如何计算曲线函数的x和y坐标 [英] How to calculate x and y coordinates of curve function

问题描述

我有一个像y = px ^ 2 + qx + r的函数。我想计算给定范围和段数以及质心位置的曲线下面积。我使用辛普森规则和梯形规则正确计算了面积。如何计算质心的位置？



我尝试过：

  public   double  f（ double  x）
 { int  p = Convert.ToInt32（textBox1.Text）; 
  int  q = Convert.ToInt32（textBox2.Text）; 
  int  r = Convert.ToInt32（textBox3.Text）; 
  double  result = p *（Math.Pow（x， 2 ））+ q * x + r; 
 返回结果; 
} 
  public   double  Calculate1（ double  a1， double  b1， int  n1）
 {
  int  i; 
  double  sum，integral; 
  double  h =（a1-b1）/ n1; 
 sum = f（a1）+ f（b1）; 
 i =  2 ; 
  while （i <  = n1）
 {
 sum = sum +  2  * f（b1 +（i  -   1 ）* h）; 
 i ++; 
} 
 integral = h * sum /  2 ; 
 返回积分; 
 
 
 
} 
  public   double 计算（ double  a， double  b， int  n）
 {
  var  h =（b  -  a）/ n; 
  var  sum =  0 。 0 ; 
  for （ var  i =  1 ; i <  = n  -   3 ; i = i +  2 ）
 sum + = f（a + i * h）; 
 sum + = f（a +（n  -   1 ）* h）; 
 sum =  4  * sum; 
  var  sum2 =  0 。 0 ; 
  for （ var  i =  2 ; i <  = n  -   4 ; i + =  2 ）
 sum2 + = f（a + i * h）; 
 sum2 + = f（a +（n  -   2 ）* h）; 
 sum2 * =  2 ; 
 sum + = sum2 + f（a）+ f（b）; 
  return  h /  3  * sum; 
} 

解决方案

从 5。整合区域的质心 [ ^ ]。

I have a function like y=px^2+qx+r. I want to calculate the area under the curve for the given range and number of segments and location of the centroid. I calculated area correctly using Simpson rule and trapezoidal rule. How can I calculate location of the centroid?

What I have tried:

public double f(double x)
       { int p =Convert.ToInt32( textBox1.Text);
         int q= Convert.ToInt32(textBox2.Text);
         int r = Convert.ToInt32(textBox3.Text);
         double result = p * (Math.Pow(x, 2)) + q * x + r;
           return result;
       }
       public double Calculate1(double a1, double b1, int n1)
       {
           int  i;
           double sum, integral;
            double h = (a1 - b1) / n1;
           sum = f(a1) + f(b1);
           i = 2;
           while (i <= n1)
           {
               sum = sum + 2 * f(b1 + (i - 1) * h);
               i++;
           }
           integral = h * sum / 2;
           return integral;



       }
       public double Calculate(double a, double b, int n)
       {
           var h = (b - a) / n;
           var sum = 0.0;
           for (var i = 1; i <= n - 3; i = i + 2)
               sum += f(a + i * h);
           sum += f(a + (n - 1) * h);
           sum = 4 * sum;
           var sum2 = 0.0;
           for (var i = 2; i <= n - 4; i += 2)
               sum2 += f(a + i * h);
           sum2 += f(a + (n - 2) * h);
           sum2 *= 2;
           sum += sum2 + f(a) + f(b);
           return h / 3 * sum;
       }

解决方案

Start from 5. Centroid of an Area by Integration[^].

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