给你写了fun_1( ),fun_2(),请自己添加另外几个被积函数。
调用方法 t=fsimp(a,b,eps,fun_i)
a,b --上下限,eps -- 迭代精度要求。
#include<stdio.h>
#include<stdlib.h>
#include <math.h>
double fun_1(double x)
{
return 1.0 + x
}
double fun_2(double x)
{
return 2.0 * x + 3.0
}
double fsimp(double a,double b,double eps, double (*P)(double))
{
int n,k
double h,t1,t2,s1,s2,ep,p,x
n=1h=b-a
t1=h*(P(a)+P(b))/2.0
s1=t1
ep=eps+1.0
while (ep>=eps)
{
p=0.0
for (k=0k<=n-1k++)
{
x=a+(k+0.5)*h
p=p+P(x)
}
t2=(t1+h*p)/2.0
s2=(4.0*t2-t1)/3.0
ep=fabs(s2-s1)
t1=t2s1=s2n=n+nh=h/2.0
}
return(s2)
}
void main()
{
double a,b,eps,t
a=0.0b=3.141592653589793238eps=0.0000001
// a definite integral by Simpson Method.
t=fsimp(a,b,eps,fun_1)
printf("%g\n",t)
t=fsimp(a,b,eps,fun_2)
printf("%g\n",t)
// ...
printf("\n Press any key to quit...")
getch()
}
C程序:定积分通用函数2007-12-15 11:23//定积分通用函数#include<stdio.h>
#include<math.h>
float f1(float x)
float f2(float x)
void process(float x0,float xf,float (*p)(float))
float jf(float x0,float xf,int n,float (*p)(float))void main()
{
printf("以 4/(1+x*x) 在(0,1)上的积分为例\n\n")
process(0,1,f1)
}float f1(float x)
{
float z
z=4/(1+x*x)
return z
}float jf(float x0,float xf,int n,float (*p)(float))
{
float z=0,sum=0
int i
for(i=0i<=n-1i++)
{
z=((xf-x0)/n)*(*p)(x0+(xf-x0)*i/n)
sum=sum+z
}
return sum
}void process(float x0,float xf,float (*p)(float))
{
int i
for(i=2i++)
{
if(fabs( jf(x0,xf,10*i,p) - jf(x0,xf,10*(i+1),p) ) <1e-6)
{
printf("它在(%f,%f)的积分=%f,计算%d次\n",x0,xf,jf(x0,xf,10*i,p),i)
break
}
}
}
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