minimumga ega deyiladi. 4)
0
x
nuqtaning shunday atrofi mavjud bo’lsaki, bu
atrofning har qanday
0
x
x
nuqtasi uchun
)
(
)
(
0
x
f
x
f
tengsizlik bajarilsa,
)
(x
f
y
funksiya
0
x
nuqtada minimumga ega deyiladi.
A)
1),3)
) 1),2),3)
D) 2),3)4)
E) hammasi
236. Ekstremumga ega bo’lshishinig zaruriy shartini toping.
A)
)
(x
f
y
funksiya
0
x
nuqtada ekstremumga ega bo’lsa,
)
(
0
x
f
y
no’lga
teng yoki u mavjud bo’lmaydi
)
)
(x
f
y
funksiya
0
x
nuqtada ekstremumga ega bo’lsa,
)
(
0
x
f
y
no’lga
teng bo’lmaydi
D)
)
(x
f
y
funksiya
0
x
nuqtada ekstremumga ega bo’lsa,
)
(
0
x
f
y
no’ldan katta bo’ladi
E)
)
(x
f
y
funksiya
0
x
nuqtada ekstremumga ega bo’lsa,
)
(
0
x
f
y
hosila
mavjud bo’lmaydi
237. Quyidagilarning qaysilarida ekstremumning yetarli shartlari to’g’ri berilgan:
1)
0
x
nuqta
)
(x
f
y
funksiyaning kritik nuqtasi bo’lib, funksiya hosilasi
ishorasi bu nuqtadan o’tishda ishorasini o’zgartirsa,
0
x
nuqta, funksiyaning
ekstremum nuqtasi, va:
)
0
x
nuqtadan chapdan o’ngga o’tishda
)
(x
f
o’z
ishorasini musbatdan manfiyga o’zgartirsa,
0
x
nuqtada funksiya maksimumga;
)
0
x
nuqtadan chapdan o’ngga o’tishda
)
(x
f
o’z ishorasini manfiydan
musbatga o’zgartirsa,
0
x
nuqtada funksiya minimumga ega bo’ladi; 2)
0
x
nuqta
)
(x
f
y
funksiyaning kritik nuqtasi bo’lib, funksiya hosilasi ishorasi bu
nuqtadan o’tishda ishorasini o’zgartirsa,
0
x
nuqta, funksiyaning ekstremum
nuqtasi, va:
)
0
x
nuqtadan chapdan o’ngga o’tishda
)
(x
f
o’z ishorasini
musbatdan manfiyga o’zgartirsa,
0
x
nuqtada funksiya minimumga
)
0
x
nuqtadan chapdan o’ngga o’tishda
)
(x
f
o’z ishorasini manfiydan musbatga
o’zgartirsa,
0
x
nuqtada funksiya maksimumga ega bo’ladi; 3)
0
x
nuqta
)
(x
f
y
funksiyaning kritik nuqtasi bo’lib, funksiya hosilasi ishorasi bu
nuqtadan o’tishda ishorasini o’zgartirmasa,
0
x
nuqta, funksiyaning ekstremum
nuqtasi, va:
)
0
x
nuqtadan chapdan o’ngga o’tishda
)
(x
f
o’z ishorasini
musbatdan manfiyga o’zgartirsa,
0
x
nuqtada funksiya maksimumga;
)
0
x
128
nuqtadan chapdan o’ngga o’tishda
)
( x
f
o’z ishorasini manfiydan musbatga
o’zgartirsa,
0
x
nuqtada funksiya minimumga ega bo’ladi
A) 1)
) 2)
D) 3)
E) hammasida
238. Quyidagilarning qaysilarida ekstremumning ikkinchi qoidasi to’g’ri berilgan:
1)
0
x
nuqtada birinchi hosila nolga teng bo’lib, ikkinchi hosila no’ldan farqli
bo’lsa,
0
x
nuqta funksiyaning ekstremum nuqtasi va :
0
)
(
0
x
f
bo’lsa,
maksimum nuqtasi;
0
)
(
0
x
f
bo’lsa, minimum nuqtasi bo’ladi; 2)
0
x
nuqtada
birinchi hosila nolga teng bo’lib, ikkinchi hosila no’ldan farqli bo’lsa,
0
x
nuqta
funksiyaning ekstremum nuqtasi va :
0
)
(
0
x
f
bo’lsa, minimum nuqtasi;
0
)
(
0
x
f
bo’lsa, minimum nuqtasi bo’ladi; 3)
0
x
nuqtada birinchi hosila
noldan farqli bo’lib, ikkinchi hosila no’lga teng bo’lsa,
0
x
nuqta funksiyaning
ekstremum nuqtasi va :
0
)
(
0
x
f
bo’lsa, maksimum
nuqtasi;
0
)
(
0
x
f
bo’lsa, minimum nuqtasi bo’ladi
A) 1)
) hammasi
D) 2)
E) 3)
239.
3
2
2
6
2
1
3
1
)
(
2
3
x
x
x
x
f
funksiyaninshg ekstremumini toping.
A)
,
10
max
min
y
6
65
)
10
max
D)
min
y
6
65
E) ekstremum yo’q
240.
4
9
6
2
11
2
4
1
)
(
2
3
4
x
x
x
x
x
f
funksiya ekstremumini
ikkinchi qoida bilan toping.
A)
0
)
3
(
min
;
25
.
0
)
2
(
max
;
0
)
1
(
min
f
f
f
)
25
.
0
)
2
(
max
;
0
)
1
(
min
f
f
D)
0
)
3
(
min
;
25
.
0
)
2
(
max
f
f
E)
0
)
3
(
min
;
0
)
1
(
min
f
f
241. Funksiyaning eng kichik va eng katta qiymatlarini topish ketma-ketligi
quyidagi raqamlarning qaysilarida to’g’ri berilgan:
1)
)
( x
f
y
funksiyaning
b
a,
kesmadagi eng kichik va eng katta qiymatlarini
topish uchun:
)
kritik nuqtalarni topamiz;
)
funksiyaning bu kritik nuqtalardagi
qiymatlarini hisoblaymiz;
)
d
bu topilgan qiymatlarni taqqoslab, eng kichigi
funksiyaning berilgan kesmadagi eng kichik qiymati, eng kattasi bu kesmadagi eng
katta qiymati ekanligini topamiz; 2)
)
( x
f
y
funksiyaning
b
a,
kesmadagi eng
kichik va eng katta qiymatlarini topish uchun:
)
kritik nuqtalarni topamiz;
)
funksiyaning bu kritik nuqtalardagi va kesmaning chetlaridagi qiymatlarini
129
hisoblaymiz;
)
d
bu topilgan qiymatlarni taqqoslab, eng kichigi funksiyaning
berilgan kesmadagi eng kichik qiymati, eng kattasi bu kesmadagi eng katta qiymati
ekanligini topamiz; 3)
)
( x
f
y
funksiyaning
b
a,
kesmadagi eng kichik va eng
katta qiymatlarini topish uchun:
)
kritik nuqtalarni topamiz;
)
funksiyaning
kesmaning chetlaridagi qiymatlarini hisoblaymiz;
)
d
bu topilgan qiymatlarni
taqqoslab, eng kichigi funksiyaning berilgan kesmadagi eng kichik qiymati, eng
kattasi bu kesmadagi eng katta qiymati ekanligini topamiz.
A) 2)
) 1)
D) hammasi
E) 3)
242.
5
2
)
(
2
4
x
x
x
f
y
funksiyaning
3
;
2
kesmadagi eng kichik va
eng katta qiymatlarini toping.
A)
69
,
4
.
.
y
y
)
4
.
y
D)
68
.
y
E) bunday qiymatlari yo’q
243. Funksiya grafigining qavariq yoki botiq bo’lishining yetarli shartlari
quyidagilarning qaysi raqamlarida to’g’ri berilgan:
1)
)
,
(
b
a
oraliqda differensiallanuvchi
)
( x
f
y
funksiyaning ikkinchi
tartibli hosilasi manfiy, ya’ni
0
)
( x
f
bo’lsa, bu oraliqda funksiya grafigi
qavariq bo’ladi; 2)
)
,
(
b
a
oraliqda differensiallanuvchi
)
( x
f
y
funksiyaning
ikkinchi tartibli hosilasi manfiy, ya’ni
0
)
( x
f
bo’lsa, bu oraliqda funksiya
grafigi botiq bo’ladi; 3)
)
,
(
b
a
oraliqda differensiallanuvchi
)
( x
f
y
funksiyaning ikkinchi tartibli hosilasi musbat, ya’ni
0
)
( x
f
bo’lsa, bu oraliqda
funksiya grafigi botiq bo’ladi.
A)
1),3)
) hammasi to’g’ri
D)
2),3)
E) 1),2)
244. Egilish nuqtalari mavjud bo’lishining yetarli sharti quyidagi raqamlarning
qaysilarida to’g’ri berilgan: 1)
0
nuqta
)
( x
f
y
funksiya uchun ikkinchi tur
kritik nuqta bo’lsa va
)
( x
f
ikkinchi tartibli hosila bu nuqtadan o’tishda ishorasni
o’zgartirmasa,
0
x
abssissali nuqta egilish nuqtasi bo’ladi; 2)
0
nuqta
)
( x
f
y
funksiya uchun ikkinchi tur kritik nuqta bo’lsa va
)
( x
f
ikkinchi
tartibli hosila bu nuqtadan o’tishda ishorasni musbatdan manfiyga o’zgartirsa,
0
x
abssissali nuqta egilish nuqtasi bo’ladi; 3)
0
nuqta
)
( x
f
y
funksiya uchun
ikkinchi tur kritik nuqta bo’lsa va
)
( x
f
ikkinchi tartibli hosila bu nuqtadan
o’tishda ishorasni o’zgartirsa,
0
x
abssissali nuqta egilish nuqtasi bo’ladi.
A) 3)
) 1)
D) hammasi
E) 2)
245.
)
( x
f
y
funksiya grafigining
b
kx
y
og’ma asimptotasi
k
va
b
parametrlarini topish, quyidagi raqamlarning qaysilarida to’g’ri berilgan:
130
1)
kx
x
f
b
x
x
f
k
x
x
)
(
lim
)
(
lim
;
2)
bx
x
f
k
x
x
f
b
x
x
)
(
lim
)
(
lim
;
3)
kx
x
f
b
x
x
f
k
x
x
)
(
lim
)
(
lim
.
A) 1)
) 2)
D) 3)
E) hammasi noto’g’ri
Integral hisob
246.
5
x
x
f
y
funksiyaning hamma boshlang’ich funksiyalarini toping.
A)
,
6
6
ixtiyoriy o’zgarmas son )
4
5
D)
105
6
6
E)
6
6
247. Aniqmas integral ta’rifi quyidagi raqamlarning qaysi larida to’g’ri berilgan: 1)
)
( x
f
funksiya biror oraliqda
)
( x
F
funksiyaning boshlang’ich funksiyasi bo’lsa,
C
x
F
)
(
(bunda
C
ixtiyoriy o’zgarmaD) funksiyalar to’plami shu oraliqda
)
( x
f
funksiyaning aniqmas integrali deyiladi;va
C
x
F
dx
x
f
)
(
)
(
simvol bilan belgilanadi; 2)
)
( x
F
funksiya biror oraliqda
)
( x
f
funksiyaning
boshlang’ich funksiyasi bo’lsa,
C
x
F
)
(
(bunda
C
ixtiyoriy o’zgarmaD)
funksiyalar to’plami shu oraliqda
)
( x
f
funksiyaning aniqmas integrali deyiladi
va
C
x
F
dx
x
f
)
(
)
(
simvol bilan belgilanadi; 3)
)
( x
F
funksiya biror oraliqda
)
( x
f
funksiyaning
boshlang’ich funksiyasi bo’lsa,
C
x
f
)
(
(bunda
C
ixtiyoriy o’zgarmas)
funksiyalar to’plami shu oraliqda
)
( x
f
funksiyaning aniqmas integrali deyiladi
va
C
x
F
dx
x
f
)
(
)
(
simvol bilan belgilanadi.
A) 2)
)1)
D)
hammasida
E) 3)
248. Aniqmas integralning xossalari quyidagi raqamlarning qaysilarida to’g’ri
berilgan:
1) aniqmas integralning hosilasi integral ostidagi funksiyaga, differensiali
esa integral ostidagi ifodaga teng, ya’ni
;
)
(
)
(
)
(
)
(
dx
x
F
dx
x
F
d
x
f
dx
x
f
131
2) biror funksiyaning hosilasidan hamda differensialidan aniqmas integral
shu funksiya bilan ixtiyoriy o’zgarmasning yig’indisiga teng, ya’ni
.
)
(
)
(
)
(
)
(
C
x
F
x
dF
C
x
f
dx
x
f
3) o’zgarmas ko’paytuvchini integral belgisi tashqarisiga chiqarish
mumkin, ya’ni
0
const
K
bo’lsa,
;
)
(
)
(
dx
x
f
K
dx
x
Kf
4) chekli sondagi funksiyalar algebraik yig’indisining aniqmas integrali,
shu funksiyalar aniqmas integrallarining ayirmasiga teng, ya’ni
dx
x
f
dx
x
f
dx
x
f
dx
x
f
x
f
x
f
)
(
)
(
)
(
)
(
)
(
)
(
3
2
1
3
2
1
.
A) 1),2),3)
) 1),2),4)
D)
hammasida
E) 2),3),4)
249. Asosiy integrallar jadvali quyidagi rim raqamlarning qaysilarida to’g’ri
berilgan:
I.
;
)
6
;
sin
cos
)
5
;
cos
sin
)
4
;
ln
1
)
3
;
)
2
;
1
,
1
)
1
1
C
e
dx
e
C
x
xdx
C
x
xdx
C
x
dx
x
C
x
dx
n
C
n
x
dx
x
x
x
n
n
II.
;
cos
1
)
4
;
arcsin
1
)
3
;
1
1
)
2
);
1
0
(
,
ln
)
1
2
2
2
2
2
C
tgx
dx
x
C
a
x
dx
x
a
C
a
x
arctg
a
dx
x
a
a
C
a
a
dx
a
x
x
III.
.
ln
)
3
;
0
,
ln
2
1
)
2
;
sin
1
)
1
2
2
2
2
2
C
k
x
x
k
x
dx
a
C
a
x
a
x
a
a
x
dx
C
ctgx
dx
x
A)
hammasida
) I.
C) II.
E) III.
250.
dx
x
x
)
9
sin
5
(
3
integralni toping.
A)
C
x
x
x
dx
x
x
9
cos
5
4
)
9
sin
5
(
4
3
)
C
x
x
dx
x
x
9
cos
5
3
)
9
sin
5
(
2
3
132
D)
C
x
x
x
dx
x
x
9
cos
5
3
)
9
sin
5
(
3
3
E)
C
x
x
x
dx
x
x
9
cos
5
4
)
9
sin
5
(
4
3
251.
dx
x
x
2
3
3
1
2
1
integralni toping.
A)
dx
x
x
2
3
3
1
2
1
=
3
)
dx
x
x
2
3
3
1
2
1
=
3
D)
dx
x
x
2
3
3
1
2
1
=
5
E)
dx
x
x
2
3
3
1
2
1
=
6
252.
x
x
dx
2
2
cos
sin
3
integralni toping.
A)
x
x
dx
2
2
cos
sin
3
=
C
ctgx
tgx
)
(
3
)
x
x
dx
2
2
cos
sin
3
=
C
tgx
ctgx
)
(
3
D)
x
x
dx
2
2
cos
sin
3
=
C
tgx
ctgx
)
(
3
E)
x
x
dx
2
2
cos
sin
3
=
C
ctgx
tgx
253.
2
5 x
dx
integralni toping.
A)
C
x
x
dx
x
dx
5
arcsin
)
5
(
5
2
2
2
)
C
x
x
dx
x
dx
5
arccos
)
5
(
5
2
2
2
133
D)
C
x
x
dx
x
dx
5
arcsin
)
5
(
5
2
2
2
E)
C
x
x
dx
x
dx
5
arcsin
)
5
(
5
2
2
2
254.
dx
x
7
)
1
3
(
integralni toping.
A)
C
x
C
t
C
t
dt
t
dx
x
24
)
1
3
(
24
8
3
1
3
)
1
3
(
8
8
8
7
7
)
C
x
C
t
C
t
dt
t
dx
x
7
)
1
3
(
7
7
7
)
1
3
(
8
8
8
7
7
C)
C
x
C
t
C
t
dt
t
dx
x
8
)
1
3
(
8
8
)
1
3
(
8
8
8
7
7
E)
C
x
dx
x
8
)
1
3
(
)
1
3
(
8
7
255.
dx
x
x
3
2
1
integralni toping.
A)
C
dt
t
xdx
x
3
2
2
3
3
2
1
1
8
3
2
1
)
C
xdx
x
3
2
2
3
2
1
1
8
3
1
D)
C
x
x
dt
t
xdx
x
3
2
2
3
3
2
1
)
1
(
8
3
2
1
E)
C
dt
t
xdx
x
3
2
2
3
3
2
1
1
8
3
2
1
256.
mxdx
cos
integralni toping.
A)
C
mx
m
mx
mxd
m
mxdx
sin
1
)
(
cos
1
cos
)
C
mx
m
mx
mxd
m
mxdx
sin
1
)
(
cos
1
cos
D)
C
mx
m
mx
mxd
m
mxdx
cos
1
)
(
cos
1
cos
E)
C
mx
m
mx
mxd
m
mxdx
cos
1
)
(
sin
1
cos
257.
x
dx
x
3
)
(ln
integralni toping.
134
A)
C
x
C
t
dt
t
x
dx
x
4
)
(ln
4
)
(ln
4
4
3
3
)
C
x
C
t
dt
t
x
dx
x
4
)
(ln
4
)
(ln
4
4
3
3
D)
C
x
C
t
dt
t
x
dx
x
3
)
(ln
3
2
3
)
(ln
2
2
3
3
E)
C
x
dt
t
x
dx
x
3
)
(ln
4
)
(ln
4
3
3
258.
xdx
e
x
cos
sin
integralni toping.
A)
C
e
x
d
e
xdx
e
x
x
x
sin
sin
sin
)
(sin
cos
)
C
e
x
d
e
xdx
e
x
x
x
sin
sin
sin
)
(sin
cos
D)
C
e
x
d
e
xdx
e
x
x
x
cos
cos
sin
)
(sin
cos
E)
C
e
x
d
e
xdx
e
x
x
x
cos
cos
sin
)
(sin
cos
259. Bo’laklab integrallash formulasini toping.
A)
vdu
uv
udv
)
vdu
uv
udv
D)
udu
uv
udv
E)
udv
uv
udv
260.
xdx
x cos
integralni toping.
A)
C
x
x
x
xdx
x
x
xdx
x
cos
sin
sin
sin
cos
)
C
x
x
x
xdx
x
x
xdx
x
cos
sin
sin
sin
cos
D)
C
x
x
x
xdx
x
x
xdx
x
cos
sin
sin
sin
cos
E)
C
x
x
xdx
x
xdx
x
cos
sin
sin
sin
cos
261. Rasional funsiyalarning sodda kasrlar ko’rinishi quyidagilarning qaysilarida
to’g’ri berilgan:
0
4
(
;
)
3
);
1
(
)
(
)
2
;
)
1
2
2
q
p
q
px
x
B
Ax
son
butun
k
a
x
A
a
x
A
k
ya’ni, kvadrat uch had haqiqiy ildizga ega emas);
1
(
)
(
)
4
2
n
q
px
x
B
Ax
n
butun son,
)
0
4
2
q
p
.
A)
hammasida
)
1),2)
D)
3),4)
E) 2),3),4)
262.
dx
a
x
A
integralni toping.
A)
C
a
x
A
dx
a
x
A
ln
)
C
a
x
dx
a
x
A
ln
135
D)
C
a
x
dx
a
x
A
ln
E)
C
a
x
dx
a
x
A
ln
263.
dx
x
x
9
2
4
integralni toping.
A)
.
3
27
9
3
9
3
2
4
C
x
arctg
x
x
dx
x
x
)
.
3
27
9
3
9
2
2
4
C
x
arctg
x
dx
x
x
D)
.
3
27
9
3
9
3
2
4
C
x
arctg
x
x
dx
x
x
E)
9
81
9
9
2
2
2
4
x
x
x
x
264.
dx
x
x
x
25
8
3
2
integralni toping.
A)
.
3
4
3
7
)
25
8
ln(
2
1
9
3
4
25
8
3
2
2
2
C
x
arctg
x
x
dt
t
t
dx
x
x
x
)
.
3
4
3
7
)
25
8
ln(
2
1
9
4
25
8
3
2
2
2
C
x
arctg
x
x
dt
t
t
dx
x
x
x
D)
.
3
4
3
7
)
25
8
ln(
2
1
25
8
3
2
2
C
x
arctg
x
x
dx
x
x
x
E)
.
3
4
3
7
)
25
8
ln(
25
8
3
2
2
C
x
arctg
x
x
dx
x
x
x
265.
6
5
1
2
2
x
x
x
rasional funksiyani sodda kasrlar yoyilmasi ko’rinishini
toping.
A)
2
3
3
5
6
5
1
2
2
x
x
x
x
x
)
2
3
3
5
6
5
1
2
2
x
x
x
x
x
D)
2
5
3
3
6
5
1
2
2
x
x
x
x
x
E)
2
3
3
5
6
5
1
2
2
x
x
x
x
x
266.
)
3
)(
2
(
x
x
dx
integralni toping.
A)
C
x
x
dx
x
x
dx
x
x
2
3
ln
3
1
2
1
)
3
)(
2
(
1
136
)
C
x
x
dx
x
x
dx
x
x
3
2
ln
3
1
2
1
)
3
)(
2
(
1
D)
C
x
x
dx
x
x
dx
x
x
2
3
ln
3
1
2
1
)
3
)(
2
(
1
E)
C
x
x
dx
x
x
dx
x
x
3
2
ln
3
1
2
1
)
3
)(
2
(
1
267.
5
2
2
x
x
dx
integralni toping.
A)
C
x
x
x
u
du
x
x
dx
5
2
)
1
(
ln
4
5
2
2
2
2
)
C
x
x
x
u
du
x
x
dx
5
2
)
1
(
ln
4
5
2
2
2
2
D)
C
x
x
x
u
du
x
x
dx
5
2
)
1
(
ln
4
5
2
2
2
2
E)
C
x
x
x
x
x
dx
5
2
)
1
(
ln
5
2
2
2
268. Trigonometrik funksiyalarning ko’paytmasini yig’indiga keltirish, quyidagi
formulalaridan qaysilari to’g’ri berilgan:
.
)
cos(
)
cos(
2
1
cos
cos
)
3
;
)
cos(
)
cos(
2
1
sin
sin
)
2
;
)
sin(
)
sin(
2
1
cos
sin
)
1
A) hammasi
) 1)
D) 2)
E) 3)
269.
xdx
x
7
cos
2
sin
integralni toping.
A)
.
9
cos
18
1
5
cos
10
1
)
5
sin
9
(sin
2
1
7
cos
2
sin
C
x
x
dx
x
x
xdx
x
)
.
9
cos
18
1
5
cos
10
1
)
5
sin
9
(sin
2
1
7
cos
2
sin
C
x
x
dx
x
x
xdx
x
D)
.
9
cos
18
1
5
cos
10
1
)
5
sin
9
(sin
2
1
7
cos
2
sin
C
x
x
dx
x
x
xdx
x
E)
.
9
cos
9
1
5
cos
5
1
)
5
sin
9
(sin
2
1
7
cos
2
sin
C
x
x
dx
x
x
xdx
x
137
270.
xdx
x
4
3
cos
sin
integralni toping.
A)
.
7
cos
5
cos
sin
cos
sin
cos
sin
7
5
4
2
4
3
C
x
x
xdx
x
x
xdx
x
)
.
7
cos
5
sin
sin
cos
sin
cos
sin
7
5
4
2
4
3
C
x
x
xdx
x
x
xdx
x
D)
.
7
sin
5
cos
sin
cos
sin
cos
sin
7
5
4
2
4
3
C
x
x
xdx
x
x
xdx
x
E)
.
7
cos
5
cos
sin
cos
sin
cos
sin
7
5
4
2
4
3
C
x
x
xdx
x
x
xdx
x
271.
dx
x
x
2
3
cos
sin
integralni toping.
A)
C
x
x
dt
t
t
x
xdx
x
dx
x
x
cos
cos
1
)
(
1
cos
sin
sin
cos
sin
2
2
2
2
2
3
)
C
x
x
dt
t
t
x
xdx
x
dx
x
x
cos
cos
1
1
cos
sin
sin
cos
sin
2
2
2
2
2
3
D)
C
x
x
dt
t
t
x
xdx
x
dx
x
x
cos
cos
1
)
(
1
cos
sin
sin
cos
sin
2
2
2
2
2
3
E)
C
x
x
dt
t
t
x
xdx
x
dx
x
x
sin
cos
1
)
(
1
cos
sin
sin
cos
sin
2
2
2
2
2
3
272. Quyidagi formulalardan qaysilari to’g’ri berilgan:
x
x
x
x
x
x
x
2
sin
2
1
cos
sin
)
3
;
2
2
cos
1
cos
)
2
;
2
2
cos
1
sin
)
1
2
2
A) hammasi
) 3)
D) 1)
E) 2)
273.
dx
x
2
sin
integralni hisoblang.
A)
C
x
x
dx
x
dx
dx
x
xdx
2
sin
4
1
2
1
2
2
cos
2
1
2
2
cos
1
sin
2
)
C
x
x
dx
x
dx
dx
x
xdx
2
sin
4
1
2
1
2
2
cos
2
1
2
2
cos
1
sin
2
D)
C
x
x
dx
x
dx
dx
x
xdx
2
sin
4
1
2
1
2
2
cos
2
1
2
2
cos
1
sin
2
E)
C
x
x
dx
x
dx
dx
x
xdx
2
sin
4
1
2
1
2
2
cos
2
1
2
2
cos
1
sin
2
274. Aniq integralning ta’rifini toping.
138
A) ta’rif.
n
i
i
i
x
c
f
1
integral yig’indining
b
a,
kesmaning
)
,...,
3
,
2
,
1
(
,
1
n
i
x
x
i
i
qismiy kesmalarga bo’linish usuliga va ularda
n
c
c
c
,
....
,
,
2
1
nuqtalarning tanlanishiga bog’liq bo’lmagan, qismiy kesmalar
eng kattasi uzunligi
0
dagi chekli limiti mavjud bo’lsa, bu limitga
)
( x
f
funksiyaning
b
a,
kesmadagi aniq integrali deyiladi va
b
a
dx
x
f
)
(
simvol bilan belgilanadi
) ta’rif.
n
i
i
i
x
c
f
1
integral yig’indining
b
a,
kesmaning
)
,...,
3
,
2
,
1
(
,
1
n
i
x
x
i
i
qismiy kesmalarga bo’linish usuliga bog’liq bo’lmagan
0
dagi chekli limiti mavjud bo’lsa, bu limitga
)
( x
f
funksiyaning
b
a,
kesmadagi aniq integrali deyiladi va
b
a
dx
x
f
)
(
simvol bilan belgilanadi
D) ta’rif.
n
i
i
i
x
c
f
1
integral yig’indining
b
a,
kesmada
n
c
c
c
,
....
,
,
2
1
nuqtalarning tanlanishiga bog’liq bo’lmagan
0
dagi chekli limiti mavjud
bo’lsa, bu limitga
)
( x
f
funksiyaning
b
a,
kesmadagi aniq integrali deyiladi va
b
a
dx
x
f
)
(
simvol bilan belgilanadi
E) ta’rif.
n
i
i
i
x
c
f
1
integral yig’indining
b
a,
kesmaning
)
,...,
3
,
2
,
1
(
,
1
n
i
x
x
i
i
qismiy kesmalarga bo’linish usuliga va ularda
n
c
c
c
,
....
,
,
2
1
nuqtalarning tanlanishiga bog’liq bo’lmagan chekli limiti mavjud
bo’lsa, bu limitga
)
( x
f
funksiyaning
b
a,
kesmadagi aniq integrali deyiladi va
b
a
dx
x
f
)
(
simvol bilan belgilanadi
275. Aniq integral quyidagi xossalardan qaysilari to’g’ri berilgan: 1) chekli
sondagi integrallanuvchi funksiyalar algebraik yig’indisining aniq integrali
qo’shiluvchilar aniq integrallarining algebraik yig’indisiga teng, ya’ni
139
;
)
(
)
(
)
(
)
(
)
(
)
(
3
2
1
3
2
1
b
a
b
a
b
a
b
a
dx
x
f
dx
x
f
dx
x
f
dx
x
f
x
f
x
f
2) o’zgarmas ko’paytuvchini aniq integral belgisidan chiqarish mumkin,
ya’ni
b
a
b
a
dx
x
f
k
dx
x
kf
)
(
)
(
;
3)
b
a,
kesmada
0
)
(x
f
bo’lsa,
b
a
dx
x
f
.
0
)
(
bo’ladi;
4)
b
a,
kesmada
)
(
)
(
x
g
x
f
tengsizlik bajarilsa,
b
a
b
a
dx
x
g
dx
x
f
)
(
)
(
bo’ladi.
A) hammasi
)
1),3)
D)
2),4)
E) 2),3)
276. Aniq integral quyidagi xossalardan qaysilari to’g’ri berilgan: 1)
b
a
c
,
kesmadagi biror nuqta bo’lsa,
b
c
c
a
b
a
dx
x
f
dx
x
f
dx
x
f
)
(
)
(
)
(
tenglik o’rinli bo’ladi; 2)
m
va
M
sonlar
)
(x
f
y
funksiyaning
b
a,
kesmadagi mos ravishda eng kichik va eng katta qiymatlari bo’lsa,
b
a
a
b
M
dx
x
f
a
b
m
)
(
)
(
)
(
tenglik o’rinli bo’ladi;
3)
a
b
b
a
dx
x
f
dx
x
f
)
(
)
(
.
A) hammasi
) 3)
D) 1)
E) 2)
277. Aniq integral quyidagi xossalardan qaysilari to’g’ri berilgan: 1)
;
0
)
(
a
a
dx
x
f
2)
b
a
b
a
b
a
dn
n
f
dt
t
f
dx
x
f
)
(
)
(
)
(
bo’ladi;
3)
)
(x
f
y
b
a,
kesmada uzluksiz bo’lsa, bu kesmada shunday bir
c
nuqta
topiladiki
)
)(
(
)
(
a
b
c
f
dx
x
f
b
a
tenglik o’rinli bo’ladi.
140
A) hammasi
) 2)
D) 1)
E) 3)
278.
x
F
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