Thread: The trapezoidal rule question

Hi,
I can't figure out where is the problem.Please help me to correct my code

Code:

#include <stdio.h>
#include <math.h>
void trap(double a,double b, int n, double *areag, double *areah);
double g(double x);
double h(double x);   
int main(void)
{
	double  areag = 0.0, areah = 0.0;
	double a1 = 0, b1 = 10;
	int n;
	for(n=2;n<=128;n*=2){
		trap(a1, b1, n, &areag, &areah);
		printf("%f %f\n", areag, areah);
	}
	return(0);
}

double g(double x){
	return(pow(x,2)*sin(x));
}
double h(double x){
	return(sqrt(4-pow(x,2)));
}
void trap(double a,double b, int n, double *areag, double *areah){
	int i, l;
	*areag = (b-a)/2*n * ( g(a) + g(b));
	for(i = 1; i<=n-1;i++)
		*areag += 2*g(i);
	*areah = (b-a)/2*n * ( h(a) + h(b));	
	for(l=1;l<=n-1;l++)
		*areah += 2*h(i);
}

Thanks for your advices.If you build and run the code, you will see it doesn't work properly.This question from a book(Problem solving and program design in C).It gives 2 equation and wants find approximating area under a curve.And adds call traps with values for n of 2, 4,8, 16, 32, 64, 128.
Output of my code is negative and -nan.
Equations are:
g(x) = x^2sinx (a = 0, b = 3.14159)
h(x) = sqrt(4-pow(x.2)) ( a =-2, b=2);

I see.I will do what you said.
Solved the problem;
The problem was in trap function :

Code:

void trap(double a,double b, int n, double *areag, double *areah){
        int i, l;
	*areag = (b-a)/2*n * ( g(a) + g(b));
	for(i = 1; i<=n-1;i++)
		*areag += 2*g( (double)i / n);
	*areah = (b-a)/2*n * ( h(a) + h(b));	
	for(l=1;l<=n-1;l++)
		*areah += 2*h( (double) l / n);
}

Originally Posted by nimitzhunter

you misunderstood the interior points of traperzoid rule.

The for loop should be:

Code:

for( something with i)
    *areag += (b-a)/n*g( a+ i*(b-a)/n)

or define the step-size like tripleA suggested

Code:

double h = (b-a)/n;
for( something in here)
    *area += h*g(a+i*h);

To keep it efficient you don't want to do all those multiplication over and over again in the for loop. The h multiplication in my suggestion can be taken out and do it after the loop.

And my previous point #2 is wrong, the multiplication of the stepping size should be taken out from the equation and applies to to the total area before returning it.

Thanks for your advices.And I guess proper code is

Code:

#include <stdio.h>
#include <math.h>
void trap(double a,double b, int n, double *areag, double *areah);
double g(double x);
double h(double x);   
int main(void)
{
	double  areag = 0.0, areah = 0.0;
	double a1 = 0, b1 = 3;
	int n;
	for(n=2;n<128;n*=2){
		trap(a1, b1, n, &areag, &areah);
		printf("%f %f\n", areag, areah);
	}
	return(0);
}

double g(double x){
	return(pow(x,2)*sin(x));
}
double h(double x){
	return(sqrt(fabs(4-pow(x,2))));
}
void trap(double a,double b, int n, double *areag, double *areah){
	int i, l;
	*areag = ((b-a)/2)*n * ( g(a) + g(b));
	for(i = 1; i<=n-1;i++)
		*areag += (b-a)/n*g( a+ i*(b-a)/n);
	*areah = (b-a)/2*n * ( h(a) + h(b));	
	for(l=1;l<=n-1;l++)
		*areah += (b-a)/n*h(( a+ l*(b-a)/n));
}