O‘zbekiston respublikasi oliy va o‘rta maxsus ta’lim vazirligi samarqand iqtisodiyot va servis instituti «oliy matematika» kafedrasi
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oliy matematika
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minimumga ega deyiladi. 4) 0 x nuqtaning shunday atrofi mavjud bo’lsaki, bu atrofning har qanday 0 x x nuqtasi uchun ) ( ) ( 0 x f x f tengsizlik bajarilsa, ) (x f y funksiya 0 x nuqtada minimumga ega deyiladi. A) 1),3) ) 1),2),3) D) 2),3)4) E) hammasi 236. Ekstremumga ega bo’lshishinig zaruriy shartini toping. A) ) (x f y funksiya 0 x nuqtada ekstremumga ega bo’lsa, ) ( 0 x f y no’lga teng yoki u mavjud bo’lmaydi ) ) (x f y funksiya 0 x nuqtada ekstremumga ega bo’lsa, ) ( 0 x f y no’lga teng bo’lmaydi D) ) (x f y funksiya 0 x nuqtada ekstremumga ega bo’lsa, ) ( 0 x f y no’ldan katta bo’ladi E) ) (x f y funksiya 0 x nuqtada ekstremumga ega bo’lsa, ) ( 0 x f y hosila mavjud bo’lmaydi 237. Quyidagilarning qaysilarida ekstremumning yetarli shartlari to’g’ri berilgan: 1) 0 x nuqta ) (x f y funksiyaning kritik nuqtasi bo’lib, funksiya hosilasi ishorasi bu nuqtadan o’tishda ishorasini o’zgartirsa, 0 x nuqta, funksiyaning ekstremum nuqtasi, va: ) 0 x nuqtadan chapdan o’ngga o’tishda ) (x f o’z ishorasini musbatdan manfiyga o’zgartirsa, 0 x nuqtada funksiya maksimumga; ) 0 x nuqtadan chapdan o’ngga o’tishda ) (x f o’z ishorasini manfiydan musbatga o’zgartirsa, 0 x nuqtada funksiya minimumga ega bo’ladi; 2) 0 x nuqta ) (x f y funksiyaning kritik nuqtasi bo’lib, funksiya hosilasi ishorasi bu nuqtadan o’tishda ishorasini o’zgartirsa, 0 x nuqta, funksiyaning ekstremum nuqtasi, va: ) 0 x nuqtadan chapdan o’ngga o’tishda ) (x f o’z ishorasini musbatdan manfiyga o’zgartirsa, 0 x nuqtada funksiya minimumga ) 0 x nuqtadan chapdan o’ngga o’tishda ) (x f o’z ishorasini manfiydan musbatga o’zgartirsa, 0 x nuqtada funksiya maksimumga ega bo’ladi; 3) 0 x nuqta ) (x f y funksiyaning kritik nuqtasi bo’lib, funksiya hosilasi ishorasi bu nuqtadan o’tishda ishorasini o’zgartirmasa, 0 x nuqta, funksiyaning ekstremum nuqtasi, va: ) 0 x nuqtadan chapdan o’ngga o’tishda ) (x f o’z ishorasini musbatdan manfiyga o’zgartirsa, 0 x nuqtada funksiya maksimumga; ) 0 x 128 nuqtadan chapdan o’ngga o’tishda ) (x f o’z ishorasini manfiydan musbatga o’zgartirsa, 0 x nuqtada funksiya minimumga ega bo’ladi A) 1) ) 2) D) 3) E) hammasida 238. Quyidagilarning qaysilarida ekstremumning ikkinchi qoidasi to’g’ri berilgan: 1) 0 x nuqtada birinchi hosila nolga teng bo’lib, ikkinchi hosila no’ldan farqli bo’lsa, 0 x nuqta funksiyaning ekstremum nuqtasi va : 0 ) ( 0 x f bo’lsa, maksimum nuqtasi; 0 ) ( 0 x f bo’lsa, minimum nuqtasi bo’ladi; 2) 0 x nuqtada birinchi hosila nolga teng bo’lib, ikkinchi hosila no’ldan farqli bo’lsa, 0 x nuqta funksiyaning ekstremum nuqtasi va : 0 ) ( 0 x f bo’lsa, minimum nuqtasi; 0 ) ( 0 x f bo’lsa, minimum nuqtasi bo’ladi; 3) 0 x nuqtada birinchi hosila noldan farqli bo’lib, ikkinchi hosila no’lga teng bo’lsa, 0 x nuqta funksiyaning ekstremum nuqtasi va : 0 ) ( 0 x f bo’lsa, maksimum nuqtasi; 0 ) ( 0 x f bo’lsa, minimum nuqtasi bo’ladi A) 1) ) hammasi D) 2) E) 3) 239. 3 2 2 6 2 1 3 1 ) ( 2 3 x x x x f funksiyaninshg ekstremumini toping. A) , 10 max min y 6 65 ) 10 max D) min y 6 65 E) ekstremum yo’q 240. 4 9 6 2 11 2 4 1 ) ( 2 3 4 x x x x x f funksiya ekstremumini ikkinchi qoida bilan toping. A) 0 ) 3 ( min ; 25 . 0 ) 2 ( max ; 0 ) 1 ( min f f f ) 25 . 0 ) 2 ( max ; 0 ) 1 ( min f f D) 0 ) 3 ( min ; 25 . 0 ) 2 ( max f f E) 0 ) 3 ( min ; 0 ) 1 ( min f f 241. Funksiyaning eng kichik va eng katta qiymatlarini topish ketma-ketligi quyidagi raqamlarning qaysilarida to’g’ri berilgan: 1) ) (x f y funksiyaning b a, kesmadagi eng kichik va eng katta qiymatlarini topish uchun: ) kritik nuqtalarni topamiz; ) funksiyaning bu kritik nuqtalardagi qiymatlarini hisoblaymiz; ) d bu topilgan qiymatlarni taqqoslab, eng kichigi funksiyaning berilgan kesmadagi eng kichik qiymati, eng kattasi bu kesmadagi eng katta qiymati ekanligini topamiz; 2) ) (x f y funksiyaning b a, kesmadagi eng kichik va eng katta qiymatlarini topish uchun: ) kritik nuqtalarni topamiz; ) funksiyaning bu kritik nuqtalardagi va kesmaning chetlaridagi qiymatlarini 129 hisoblaymiz; ) d bu topilgan qiymatlarni taqqoslab, eng kichigi funksiyaning berilgan kesmadagi eng kichik qiymati, eng kattasi bu kesmadagi eng katta qiymati ekanligini topamiz; 3) ) (x f y funksiyaning b a, kesmadagi eng kichik va eng katta qiymatlarini topish uchun: ) kritik nuqtalarni topamiz; ) funksiyaning kesmaning chetlaridagi qiymatlarini hisoblaymiz; ) d bu topilgan qiymatlarni taqqoslab, eng kichigi funksiyaning berilgan kesmadagi eng kichik qiymati, eng kattasi bu kesmadagi eng katta qiymati ekanligini topamiz. A) 2) ) 1) D) hammasi E) 3) 242. 5 2 ) ( 2 4 x x x f y funksiyaning 3 ; 2 kesmadagi eng kichik va eng katta qiymatlarini toping. A) 69 , 4 . . y y ) 4 . y D) 68 . y E) bunday qiymatlari yo’q 243. Funksiya grafigining qavariq yoki botiq bo’lishining yetarli shartlari quyidagilarning qaysi raqamlarida to’g’ri berilgan: 1) ) , ( b a oraliqda differensiallanuvchi ) (x f y funksiyaning ikkinchi tartibli hosilasi manfiy, ya’ni 0 ) (x f bo’lsa, bu oraliqda funksiya grafigi qavariq bo’ladi; 2) ) , ( b a oraliqda differensiallanuvchi ) (x f y funksiyaning ikkinchi tartibli hosilasi manfiy, ya’ni 0 ) (x f bo’lsa, bu oraliqda funksiya grafigi botiq bo’ladi; 3) ) , ( b a oraliqda differensiallanuvchi ) (x f y funksiyaning ikkinchi tartibli hosilasi musbat, ya’ni 0 ) (x f bo’lsa, bu oraliqda funksiya grafigi botiq bo’ladi. A) 1),3) ) hammasi to’g’ri D) 2),3) E) 1),2) 244. Egilish nuqtalari mavjud bo’lishining yetarli sharti quyidagi raqamlarning qaysilarida to’g’ri berilgan: 1) 0 nuqta ) (x f y funksiya uchun ikkinchi tur kritik nuqta bo’lsa va ) (x f ikkinchi tartibli hosila bu nuqtadan o’tishda ishorasni o’zgartirmasa, 0 x abssissali nuqta egilish nuqtasi bo’ladi; 2) 0 nuqta ) (x f y funksiya uchun ikkinchi tur kritik nuqta bo’lsa va ) (x f ikkinchi tartibli hosila bu nuqtadan o’tishda ishorasni musbatdan manfiyga o’zgartirsa, 0 x abssissali nuqta egilish nuqtasi bo’ladi; 3) 0 nuqta ) (x f y funksiya uchun ikkinchi tur kritik nuqta bo’lsa va ) (x f ikkinchi tartibli hosila bu nuqtadan o’tishda ishorasni o’zgartirsa, 0 x abssissali nuqta egilish nuqtasi bo’ladi. A) 3) ) 1) D) hammasi E) 2) 245. ) (x f y funksiya grafigining b kx y og’ma asimptotasi k va b parametrlarini topish, quyidagi raqamlarning qaysilarida to’g’ri berilgan: 130 1) kx x f b x x f k x x ) ( lim ) ( lim ; 2) bx x f k x x f b x x ) ( lim ) ( lim ; 3) kx x f b x x f k x x ) ( lim ) ( lim . A) 1) ) 2) D) 3) E) hammasi noto’g’ri Integral hisob 246. 5 x x f y funksiyaning hamma boshlang’ich funksiyalarini toping. A) , 6 6 ixtiyoriy o’zgarmas son ) 4 5 D) 105 6 6 E) 6 6 247. Aniqmas integral ta’rifi quyidagi raqamlarning qaysi larida to’g’ri berilgan: 1) ) (x f funksiya biror oraliqda ) (x F funksiyaning boshlang’ich funksiyasi bo’lsa, C x F ) ( (bunda C ixtiyoriy o’zgarmaD) funksiyalar to’plami shu oraliqda ) (x f funksiyaning aniqmas integrali deyiladi;va C x F dx x f ) ( ) ( simvol bilan belgilanadi; 2) ) (x F funksiya biror oraliqda ) (x f funksiyaning boshlang’ich funksiyasi bo’lsa, C x F ) ( (bunda C ixtiyoriy o’zgarmaD) funksiyalar to’plami shu oraliqda ) (x f funksiyaning aniqmas integrali deyiladi va C x F dx x f ) ( ) ( simvol bilan belgilanadi; 3) ) (x F funksiya biror oraliqda ) (x f funksiyaning boshlang’ich funksiyasi bo’lsa, C x f ) ( (bunda C ixtiyoriy o’zgarmas) funksiyalar to’plami shu oraliqda ) (x f funksiyaning aniqmas integrali deyiladi va C x F dx x f ) ( ) ( simvol bilan belgilanadi. A) 2) )1) D) hammasida E) 3) 248. Aniqmas integralning xossalari quyidagi raqamlarning qaysilarida to’g’ri berilgan: 1) aniqmas integralning hosilasi integral ostidagi funksiyaga, differensiali esa integral ostidagi ifodaga teng, ya’ni ; ) ( ) ( ) ( ) ( dx x F dx x F d x f dx x f 131 2) biror funksiyaning hosilasidan hamda differensialidan aniqmas integral shu funksiya bilan ixtiyoriy o’zgarmasning yig’indisiga teng, ya’ni . ) ( ) ( ) ( ) ( C x F x dF C x f dx x f 3) o’zgarmas ko’paytuvchini integral belgisi tashqarisiga chiqarish mumkin, ya’ni 0 const K bo’lsa, ; ) ( ) ( dx x f K dx x Kf 4) chekli sondagi funksiyalar algebraik yig’indisining aniqmas integrali, shu funksiyalar aniqmas integrallarining ayirmasiga teng, ya’ni dx x f dx x f dx x f dx x f x f x f ) ( ) ( ) ( ) ( ) ( ) ( 3 2 1 3 2 1 . A) 1),2),3) ) 1),2),4) D) hammasida E) 2),3),4) 249. Asosiy integrallar jadvali quyidagi rim raqamlarning qaysilarida to’g’ri berilgan: I. ; ) 6 ; sin cos ) 5 ; cos sin ) 4 ; ln 1 ) 3 ; ) 2 ; 1 , 1 ) 1 1 C e dx e C x xdx C x xdx C x dx x C x dx n C n x dx x x x n n II. ; cos 1 ) 4 ; arcsin 1 ) 3 ; 1 1 ) 2 ); 1 0 ( , ln ) 1 2 2 2 2 2 C tgx dx x C a x dx x a C a x arctg a dx x a a C a a dx a x x III. . ln ) 3 ; 0 , ln 2 1 ) 2 ; sin 1 ) 1 2 2 2 2 2 C k x x k x dx a C a x a x a a x dx C ctgx dx x A) hammasida ) I. C) II. E) III. 250. dx x x ) 9 sin 5 ( 3 integralni toping. A) C x x x dx x x 9 cos 5 4 ) 9 sin 5 ( 4 3 ) C x x dx x x 9 cos 5 3 ) 9 sin 5 ( 2 3 132 D) C x x x dx x x 9 cos 5 3 ) 9 sin 5 ( 3 3 E) C x x x dx x x 9 cos 5 4 ) 9 sin 5 ( 4 3 251. dx x x 2 3 3 1 2 1 integralni toping. A) dx x x 2 3 3 1 2 1 = 3 ) dx x x 2 3 3 1 2 1 = 3 D) dx x x 2 3 3 1 2 1 = 5 E) dx x x 2 3 3 1 2 1 = 6 252. x x dx 2 2 cos sin 3 integralni toping. A) x x dx 2 2 cos sin 3 = C ctgx tgx ) ( 3 ) x x dx 2 2 cos sin 3 = C tgx ctgx ) ( 3 D) x x dx 2 2 cos sin 3 = C tgx ctgx ) ( 3 E) x x dx 2 2 cos sin 3 = C ctgx tgx 253. 2 5 x dx integralni toping. A) C x x dx x dx 5 arcsin ) 5 ( 5 2 2 2 ) C x x dx x dx 5 arccos ) 5 ( 5 2 2 2 133 D) C x x dx x dx 5 arcsin ) 5 ( 5 2 2 2 E) C x x dx x dx 5 arcsin ) 5 ( 5 2 2 2 254. dx x 7 ) 1 3 ( integralni toping. A) C x C t C t dt t dx x 24 ) 1 3 ( 24 8 3 1 3 ) 1 3 ( 8 8 8 7 7 ) C x C t C t dt t dx x 7 ) 1 3 ( 7 7 7 ) 1 3 ( 8 8 8 7 7 C) C x C t C t dt t dx x 8 ) 1 3 ( 8 8 ) 1 3 ( 8 8 8 7 7 E) C x dx x 8 ) 1 3 ( ) 1 3 ( 8 7 255. dx x x 3 2 1 integralni toping. A) C dt t xdx x 3 2 2 3 3 2 1 1 8 3 2 1 ) C xdx x 3 2 2 3 2 1 1 8 3 1 D) C x x dt t xdx x 3 2 2 3 3 2 1 ) 1 ( 8 3 2 1 E) C dt t xdx x 3 2 2 3 3 2 1 1 8 3 2 1 256. mxdx cos integralni toping. A) C mx m mx mxd m mxdx sin 1 ) ( cos 1 cos ) C mx m mx mxd m mxdx sin 1 ) ( cos 1 cos D) C mx m mx mxd m mxdx cos 1 ) ( cos 1 cos E) C mx m mx mxd m mxdx cos 1 ) ( sin 1 cos 257. x dx x 3 ) (ln integralni toping. 134 A) C x C t dt t x dx x 4 ) (ln 4 ) (ln 4 4 3 3 ) C x C t dt t x dx x 4 ) (ln 4 ) (ln 4 4 3 3 D) C x C t dt t x dx x 3 ) (ln 3 2 3 ) (ln 2 2 3 3 E) C x dt t x dx x 3 ) (ln 4 ) (ln 4 3 3 258. xdx e x cos sin integralni toping. A) C e x d e xdx e x x x sin sin sin ) (sin cos ) C e x d e xdx e x x x sin sin sin ) (sin cos D) C e x d e xdx e x x x cos cos sin ) (sin cos E) C e x d e xdx e x x x cos cos sin ) (sin cos 259. Bo’laklab integrallash formulasini toping. A) vdu uv udv ) vdu uv udv D) udu uv udv E) udv uv udv 260. xdx x cos integralni toping. A) C x x x xdx x x xdx x cos sin sin sin cos ) C x x x xdx x x xdx x cos sin sin sin cos D) C x x x xdx x x xdx x cos sin sin sin cos E) C x x xdx x xdx x cos sin sin sin cos 261. Rasional funsiyalarning sodda kasrlar ko’rinishi quyidagilarning qaysilarida to’g’ri berilgan: 0 4 ( ; ) 3 ); 1 ( ) ( ) 2 ; ) 1 2 2 q p q px x B Ax son butun k a x A a x A k ya’ni, kvadrat uch had haqiqiy ildizga ega emas); 1 ( ) ( ) 4 2 n q px x B Ax n butun son, ) 0 4 2 q p . A) hammasida ) 1),2) D) 3),4) E) 2),3),4) 262. dx a x A integralni toping. A) C a x A dx a x A ln ) C a x dx a x A ln 135 D) C a x dx a x A ln E) C a x dx a x A ln 263. dx x x 9 2 4 integralni toping. A) . 3 27 9 3 9 3 2 4 C x arctg x x dx x x ) . 3 27 9 3 9 2 2 4 C x arctg x dx x x D) . 3 27 9 3 9 3 2 4 C x arctg x x dx x x E) 9 81 9 9 2 2 2 4 x x x x 264. dx x x x 25 8 3 2 integralni toping. A) . 3 4 3 7 ) 25 8 ln( 2 1 9 3 4 25 8 3 2 2 2 C x arctg x x dt t t dx x x x ) . 3 4 3 7 ) 25 8 ln( 2 1 9 4 25 8 3 2 2 2 C x arctg x x dt t t dx x x x D) . 3 4 3 7 ) 25 8 ln( 2 1 25 8 3 2 2 C x arctg x x dx x x x E) . 3 4 3 7 ) 25 8 ln( 25 8 3 2 2 C x arctg x x dx x x x 265. 6 5 1 2 2 x x x rasional funksiyani sodda kasrlar yoyilmasi ko’rinishini toping. A) 2 3 3 5 6 5 1 2 2 x x x x x ) 2 3 3 5 6 5 1 2 2 x x x x x D) 2 5 3 3 6 5 1 2 2 x x x x x E) 2 3 3 5 6 5 1 2 2 x x x x x 266. ) 3 )( 2 ( x x dx integralni toping. A) C x x dx x x dx x x 2 3 ln 3 1 2 1 ) 3 )( 2 ( 1 136 ) C x x dx x x dx x x 3 2 ln 3 1 2 1 ) 3 )( 2 ( 1 D) C x x dx x x dx x x 2 3 ln 3 1 2 1 ) 3 )( 2 ( 1 E) C x x dx x x dx x x 3 2 ln 3 1 2 1 ) 3 )( 2 ( 1 267. 5 2 2 x x dx integralni toping. A) C x x x u du x x dx 5 2 ) 1 ( ln 4 5 2 2 2 2 ) C x x x u du x x dx 5 2 ) 1 ( ln 4 5 2 2 2 2 D) C x x x u du x x dx 5 2 ) 1 ( ln 4 5 2 2 2 2 E) C x x x x x dx 5 2 ) 1 ( ln 5 2 2 2 268. Trigonometrik funksiyalarning ko’paytmasini yig’indiga keltirish, quyidagi formulalaridan qaysilari to’g’ri berilgan: . ) cos( ) cos( 2 1 cos cos ) 3 ; ) cos( ) cos( 2 1 sin sin ) 2 ; ) sin( ) sin( 2 1 cos sin ) 1 A) hammasi ) 1) D) 2) E) 3) 269. xdx x 7 cos 2 sin integralni toping. A) . 9 cos 18 1 5 cos 10 1 ) 5 sin 9 (sin 2 1 7 cos 2 sin C x x dx x x xdx x ) . 9 cos 18 1 5 cos 10 1 ) 5 sin 9 (sin 2 1 7 cos 2 sin C x x dx x x xdx x D) . 9 cos 18 1 5 cos 10 1 ) 5 sin 9 (sin 2 1 7 cos 2 sin C x x dx x x xdx x E) . 9 cos 9 1 5 cos 5 1 ) 5 sin 9 (sin 2 1 7 cos 2 sin C x x dx x x xdx x 137 270. xdx x 4 3 cos sin integralni toping. A) . 7 cos 5 cos sin cos sin cos sin 7 5 4 2 4 3 C x x xdx x x xdx x ) . 7 cos 5 sin sin cos sin cos sin 7 5 4 2 4 3 C x x xdx x x xdx x D) . 7 sin 5 cos sin cos sin cos sin 7 5 4 2 4 3 C x x xdx x x xdx x E) . 7 cos 5 cos sin cos sin cos sin 7 5 4 2 4 3 C x x xdx x x xdx x 271. dx x x 2 3 cos sin integralni toping. A) C x x dt t t x xdx x dx x x cos cos 1 ) ( 1 cos sin sin cos sin 2 2 2 2 2 3 ) C x x dt t t x xdx x dx x x cos cos 1 1 cos sin sin cos sin 2 2 2 2 2 3 D) C x x dt t t x xdx x dx x x cos cos 1 ) ( 1 cos sin sin cos sin 2 2 2 2 2 3 E) C x x dt t t x xdx x dx x x sin cos 1 ) ( 1 cos sin sin cos sin 2 2 2 2 2 3 272. Quyidagi formulalardan qaysilari to’g’ri berilgan: x x x x x x x 2 sin 2 1 cos sin ) 3 ; 2 2 cos 1 cos ) 2 ; 2 2 cos 1 sin ) 1 2 2 A) hammasi ) 3) D) 1) E) 2) 273. dx x 2 sin integralni hisoblang. A) C x x dx x dx dx x xdx 2 sin 4 1 2 1 2 2 cos 2 1 2 2 cos 1 sin 2 ) C x x dx x dx dx x xdx 2 sin 4 1 2 1 2 2 cos 2 1 2 2 cos 1 sin 2 D) C x x dx x dx dx x xdx 2 sin 4 1 2 1 2 2 cos 2 1 2 2 cos 1 sin 2 E) C x x dx x dx dx x xdx 2 sin 4 1 2 1 2 2 cos 2 1 2 2 cos 1 sin 2 274. Aniq integralning ta’rifini toping. 138 A) ta’rif. n i i i x c f 1 integral yig’indining b a, kesmaning ) ,..., 3 , 2 , 1 ( , 1 n i x x i i qismiy kesmalarga bo’linish usuliga va ularda n c c c , .... , , 2 1 nuqtalarning tanlanishiga bog’liq bo’lmagan, qismiy kesmalar eng kattasi uzunligi 0 dagi chekli limiti mavjud bo’lsa, bu limitga ) (x f funksiyaning b a, kesmadagi aniq integrali deyiladi va b a dx x f ) ( simvol bilan belgilanadi ) ta’rif. n i i i x c f 1 integral yig’indining b a, kesmaning ) ,..., 3 , 2 , 1 ( , 1 n i x x i i qismiy kesmalarga bo’linish usuliga bog’liq bo’lmagan 0 dagi chekli limiti mavjud bo’lsa, bu limitga ) (x f funksiyaning b a, kesmadagi aniq integrali deyiladi va b a dx x f ) ( simvol bilan belgilanadi D) ta’rif. n i i i x c f 1 integral yig’indining b a, kesmada n c c c , .... , , 2 1 nuqtalarning tanlanishiga bog’liq bo’lmagan 0 dagi chekli limiti mavjud bo’lsa, bu limitga ) (x f funksiyaning b a, kesmadagi aniq integrali deyiladi va b a dx x f ) ( simvol bilan belgilanadi E) ta’rif. n i i i x c f 1 integral yig’indining b a, kesmaning ) ,..., 3 , 2 , 1 ( , 1 n i x x i i qismiy kesmalarga bo’linish usuliga va ularda n c c c , .... , , 2 1 nuqtalarning tanlanishiga bog’liq bo’lmagan chekli limiti mavjud bo’lsa, bu limitga ) (x f funksiyaning b a, kesmadagi aniq integrali deyiladi va b a dx x f ) ( simvol bilan belgilanadi 275. Aniq integral quyidagi xossalardan qaysilari to’g’ri berilgan: 1) chekli sondagi integrallanuvchi funksiyalar algebraik yig’indisining aniq integrali qo’shiluvchilar aniq integrallarining algebraik yig’indisiga teng, ya’ni 139 ; ) ( ) ( ) ( ) ( ) ( ) ( 3 2 1 3 2 1 b a b a b a b a dx x f dx x f dx x f dx x f x f x f 2) o’zgarmas ko’paytuvchini aniq integral belgisidan chiqarish mumkin, ya’ni b a b a dx x f k dx x kf ) ( ) ( ; 3) b a, kesmada 0 ) (x f bo’lsa, b a dx x f . 0 ) ( bo’ladi; 4) b a, kesmada ) ( ) ( x g x f tengsizlik bajarilsa, b a b a dx x g dx x f ) ( ) ( bo’ladi. A) hammasi ) 1),3) D) 2),4) E) 2),3) 276. Aniq integral quyidagi xossalardan qaysilari to’g’ri berilgan: 1) b a c , kesmadagi biror nuqta bo’lsa, b c c a b a dx x f dx x f dx x f ) ( ) ( ) ( tenglik o’rinli bo’ladi; 2) m va M sonlar ) (x f y funksiyaning b a, kesmadagi mos ravishda eng kichik va eng katta qiymatlari bo’lsa, b a a b M dx x f a b m ) ( ) ( ) ( tenglik o’rinli bo’ladi; 3) a b b a dx x f dx x f ) ( ) ( . A) hammasi ) 3) D) 1) E) 2) 277. Aniq integral quyidagi xossalardan qaysilari to’g’ri berilgan: 1) ; 0 ) ( a a dx x f 2) b a b a b a dn n f dt t f dx x f ) ( ) ( ) ( bo’ladi; 3) ) (x f y b a, kesmada uzluksiz bo’lsa, bu kesmada shunday bir c nuqta topiladiki ) )( ( ) ( a b c f dx x f b a tenglik o’rinli bo’ladi. |
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