洛谷T219056工科数学分析-优快云博客

本文通过C++代码展示了使用数值积分方法求解不同函数的积分过程，包括倒数函数、对数函数及复杂复合函数等。采用梯形法则进行估算，并比较了不同函数在指定区间内的积分结果。

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#include<iostream>
#include<cmath>
using namespace std;
int main()
{
 int a, b;
 double sum,aa;
 cin >> a >> b;
 int n = 30000;
 double h =1.0* (b - a) / n;

 sum = (1.0 / a + 1.0 / b) / 2;
 for (int i = 1; i < n; i++)
  sum = sum + (1.0 / (a + i * h));
 sum = sum * h;
 printf("%.6f\n", sum);

 sum = (log(a) + log(b)) / 2;
 for (int i = 1; i < n; i++)
  sum = sum + (log(a+i*h));
 sum = sum * h;
 printf("%.6f\n", sum);

 
 sum = (a*sin(a*a)*sin(a*a) * sin(a * a) * sin(a * a) * sin(a * a) + b * sin(b * b) * sin(b * b) * sin(b * b) * sin(b * b) * sin(b * b)) / 2;
 for (int i = 1; i < n; i++)
 {
   aa = a + i * h;
  sum = sum + (aa * sin(aa * aa) * sin(aa * aa) * sin(aa * aa) * sin(aa * aa) * sin(aa * aa));
 }
  
 sum = sum * h;
 printf("%.6f\n", sum);

 sum = (pow(3 + 2 * a * a, 0.5) + pow(3 + 2 * b * b, 0.5)) / 2;
 for (int i = 1; i < n; i++)
  sum = sum + (pow(3 + 2 * (a + i * h) * (a + i * h), 0.5));
 sum = sum * h;
 printf("%.6f\n", sum );
 
 sum = (atan((a-3)*(a-3)) , atan((b - 3) * (b - 3))) / 2;
 for (int i = 1; i < n; i++)
  sum=sum+ atan((a+i*h - 3) * (a+i*h - 3));
 sum = sum * h;
 printf("%.6f\n", sum+0.000001);
 return 0;
}