|
1-shakl
1
O
y
2
2
2
2
1
x
1
x
1
x
y
_
_
x
.
.
.
1
2
1
2
1
10
.
8
o
Funksiyani qavariqlikka va botiqlikka tekshiramiz va egilish nuqtalarini
topamiz.
4
2
2
)
1
(
)
1
2
)(
1
(
2
)
1
)(
2
2
(
)
(
x
x
x
x
x
x
x
f
0
)
(
,
)
1
(
4
3
x
f
x
Ikkinchi tartibli hosila
1
3
x
nuqtada mavjud emas.
y
hosilaning ishorasi bu
nuqtadan chapda manfiy va o‘ngda musbat.
Demak, funksiyaning grafigi
)
1
;
(
intervalda qavariq,
)
;
1
(
intervalda
botiq bo‘ladi. Funksiya grafigining egilish nuqtasi yo‘q.
o
o
8
1
bandlardagi tekshirishlar asosida funksiya grafigini chizamiz (1-shakl).
2-MAVZU. BIR O‘ZGARUVCHI FUNKSIYASINING
INTEGRAL HISOBI
5-masala.
Aniqmas integrallarni toping:
1)
dx
x
x
x
x
x
)
5
2
)(
1
(
5
7
4
2
2
; 2)
dx
x
x
x
cos
1
cos
3
sin
2
;
3)
dx
x
x
x
x
3
6
3
2
.
3
3
3
)
3
(
; 4)
.
)
1
(
12
5
3
2
4
dx
x
x
x
Yechish.
1) Integral ostidgi funksiya to‘g‘ri kasrdan iborat. Kasrning
maxrajidagi
5
2
2
x
x
kvadrat uchhad ko‘paytuvchilarga ajralmaydi, chunki
.
0
4
4
2
q
p
U holda kasrni
5
2
1
)
5
2
)(
1
(
5
7
4
2
2
2
x
x
C
Bx
x
A
x
x
x
x
x
ko’rinishda yozib olamiz.
Tenglikning chap va o‘ng tomonlarini umumiy maxrajga keltiramiz va suratlarni
tenglashtiramiz:
).
1
)(
(
)
5
2
(
5
7
4
2
2
x
C
Bx
x
x
A
x
x
C
B
A
,
,
koeffitsiyentlarni topamiz:
.
5
5
:
,
4
:
,
8
16
:
1
0
2
C
A
x
B
A
x
A
x
11
Bundan
.
5
,
2
,
2
C
B
A
Shunday qilib,
5
2
)
5
2
(
|
1
|
ln
2
5
2
5
2
1
2
)
5
2
)(
1
(
5
7
4
2
2
2
2
2
x
x
x
x
d
x
dx
x
x
x
x
dx
dx
x
x
x
x
x
.
2
1
2
3
|
5
2
|
ln
|
1
|
ln
2
2
)
1
(
)
1
(
3
2
2
2
C
x
arctg
x
x
x
x
x
d
2) Integralda almashtirishlar bajaramiz:
.
3
cos
1
sin
1
3
cos
1
sin
1
cos
3
3
cos
1
cos
3
sin
2
1
C
I
x
dx
x
x
dx
dx
x
x
x
dx
x
x
x
1
I
integralni universal trigonometrik o‘rniga qo‘yish orqali ratsionallashtiramiz:
arctgt
x
t
dt
dx
t
t
x
t
t
x
x
tg
t
dx
x
x
I
,
1
2
,
1
1
cos
,
1
2
sin
,
2
cos
1
sin
1
2
2
2
2
1
2
2
2
2
1
2
1
1
1
1
2
1
t
dt
t
t
t
t
2
2
2
2
2
1
)
1
(
1
2
1
2
1
t
t
d
t
t
tdt
dt
dt
t
t
t
.
2
cos
ln
2
2
2
1
ln
2
|
1
|
ln
2
2
x
x
tg
x
tg
x
tg
t
t
Demak,
.
2
cos
ln
2
2
3
cos
1
cos
3
sin
2
C
x
x
tg
x
dx
x
x
x
3)
6
3
t
x
belgilash kiritamiz, chunki
6
)
6
,
3
,
2
(
EKUK
.
Bundan
.
6
,
3
5
6
dt
t
dx
t
x
U holda
dt
t
t
t
t
t
dx
x
x
x
x
5
2
3
4
3
6
3
2
6
.
3
3
3
)
3
(
dt
t
t
t
dt
t
t
t
)
1
(
6
1
1
6
2
4
4
3
.
)
3
(
5
6
)
3
(
7
6
5
6
7
6
6
5
6
7
5
6
7
C
x
x
x
C
t
t
t
12
4). Integral ostidagi funksiyani standart shaklda yozib olamiz:
.
1
3
2
4
1
12
17
x
x
Demak,
.
3
2
,
4
1
,
12
17
p
n
m
Bundan
.
1
1
p
n
m
Chebishevning uchinchi o‘rniga qo‘yishidan foydalanamiz:
3
4
1
4
1
1
t
x
x
yoki
.
1
)
1
(
3
4
1
t
x
Bundan
,
1
3
1
4
4
x
x
t
,
)
1
(
4
3
t
x
.
)
1
(
12
5
3
2
dt
t
t
dx
U holda
dt
t
t
t
t
t
dx
x
x
x
5
3
2
3
2
1
3
3
3
17
2
12
5
3
2
4
)
1
(
)
)
1
(
(
)
1
(
12
)
1
(
dt
t
dt
t
t
4
2
2
5
3
2
3
17
2
12
)
1
(
12
.
1
5
12
5
12
3
5
4
4
5
C
x
x
C
t
6-masala.
Aniq integrallarni hisoblang:
1)
9
0
2
3
cos
x
xdx
; 2)
2
2
6
8
.
cos
sin
2
xdx
x
Yechish.
1) Aniq integralni bo‘laklab integrallash usuli bilan hisoblaymiz:
9
0
9
0
2
9
0
2
3
3
1
3
3
1
3
3
1
,
3
cos
,
,
3
cos
xdx
tg
x
xtg
x
tg
v
x
dx
dv
dx
du
x
u
x
xdx
|
0
cos
|
ln
3
cos
ln
9
1
27
3
|
3
cos
|
ln
9
1
0
3
9
3
1
9
0
x
tg
2
ln
3
3
27
1
1
ln
2
1
ln
9
1
27
3
.
13
2) Integral ostidagi funksiyaning darajasini pasaytiramiz:
2
2
2
2
2
2
4
2
4
2
6
8
)
cos
sin
2
(
)
sin
2
(
16
)
cos
sin
2
)(
sin
2
(
2
cos
sin
2
x
x
x
x
x
x
x
x
x
x
x
x
x
2
sin
)
2
cos
2
cos
2
1
(
16
2
sin
)
2
cos
1
(
16
2
2
2
2
x
x
x
x
x
2
cos
2
sin
16
2
sin
2
cos
32
2
sin
16
2
2
2
2
2
2
2
)
2
cos
2
sin
2
(
4
2
sin
2
cos
32
)
2
sin
2
(
8
x
x
x
x
x
)
8
cos
1
(
2
2
sin
2
cos
32
4
cos
8
8
2
x
x
x
x
.
2
cos
2
sin
32
8
cos
2
4
cos
8
10
2
x
x
x
x
Integralni hisoblaymiz:
2
2
2
2
2
2
2
6
8
2
cos
2
sin
32
8
cos
2
4
cos
8
10
cos
sin
2
xdx
x
xdx
xdx
dx
xdx
x
2
2
2
2
2
)
2
(sin
2
sin
16
8
8
sin
2
4
4
sin
8
10
x
xd
x
x
x
.
5
3
2
sin
16
0
0
2
10
2
3
x
7-masala.
Berilgan
l
egri chiziqning ko‘rsatilgan o‘q atrofida aylanishidan
hosil bo‘lgan sirt yuzasini hisoblang:
:
l
t
y
t
x
3
3
sin
5
,
cos
5
astroidaning
0
t
dan
2
t
gacha qismi,
.
Oy
Yechish.
),
(
t
x
),
(
t
y
t
parametrik tenglamalar bilan berilgan
egri chiziqning
Oy
o‘q atrofida aylanishidan hosil bo‘lgan jism sirti yuzasi
dt
t
t
t
)
(
)
(
)
(
2
2
2
formula bilan hisoblanadi.
t
y
t
x
3
3
sin
5
,
cos
5
astroidaning
2
0
t
Oy
o‘q atrofida aylanishidan
hosil bo‘lgan sirt yuazini hisoblaymiz: (2-shakl).
2
0
2
2
2
2
3
)
cos
sin
15
(
)
sin
cos
15
(
cos
5
2
dt
t
t
t
t
t
2
0
3
2
0
2
2
2
3
sin
cos
cos
150
)
sin
(cos
)
sin
(cos
cos
150
tdt
t
t
dt
t
t
t
t
t
14
2-shakl
O
5
x
y
5
2
0
2
0
5
4
2
0
4
.
30
5
cos
150
)
(cos
cos
150
sin
cos
150
t
t
td
tdt
t
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