如何在apache commons math3库中计算函数的集成？ [英] How to compute integration of a function in apache commons math3 library?

问题描述

我正在尝试集成一个非常简单的功能。积分（x.dx）。我没有得到1的答案，我得到的答案为0或0.5，当我包含从0到1的限制时。有什么我误解了apache commons库中集成的实现吗？

I am trying to integrate a very simple function. integral(x.dx). Instead of getting an answer of 1, I am getting an answer as 0, or 0.5 when I include limits from 0 to 1. Is there something I misunderstand about the implementation of the integration in apache commons library?

import org.apache.commons.math3.analysis.integration.*;
import org.apache.commons.math3.analysis.polynomials.*;

public static void main(String args[])
{
    SimpsonIntegrator simpson = new SimpsonIntegrator();
    TrapezoidIntegrator trapezoid = new TrapezoidIntegrator();
    double[] vector = new double[2];
    vector[0] = 0;
    vector[1] = 1;

    PolynomialFunction f = new PolynomialFunction(vector);
    UnivariateFunction uf = (UnivariateFunction)new PolynomialFunction(vector);
    System.out.println("To String " + uf.toString());
    System.out.println("Degree: " + f.degree());

    double i = simpson.integrate(10, uf, -Float.MAX_VALUE, Float.MAX_VALUE);
    double j = trapezoid.integrate(10, uf, 0, 1);
    System.out.println("Simpson integral : " + i);        
    System.out.println("Trapezoid integral : " + j);        
}
/*** OUTPUT 
To String x
Degree: 1
Simpson integral : 0.0
Trapezoid integral : 0.5
***/

推荐答案

我认为这是按预期运作的。您正在积分的函数是斜率1的直线。

I think this is functioning as expected. The function you are integrating is the straight line of slope 1.

在0和1之间，您获得的面积为0.5。在所有空间中，上方和下方的积分抵消为0。

Between 0 and 1 you get an area of 0.5. Over all of the space the integrals above and below cancel out to give 0.

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