```
/********************************************************************
/
/ noname.c
/
/ A program in C to calculate the area under a curve by using the
/ 5th Degree Open Newton-Cotes Formula, which has no name
/
/ Paul Soper
/
/ March 2, 2018
/
/*******************************************************************/
#include "stdio.h"
/********************************************************************
/
/ double f(double x)
/
/ We need to define the curve by an algebraic function. We will use
/ a polynomial here, but the function could be any continuous
/ function
/
/*******************************************************************/
double f(double x)
{
/* use 5x^3 - 12x^2 + 7x - 3 as the polynomial function */
return (5.0*x*x*x - 12.0*x*x + 7.0*x - 3.0);
}
int main()
{
int n = 10; /* n = the number of segments */
int i = 0; /* a counter */
double area = 0; /* this will hold the area under the curve
as we sum over the areas of each segment */
/* We will be calculating the area under the curve y = f(x)
from x = a to x = b, using n segments, and using the
fifth degree open Newton-Cotes method */
/* x1 and x2 are the endpoints for the whole calculation */
double x1 = 2.0;
double x2 = 6.0;
double width = (x2 - x1)/(float)n;
/* a and b are the endpoints for each segment */
double a;
double b;
for (i = 0; i < n; ++i)
{
a = x1 + i * width;
b = x1 + (i + 1) * width;
area = area + ((b-a)/24.0) * (11.0 * f(a + 1.0 * (b - a)/5.0) +
1.0 * f(a + 2.0 * (b - a)/5.0) +
1.0 * f(a + 3.0 * (b - a)/5.0) +
11.0 * f(a + 4.0 * (b - a)/5.0));
}
printf ("Area = %16.10fn", area);
/* We can easily check our answer, because f is just a
polynomial and is easily integrated, so we can solve the problem
analytically rather than numerically.
The integral is g(x) = (5/4)x^4 - (12/3)x^3 + (7/2)x^2 -3x + C
The definite integral (which is the area under the curve), is
g(b) - g(a).
In our case that is 864 - (-4) = 868.0 */
printf ("Analytic solution: area = 868.00n");
return (0);
}
```

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