egri chiziqli trapetsiya

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aim.uz egri chiziqli trapetsiyaning yuzini reja: 1.egri chiziqli trapetsiyaning yuzi. 2.misollar yechish. kesmada aniqlangan funksiya qaralayotgan bo’lsin. kesmani 4 ta ushbu ko’rinishdagi bo’lakchalarga bo’lib, uzunligi bo’lgan xar bir bo’lakchada ixtiyoriy nuqta tanlaymiz. ushbu ko’rinishdagi yig’indiga funksiyaning kesmadagi integral yig’indisi deyiladi. 2. agar chekki ilmiy mavjud bo’lsa, u xolda i soniga funksiyadan kesmada olingan aniq integral deyiladi va quyidagicha yoziladi: a – integrallashning quyi chegarasi b – integralllashni yuqori chegarasi. 3. agar funksiya temada uzluksiz bo’lsa, u xolda, i chekli limit mavjud bo’ladi. bu limit kesmani bo’lakchalargabo’lish usulidan va nuqtani tanlashdan bog’liq bo’lmaydi. 4. agar kesmada funksiya manfiy bo’lmasa, u xolda aniq integral ushbu chiziqlar bilan chegaralangan egri chiziqli trapesiyaning yuzini ifodalaydi (1-shakl.) misol. integralni integral yig’indidan foydalanib hisoblang. yechish. bu erda embed equation.3 kesmani n ta teng bo’lakchalarga bo’lamiz. u xolda bo’ladi. sifatida ni tanlaymiz. natijada: embed equation.3 embed equation.3 hosil bo’ladi va bo’lishi uchun intilishi ta`rifga ko’ra: ma`lumki, shunga …

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inishda bo’ladi: (8) shuningdek, quyi chegarasi o’zgaruvchidan iborat bo’lgan aniq integral ifodasi esa quyidagicha bo’ladi: (9) aniq integralni hisoblashda quyidagi bosqich ishlari ketma – ket bajariladi: 1. quyidagi aniqmas integral topiladi: 2. ning dagi qiymati topiladi, ya’ni 3. ning dagi qiymati hisoblanadi, ya’ni topiladi. nazorat savollari: 1. oraliqlarda chegarallangan figurallarning yuzini topish formulalarni toping? 2. integral deb nimaga aytiladi? 3. aniq integralning xossalari? misollar. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. tayanch so’z lar. boshlang’ich funksiya intregral aniqmas integral egri chiziqli trapetsiyaning yuzi adabiyotlar: “algebra va matematik analiz asoslari” i-ii qism a.u. abduhamidov, h.a. nasimov ,u.m.nosirov, j.h.husanov.sh.o.alimov alimov “algebra va analiz asoslari” aim.uz _1555603126.unknown _1555603142.unknown _1555603158.unknown _1555603166.unknown _1555603174.unknown _1555603178.unknown _1555603180.unknown _1555603182.unknown _1555603183.unknown _1555603181.unknown _1555603179.unknown _1555603176.unknown _1555603177.unknown _1555603175.unknown _1555603170.unknown _1555603172.unknown _1555603173.unknown _1555603171.unknown _1555603168.unknown _1555603169.unknown _1555603167.unknown _1555603162.unknown _1555603164.unknown _1555603165.unknown _1555603163.unknown _1555603160.unknown _1555603161.unknown _1555603159.unknown _1555603150.unknown _1555603154.unknown _1555603156.unknown _1555603157.unknown _1555603155.unknown _1555603152.unknown _1555603153.unknown _1555603151.unknown _1555603146.unknown _1555603148.unknown _1555603149.unknown _1555603147.unknown _1555603144.unknown _1555603145.unknown …

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_1555603117.unknown _1555603122.unknown _1555603124.unknown _1555603125.unknown _1555603123.unknown _1555603119.unknown _1555603121.unknown _1555603118.unknown _1555603113.unknown _1555603115.unknown _1555603116.unknown _1555603114.unknown _1555603111.unknown _1555603112.unknown _1555603110.unknown _1555603101.unknown _1555603105.unknown _1555603107.unknown _1555603108.unknown _1555603106.unknown _1555603103.unknown _1555603104.unknown _1555603102.unknown _1555603097.unknown _1555603099.unknown _1555603100.unknown _1555603098.unknown _1555603095.unknown _1555603096.unknown _1555603094.unknown [ ] b a ; [ ] b a ; b x x x x x a = < < < < < = - 4 1 4 2 1 0 ... 1 - - = d i i i x x x x å - d = 4 1 ) ( i i x f g x ( ) x f [ ] b a ; i x f i i x i = d å = ® d 4 1 0 max ) ( lim x ( ) x f [ ] b a ; ò = b a dx x f i ) ( ( ) x f [ ] b a ; [ ] b a …

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( ) . a f b f x f dx x f b a b a - = = ò ( ) ( ) ( ) ( ) . a f x f t f dt t f x a x a - = = ò ( ) ( ) ( ) ( ) . x f b f t f dt t f b x b x - = = ò ( ) ò b a dx x f ( ) ( ) ò + = . c x f dx x f ( ) c x f + b x = ( ) . c b f + ( ) c x f + a x = ( ) . c a f + ( ) [ ] ( ) [ ] ( ) ( ) a f b f c a f c b f - = …

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aim.uz egri chiziqli trapetsiyaning yuzini reja: 1.egri chiziqli trapetsiyaning yuzi. 2.misollar yechish. kesmada aniqlangan funksiya qaralayotgan bo’lsin. kesmani 4 ta ushbu ko’rinishdagi bo’lakchalarga bo’lib, uzunligi bo’lgan xar bir bo’lakchada ixtiyoriy nuqta tanlaymiz. ushbu ko’rinishdagi yig’indiga funksiyaning kesmadagi integral yig’indisi deyiladi. 2. agar chekki ilmiy mavjud bo’lsa, u xolda i soniga funksiyadan kesmada olingan aniq integral deyiladi va quyidagicha yoziladi: a – integrallashning quyi chegarasi b – integralllashni yuqori chegarasi. 3. agar funksiya temada uzluksiz bo’lsa, u xolda, i chekli limit mavjud bo’ladi. bu limit kesmani bo’lakchalargabo’lish usulidan va nuqtani tanlashdan bog’liq bo’lmaydi. 4. agar kesmada funksiya manfiy bo’lmasa, u xolda aniq integral ush...

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