nyuton - leybnits formulasi. xosmas integral.

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nyuton - leybnits formulasi nyuton - leybnits formulasi. xosmas integral. reja: 1. nyuton - leybnits formulasi. 2. xosmas integralni hisoblash. 3. bo’laklab integrallash yordamida aniq integralni hisoblash. nyuton - leybnits formulasi. aniq integral hisobining asosiy formulasi teorema: agar f(x) funksiya f(x) funksiyaning [a;b]kesmadagi boshlang’ich funksiyasi bo’lsa, u holda aniq integral boshlang’ich funksiyaning integrallash oralig’idagi orttirmasiga teng, ya`ni misol: aniq integralda o’zgaruvchini almashtirish integral berilgan bo’lsin, bunda f(x) funksiya [a,b] kesmada uzluksiz funksiya. x=((t) formula bo’yicha yangi t o’zgaruvchini kiritamiz. teorema: agar 1) ((()=a ; ((()=b ya`ni x=((t) funksiya ( va ( ni mos ravishda a va b ga o’tkazsa; 2) ((t) va (1(t) funksiyalar ham [(, (] kesmada uzluksiz funksiyalar bo’lsa, 3) f (((t)) funksiya [(, (] kesmada aniqlangan hamda uzluksiz bo’lsa, u holda quyidagi formula (1) o’rinli bo’ladi. isbot: agar f(x) funksiya f(x) ning boshlang’ich funksiyasi bo’lsa, u holda f(((t)) funksiya esa f(((t)) funksiyaning boshlang’ich funksiyasi bo’ladi. endi (1) …

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$ligidan tengliklar, ular mabjud bo`lgan taqdirda, o`rinlidir (12.7) ga asosan 1) ko`rinishda yozib olamiz. u holda a) yaqinlanuvchi. demak, yuqoridagi eslatmaga ko`ra va 2 - teoremaga asosan ning yaqinlanuvchi ekanligi kelib chiqadi. b) , ham yaqinlashuvchidir (yuqoridagidek ko`rsatiladi). demak, yaqinlashuvchidir. 2) a) , - yaqinlashuvchi (yuqoridagidek ko`rsatiladi), demak, yaqinlashuvchi b) va yaqinlashuvchi 2-misolga qarang, bundan 12.4- teoremaga asosan ning yaqinlashuvchiligi kelib chiqadi. demak, yaqinlashuvchi. shunday qilib, xosmas integralning yaqinlashuvchi ekanligi kelib chiqadi. > with(integrationtools): xi2 := int(1/sqrt(abs(x*(x^2-1))), x=-infinity..infinity); shape \* mergeformat > split(xi2, [-1, 0, 1]); shape \* mergeformat > xi2:=value(%); > evalf(xi2,5); 5-misol. xosmas integralning yaqinlashuvchiligi tekshirilsin (((r). yechish. bu yerda (((-(;1), (=1 va (((1;() bo`lgan uch holni ajratamiz. 1) ((1 bo`lsin, u holda oxirgi limit ( 1 bo`lsa, , ya`ni, - xosmas integral uzoqlashuvchi. 2) , ya`ni xosmas integral uzoqlashuvchi ekan. demak, -xosmas integral ( restart; > with(plots): f:=x->8/(x^2+4): > plot({f(x)}, x=-6..6, y=0..2,color=red, style=line, thickness=2, title=`yuza`); > xi1:=int( …$

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$ng ox o`qi atrofida aylanishdan hosil bolgan jisim xajmini hisoblang. yechish. aylanish xajmi elimenti: . izlanayotgan jism qiymatini chegarasi bo`lgan quyidagi integralga teng: 1)grafigini quyidagich quramiz: > restart; > with(plots): warning, the name changecoords has been redefined > implicitplot(y=2*(1/x-1/(x^2)), x=0..6, y=-1..1,color= blue, thickness=2); 2)jisim xajmini 2 xil usulda hisoblaymiz. a) formula bo`yicha: > xi4:=4*pi*int((1/x-1/(x^2))^2,x=1..infinity); shape \* mergeformat > xi4:=4*pi*int((1/x-1/(x^2))^2,x=1..infinity); b) volumeofrevolution buyrug`i bo`yicha jisim xajmini [1,6] dagi qismi: > restart; with(plots): with(student[calculus1]): > volumeofrevolution((x-1)/x^2,x=1..6,output=plot); > volumeofrevolution(2*(x-1)/x^2,x=1..6, output=integral); shape \* mergeformat > value(%); xi2 := ò k n n 1 | x ( x 2 k 1 ) | d x ∫ k n k 1 1 | x ( x 2 k 1 ) | d x c � k 1 0 1 | x ( x 2 k 1 ) | d x c � 0 1 1 | x ( x 2 k 1 ) | d x …$

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own _1405583394.unknown _1405583392.unknown _1011503632.unknown _1011503638.unknown _1011503641.unknown _1185106319.unknown _1011503639.unknown _1011503635.unknown _1011503637.unknown _1011503634.unknown _1011503625.unknown _1011503627.unknown _1011503630.unknown _1011503626.unknown _1011503622.unknown _1011503623.unknown _1011503621.unknown f x dx a b ( ) ò f x dx f x f b f a a b a b ( ) ( ) ( ) ( ) | ò = = - ( ) ( ) x dx x 3 1 2 4 1 2 4 4 4 1 4 2 1 1 4 16 1 1 4 15 15 4 ò = = - = - = = | ; f x dx a b ( ) ò ò ò ¢ = b a j j dt t t f dx x f b a )) ( )) ( ( ) ( ) ( t j ¢ ( ) ( ) ( ) f x dx f x f b f a f t t dt f t f …

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x x x dx x x dx x x dx ò ò ò ò - ¥ - - +¥ - = - - = - 1 1 0 2 0 1 2 1 2 2 ) 1 ( ) 1 ( ; ) 1 ( ) 1 ( x x dx x x dx x x dx x x dx ò ò ò - + - = - 1 0 1 5 , 0 2 5 , 0 0 2 2 ) 1 ( ) 1 ( ) 1 ( x x dx x x dx x x dx x x x x x x x 1 75 , 0 1 ) 1 ( 1 0 75 , 0 ) 1 ( ], 5 , 0 ; 0 ( 2 2 × £ - = - = - +¥ ® +¥ ® x x x x x x ò ò …

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nyuton - leybnits formulasi nyuton - leybnits formulasi. xosmas integral. reja: 1. nyuton - leybnits formulasi. 2. xosmas integralni hisoblash. 3. bo’laklab integrallash yordamida aniq integralni hisoblash. nyuton - leybnits formulasi. aniq integral hisobining asosiy formulasi teorema: agar f(x) funksiya f(x) funksiyaning [a;b]kesmadagi boshlang’ich funksiyasi bo’lsa, u holda aniq integral boshlang’ich funksiyaning integrallash oralig’idagi orttirmasiga teng, ya`ni misol: aniq integralda o’zgaruvchini almashtirish integral berilgan bo’lsin, bunda f(x) funksiya [a,b] kesmada uzluksiz funksiya. x=((t) formula bo’yicha yangi t o’zgaruvchini kiritamiz. teorema: agar 1) ((()=a ; ((()=b ya`ni x=((t) funksiya ( va ( ni mos ravishda a va b ga o’tkazsa; 2) ((t) va (1(t) funksiyalar ham [(, (] kesma...

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