标题直接上代码了，根据代码走就行了

说明：这里讲龙贝格算法的步骤分开实现了，

• *可以将步骤合一实现，在求Tn的后面，每增加一个Tn 就可以增加一个Sn每增加一个Sn就可以增加一个Rn，由于前面两项后面产生一项，可以对每一个序列进行一个flag标记，初始标记为0当标记大于1
  的时候，每增加一个前面的序列值，后面就会增加一个，依次类推，写成一个在不超过精度的最简算法，节省空间，一步达成，这里分开写

**

修改函数，和区间可以实现任意函数的定积分

引用注明来源，请勿抄袭！！！

package 龙贝格算法;import java.util.ArrayList;
import java.util.Scanner;/*** @author liqiming* @date 2021-10-28 -18:33*/
public class Rebom {//math.e xpublic static void main(String[] args){Scanner scanner = new Scanner(System.in);System.out.println("请输入Tn的长度");int n = scanner.nextInt();double[] orgin = {0,1};double[] Tn = new double[n];double[] Sn = new double[n-1];double[] Cn = new double[n-2];double[] Rn = new double[n-3];ComplexTrapezoids(orgin, Tn);Sn=Sn(Tn);Cn= Cn(Sn);Rn = Rn(Cn);System.out.println("Tn序列");for (int i = 0; i < Tn.length; i++) {System.out.println(Tn[i]);} System.out.println("Sn序列");for (int i = 0; i < Sn.length; i++) {System.out.println(Sn[i]);} System.out.println("Cn序列");for (int i = 0; i < Cn.length; i++) {System.out.println(Cn[i]);} System.out.println("Rn序列");for (int i = 0; i < Rn.length; i++) {System.out.println(Rn[i]);}System.out.println("4/（1+x*x）在 0 -1 上的积分最终是"+Rn[Rn.length-1]);}/**** @param orgin 需要传入的原来的数组，里面规定了要积分的上下限* @param Tn     Tn里面是根据*/public  static  void ComplexTrapezoids(double[]orgin,double []Tn) {Tn[0]=(Fun(orgin[0])+Fun(orgin[1]))*(orgin[1]-orgin[0])/2;ArrayList<double[]> arrayList = MiddleInsertDot(orgin);for (int i = 1; i <Tn.length; i++) {double newadd=0;//设置二分之一的步长为hdouble h=(arrayList.get(1)[1]-arrayList.get(1)[0]);for (int i1 = 0; i1 < arrayList.get(0).length; i1++) {//----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------newadd += Fun(arrayList.get(0)[i1])*h;}Tn[i]=(Tn[i-1]/2+newadd);//更新数组arrayList = MiddleInsertDot(arrayList.get(1));}}/**** @param doubles 该参数是给定的一个数组的插值序列，是原初的x的坐标位置* @return 返回的是一个集合，该集合的第一项是新产生点的数组，第二项是扩大二倍之后的数组*/public static ArrayList<double[]>  MiddleInsertDot(double [] doubles) {double [] value =new double [doubles.length*2-1];double []returnvalue = new double[doubles.length-1];//新产生的值被保存了for (int i = 0; i < doubles.length-1; i++) {returnvalue[i]=(doubles[i]+doubles[i+1])/2;}
//for (int i = 0; i < doubles.length; i++) {value[i*2]=doubles[i];}for (int i = 1; i < doubles.length; i++) {value[i*2-1]=returnvalue[i-1];}ArrayList<double[]> arrayList = new ArrayList<>();arrayList.add(returnvalue);arrayList.add(value);return arrayList;}// 返回函数值/**** @param x 传入的自变量* @return  返回函数值*/public static  double Fun(double x) {double y=0;y=4/(1+x*x);//需要修改的函数部分实现任意函数积分return y;}public static double[] Sn(double [] Tn ){double [] Sn=new double[Tn.length-1];for (int i = 0; i < Sn.length; i++) {Sn[i] =(4*Tn[i+1]-Tn[i])/3;}return Sn;}public static double[] Cn(double [] Sn ){double [] Cn=new double[Sn.length-1];for (int i = 0; i < Sn.length-1; i++) {Cn[i] =(4*4*Sn[i+1]-Sn[i])/15;}return Cn;}public static double[]Rn(double [] Cn ){double [] Rn=new double[Cn.length-1];for (int i = 0; i < Cn.length-1; i++) {Rn[i] =(4*4*4*Cn[i+1]-Cn[i])/63;}return Cn;}}