Simpsons 3/8th rule

    static double mathFunc(double x) {
        // function = e^x + x^3
        return Math.pow(Math.E, x)+Math.pow(x, 3);
    }

    // Simpsons 3/8th rule
    // a is begging of interval b is the end
    // n is the amount of times we are splitting
    //that function into smaller segments that are gonna be added together
    static double simpsons38th(double a, double b, double n) {
        double result = 0;
        double h = (b-a)/n;
        result += (mathFunc(a)+mathFunc(b));

        for(int i = 0; i < n ; i++) {
            double x;
            if(i%3 != 0) {
                x = (a + i * h);
                result +=  3.0 * mathFunc(x);
                //System.out.println(i);
            }
        }

        for(int i = 0; i < n/3  ; i++) {
            double x = (a + i * h * 3);
            result +=  2.0 * mathFunc(x);
            //System.out.println(i*3);
        }

        return (3.0 * h * result) / 8.0 ;
    }

    public static void main(String[] args) {
        System.out.println("f(x) = e^x + x^3");
        double a = 0, b = 1;
        int n = 10000;
        System.out.println("Simpson 3/8th    = " + simpsons38th(a, b, n));
    }