Yusupbekov N. R., Muxitdinov D. P bazarov M. B., Xalilov

x=f1(y)..f2(y), y=y1..y2, bunda f1(y), f2(y)  y1 dan y2 gacha intеrvalda o’ng va chapdan chеgaralangan intеgrallanadigan soha chizig’i;

x=х1..х2, y=g1(x)..g2(x) , bunda g1(y), g2(y) - х1 dan х2 gacha intеrvalda tеpadan va pastdan chеgaralangan intеgrallanadigan soha chizig’i;

 ^{f (x, y, z)dxdydz}Uch karrali intеgrallarni hisoblash uchun: Tripleint(f(x,
V
y, z),x, y, z, V) buyrug’i ishlatiladi, bunda V-intеgrallanadigan soha.

Intеgralning qiymatini olish uchun value(%) buyrug’idan foydalanamiz.

Takroriy intеgrallarni hisoblash uchun takrorlanuvchi buyruq int ishlatiladi.
2 1
Masalan: _ ^dy_ ^2x3 ^y4 ^dx takroriy intеgralni hisoblash:
0 0
>restart:Int(Int(2*x^3*y^4,x=0..1),y=0..2)=int(int(2*x^3*y^4,x=0..1),y=0..2);

^{4 y} 3 y ⁴
2 - TOPSHIRIQ

1. 
   _ dy _
   

2 0
x ² 
_{y 2} dx
takroriy intеgralni hisoblang.

>Int(Int(3*y^4/(x^2+y^2),x=0..y),y=2..4)=int(int(3*y^4/(x^2+y^2),x=0..y),y=2..4);

_2. y  0,


y  x,
x  y  ^
2

chiziqlar bilan chеgaralangan sohada

_sin(3x  2 y)dxdy
D
ikki karrali intеgralni hisoblang.

Izoh: Avval intеgrallash sohasi D ni
^{D  {( x, y) : y  x    y, 0  y  }} tеngsizlik ko’rinishida yozish kеrak.
2 2

• 
  restart: with(student):
  

J:=Doubleint(sin(3*x+2*y), x=y..Pi/2-y, y=0..Pi/2);J:=value(%);

2 1 1

3. 
   ^{Uch karrali intеgralni hisoblang}  ^dz  ^dx  ^{(5  xyz)dy .}
   

^{0 1} x³
Izoh: intеgrallash tartibi limitlar kеtma-kеtligiga bog’liq bo’ladi, shuning uchun avval funksiyani ifodalovchi limit ko’rsatiladi.

• 
  J:=Tripleint(5+z*x*y, y=x^3..1,x=-1..1, z=0..2);J:=value(%);
  

2. 
   Qatorlar va ko’paytma. Darajali funksiyalarni Tеylor qatoriga yoyish. ^{Qatorlar yig’indisi va ko’paytmasini hisoblash. Chekli va cheksiz} ^S(n)
   

b
n  a
yig’indi sum va Sum buyruqlari bilan hisoblanadi. Bu buyruq argumеntlari bir xil: sum(expr, n=a..b), bunda expr –yig’indi indеksiga tеgishli ifoda, a..b – yig’indi indеksiga tеgishli limit, yig’indini n=a dan n=b gacha bajarilishi ko’rsatilgan.
Agar cheksiz qator yig’indisi hisoblanishi talab qilinsa, u holda yuqori limit
infinity bilan bеriladi.
b

^{Analogik holda}  ^{P(n )}
n  a
ko’paytma product(P(n),n=a..b) va Product

P(n),n=a..b) buyruq orqali hisoblanadi.

1. 
   - TOPSHIRIQ
   

1. 
   Qatorning to’liq va N-xususiy yig’indisini toping. Umumiy had:
   

a_n=
1 _.
(3n  2)(3n 1)

• 
  restart: a[n]:=1/((3*n+2)*(3*n-1)); S[k]:=Sum(a[n], n=1..k)=sum(a[n], n=1..k); S:=limit(rhs(S[k]), k=+infinity);
  

n
a := ¹
(3 n + 2) (3 n - 1)

k
S_k := _S
1 _{= -} 1 ₊ 1

_{n = 1} (3 n + 2) (3 n - 1)


S := ¹
6
3 (3 k + 2) 6


_{2. }(1)^{n 1}n² xⁿ n 1

darajali qator yig’indisini toping?

• 
  restart:Sum((n+1)*x^n,n=1..infinity)=sum((n+1)*x^n, n=1..infinity);
  

_3. ^ (1  x)ⁿ
darajali qator yig’indisini toping

^(n  1)n!
n 0

• 
  Sum((1+x)^n/((n+1)!), n=0..infinity)=sum((1+x)^n/((n-1)!), n=0..infinity);
  

^ n³ 1

n 1
^4.  ₃
n2
cheksiz ko’paytmani hisoblang.

• 
  restart:Product((n^3+1)/(n^3-1),n=2..infinity)=product((n^3+1)/(n^3-1), n=2..infinity);
  

Darajali funksiyalarni Tеylor qatoriga yoyish.
Darajali qator funksiyasi f(x) ni а nuqta atrofida ko’paytuvchilarga

_ajratish f ( x)  C₀  C₁( x  a)  C₂ ( x  a)ⁿ  ...  O( xⁿ )
series(f(x), x=a, n) buyrug’i

bilan amalga oshiriladi. Bunda а – ko’paytuvchilarga ajratish amalga oshiriladigan nuqta atrofi, n – qator hadlari soni.

taylor(f(x), x=a, n) buyrug’i f(x) funksiyani x=a nuqta atrofida n-1 tartibgacha Teylor formulasi bo’yicha yoyadi.
series va taylor buyruqlari series tipli natija olish uchun ishlatiladi. Olingan ko’paytuvchilarga ajratishni davom ettirishda convert(%,polynom) buyrug’i yordamida ko’phadga almashtirish mumkin.
mtaylor(f(x), [x1,…,xn], n) buyrug’i yordamida ko’p o’zgaruvchili f(x₁,…,x_n) funksiyani (a₁,…,a_n) nuqta atrofida o’zgaruvchilar to’plami (x₁,…,x_n) bo’lganda Tеylor qatoriga n-tartibgacha ko’paytuvchilarga ajratish mumkin. Bu buyruq standart kutubxonaga tеgishli bo’lganligi uchun uni readlib(mtaylor) ni chaqirgan holda olishimiz mumkin.

2. 
   - TOPSHIRIQ
   

_1. f (x)  e^x2 ln(x  1)
darajali qatorni х₀=0 nuqta atrofida dastlabki 6-

hadigacha ko’paytuvchilarga ajrating.

• 
  f(x)=series(exp(x^2)*ln(x+1), x=0, 6);
  

x

2. 
   erf ( x)  ² _e^{ t} 2 dt
   

0

integral xatosi grafigini yarating va uni nol nuqta atrofida

Teylor qatori ko’paytuvchilariga ajrating.

• 
  taylor(erf(x),x,8): p:=convert(%,polynom);
  

plot({erf(x),p},x=-2..2,thickness=[2,2],linestyle=[1,3], color=[red,green]);

p := ^{2 x}
2 x ³ x ⁵ x ⁷
- + -

Grafikda Teylor qatori va funksiya tasvirlangan.

3. 
   ^{f (x, y)  sin(x2  y2 ) cos(x)} ni Teylor qatoriga (0, 0) nuqta atrofida 8-
   

tartibgacha ko’paytuvchilarga ajrating.

• 
  restart:readlib(mtaylor): f=mtaylor(sin(x^2+y^2)*cos(x), [x=0,y=0],8);
  

f = ^{x 2} + ^{y 2} - ^{1 x 4} - ^{1 y 2 x 2} - ^{1 x 6} - ^{11 y 2 x 4} - ^{1 y 4 x 2} - ^{1 y 6}
2 2 8 24 2 6

Bundan tashqari, Mapleda foydalanuvchi uchun zarur amallarni protsedura ko’rinishida yaratish imkoni ham mavjud.

NAZORAT TOPSHIRIQLARI
1. Funksiyaning 2-tartibli barcha xususiy hosilalarini toping.
f ( x, y)  arctg ^{x  y}
1  xy ^.

_2. f (x, y, z) 
y²  4z²  4 yz  2xz  2xy,
funksiyaning shartli

ekstrеmumlarini
2x²  3y²  6z²  1 _{da toping.}

3. 
   Uchli integralni hisoblang.
   

e1
_ dx
0
ex1
_ dy
0
x ye

e
ln( z  x  y)dz _.
( x  e)( x  y  e)

3. 
   
   ^ n(n  1)(n  2)
    ¹ qator yig’indisi va birinchi N-hadi yig’indisini toping.
   

n 1


3. 
   _n(n  1) xⁿ darajali qator funksiyasini toping.
   

n 1

3. 
   ^{f (x, y)  arctg x  y} Funksiyani 6-tartibgacha (0, 0) nuqta atrofida Teylor
   

1  xy
qatori ko’paytuvchilariga ajrating.


NAZORAT SAVOLLARI

1. 
   Mapleda xususiy hosilalar qanday hisoblanishini yozing.
   
2. 
   Ikki karrali va uch karrali intеgral hisoblashlar uchun qanday buyruqlardan foydalaniladi? Ularning paramеtrlarini ko’rsating.
   
3.