Ushbu o‘quv-uslubiy materiallar qurilish ta’lim yo‘nalishlarida sirtdan
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Oliy matmatika 2 kurs
- Bu sahifa navigatsiya:
- 2-MAVZU. BIR NECHA O‘ZGARUVCHI FUNKSIYASINING INTEGRAL HISOBI
- 4-masala.
- 3-MAVZU. SONLI VA FUNKSIONAL QATORLAR
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Funksiyaning hisoblangan qiymatlarini taqqoslaymiz. Demak,
16 ) 0 , 4 ( z z katta eng va
. 0 ) 0 , 0 ( z Z kichik eng
2-MAVZU. BIR NECHA O‘ZGARUVCHI FUNKSIYASINING INTEGRAL HISOBI
3-masala. Karrali integrallarni hisoblang 1)
dxdy y x ) 3 ( 2 , , 0 , 1 , 1 : 2 x x y y x D
2) . 3 1 , 1 0 , 1 0 : , ) 2 3 ( 2
z y x V dxdydz z y x V
keltirilgan. Agar ichki integrallash
bo‘yicha va tashqi integrallash x bo‘yicha bajarilsa berilgan ikki karrali integral bitta takroriy integral bilan ifodalanadi. Integralni hisoblaymiz: D x x dy y x dx dxdy y x 1 0 1 1 2 2 2 ) 3 ( ) 3 (
1 0 1 1 2 2 2 2 3 x x y y x
x x x x x x x x 1 0 2 4 2 2 4 3 2 ) 1 2 2 1 ( 2 3
x x x x x x 1 0 4 2 4 3 2 ) 6 3 9 2 2 4 ( 2 1 dx x x x x 1 0 4 3 2 ) 6 5 2 13 ( 2 1
. 12 1 3 2 1 3 13 2 1 1 0 2 5 4 3 x x x x
2) . 3 1 , 1 0 , 1 0 : , ) 2 3 ( 2
z y x V dxdydz z y x V
Berilgan to‘g‘ri burchakli parllelopiped uchun topamiz:
dz z y x dy dx dxdydz z y x 1 0 1 0 3 1 2 2 ) 2 3 ( ) 2 3 (
2 -shakl.
O
1 1
1
0
1
y x
1 2 x y
11
dy z z y x dx 1 0 1 0 3 1 2 2 2 ) 2 3 ( dy y x dx 1 0 1 0 2 ) 1 2 3 ( 4 1 0 1 0 1 0 3 2 1 0 2 2 12 ) 2 ( 4 ) 2 3 ( 4 ) ) 1 3 (( 4 x x dx x dx y y x .
1)
dl y 2 , ) cos
1 ( 2 ), sin
( 2 : t y t t x L sikloidaning bir arkasi; 2) L dy x dx y 2 2 , t b y t a x L sin
, cos
: ellipsning soat strelkasi yo‘nalishida aylanib o‘tishdagi yuqori yoyi. Yechish. 1) Sikloidaning parametrik tenglamasidan topamiz: , sin 2 ), cos 1 ( 2 t y t x t t . cos 1 2 2 sin 4 ) cos 1 ( 4 2 2 dt t dt t t dl U holda
2 0 cos 1 2 2 ) cos 1 ( 2 2 2
t t dl y L
2 0 2 0 . 2 8 ) sin ( 2 4 ) cos 1 ( 2 4 t t dt t
2) Ellipsning parametrik tenglamasiga ko‘ra , sin tdt a dx . costdt b dy
Bunda soat strelkasi yo‘nalishida t parametr dan
0 gacha o‘zgaradi. U holda 0 2 2 2 2 2 2 ) sin
cos cos
sin (
dt t tb a t ta b dy x dx y L
0 2 2 ) (cos
) cos
1 (
t d t a b 0 2 2 ) (sin ) sin
1 (
t d t b a
. 3 4 sin 3 1 sin cos 3 1 cos 2 0 3 2 0 3 2
t t b a t t a b
5-masala. Birinchi tur sirt integralini hisoblang, bu yerda D
tekislikning koordinata tekisliklari bilan ajratilgan qismi. . 4 2 2 : , ) 4 4 ( z y x D d z y x
12
Yechish.Tekislik tenglamasidan topamiz: . 2 , 2 , 2 2 4 y x z z y x z
U holda . 3 1 2 2
dxdy z z d y x
Sirt integralini xy soha bo‘yicha ikki karrali integralni hisoblashga keltiramiz, bu yerda
sirtning
uchburchak (3-shakl).
d z y x ) 4 4 (
dxdy y x y x 3 ) 8 8 16 4 (
2 0 2 0 ) 9 4 16 ( 3
dy y x dx 2 0 2 0 2 2 9 4 16 3 dx y y x x
dx x x x 2 0 2 ) 2 ( 9 ) 4 16 ( ) 2 ( 3
2 0 ) 14 )( 2 ( 2 3 dx x x
44 3 6 28 2 3 12 28 2 3 2 0 3 2 2 0 2
x x dx x x .
) ,
( z y x u u
funksiyaning 1 M nuqtadagi 2 1
M
vektor yo‘nalishidagi hosilasini toping. ). 4 ; 5 ; 3 ( ), 1 ; 1 ; 1 ( ), 1 ln( 2 1 2 2 M M z y x u
2 1
M vektor yo‘nalishidagi l birlik vektorning yo‘naltiruvchi kosinuslarini topamiz: }, 3 ; 6 ; 2 { 2 1 M M
, 7 3 7 6 7 2 7 3 6 2 2 1 2 1 0 k j i k j i M M M M l 7 3 cos , 7 6 cos
, 7 2 cos
. ) 1 ln(
2 2
y x u funksiya xususiy hosilalarining ) 1 ; 1 ; 1 ( 1
nuqtadagi 3-shakl. z
2
2
4 C 13
qiymatlarini topamiz: , 4 1 1 1 0 0 2 2
M z y x x u
, 2 1 1 2 0 0 2 2 M M z y x y y u
. 2 1 1 2 0 0 2 2
M z y x z z u
U holda . 7 1 7 3 2 1 7 6 2 1 7 2 4 1
u
3-MAVZU. SONLI VA FUNKSIONAL QATORLAR
7-masala. 2. Qatorni yaqinlashishga tekshiring: 1) 1 3 4 2 sin 1
n n ; 2)
1 )! 3 (
n n n ; 3) 1 2 5 1 5 3 1
n n n n ; 4)
1 1 3 5 ) 1 ( n n n n
tekshiramiz. Etalon qator sifatida umumiy hadi
bo‘lgan yaqinlashuvchi qatorni olamiz. Berilgan va etalon qatorlar hadlari nisbatlarining limitini topamiz:
3 4 2 sin lim
3 4 2 sin 1 lim lim n n n n n n b a n n n n n
1 3 4 2 lim 3 4 2 3 4 2 sin
3 4 2 lim
n n n n n n n . Demak, taqqoslashning limit alomatiga ko‘ra berilgan qator yaqinlashadi. 14
2) Berilgan qatorda n n n n a )! 3 ( , 1 1 ) 1 ( )! 4 ( n n n n a .
U holda . 1 1 1 1 1 lim
1 1 4 lim )! 3 ( ) 1 ( )! 4 ( lim
lim 1 1
n n n n n n n n n a a n n n n n n n n n n
Demak, Dalamber alomatiga ko‘ra qator yaqinlashadi. 3) Qatorni yaqinlashishga Koshining ildiz alomati bilan tekshiramiz: . 1 3 5 1 1 lim
3 1 5 1 5 3 1 lim
5 1 5 3 1 lim lim 5 5 1 5 2 e n n n n n a n n n n n n n n n n n
Demak, qator yaqinlashadi. 4) Qatorning yoyilmasini yozamiz: ...
3 5 ) 1 ( ... 81 9 27 8 9 7 3 6 3 5 ) 1 ( 1 1 1
n n n n n n
Demak, qator ishora almashinuvchi. Bu qator hadlarining absolut qiymatlaridan tashkil topgan 1 3 5
n n qatorni Dalamber alomati bilan yaqinlashishga tekshiramiz: . 1 3 1 5 6 lim
3 1 5 3 3 6 lim lim
1 1 n n n n a a n n n n n n n
1 3 5 n n n qator yaqinlashadi. Demak, berilgan qator absolut yaqinlashadi.
. ) 1 ( 3 1 3 n n n n x
Berilgan qator uchun 3 1 1 3 1 3 , 3
a n a n n n n .
15
Bundan . 3 1 3 1 3 lim
lim 1 3 3 1
n n n n n n n a a R
Demak, qator 3 1 1 ; 3 1 1 , ya’ni 3 4 ; 3 2 oraliqda yaqinlashadi. Intervalning chegaraviy nuqtalarida tekshiramiz.
3 2 x da qator
1 3 ) 1 ( n n n ko‘rinishni oladi. Leybnits alomatiga ko‘ra 1)
...; 3 1 2 1 1 3 3 2) 0 1 lim 3 n n . Demak, qator 3 2 x da yaqinlashadi.
3
x da qator
3 1
n ko‘rinishini oladi. Bu qator uzoqlashuvchi. Shunday qilib, qatorning yaqinlashish sohasi 3 4 ; 3 2 dan iborat.
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