Simpsons Rule Problem

[askinne2]

Hey guys, I have looked around and am having trouble figuring out why my program is not integrating the function correctly. Here is my program and a sample of the output.

#include <stdio.h>  //for use of the standard i/o functions
#include <stdlib.h> //for use of the standard library functions
#include <math.h>   //for use of the math functions

double function(double x) //prototype for passing a number into the function
{
    return exp(-x*x); //the hardcoded function e to the -x squared
}

double simpson(double a, double b, int c) //passing variables into the simpson function for evaluating simpsons rule 
//a is the integrate to, b is the integrate from
//and c is the number of subdivisions
{
       double dx = 1;  //intial value of dx is 1 for the first pass of the for loop
       int i = 0;   //this variable is used for a 'for' loop as a counter and index position in the array
       int n = 1;  //this will be used as the multiplier to dx which increments based on the number of subdivisions
       int j = 1;
       int arr_length = ++c;  //this array will hold the number of values calculated for the integration
       double odd_sum = 0;
       double even_sum = 0;
       double total_sum = 0;
       double answer_sum = 0;
       double first_sum = 0;
       double last_sum = 0;
       
       dx = (b - a)/(c);
       
       double arr_nums[arr_length]; //intialization of the array of numbers which will be summated for the integration
       
       for(i = 1; i < --c; i++) //runs a for loop to evaluate the areas of the subdivisions of the function
       {
           arr_nums[i] = function(a + (dx * j));
           j++;
       }
       
       if(n % 2 != 0 & n != c)
       {
           odd_sum = ((arr_nums[n]) * 4) + odd_sum;
           n++;
           n++;
       }
           
       n = 2;
       
       if(n % 2 == 0)
       {
            even_sum = ((arr_nums[n]) * 2) + even_sum;
            n++;
            n++;
       }
           
       first_sum = function(a);
       last_sum = function(b);
           
       for(i = 0; i < c; i++)
       {
            total_sum = arr_nums[i] + total_sum;
       }
       
       total_sum = total_sum + first_sum + last_sum;
           
       answer_sum = (dx / 3) * total_sum;
           
       return answer_sum; 
}

int main()
{
    double integrate_from;
    double integrate_to;
    int num_of_subdivisions;
    int temp_holder;
    double answer;
    
    printf("Integrate from: \n");
    scanf("%lf", &integrate_from);
    printf("Integrate to: \n");
    scanf("%lf", &integrate_to);
    printf("Number of subdivisions (must be a multiple of 2): \n");
    scanf("%d", &num_of_subdivisions);
    
    if(num_of_subdivisions % 2 != 0)
    {
         printf("That is not a multiple of 2");          
         exit(1);
    }
        
    else
    {
         // answer = simpson(integrate_from, integrate_to, num_of_subdivisions);
         answer = simpson(integrate_from, integrate_to, num_of_subdivisions);
         printf("%.10f\nThe integration is: ", answer);        
    }
    
}

Sample output:

C:\Users\askinne2\Desktop\Fall 2010\CS\Labs>simpson
Integrate from:
0
Integrate to:
3.14
Number of subdivisions (must be a multiple of 2):
6
0.3386618454

[rags_to_riches]

Warnings generated:

simp.c: In function ‘simpson’:
simp.c:28: warning: ISO C90 forbids variable-size array ‘arr_nums’
simp.c:28: warning: ISO C90 forbids mixed declarations and code
simp.c:36: warning: suggest parentheses around comparison in operand of &
simp.c: In function ‘main’:
simp.c:72: warning: unused variable ‘temp_holder’
simp.c:95: warning: control reaches end of non-void function

Of particular interest is the one on line 36. What is the correct syntax for boolean and?

if(n % 2 != 0 & n != c)

What's the expected result, BTW? I left my calculus far behind me.