84
85
Berilgan
f
(
x
) funksiya uchun uning shunday boshlang‘ich funk-
siyasini topingki, bu boshlang‘ich funksiyaning grafigi
y
to‘g‘ri chiziq
bilan faqat bitta umumiy nuqtaga ega bo‘lsin (
8 – 9
):
8.
1)
f
(
x
) = 4
x
+8,
y
= 3;
2)
f
(
x
) = 3–
x
,
y
= 7,
3)
f
(
x
) = 4,5
x
+9,
y
= 6,8; 4)
f
(
x
) = 2
x
–6,
y
= 1.
9*.
f
(
x
)=
ax
+
b
,
y=k
.
Ko‘rsatma:
2
( )
,
2
ax
F x
bx C
=
+
+
masala shartidan va
2
2
ax
bx C k
+
+ =
kvadrat tenglamadan
C
ni toping.
2
2
2
2
2
ak b
b
C
k
a
a
+
=
= +
bo‘ladi.
10*.
f
(
x
) uchun uning shunday boshlang‘ich funksiyasini topingki, bu
boshlang‘ich funksiyaning grafigi ko‘rsatilgan nuqtalardan o‘tsin:
1)
3
16
( )
,
f x
x
′
=
A
(1; 10) va
B
(4; –2);
2)
4
54
( )
,
f x
x
′
=
A
(–1; 4) va
B
(3; 4);
3)
f
′(
x
)=6
x
,
A
(1;6) va
B
(3;30);
4)
f
′(
x
) = 20
x
3
;
A
(1;9) va
B
(–1;7).
Ko‘rsatma:
Berilgan
( )
f x
′
bo‘yicha
f
(
x
)
+C
1
topiladi. So‘ngra
f
(
x
)
+C
1
uchun boshlang‘ich funksiyasi
1
2
( )
( )
F x
f x dx C x C
=
+
+
∫
topiladi.
Berilgan nuqtalar koordinatalarini oxirgi tenglikka qo‘yib,
C
1
va
C
2
sonlarni topish uchun chiziqli tenglamalar sistemasiga kelinadi.
11*.
Berilgan
f
(
x
) funksiya uchun uning shunday boshlang‘ich funk siya-
sini topingki, bu boshlang‘ich funksiyaning grafigi bilan
f
(
x
) hosila -
sining grafigi abssissasi ko‘rsatilgan nuqtada kesishsin:
1)
1
3
( ) (3
2) ,
f x
x
=
−
0
1;
x
=
2)
1
4
( ) (4
5) ,
f x
x
=
+
0
1
x
= −
;
3)
1
7
( ) (7
5) ,
f x
x
=
−
0
1;
x
=
4)
1
( ) (
) ,
k
f x
kx b
=
+
0
1
.
b
x
k
−
=
84
85
12.
Berilgan
f
(
x
) funksiya uchun ko‘rsatilgan nuqtadan o‘tuvchi bosh-
lang‘ich funksiyani toping:
1)
5
( )
,
2
f x
x
=
−
A
(3; 7); 2)
3
( )
,
1
f x
x
=
+
A
(0; 1);
3)
( ) cos ,
f x
x
=
( ; 8);
2
A
π
4)
( ) sin ,
f x
x
=
( ; 1 0).
A
π
13.
F
(
x
) funksiya son o‘qida
f
(
x
) funksiyaning boshlang‘ich funksiyasi
ekanini ko‘rsating:
1)
( )
,
x
k
F x
k e
= ⋅
( )
,
x
k
f x
e
=
0;
k
≠
2)
F
(
x
)
=C+
sin
kx, f
(
x
)
=k∙
cos
kx, C
– o‘zgarmas son;
3)
F
(
x
)
=C+
cos
kx, f
(
x
)
=–k∙
sin
kx, C
– o‘zgarmas son;
4)
1
( )
sin(5 12),
5
F x
x
=
+
( ) cos(5 12)
f x
x
=
+
.
14.
f
(
x
) funksiyaning ko‘rsatilgan nuqtadan o‘tuvchi boshlang‘ich
funk siyasini toping:
1)
( ) sin 3 ,
f x
x
=
1
( ; );
3 3
A
π
2)
( ) cos5 ,
f x
x
=
4
( ; );
2 5
A
π
3)
( ) cos ,
2
x
f x
=
( ; 1 );
3
A
π
4)
( ) sin ,
3
x
f x
=
9
( ; ).
2
A
π
15.
f
(
x
) funksiya uchun uning berilgan tenglamalar sistemasining yechimi
(
x
0
;
y
0
) nuqtadan o‘tuvchi boshlang‘ich funksiyasini toping:
1)
f
(
x
)
=
3
x
2
;
2
2
2
2
log
log
3,
4log
log
2.
x
y
x
y
+
=
−
=
;
2)
f
(
x
)
=
4
x
3
;
5 5
30,
3 5 2 5 15.
x
y
x
y
+
=
⋅ − ⋅
=
;
3)
f
(
x
)
=
cos
x
;
3 ,
2
4
3
;
x y
x
y
π
+ =
−
= −π
4)
1
( )
5
;
f x
x e
=
+
;
2 3
4,
3 2 3
0.
x
y
x
y
+
=
⋅ −
=
86
87
40–43
INTEGRALLAR JADVALI. INTEGRALLASHNING
ENG SODDA QOIDALARI
Integrallar jadvalini
hosilalar jadvali yordamida tuzish mumkin.
№
Funksiya
f(x)
Boshlang‘ich funksiya
F(x)+C
1
x
p
, p
≠–1
1
1
p
x
C
p
+
+
+
2
1/
x
ln׀
x
׀
+C
3
e
x
e
x
+C
4
sin
x
– cos
x+C
5
cos
x
sin
x+C
6
(
) ,
p
kx b
+
1,
p
≠ −
0
k
≠
1
(
)
(
1)
p
kx b
C
k p
+
+
+
+
7
1 ,
kx b
+
0
k
≠
11 (
)
n kx b C
k
+ +
ln׀
kx+b
׀
+C
8
e
kx+b
,
0
k
≠
1
kx b
e
C
k
+
+
9
sin(
),
kx b
+
0
k
≠
1 cos(
)
kx b C
k
−
+ +
10
cos(
),
kx b
+
0
k
≠
1 sin(
)
kx b C
k
−
+ +
11
1/cos
2
x
tg
x + C
12
1/sin
2
x
– ctg
x+C
13
a
x
ln
x
a
C
a
+
14
2
1
1
x
+
arctg
x + C
15
f
(
kx+b
)
1 sin(
)
kx b C
k
−
+ +
F
(
kx+b
)
+C
16
f
(
g
(
x
))
g
(
x
)
F
(
g(x
))
+C
86
87
Biror
X
oraliqda aniqlangan
F
(
x
) funksiya
f
(
x
) funksiyaning boshlan-
g‘ich funksiyasi bo‘lishi uchun ikkala
F
(
x
) va
f
(
x
) funksiya ham ayni
shu
X
oraliqda aniqlangan bo‘lishi kerak.
Masalan,
1
5 8
x
−
funksiyaning 5
x
– 8 > 0 , ya’ni
x
>1,6 oraliqdagi integ-
rali, jadvalga muvo
fiq,
1 ln(5 –8) C
5
x
+
С
ga teng.
Differensiyalash qoidalaridan foydalanib,
integrallash qoidalarini
bayon qilish mumkin.
F
(
x
) va
G
(
x
) funksiyalar biror oraliqda, mos ravishda,
f
(
x
) va
g
(
x
)
funksiyalarning boshlang‘ich funksiyalari bo‘lsin. Ushbu qoidalar
o‘rinlidir:
1-qoida:
a∙F
(
x
) funksiya
a∙f
(
x
) funksiyaning boshlang‘ich funksiyasi
bo‘ladi, ya’ni
( )
( )
.
a f x dx a F x C
⋅
= ⋅
+
∫
2-qoida:
F
(
x
)
±
G
(
x
) funksiya
f
(
x
)
±
g
(
x
) funksiyaning boshlang‘ich
funksiyasi bo‘ladi, ya’ni:
∫(
f
(
x
) ±
g
(
x
))
dx
=∫
f
(
x
)
dx
± ∫
g
(
x
)
dx
=
F
(
x
) ±
G
(
x
)+
C.
1-misol.
f
(
x
)
=
5sin(3
x+
2) funksiyaning integralini toping.
Bu funksiyaning integralini 1-qoida va integrallar jadvalining
9-bandiga muvofiq topamiz:
( )
5sin(3
2)
5 sin(3
2)
1
5
5 (
cos(3
2))
cos(3
2)
,
3
3
f x dx
x
dx
x
dx
x
C
x
C
=
+
=
+
=
= ⋅ −
+
+ = −
+ +
∫
∫
∫
chunki integrallar jadvaliga ko‘ra
1
sin(3
2)
cos(3
2)
.
3
x
dx
x
C
+
= −
+ +
∫
Javob:
5 cos(3 2) .
3
x
C
−
+ +
▲
88
89
2-misol.
f
(
x
)
=
8
x
7
+2cos2
x
funksiyaning integralini toping.
Bu funksiyaning integralini 1- va 2-qoidalar hamda integrallar
jadvalining 1- va 10-bandiga muvofiq
topamiz:
7
7
2
8
8
( )
(8
cos 2 )
8
2 cos
1
1
8
2 sin 2
sin 2
8
2
f x dx
x z
x dx
x dx
xdx
x
x C x
x C
=
+
=
+
=
= ⋅
+ ⋅
+ =
+
+
∫
∫
∫
∫
2cos2
x
)
dx
=
7
7
2
8
8
( )
(8
cos 2 )
8
2 cos
1
1
8
2 sin 2
sin 2
8
2
f x dx
x z
x dx
x dx
xdx
x
x C x
x C
=
+
=
+
=
= ⋅
+ ⋅
+ =
+
+
∫
∫
∫
∫
cos2
xdx
7
7
2
8
8
( )
(8
cos 2 )
8
2 cos
1
1
8
2 sin 2
sin 2
8
2
f x dx
x z
x dx
x dx
xdx
x
x C x
x C
=
+
=
+
=
= ⋅
+ ⋅
+ =
+
+
∫
∫
∫
∫
.
Javob:
8
sin 2
.
x
x C
+
+
▲
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