Ggtcf

I am trying to compute the integral of the function f(x)=(1-x^2)^(1/2) from x=0 to x=1. The answer should be approximately pi/4. I am currently getting 2.

My current implementation of the trapezoidal rule is the following:

double
def_integral(double *f, double *x, int n)

 double F;
 for (int i = 0 ; i < n ; i++) 
 F += 0.5 * ( x[i+1] - x[i] ) * ( f[i] + f[i+1] );
 
 return F;

I'm creating N divisions to approximate the area under the curve between x_1=0 and x_N=1 by looping through i to N with x_i = i / N.

int
main(int argc, char **argv)

 int N = 1000;
 double f_x[N];
 double x[N];

 for (int i = 0 ; i <= N ; i++) 
 double x = i * 1. / N;
 f_x[i] = sqrt(1. - pow(x, 2.));
 //printf("%.2f %.5fn", x, f_x[i]); //uncomment if you wanna see function values
 

 double F_x = def_integral(f_x, x, N);

 printf("The integral is %gn", F_x);

The result of 2 that I am currently getting should be dependent on the number of N division, however, no matter if I make N=10000 or N=100, I still get 2.

Any suggestions?

In this for loop, you forgot updatin array x as well.

for (int i = 0 ; i <= N ; i++) 
 double x = i * 1. / N;
 f_x[i] = sqrt(1. - pow(x, 2.));
 //printf("%.2f %.5fn", x, f_x[i]); //uncomment if you wanna see function values

So, for loop should be replaced by

for (int i = 0 ; i <= N ; i++) 
 double xi = i * 1. / N;
 x[i] = xi;
 f_x[i] = sqrt(1. - pow(xi , 2.));
 //printf("%.2f %.5fn", x, f_x[i]); //uncomment if you wanna see function values

In your main code, you call def_integral with a double (x) and in the function an array of x (double * x) is expected. Perhaps (it is what I suppose), the problem comes from the fact you formula needs x(i+1)-x(i) but you use a constant step. Indeed, x(i+1)-x(i)=step_x is constant so you do not need each x(i) but only value : 1./N

Other remark, with a constant step, your formula could be simplified to :
F_x=step_x* ( 0.5*f_x(x0)+ f_x(x1)+...+f_x(xn-1)+ 0.5*f_x(xn) ) . It helps to simplify the code and to write a better efficient one.
Everything is commented in the code above. I hope it could help you.
Best regards.

#include <stdio.h>
#include <math.h>

double
def_integral(double *f, double step_x, int n)

 double F;
 for (int i = 0 ; i < n ; i++) 
 F += 0.5 * ( step_x ) * ( f[i] + f[i+1] );
 
 return F;


int main()

 int N = 1001; // 1001 abscissas means 1000 intervalls (see comment on array size and indices)
 double f_x[N]; // not needed for the simplified algorithm
 double step_x = 1. / N; // x(i+1)-x(i) is constant

 for (int i = 0 ; i < N ; i++) // Note : i<N and not i<=N
 double xi = i * step_x; // abscissa calculation
 f_x[i] = sqrt((1. - xi )*(1. + xi )); // cf chux comment
 

 double F_x = def_integral(f_x, step_x, N);
 printf("The integral is %.10gn", F_x);

// simplified algorithm
// F_x=step_x*( 0.5*f_x(x0)+f_x(x1)+...+f_x(xn-1)+0.5f_x(xn) )
 double xi;
 xi=0; // x(0)
 F_x=0.5*sqrt((1. - xi )*(1. + xi ));
 for (int i=1 ; i<=N-1 ; i++) 
 xi=step_x*i;
 F_x+=sqrt((1. - xi )*(1. + xi ));
 
 xi=step_x*N;
 F_x+=0.5*sqrt((1. - xi )*(1. + xi ));
 F_x=step_x*F_x;
 printf("The integral is %.10gn", F_x);