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2 √ x+ 3 2 x √ x+C D) 2 5 x 5/2 − 3 2 x 3/2 +C 11. Ushbu f (x) = x − 1 √ x funksiyaning boshlang’ich funksiyasini umumiy ko’rinishini toping. A) 2 3 x √ x − 2 √ x + C B) 3 2 x √ x − 2 √ x + C C) 2 3 x √ x + 2 √ x + C D) 2 3 x √ x − 1 2 √ x + C 12. (00-3-70) Ushbu f (x) = 2x − 1 x 2 − cos 2x funksiyaning boshlang’ich funksiyasini toping. A) x 2 + 1 x + 1 2 sin 2x + C B) x 2 − 1 x + 1 2 sin 2x + C C) x 2 + 1 x − 1 2 sin 2x + C D) x 2 + 1 x − sin 2x + C Yechish: Boshlang’ich funksiyalarni hisoblashn- ing 1 va 12-qoidasidan foydalanib Z f (x)dx = x 2 + x −1 − Z cos 2xdx (2) ni olamiz. 4 va 14-qoidalardan foydalanib Z cos 2xdx = 1 2 sin 2x + C ekanligini olamiz. Bu ifodani (2) ga qo’yib Z f (x)dx = x 2 + 1 x − 1 2 sin 2x + C ni olamiz. Javob: (C). 13. (98-8-31) Ushbu y = 2 e x funksiyaning boshlang’ich funksiyasini toping. A) 2 e x + C B) 2 ln x + C C) e −x + C D) −2e −x + C 14. (96-1-32) Ushbu f (x) = 1 − 1 cos 2 3x funksiya uchun boshlang’ich funksiyasining umu- miy ko’rinishini toping. A) x + 1 3 ctgx + C B) x − 1 3 tgx + C C) x − 1 3 tg3x + C D) tg3x + C 15. (96-3-31) Ushbu f (x) = 2 sin 3x funksiya uchun boshlang’ich funksiyaning umumiy ko’rinishini to- ping. A) − 2 3 cos 3x + C B) 2 3 cos 3x + C C) − 3 2 sin 2x + C D) 3 2 sin 2x + C 16. (96-7-32) f (x) = 2 cos 2 x funksiya boshlang’ich funksiyasining umumiy ko’rinishini ko’rsating. A) 2 sin 2 x + C B) x + 1 2 sin 2x + C C) 2 3 cos 3 x + C D) 2x − 1 2 sin 2x + C Yechish: Darajani pasaytirish formulasi 2 cos 2 x = 1 + cos 2x dan hamda 4 va 14-qoidalardan foy- dalanib Z f (x)dx = Z (1 + cos 2x)dx = x + 1 2 sin 2x + C ni olamiz. Javob: (B). 183 17. (97-5-35) Ushbu f (x) = sin 2 x funksiyaning bosh- lang’ich funksiyasini toping. A) − 1 2 x + 1 4 sin 2x + C B) 1 2 x − 1 4 sin 2x + C C) 1 4 sin 2x + C D) − 1 4 sin 2x + C 18. (96-10-34) Ushbu f (x) = 1 + 1 sin 2 4x funksiya boshlang’ich funksiyasining umumiy ko’rinishini toping. A) x − 1 4 ctg4x + C B) x + 1 4 tg4x + C C) x − ctg4x + C D) x + 1 4 ctgx + C 19. (96-11-32) Ushbu f (x) = 3 sin 2x funksiya uchun boshlang’ich funksiyasining umumiy ko’rinishini ko’rsating. A) − 3 2 · cos 2x + C B) − 2 3 · cos 2x + C C) 3 2 · sin 2x + C D) − 3 2 · sin 2x + C 20. (96-12-82) f (x) = x 2 funksiyaning (3; 2) nuq- tadan o’tuvchi boshlang’ich funksiyasini toping. A) x 3 3 + 7 B) x 3 3 − 7 C) 2x − 4 D) 2x + 4 Yechish: 1-qoidadan foydalanib Z f (x)dx = Z x 2 dx = 1 3 x 3 + C = F (x) ni olamiz. F funksiyaning grafigi (3; 2) nuqtadan o’tishidan foydalansak F (3) = 1 3 3 3 + C = 2 teng- likni olamiz. Bu yerdan C = −7 ekanligi kelib chiqadi. Javob: x 3 3 − 7 (B). 21. (96-13-25) f (x) = x − x 2 2 funksiyaning (6; 0) nuq- tadan o’tuvchi boshlang’ich funksiyasini toping. A) 1 − x + 5 B) 1 − x − 5 C) x 2 2 − x 3 6 − 18 D) x 2 2 − x 3 6 + 18 22. (97-6-23) Agar F 0 (x) = 2x−1 va F (1) = 2 bo’lsa, F (x) ni toping. A) F (x) = 3x 2 − 3x + 2 B) F (x) = x 2 − x + 2 C) F (x) = x 2 +x D) F (x) = x 2 2 − x + 2 1 2 23. (97-7-32) Ushbu f (x) = sin x · cos 2x funksiya boshlang’ich funksiyasining umumiy ko’rinishini ko’rsating. A) 1 3 sin 3x + 1 2 sin x + C B) 1 2 cos x − 1 3 cos 3x + C C) 1 2 cos x − 1 6 cos 3x + C D) − cos x · sin 2x + C 24. (97-10-32) Quyidagilardan qaysi biri f (x) = sin 2x· cos x funksiya boshlang’ich funksiyasining umu- miy ko’rinishi? A) − 1 2 cos 2x · sin x + C B) 1 6 cos 3x + 1 2 cos x + C C) − 1 6 cos 6x − 1 6 cos x + C D) − 1 2 cos x − 1 6 cos 3x + C 25. (98-1-31) Ushbu y = e 1−3x funksiyaning bosh- lang’ich funksiyasini ko’rsating. A) −3e x + C B) e 1−3x + C C) −3e 1−3x + C D) − 1 3 e 1−3x + C 26. (96-6-47) Quyidagi funksiyalarning qaysi biri uchun F (x) = 2 cos x + sin x + C funksiya boshlang’ich funksiya bo’ladi? A) f (x) = −2 sin x − cos x B) f (x) = 2 sin x + cos x C) f (x) = −2 sin x + cos x D) f (x) = 2 sin x − cos x Yechish: F funksiyaning hosilasini hisoblaymiz: F 0 (x) = −2 sin x + cos x = f (x). Javob: f (x) = −2 sin x + cos x (C). 27. (98-2-43) Ushbu F (x) = e x − 1 3 sin 3x + ctgx + C funksiya quyidagi funksiyalardan qaysi birining boshlang’ich funksiyasi? A) f (x) = e x − cos 3x − 1 sin 2 x B) f (x) = e x + cos 3x − 1 sin 2 x C) f (x) = e x − cos 3x + 1 sin 2 x D) f (x) = e x + cos 3x + 1 sin 2 x 28. (96-6-48) Agar y = f (x) funksiyaning boshlang’ich funksiyasi F (x) bo’lsa, 2f (2x) funksiyaning bosh- lang’ich funksiyasini toping. A) 2F (2x) B) 1 2 F (2x) C) F (2x) D) 2F (x) 29. (98-9-41) F (x) = 2 cos 2x + sin x + C funksiya quyidagi funksiyalardan qaysi birining boshlang’ich funksiyasi hisoblanadi? A) −4 sin 2x − cos x B) 4 sin x + cos x C) −2 sin 2x + cos x D) −4 sin 2x + cos x 30. (99-2-43) F (x) = 1 2 x 2 + cos x + C funksiya y = f (x) funksiyaning boshlang’ich funksiyasi. y = f (x) funksiyaning hosilasini toping. A) 2 cos 2 x 2 B) 2 sin 2 ( π 4 − x 2 ) C)1 + 2 · cos x D) 2 · sin 2 x 2 31. (99-3-59) Ushbu f (x) = x + ctg 2 x funksiyaning boshlang’ich funksiyasini toping. A) x 2 2 + 1 3 ctg 3 x + C B) x 2 2 − 1 3 ctg 3 x + C C) x 2 2 − x − ctgx + C D) x 2 2 − x + ctgx + C 184 32. (01-12-50) Ushbu f (x) = (ln sin x+1)·cos x funksiya uchun boshlang’ich funksiyani toping. A) cos x · ln sin x + C B) sin x · ln sin x + C C) sin x · ln cos x + C D) x + ln sin x + C 33. (02-3-51) f (x) = (tgx + ctgx) 2 funksiyaning boshlang’ich funksiyasini toping. A) tgx − ctgx + C B) tgx − ctgx + 2x + C C) tgx − ctgx + 4x + C D) tgx − ctgx − 4x + C 34. (99-8-41) Ushbu f (x) = 3x 2 − 2 funksiyaning boshlang’ich funksiyalaridan qaysi birining grafigi M (2; 4) nuqtadan o’tadi? A) F (x) = x 3 − 2x B) F (x) = x 3 − 2x + 1 C) F (x) = x 3 − 2x + 5 D) F (x) = x 3 − 2x + 8 35. (99-10-45) Ushbu f (x) = 2 cos 2 ( x 2 ) funksiyaning M (0; 3) nuqtadan o’tadigan boshlang’ich funksiyasini toping. A) F (x) = x − sin x + 3 B) F (x) = −x + sin x + 3 C) F (x) = x + sin x + 3 D) F (x) = x + cos x + 3 36. (02-10-32) f (x) = 6x 2 − 6x + 7 funksiyaning M (1; 0) nuqtadan o’tuvchi boshlang’ich funksiya- sini ko’rsating. A) 2x 3 − 3x 2 + 7x − 6 B) 6x 2 − 6x C) 6x 3 − 6x 2 + 7x − 7 D) 3x 3 − 3x 2 + 7x − 7 37. (01-1-36) f (x) = 3x 2 − 2 cos(2x + π 3 ) funksiyaning, grafigi koordinata boshidan o’tuvchi boshlang’ich funksiyasini toping. A) x 3 − 1 2 sin(2x + π 3 ) − √ 3 2 B) 3x 3 − sin 2x − √ 3 2 C) x 3 − sin x + 1 2 D) x 3 − sin(2x + π 3 ) + √ 3 2 38. (01-4-24) f (x) = 1 x funksiyaning, grafigi (e; 2) nuq- tadan o’tuvchi boshlang’ich funksiyasini toping. A) 2 ln |x| B) 3 − ln |x| C) e ln |x| D) ln |x| + 1 39. (01-7-51) f (x) = 1 √ x − 2 funksiyaning, grafigi A(3; 5) nuqtadan o’tuvchi boshlang’ich funksiyasini to- ping. A) √ x − 2 + 4 B) 2 √ x − 2 + 3 C) √ x − 2 + 3 D) 2 √ x − 2 + 4 40. (01-8-30) Agar F 0 (x) = e −3x va F (1) = 0 bo’lsa, F (x) ni toping. A) −3e −3x + 1 B) − 1 3 e −3x + 1 3 C) 1 3 e −3x + e D) − 1 3 e −3x + 1 3 e −3 Yechish: 11-qoidaga ko’ra, F (x) = Z F 0 (x)dx = − 1 3 e −3x + C. (3) Endi F (1) = 0 shartdan foydalanamiz: 0 = − 1 3 e −3 + C. Bu yerdan C = 1 3 e −3 topib, uni (3) ga qoy- amiz, natijada D) javobni olamiz. Javob: (D). 41. (01-1-37) Agar F 0 (x) = e x + sin 2x va F (0) = 3, 5 bo’lsa, F (x) ini toping. A) e x − 1 2 cos 2x + 3 B) e x − 1 2 cos 2x + 4 C) e x − cos 2x + 4, 5 D) e x − cos x + 3 42. (01-11-41) Agar F 0 (x) = 3x 2 − 2x va F (0) = 4 bo’lsa, F (x) ni toping. A) F (x) = x 4 +2x 2 −4 B) F (x) = x 4 −2x 2 +4 C) F (x) = x 4 − x 2 − 4 D) F (x) = x 3 − x 2 + 4 43. (97-11-23) Agar F 0 (x) = x − 4, F (2) = 0 bo’lsa, F (x) ni toping. A) F (x) = x 2 − 2x B) F (x) = x 2 − 4x + 4 C) F (x) = 2x 2 −4x D) F (x) = 1 2 x 2 − 4x + 6 44. (02-2-31) Agar f 0 (x) = 6x 2 − 3x + 5 va f (4) = 130 bo’lsa, f (0) =? A) 6 B) 4 C) −4 D) −6 14.4.1 Aniq integral Aniq integralning ta’rifiga to’xtalmaymiz, ammo uning xossalari va tadbiqlarini qarab chiqamiz. f funksiyadan [a; b] kesma bo’yicha olingan aniq integral quyidagicha belgilanadi: Z b a f (x)dx. Aniq integral tushunchasi [a; b] kesma f funksiya grafigi va abssissalar o’qi bilan chegaralangan geometrik figura yuzasini hisoblash masalasi bilan uzviy bog’liqdir. Faraz qilaylik, F funksiya f funksiyaning [a; b] kesmadagi boshlang’ich funksiyasi bo’lsin, ya’ni F 0 (x) = f (x), x ∈ [a; b]. Endi integral hisobning asosiy formulasi − Nyuton- Lebnist formulasini keltiramiz: 1. Nyuton-Lebnist formulasi: Z a b f (x)dx = F (x)| b a = F (b) − F (a). (14.4) Bizga [a; b] kesmada aniqlangan manfiymas f funksiya berilgan bo’lsin. Yuqoridan f funksiyaning grafigi, quyi- dan abssissalar o’qi va yon tomonlardan x = a hamda x = b vertikal to’g’ri chiziqlar bilan chegaralangan figura egri chiziqli trapetsiya (14.3-chizma) deyiladi. Egri chiziqli trapetsiya yuzasi S uchun quyidagi tenglik o’rinli: S = Z a b f (x)dx = F (b) − F (a). (14.5) 185 1. (98-8-32) Hisoblang. Z π − π 2 | cos x|dx A) 1 B) 3 C) −1 D) 4 Yechish: − π 2 ≤ x ≤ π 2 da cos x ≥ 0 ekanidan bu oraliqda | cos x| = cos x tenglik, π 2 ≤ x ≤ π da esa cos x ≤ 0 ekanidan bu oraliqda | cos x| = − cos x tenglik o’rinli ekani kelib chiqadi. Shu sababli berilgan integralni ikkita integralga ajratib hisoblaymiz. Z π − π 2 | cos x|dx = Z π 2 − π 2 cos xdx − Z π π 2 cos xdx = = sin x| π 2 − π 2 − sin x| π π 2 = sin π 2 − sin(− π 2 )− − sin π + sin π 2 = 1 + 1 − 0 + 1 = 3 Javob: 3 (B). 2. (97-6-63) Hisoblang. Z 3 −2 |3 − x|dx A) 9 B) 8 C) 4 D) 12,5 3. (96-1-31) Integralni hisoblang. Z π 2 π 3 sin xdx A) √ 3 2 B) √ 2 2 C) 1 2 D) − √ 2 4. (96-6-49) Integralni hisoblang. Z e 2 −1 0 dx x + 1 A) 3 B) 2 C) −2 D) −3 5. (96-7-31) Hisoblang. Z 2 0 (1 − 2x) 2 dx A) 4 1 2 B) −3 1 3 C) 9 D) 4 2 3 6. (97-3-31) Hisoblang. Z 1 0 (3x − 1) 2 dx A) 3 B) 1 C) − 1 3 D) 7 9 7. (96-9-82) Hisoblang. Z π 4 0 sin 2xdx A) 1 2 B) −1 C) − 1 2 D) 1 Yechish: Nyuton-Lebnist formulasiga ko’ra, Z π 4 0 sin 2xdx = − 1 2 cos 2x| π 4 0 = − 1 2 · 0 + 1 2 = 1 2 . Javob: 1 2 (A). 8. (96-10-33) Integralni hisoblang. Z π 2 π 3 cos 2xdx A) 1 2 B) − √ 3 4 C) 0 D) √ 3 4 9. (97-1-22) Integralni hisoblang. Z 0 − π 2 cos 3xdx A) 1 3 B) 0 C) − 1 3 D) 2 3 10. (97-6-22) Hisoblang. Z − π 4 π 4 cos 2xdx A) 0 B) −2 C) −1 D) √ 2 11. (97-8-49) Hisoblang. Z π 2 π 4 (1 + ctg 2 x)dx A) √ 3 3 B) 1 C) √ 3 − 1 D) −1 12. (97-7-31) Hisoblang. Z 0 −1 (2x + 1) 2 dx A) 1 6 B) 2 3 C) 1 D) 1 3 13. (97-10-31) Hisoblang. Z 0 −1 (1 + 3x) 2 dx A) 1 B) −1 C) 7 9 D) − 1 3 186 14. (97-11-22) Integralni hisoblang. Z π 2 0 sin 5xdx A) 1 5 B) − 2 5 C) 1 D) −1 15. (98-4-43) a ning qanday qiymatlarida Z 2 0 (t − log 2 a)dt = 2 log 2 2 Download 1.09 Mb. Do'stlaringiz bilan baham: |
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