梯形法则问题

发布于 2024-11-03 10:41:41 字数 1065 浏览 8 评论 0原文

如果你构建并运行代码，你会发现它无法正常工作。这个问题来自一本书（C语言中的问题解决和程序设计）。它给出了2个方程，并想要找到曲线下的近似面积。并添加了调用陷阱n 的值为 2, 4,8, 16, 32, 64, 128。

我的代码的输出是负数和 -nan。方程为：

g(x) = x^2sinx (a = 0, b = 3.14159)
h(x) = sqrt(4-pow(x.2)) ( a =-2, b=2);

代码为：

#include <stdio.h>
#include <math.h>
void trap(double a,double b, int n, double *areag, double *areah);
double g(double x);
double h(double x);   
int main(void)
{
    double  areag = 0.0, areah = 0.0;
    double a1 = 0, b1 = 10;
    int n;
    for(n=2;n<=128;n*=2){
        trap(a1, b1, n, &areag, &areah);
        printf("%f %f\n", areag, areah);
    }
    return(0);
}

double g(double x){
    return(pow(x,2)*sin(x));
}
double h(double x){
    return(sqrt(4-pow(x,2)));
}
void trap(double a,double b, int n, double *areag, double *areah){
    int i, l;
    *areag = (b-a)/2*n * ( g(a) + g(b));
    for(i = 1; i<=n-1;i++)
        *areag += 2*g(i);
    *areah = (b-a)/2*n * ( h(a) + h(b));    
    for(l=1;l<=n-1;l++)
        *areah += 2*h(i);
}

原文

If you build and run the code, you will see it doesn't work properly.This question from a book(Problem solving and program design in C).It gives 2 equation and wants find approximating area under a curve.And adds call traps with values for n of 2, 4,8, 16, 32, 64, 128.

Output of my code is negative and -nan.
Equations are:

g(x) = x^2sinx (a = 0, b = 3.14159)
h(x) = sqrt(4-pow(x.2)) ( a =-2, b=2);

And the code is:

#include <stdio.h>
#include <math.h>
void trap(double a,double b, int n, double *areag, double *areah);
double g(double x);
double h(double x);   
int main(void)
{
    double  areag = 0.0, areah = 0.0;
    double a1 = 0, b1 = 10;
    int n;
    for(n=2;n<=128;n*=2){
        trap(a1, b1, n, &areag, &areah);
        printf("%f %f\n", areag, areah);
    }
    return(0);
}

double g(double x){
    return(pow(x,2)*sin(x));
}
double h(double x){
    return(sqrt(4-pow(x,2)));
}
void trap(double a,double b, int n, double *areag, double *areah){
    int i, l;
    *areag = (b-a)/2*n * ( g(a) + g(b));
    for(i = 1; i<=n-1;i++)
        *areag += 2*g(i);
    *areah = (b-a)/2*n * ( h(a) + h(b));    
    for(l=1;l<=n-1;l++)
        *areah += 2*h(i);
}

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

一袭水袖舞倾城 2024-11-10 10:41:41

我不确定它的意图是什么，因为你没有解释它应该如何工作，但这部分是取负数的平方根：

sqrt(4-pow(x,2))

啊，现在我明白这就是你想要积分的函数。问题是你需要将集成的范围分成小块，而不是在更大的范围内集成。尝试

*areah += 2*h( (double) i / n);

I'm not sure what's intended, because you didn't explain how it is supposed to work, but this part is taking the square root of negative numbers:

sqrt(4-pow(x,2))

Ah, now I see that is the function you want to integrate. The problem is that you need to divide the range of integration into small pieces, rather than integrating over a wider range. Try