| 1 |
|
|
#include<stdio.h> |
|
|
| 2 |
|
|
#include<conio.h> |
|
|
| 3 |
|
|
#include<math.h> |
|
|
| 4 |
|
|
float f(float x) |
|
|
| 5 |
|
|
{ |
|
|
| 6 |
|
|
return(1/(1+pow(x,2))); |
|
|
| 7 |
|
|
} |
|
|
| 8 |
|
|
void main() |
|
|
| 9 |
|
|
{ |
|
|
| 10 |
|
|
int i,n; |
|
|
| 11 |
|
|
float xn,x0,h,y[20],se,so,ans,x[20]; |
|
|
| 12 |
|
|
printf("\n Enter the values of xn,x0,h:\n"); |
|
|
| 13 |
|
|
scanf("%f%f%f",&x0,&xn,&h); |
|
|
| 14 |
|
|
n=(xn-x0)/h; |
|
|
| 15 |
|
|
if(n%2==1) |
|
|
| 16 |
|
|
{ |
|
|
| 17 |
|
|
n=n+1; |
|
|
| 18 |
|
|
} |
|
|
| 19 |
|
|
h=(xn-x0)/n; |
|
|
| 20 |
|
|
printf("\nrefined_value _of n and h are:%d %f\n",n,h); |
|
|
| 21 |
|
|
printf("\n Y values \n"); |
|
|
| 22 |
|
|
for(i=0; i<=n; i++) |
|
|
| 23 |
|
|
{ |
|
|
| 24 |
|
|
x[i]=x0+i*h; |
|
|
| 25 |
|
|
y[i]=f(x[i]); |
|
|
| 26 |
|
|
printf("\n%f\n",y[i]); |
|
|
| 27 |
|
|
} |
|
|
| 28 |
|
|
so=0; |
|
|
| 29 |
|
|
se=0; |
|
|
| 30 |
|
|
for(i=1; i<n; i++) |
|
|
| 31 |
|
|
{ |
|
|
| 32 |
|
|
if(i%2==1) |
|
|
| 33 |
|
|
{ |
|
|
| 34 |
|
|
so=so+y[i]; |
|
|
| 35 |
|
|
} |
|
|
| 36 |
|
|
else |
|
|
| 37 |
|
|
{ |
|
|
| 38 |
|
|
se=se+y[i]; |
|
|
| 39 |
|
|
} |
|
|
| 40 |
|
|
} |
|
|
| 41 |
|
|
ans=h/3*(y[0]+y[n]+4*so+2*se); |
|
|
| 42 |
|
|
printf("\nFinal_integration is %f",ans); |
|
|
| 43 |
|
|
getch(); |
|
|
| 44 |
|
|
} |
|
|