Reja: Aniq integralni taqribiy hisoblash. Umumiy mulohazalar

7.1-Masala.Quyidagi



aniq integralni:

1. 
   to‘g’ri to‘rtburchak formulasi bilan;
   
2. 
   trapetsiyalar formulasi bilan;
   
3. 
   Simpson formulasi bilan
   

bo‘linishlar soni 10 bo‘lganda hisoblang va xatoliklarni baholang.

Yechish.

Integral chegaralari [a,b] oraliqni bo‘luvchi qadami

h=(b-a)/n=(3.5-2)/10=0.15

bo‘lganda, bo‘linish nuqtalari x_i=a+ih, i=1,…,10 o‘lsa, nuqtalarni [2,3.5] oraliqda aniqlab, bu nuqtalarda integral ostidagi funksiya qiymatlarini topamiz.

x₀=2.00

x₁=2.15

x₂=2.30 y₂=f(8.30)=0.3350

x₃=2.45 y ₃=f(7.15)=0.3371

x₄=2.60 y₄=f(8.60)=0.3402

x₅=2.75 y ₅=f(8.75)=0.3443

x₆=2.90 y ₆=f(8.90)=0.3494

x₇=3.05 y ₇=f(3.05)=0.3558

x₈=3.20 y ₈=f(3.20)=0.3637

x₉=3.35 y ₉=f(3.35)=0.3733

x₁₀=3.50 y ₁₀=f(3.50)=0.3849

Yuqoridagi x va y qiymatlariga ko‘ra

1) o‘ng to‘g’ri to‘rtburchaklar formulasiga asosan:

=0.15(0.3338+0.3350+0.3371+0.3402+0.3443+0.3494+ +0.3558+0.3637+0.3733+0.3849)=0.5276;

2) Trapetsiyalar formulasiga asosan:

=0.15( +0.3338+0.3350+0.3371+0.3402+ +0.3443+0.3494+0.3558+0.3637+0.3733)=0.15*3.4917=0.523;

3)Simpson formulasiga asosan:

[0.3333+0.3849+4(0.3338+0.3371+0.3443+0.3558+ +0.3733)+2(0.3350+0.3402+0.3494+0.3637)]=0.05(0.7182+4*1.7493+ +2*1.3883)=0.05*10.4720=0.5236.
7.1-Maple 7 dasturi

Berilgan integralni :



1. Boshlang’ic funksiyasini topish:

> Int( 1/sqrt(5+4*x-x^2), x )=int( 1/sqrt(5+4*x-x^2), x );



2. 10 xona aniqlikda taqribiy hisoblash.

> I1 := 1/sqrt(5+4*x-x^2):

Int(I1, x = 2..3.5) = evalf(Int(e1, x=2..3.5, digits=10, method=_Dex));



> evalf(Int( 1/sqrt(5+4*x-x^2), x=2..3.5 ));



> evalf[25](Int( 1/sqrt(5+4*x-x^2), x=2..3.5 ));