Unit 4.3 Simpson's 3/8 Integration

#include <stdio.h>
#include <math.h>
#define f(x) 1/(1+x*x)
int main()
{
    float b,a,h;
    int n;
    printf("Enter the values of upper limit,lower limit and h\n");
    scanf("%f %f %f",&b,&a,&h);
    n = (b-a)/h +1 ;
    float x[n];
    for(int i=0 ; i<n ; i++){
        x[i] = a+ (i*h);
    }
    float div_3=0,nondiv_3=0;
    for(int i=1 ; i<n-1 ; i++){
        if(i%3==0)
            div_3 += f(x[i]);
        else
            nondiv_3 += f(x[i]);
    }
    float ex_ord = f(x[0]) + f(x[n-1]); // extreme ordinates
    float ans = (3*h)/8 *  ( ex_ord + (3*nondiv_3) + (2*div_3) ) ;
    printf ("\nFinal answer is %f",ans);

    return 0;
}
/*
Enter the values of upper limit,lower limit and h
1 0 0.166666
Final answer is 0.785394
*/

// Algorithm

define function
read a,b and h , where a lower , b upper limit and h size of interval
n = (b-a)/h + 1
declare float array x of size of n

start a loop from i=0 to n-1
    x[i] = a + (i*h)
end i loop

calulating the sum of ordinate divisible by 3 and not divisible by 3
declare div_3 = 0 and nondiv_3 = 0
start loop from i=1 to n-2
    check if i%3==0
    if yes
    then
        div_3 = div_3 + f(x[i])
    if no
    else
        nondiv_3 = nondiv_3 + f(x[i])
end i loop

ex_ord = f(x[0]) + f(x[n-1]) , sum of extreme ordinates

ans = (3*h)/8 *  ( ex_ord + (3*nondiv_3) + (2*div_3) )

print ans
stop