在c++下用梯形法求函数e^(-x的平方）在0到1上的积分

#include <iostream>
#include <cmath>
using std::cout;
using std::endl;

double fun(double x){
double p=x*x*-1;
return exp(p);
}

int main(){
double result=0;
double x=0;
double h=1e-6;
while(x<1){
result+=fun(x);
x+=h;
}
result*=h;
cout<<result<<endl;
return 0;
}

答案应该是这样的，h的值取再小，小数点后面6为的值都是不变的。你检查一下。

#include <iostream>
#include <cmath>
using namespace std;

typedef double(*pf)(double);
double integral(pf,double a,double b);
double fun(double x){return fabs(-x);}
int main()
{
cout << integral(fun,0,1) << endl;
return 0;
}
double integral(pf f,double a,double b)
{
int i,n=100;
double x1,x2,sum=0,dx=(b-a)/n;
for(i=0;i<n-1;++i)
{
x1=f(a+i*dx);
x2=f(a+(i+1)*dx);