aniq integral 2

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1576220891.doc [ ] a b , f x ( ) ³ 0 [ ] a b , a x x x x b n = > > = 0 1 . . . . d x x x i i i = - b bo`lsa =- bo`ladi, chunki bo`lib bo`ladi. agar a=b bo`lsa =0 bo`ladi. integral o`zgaruvchini har qanday harf bilan belgilash mumkin: = = = 3. aniq integralning xossalari. 1-xossa. o`zgarmas ko`paytuvchini aniq integral belgisidan tashqari chiqarish mumkin: a (a-o`zgarmas son ) isboti. = = = a keyingi xossalari ham aniq integralning ta`rifidan va limitning xossalaridan foydalanib osongina isbotlanadi. shuning uchun biz ularning isbotini o`quvchilarga havola qilib, ba`zilarining geometrik ma`nosini ko`rsatib ketamiz. 2-xossa. 3-xossa. agar da f(x) va funksiyalar f(x) tengsizlikni qanoatlantirsalar, u holda munosabat o`rinli bo`ladi. isboti. geometrik nuqtai nazardan ko`raylik. agar f(x)>0 , >0 bo`lib, f(x) bo`lsa u holda aa1b1b egri chiziqli trapesiyaning yuzi aa2b2b egri chiziqli …

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tegral bo`lsin biz bilamizki integral o`zgaruvchini istalgan harf bilan belgilash mumkin, shuning uchun = deylik. biz bilamizki agar f(x) funksiya [a,b] da integrallanuvchi bo`lsa, [a,x] da ( ) ya`ni ) ham integrallanuvchi bo`lib, integral qiymati x ning funksiyasi bo`ladi. (1) (1) integral acdx egri chiziqli trapesiyaning yuzini ifodalaydi. agar [a,b] da x o`zgarsa ham o`zgarishi ravshan. y d c 0 a x ( x+(x b x teorema. agar f(u) funksiya u= nuqtada uzluksiz bo`lsa, u holda integraldan yuqori chegarasi bo`yicha olingan hosila integral ostidagi funksiyaga teng bo`lib, unda integrallash o`zgaruvchisi o`rniga integralning yuqori chegarasi qo`yiladi, ya`ni yoki =f(x) bo`ladi. isboti. x ga orttirma bersak funksiya ham orttirma oladi: . agar integralga o`rta qiymat xaqidagi teoremani tatbiq qilsak ( ) bo`lib, bo`ladi. hosilaning ta`rifiga ko`ra chunki da demak bo`lgani uchun da ravshan. shunday qilib bu isbot qilingan, matematik analiz kursining asosiy teoremalarining biri bo`lgan teoremadan ko`rinadiki [a,b] da uzluksiz bo`lgan har …

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n va uzluksiz hosilaga ega bo`lsin. 2. yangi o`zgaruvchi t [c,d] kesmada o`zgarganda funksiyaning qiymati dan chiqmasin. 3. bo`lsin, bu holda (3) tenglik o`rinli bo`ladi. isboti. f(x) funksiya da uzluksiz bo`lgani uchun uning shu kesmadagi boshlang`ich funksiyasini f(x) desak u holda nyuton-leybnis formulasiga ko`ra (4) tenglik o`rinli bo`ladi. agar desak f(x)=f( ) funksiyaning funksiya uchun boshlang`ich funksiya ekanligini ko`rish qiyin emas. haqiqatan chunki desak [ ]=f[ ]. demak nyuton-leybnis formulasiga ko`ra =f(b)-f(a)=( (4) ga ko`ra )= misol yechganda integral ostidagi funksiya yuqoridagi shartlarni qanoatlantirishi shart. misol. 1) 2. 6. aniq integralni bo`laklab integrallash. teorema. agar u(x), v(x) funksiyalar kesmada uzluksiz bo`lgan va hosilalarga ega bo`lsa, u holda (1) formula o`rinli bo`ladi. isboti. haqiqatan [u(x) v(x)]`=u`(x)v(x)+u(x)v`(x) dan ko`rinadiki u(x) v(x) funksiya u`(x)v(x)+u(x)v`(x) funksiya uchun da boshlang`ich funksiya, bu holda nyuton-leybnis formulasiga ko`ra yoki yoki bu yerdagi asosiy maqsad integralda u va dv deb shunday tanlash kerakki natijada integral integralga nisbatan ancha …

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1576220891.doc [ ] a b , f x ( ) ³ 0 [ ] a b , a x x x x b n = > > = 0 1 . . . . d x x x i i i = - b bo`lsa =- bo`ladi, chunki bo`lib bo`ladi. agar a=b bo`lsa =0 bo`ladi. integral o`zgaruvchini har qanday harf bilan belgilash mumkin: = = = 3. aniq integralning xossalari. 1-xossa. o`zgarmas ko`paytuvchini aniq integral belgisidan tashqari chiqarish mumkin: a (a-o`zgarmas son ) isboti. = = = a keyingi xossalari ham aniq integralning ta`rifidan va limitning xossalaridan foydalanib osongina isbotlanadi. shuning uchun biz ularning isbotini o`quvchilarga havola qilib, ba`zilarining geometrik ma`nosini ko`rsatib ketamiz. 2-xossa. 3-xossa. agar da f(x) va funksiyalar …

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