题目内容:已知f(x)=1/(1+x*x),编写函数用梯形法计算f(x)在区间[a,b]的积分。 输入格式:输入a和b,a<b,均为double型 输出格式:输出为double型 输入样例:0.5 1 输出样例:0.321751 c++
时间: 2024-05-12 10:15:22 浏览: 210
以下是使用C++编写的函数代码:
```cpp
#include <iostream>
using namespace std;
double f(double x) {
return 1.0 / (1 + x * x);
}
double trapezoid(double a, double b, int n) {
double h = (b - a) / n;
double sum = (f(a) + f(b)) / 2;
for (int i = 1; i < n; i++) {
double x = a + i * h;
sum += f(x);
}
return sum * h;
}
int main() {
double a, b;
cin >> a >> b;
double result = trapezoid(a, b, 100000); //将区间[a,b]分成100000个小区间计算
cout << result << endl;
return 0;
}
```
函数`f(x)`实现了题目中给定的函数。`trapezoid`函数接受三个参数:区间左端点`a`、区间右端点`b`以及小区间数目`n`。函数返回使用梯形法计算得到的积分值。
在`trapezoid`函数中,我们首先计算出每个小区间的宽度`h`,然后计算出端点处的函数值,最后使用循环计算出中间各点处的函数值并求和。最终将得到的积分值乘以小区间宽度即可得到最终结果。本题中我们将区间分成了100000个小区间进行计算,可以根据需要调整小区间数目以获得更高的计算精度。
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