1-topshiriq varianlari Berilgan integralni Simpson hamda Monte-Karlo usulida hisoblang. Oraliqni bo’linish soni N, hamda sinovlar soni m ko’rsatilgan

1-topshiriq varianlari
Berilgan integralni Simpson hamda Monte-Karlo usulida hisoblang. Oraliqni bo’linish soni N , hamda sinovlar soni M ko’rsatilgan.
𝑐𝑜𝑠(𝑥 ² + 1) ∙ (𝑥 ³ + 3𝑥 + 1 )^1/3;
[0;2]

#include
#include
using namespace std;
int main()
{int a, b, n;
cout<<"Aniq integral chegaralarini kiriting:"<cout<<"a="; cin>>a;
cout<<"b="; cin>>b;
cout<<"Bo'laklashlar sonini kiriting:"<cout<<"n="; cin>>n;
int m=n/2;
double h=1.0*(b-a)/(2*m);
float y=cos(a*a+1)*pow((a*a*a+3*a+1),1/3)+cos(b*b+1)*pow((b*b*b+3*b+1),1/3);
double y1=0;
for(int k=1; k<=m-1; k++)
{
y1=y1+cos((a+2*k*h)*(a+2*k*h)+1)*pow((pow((a+2*k*h),3)+3*(a+2*k*h)+1),1/3);
}
double y2=0;
for(int k=1; k<=m; k++)
{
y2=y2+cos((a+(2*k-1)*h)*(a+(2*k-1)*h)+1)*pow((pow((a+(2*k-1)*h),3)+3*(a+(2*k-1)*h)+1),1/3);
}
double S=(h*1.0/3)*(y+2*y1+4*y2);
cout<<"Simpson formulasi uchun natija quyidagicha: "<<"S="<return 0;
}

.2. Algebraik va transtsendent tenglamalarni yechishda oraliqni teng ikkiga bo’lish, iteratsiya usullari.

#include
#include
using namespace std;
float fx(float x){
float y=2*x*x*x+2*x-4;
return y;
}
int main(){
float a,b;
cout<<" A va b oraliqni kiriting "<cout<<"a = ";cin>>a;
cout<<"b = ";cin>>b;
if(fx(a)*fx(b)<0){
cout<<"Funksiya yechimga ega "<int i=0,k=1;
float c;
while(k==1){
i++;
c=(a+b)/2;
if(fx(a)*fx(c)<0){
b=c;
}
if(fx(c)*fx(b)<0){
a=c;
}
if(fabs(b-a)<0.001){
cout<<"x= "<k=0;
}
}
}
else{
cout<<" Bu oraliqda yechimga ega emas "< }}

#include

#include
using namespace std;
float fx(float x){
float y=2*x*x*x*x-log(x)-3;
return y;
}
int main(){
float a,b;
cout<<" A va b oraliqni kiriting "<cout<<"a = ";cin>>a;
cout<<"b = ";cin>>b;
if(fx(a)*fx(b)<0){
cout<<"Funksiya yechimga ega "<int i=0,k=1;
float c;
while(k==1){
i++;
c=(a+b)/2;
if(fx(a)*fx(c)<0){
b=c;
}
if(fx(c)*fx(b)<0){
a=c;
}
if(fabs(b-a)<0.001){
cout<<"x= "<k=0;
}
}
}
else{
cout<<" Bu oraliqda yechimga ega emas "< }}

3-Topshiriq.
Tenglamalarni yechishda Nyuton va vatarlar usullari. Yaqinlashish tezligi.


7) 𝑎) 2 − 𝑥 = ln 𝑥 ;
#include
#include
using namespace std;
float fx(float x){
float y=2-x-log(x);
return y;
}
int main(){
float a,b;
cout<<" A va b oraliqni kiriting "<cout<<"a = ";cin>>a;
cout<<"b = ";cin>>b;
if(fx(a)*fx(b)<0){
cout<<"Funksiya yechimga ega "<int i=0,k=1;
float c;
while(k==1){
i++;
c=(a+b)/2;
if(fx(a)*fx(c)<0){
b=c;
}
if(fx(c)*fx(b)<0){
a=c;
}
if(fabs(b-a)<0.001){
cout<<"x= "<k=0;
}
}
}
else{
cout<<" Bu oraliqda yechimga ega emas "< }}

𝑏) 𝑥 ³ − 0.2𝑥 ² + 0.4𝑥 − 1.4 = 0.

#include
#include
using namespace std;
float fx(float x){
float y=x*x*x-0.2*x*x+0.4*x-1.4;
return y;
}
int main(){
float a,b;
cout<<" A va b oraliqni kiriting "<cout<<"a = ";cin>>a;
cout<<"b = ";cin>>b;
if(fx(a)*fx(b)<0){
cout<<"Funksiya yechimga ega "<int i=0,k=1;
float c;
while(k==1){
i++;
c=(a+b)/2;
if(fx(a)*fx(c)<0){
b=c;
}
if(fx(c)*fx(b)<0){
a=c;
}
if(fabs(b-a)<0.001){
cout<<"x= "<k=0;
}
}
}
else{
cout<<" Bu oraliqda yechimga ega emas "< }}