O‘zbekiston respublikasi raqamli texnologiyalar vazirligi muhammad al-xorazmiy nomidagi toshkent axborot texnologiyalari universiteti teleradio eshittirish va

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.