + 24) = 12. 93
Simpson 1/3 = 0.5/3 (24 + 4(21) + 9) = 19.5
Total = 32.43

b-)
At/4 = 32.43/4 = 8.10 autos

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