1.52804 9.63106 5.69143 4.58413 12.22434 3.34052
3.91932 12.84398 1.66968 11.75795 31.35454 4.09430
0.64177 5.21029 3.87952 1.92530 5.13413 2.41910
N1= 10
Y=1
2.82219 17.06957 3.47711 4.46002
3.45048 25.29536 2.24122 5.36151
3.84781 27.88627 0.33742 5.63330
3.79823 11.64860 3.14921 3.93024
1.87480 9.54129 3.73618 3.37877
2.59128 8.44137 1.67605 3.32154
3.31921 26.38727 3.53498 5.45036
3.94437 29.15086 1.36120 5.75285
1.52804 9.63106 5.69143 3.34052
3.91932 12.84398 1.66968 4.09430
N2= 8
xtT(i) ytT(i) ztT(i) FNF(xtT(I), ytT(I))
2.82219 17.06957 3.47711 4.46002
3.45048 25.29536 2.24122 5.36151
3.84781 27.88627 0.33742 5.63330
3.79823 11.64860 3.14921 3.93024
1.87480 9.54129 3.73618 3.37877
2.59128 8.44137 1.67605 3.32154
3.31921 26.38727 3.53498 5.45036
3.94437 29.15086 1.36120 5.75285
w= 128
S1= 307.2
Yuqoridan ko`rinib turibdiki aniq yechim y=162.1 va biz Basic tilida hisoblab topgan yechimimiz s1=307.2. endi xatolikni baholab olamiz:
=(s1-y)/y=(307.2-162.1)/162.1=0.90=90% Demak, xatolik 90% , bundan ko`rinib turibdiki, biz random orqali tanlab olgan ixtiyoriy x, y o`zgaruvchilarimiz hisoblashda juda katta xatolikni keltirib chiqarar ekan, dasturdan olingan natija hisob-kitobimizni umuman qanoatlantirmaydi desak mubolag`a bo`lmaydi. Shu sababli ham o`zgaruvchilarni tanlashda alohida e`tibor qaratish lozim.
3-masala. V = {1x3, 0yx, x+yzx+2y integrallash sohasi asosida
integralni (2.41) formula asosida hisoblang.
(2.41) formula dagi o‘zgaruvchilarni berilgan misoldagi qiymatlari a=1, b=3, s=0, d=3, y=1, h=9
Bu holda quyidagicha almashtirish qilamiz:
x=1+2, u=3, z=1+8, n=24
Tasodifiy sonlar jadvaldan 80 ta qiymat olamiz (N=20). Jadvaldan yi (i = 1, 2, .. 20) shartni qanoatlantiruvchi ui larni topamiz, ular 11 ta, bularga mos keluvchi larni aniqlaymiz ular 3 ga teng, bu 3 ta songa mos keluvchi yi ni yiUi shartga asosan topamiz u n=1.
Demak,
Berilgan integral aniq yechimi:
Xatolikni baholaymiz:
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