>>707
#include <stdio.h>
#define N 10000
#define XS 0.0
#define XE 1.0
double f(double x) {return 1.0 / (1.0 + x*x);}
int main()
{
double x1, x2, s, dx = (XE - XS) / N;
int i;
s = 0.0;
for (i=0; i<N; i++) {
x1 = XS + i*dx;
x2 = x1 + dx;
s += f(x1) + f(x2);
}
s = s * dx / 2.0;
printf("ans = %12.9f\n", s);
return 0;
}