These functions will require you to tweak them if you go changing them around, and no guarantee is made for their accuracy. I have not systematically tested them for anything and do not warranty them for any use other than possibly to double check a homework assignment or two.

The first program I tied I got amazingly close to the results in my calculus book.

`.#include <stdio.h> #include <stdlib.h> // return x squared double f (double x){ return (x * x ); } int main (){ long double b = 1; // interval begin long double e = 2; // interval end long double n = 1000000; // number of steps., denomenator long double r = 0; // result long double c=(e-b); // The change in x per unit long double ca=c/n/2; // The multiplier for final result long double ba = b * n + c; // the begining numerator long double ea = e * n; // the ending numerator for (;ba<ea; ba+=c) r += f(ba/n); // sum all the intervals r *= 2; // double all the interval values. r += f(b) + f(e); //handle the first and last interval r *= ca; // apply the final multiplier printf("Answer is %.20Lf \n", r); }`

`Save the above as program.c`

Compile it with gcc program.c -o name

Then run it with ./name

`I got this as my answer:`

`Answer is 2.33333333333350000744`

The following code is similar to the above code, but I changed the square function to return the result of a polynomial. I don't have a good way of running this as

```
``````
#include <stdio.h>
#include <stdlib.h>
// return a polynomial
double f (double x){
return (2 * x * x +4*x -5);
}
int main (){
long double b = 1; // interval begin
long double e = 2; // interval end
long double n = 1000000; // number of steps., denomenator
long double r = 0; // result
long double c=(e-b); // The change in x per unit
long double ca=c/n/2; // The multiplier for final result
long double ba = b * n + c; // the begining numerator
long double ea = e * n; // the ending numerator
for (;ba<ea; ba+=c)
r += f(ba/n); // sum all the intervals
r *= 2; // double all the interval values.
r += f(b) + f(e); //handle the first and last interval
r *= ca; // apply the final multiplier
printf("Answer is %.20Lf \n", r);
}
```

This gave me a result of 5.66666666666699991731 but I am not sure how to check if that is right. I will integrate the function, and then calculate for the interval using those numbers to see if it gives similar results.

I am not sure if this program will work with fractional intervals, will require additional testing to see if that works.

A more accurate method is to use Simpson's rule, Approximating using Parabolas, which I will attempt to perform in the next few days.

**I updated the interval calculation to work right when the range is not an interval of 1.**

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