![ANIQ INTEGRALNI TAQRIBIY HISOBLASH USULLARI Reja: Kirish To`g`ri to`rt burchaklar formulasi. Trapetsiyalar formulasi. Simpson yoki parabola formulasi.](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img0.jpg)
ANIQ INTEGRALNI TAQRIBIY HISOBLASH USULLARI
Reja:
- Kirish
- To`g`ri to`rt burchaklar formulasi.
- Trapetsiyalar formulasi.
- Simpson yoki parabola formulasi.
![Tayanch iboralar 1. Integral 2. Boshlang`ich funksiya, 3. Integral chegarasi, 4. Bo`linish qadami, 5. To’g’ri to’rtburchak formulasi 6. Trapetsiyalar formulasi, 7. Simpson formulasi](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img1.jpg)
- Tayanch iboralar
1. Integral 2. Boshlang`ich funksiya, 3. Integral chegarasi, 4. Bo`linish qadami, 5. To’g’ri to’rtburchak formulasi 6. Trapetsiyalar formulasi, 7. Simpson formulasi
![KIRISH Agar [ a; b ] kesmada f(x) 0 bo`lsa, u xolda ning qiymati son jixatidan y = f(x) funktsiyani grafigi hamda x=a, x=b, to`g’ri chiziqlar bilan chegaralangan shakl (figura) ning yuziga teng (1-rasm). Agar [a;b] kesmada f(x) 0 bo`lsa, integralning qiymati yuqorida keltirilgan shaklning teskari ishora bilan olingan yuziga teng (2-rasm). Shunday kilib aniq integralni hisoblash deganda biror shaklning yuzini hisoblash tushuniladi. Quyida aniq integralni hisoblash uchun ba`zi taqribiy formulalar bilan tanishib chiqamiz.](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img2.jpg)
- KIRISH
- Agar [ a; b ] kesmada f(x) 0 bo`lsa, u xolda ning qiymati son jixatidan y = f(x) funktsiyani grafigi hamda x=a, x=b, to`g’ri chiziqlar bilan chegaralangan shakl (figura) ning yuziga teng (1-rasm). Agar [a;b] kesmada f(x) 0 bo`lsa, integralning qiymati yuqorida keltirilgan shaklning teskari ishora bilan olingan yuziga teng (2-rasm).
Shunday kilib aniq integralni hisoblash deganda biror shaklning yuzini hisoblash tushuniladi. Quyida aniq integralni hisoblash uchun ba`zi taqribiy formulalar bilan tanishib chiqamiz.
![Faraz kilaylik, bizdan aniq integralning taqribiy qiymatini topish talab etilsin. x 0 , x 1 , x 2 , . . . x n nuqtalar yordamida [a; b] kesmani N ta teng bulakchalarga bo`lamiz. Har bir bulakchaning uzunligi . Bo’linish nuqtalari esa: x 0 = a; x 1 = a + h; x 2 = x + 2h; x 3 = a+3h …; x n-1 = a+(N-1)h; x N = b](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img3.jpg)
Faraz kilaylik, bizdan
aniq integralning taqribiy qiymatini topish talab etilsin.
x 0 , x 1 , x 2 , . . . x n nuqtalar yordamida [a; b] kesmani
N ta teng bulakchalarga bo`lamiz.
Har bir bulakchaning uzunligi
.
Bo’linish nuqtalari esa:
x 0 = a; x 1 = a + h; x 2 = x + 2h;
x 3 = a+3h …; x n-1 = a+(N-1)h; x N = b
![Bu nuqtalarni tugun nuqtalar deb ataladi. f(x) funktsiyaning tugun nuqtalaridagi qiymatlari orqali aniqlanadi.](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img4.jpg)
Bu nuqtalarni tugun nuqtalar deb ataladi.
f(x) funktsiyaning tugun nuqtalaridagi qiymatlari
orqali aniqlanadi.
![[a,b] kesmani bo`lish natijasida hosil bo`lgan barcha to’rtburchaklarning yuzini hisoblab, ularni jamlab (1) Bu yerda to`g’ri turtburchak yuzini hisoblashda uning chap tomon ordinatasi olindi. Agar o’ng tomon ordinatami olsak ham shunday formulaga ega bo`lamiz: (2)](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img5.jpg)
[a,b] kesmani bo`lish natijasida hosil bo`lgan barcha to’rtburchaklarning yuzini hisoblab, ularni jamlab
(1)
Bu yerda to`g’ri turtburchak yuzini hisoblashda uning chap tomon ordinatasi olindi.
Agar o’ng tomon ordinatami olsak ham shunday formulaga ega bo`lamiz:
(2)
![Misol. To`g’ri turtburchaklar formulalari yordamida integralning taqribiy qiymatlari topilsin. Echish. Bu yerda a=0; b=1; n=10; h=(b- a)/n=0,1. x 0 =a=0; x 1 =a+h=0,1; x 2 =a+2h=0,2; x 3 =a+3h=0,3 x 4 =a+4h=0,9 … x 9 =a+9h=0,9; x 10 =b=1](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img6.jpg)
Misol. To`g’ri turtburchaklar formulalari yordamida
integralning taqribiy qiymatlari topilsin.
Echish. Bu yerda a=0; b=1; n=10; h=(b- a)/n=0,1.
x 0 =a=0; x 1 =a+h=0,1; x 2 =a+2h=0,2; x 3 =a+3h=0,3
x 4 =a+4h=0,9 … x 9 =a+9h=0,9; x 10 =b=1
![dan dan Ma`lumki , aniq yechim chap va o’ng formulalar orqali topilgan yechimlar orasida yotadi.](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img7.jpg)
- dan
- dan
Ma`lumki ,
aniq yechim chap va o’ng formulalar orqali topilgan yechimlar orasida yotadi.
![Trapetsiyalar formulasi Bu xulosalarni nazarga olgan xolda (1) va (2) formulalar xadlarini moc ravishda ko’shib o’rta arifmetigini olsak, quyidagi ifoda hosil bo`ladi: (3) (3) formula trapetsiyalar formulasi deb ataladi. Bu formula yordamida topilgan integralning taqribii qiymatining aniqligini oshirish uchun bulinish nuqtalari soni n» ni ikki, uch va x.k. marta oshirish kerak bo`ladi. Albatta bunda ham hisoblash xajmi bir necha marotaba oshadi.](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img8.jpg)
Trapetsiyalar formulasi
Bu xulosalarni nazarga olgan xolda (1) va (2) formulalar xadlarini moc ravishda ko’shib o’rta arifmetigini olsak, quyidagi ifoda hosil bo`ladi:
(3)
(3) formula trapetsiyalar formulasi deb ataladi. Bu formula yordamida topilgan integralning taqribii qiymatining aniqligini oshirish uchun bulinish nuqtalari soni n» ni ikki, uch va x.k. marta oshirish kerak bo`ladi. Albatta bunda ham hisoblash xajmi bir necha marotaba oshadi.
![Misol. Bo`linishlar soni 10 bo`lganda ε=0,001 aniqlikda hisoblang. Yechish. h=(b-a)/n=(3,5-2)/10=0,15 bo`linish nuqtalari x i =a+ih, i =1,.., 10 x 0 =2,00 x 1 =2,15 x 2 =2,30 y 2 =f(2,30)=0,3350 x 3 =2,45 y 3 =f(2,45)=0,3371 x 4 =2,60 y 4 =f(2,60)=0,3402 x 5 =2,75 y 5 =f(2,75)=0,3443 x 6 =2,90 y 6 =f(2,90)=0,3494 x 7 =3,05 y 7 =f(3,05)=0,3558 x 8 =3,20 y 8 =f(3,20)=0,3637 x 9 =3,35 y 9 =f(3,35)=0,3733 x 10 =3,50 y 10 =f(3,50)=0,3849](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img9.jpg)
Misol.
Bo`linishlar soni 10 bo`lganda ε=0,001 aniqlikda hisoblang.
Yechish.
h=(b-a)/n=(3,5-2)/10=0,15
bo`linish nuqtalari x i =a+ih, i =1,.., 10
x 0 =2,00
x 1 =2,15
x 2 =2,30 y 2 =f(2,30)=0,3350
x 3 =2,45 y 3 =f(2,45)=0,3371
x 4 =2,60 y 4 =f(2,60)=0,3402
x 5 =2,75 y 5 =f(2,75)=0,3443
x 6 =2,90 y 6 =f(2,90)=0,3494
x 7 =3,05 y 7 =f(3,05)=0,3558
x 8 =3,20 y 8 =f(3,20)=0,3637
x 9 =3,35 y 9 =f(3,35)=0,3733
x 10 =3,50 y 10 =f(3,50)=0,3849
![(3) formulaga ko’ra](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img10.jpg)
(3) formulaga ko’ra
![Simpson yoki parabola formulasi . [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(1,0515+4*1,7443+ +2*1,3883)=0,05*10,8013= 0,54265 .](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img11.jpg)
Simpson yoki parabola formulasi .
[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(1,0515+4*1,7443+ +2*1,3883)=0,05*10,8013= 0,54265 .
![Этиборингиз учун рахмат](https://fsd.videouroki.net/html/2020/04/06/v_5e8b2153d2f9d/img12.jpg)
Этиборингиз учун рахмат