公式pai=4*∫10dx/(1+x2)
把[0.1]分成10等份,用两中方法分别计算
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class work2 { public static void main (String args[]){ double h = 0.1;//参数h double m1 =0 ;//复化梯形求和求和 double n =10; double I=0;//最后结果 double H_2 =0.2; double m =5 ;//由于n=2m.n=10,所以m=5 double s1=0;//复化抛物线第一个数组求和 double s2=0;//复化抛物线第二个数组求和 double ss[] = new double[10];//复化梯形公式计算中的求和单元 double ss1[]=new double[5];//复化抛物线中用到的第一个数组 double ss2[]=new double[4];//复化抛物线中用到的第二个数组
double k[]={0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9};//10等分[0，1] for(int i=0;i<10;i++){ ss[i] = 1/(1+k[i]*k[i]); } //teat pass for(int i =0;i<10;i++){ m1 = m1 +ss[i]; } //test pass I = H_2*(0.5+2*m1);//0.5是f(b)
// --------------------test code System.out.println("===复化梯形公式计算(包含测试及结果）==="); System.out.println("复化梯形公式单项结果"); for(int i =0;i<10;i++){ System.out.println(ss[i]);
} System.out.println("结果求和 m is "+ m1); System.out.println("结果是 "+ I); System.out.println("==================================");// ------------------------------------------------ //==开始复化抛物线公式计算====== System.out.println("复化抛物线公式计算"); //由于抛物线运算对数组的操作比较复杂，所以我重新定义数组k[] double k_2[] = {0,0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9}; //这样数组k就和实际中的对应起来了，哦哈哈 for(int i =1;i<=m;i++){ int n1 = 2*i-1; ss1[i-1] = 1/(1+k_2[n1]*k_2[n1]); System.out.println(ss1[i-1]);//测试计算是否正确, } System.out.println("运算正确"); System.out.println("第二个数组计算"); for(int i = 1;i<=(m-1);i++){ int n1 = 2*i; ss2[i-1]=1/(1+k_2[n1]*k_2[n1]); System.out.println(ss2[i-1]);//测试计算是否正确, } System.out.println("运算正确"); //对2个数组求和 for(int i =0;i<5;i++){ s1 = s1 +ss1[i]; } System.out.println("ss1求和是"+ s1); for(int i =0;i<4;i++){ s2 = s2 +ss2[i]; } System.out.println("ss2求和是"+ s2); System.out.println("运算正确"); I = (h/3)*(1+0.5+4*s1+2*s2); System.out.println("结果是 "+ I); //抛物线法不知道正不正确，书上没有答案 } } ===========================================
