在 C 中集成复合函数

Question

我正在尝试编辑使用梯形规则来集成函数的代码。我的问题是该程序似乎不喜欢我有不止一个涉及 x 的术语这一事实。

$f(x)=\frac{\beta\gamma}{\delta}e^{-\gamma a}\lr{1-e^{-\delta a}}e^{- \frac{\alpha\beta \mathbf{S}}{\delta}\lr{1-e^{-\delta a}}}$

这是我想近似计算的积分图片，因为乳胶似乎不起作用：

我从使用梯形法则的 C（第 2 版）数值食谱中获取了代码：

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

double d = 0.7, b = 2.5, g = 1.1, a=0.3; /*One species case: delta, beta, gamma, alpha */
float s1=0.564029; /*s value*/
double exp(double x); /*defining e^x*/

/* Define function here */

double f (double x) { double myresult = b*g/d*exp(-g*x)*(1-exp(-d*x))*exp(-a*b*s1/d*(1-exp(-d*x))); return myresult; }

int main()
{
 float lower, upper, integration=0.0, stepSize, k;
 int i, subInterval;
system("clear");
 /* Input */
 printf("Enter lower limit of integration: ");
 scanf("%f", &lower);
 printf("Enter upper limit of integration: ");
 scanf("%f", &upper);
 printf("Enter number of sub intervals: ");
 scanf("%d", &subInterval);

 /* Calculation */
 /* Finding step size */
 stepSize = (upper - lower)/subInterval;

 /* Finding Integration Value */
 integration = f(lower) + f(upper);
 for(i=1; i<= subInterval-1; i++)
 {
  k = lower + i*stepSize;
  integration = integration + 2 * f(k);
 }
 integration = integration * stepSize/2;
 printf("\nIntegral of given function for given values is: %.3f", integration);
 getchar();
 return 0;
}

这段代码给我错误，我相信是因为我的#Define f(x) 有不止一项涉及 'x'。

错误：

constant.c:33:18: error: expected ')'
 integration = f(lower) + f(upper);
                 ^
constant.c:33:16: note: to match this '('
 integration = f(lower) + f(upper);
               ^
constant.c:13:37: note: expanded from macro 'f'
#define f(x)  b*g/d*exp(-g*x)*(1-exp(-d x))*exp(-a*b*s1/d*(1-exp(-d x)))
                                    ^
constant.c:33:18: error: expected ')'
 integration = f(lower) + f(upper);
                 ^
constant.c:33:16: note: to match this '('
 integration = f(lower) + f(upper);
               ^
constant.c:13:65: note: expanded from macro 'f'
#define f(x)  b*g/d*exp(-g*x)*(1-exp(-d x))*exp(-a*b*s1/d*(1-exp(-d x)))
                                                                ^
constant.c:33:29: error: expected ')'
 integration = f(lower) + f(upper);
                            ^
constant.c:33:27: note: to match this '('
 integration = f(lower) + f(upper);
                          ^
constant.c:13:37: note: expanded from macro 'f'
#define f(x)  b*g/d*exp(-g*x)*(1-exp(-d x))*exp(-a*b*s1/d*(1-exp(-d x)))
                                    ^
constant.c:33:29: error: expected ')'
 integration = f(lower) + f(upper);
                            ^
constant.c:33:27: note: to match this '('
 integration = f(lower) + f(upper);
                          ^
constant.c:13:65: note: expanded from macro 'f'
#define f(x)  b*g/d*exp(-g*x)*(1-exp(-d x))*exp(-a*b*s1/d*(1-exp(-d x)))
                                                                ^
constant.c:37:37: error: expected ')'
  integration = integration + 2 * f(k);
                                    ^
constant.c:37:35: note: to match this '('
  integration = integration + 2 * f(k);
                                  ^
constant.c:13:37: note: expanded from macro 'f'
#define f(x)  b*g/d*exp(-g*x)*(1-exp(-d x))*exp(-a*b*s1/d*(1-exp(-d x)))
                                    ^
constant.c:37:37: error: expected ')'
  integration = integration + 2 * f(k);
                                    ^
constant.c:37:35: note: to match this '('
  integration = integration + 2 * f(k);
                                  ^
constant.c:13:65: note: expanded from macro 'f'
#define f(x)  b*g/d*exp(-g*x)*(1-exp(-d x))*exp(-a*b*s1/d*(1-exp(-d x)))
                                                                ^
6 errors generated.

如果我只有

#define f(x)  b*g/d*exp(-g*x)

然后它确实给出了该项的正确近似值。

我试过分别定义这三个部分，即

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

double d = 0.7, b = 2.5, g = 1.1, a=0.3; /*One species case: delta, beta, gamma, alpha */
float s1=0.564029; /*s value*/
double exp(double x); /*defining e^x*/

/* Define function here */

#define f(x)  b*g/d*exp(-g*x)
#define g(x) (1-exp(-d x))
#define h(x) exp(-a*b*s1/d*(1-exp(-d x)))

int main()
{
 float lower, upper, integration=0.0, stepSize, k;
 int i, subInterval;
system("clear");
 /* Input */
 printf("Enter lower limit of integration: ");
 scanf("%f", &lower);
 printf("Enter upper limit of integration: ");
 scanf("%f", &upper);
 printf("Enter number of sub intervals: ");
 scanf("%d", &subInterval);

 /* Calculation */
 /* Finding step size */
 stepSize = (upper - lower)/subInterval;

 /* Finding Integration Value */
 integration = f(lower)*g(lower)*h(lower) + f(upper)*g(upper)*h(upper);
 for(i=1; i<= subInterval-1; i++)
 {
  k = lower + i*stepSize;
  integration = integration + 2 * f(k)*g(k)*h(k);
 }
 integration = integration * stepSize/2;
 printf("\nIntegral of given function for given values is: %.3f", integration);
 getchar();
 return 0;
}

但这也给我一个错误。

Answer 1

只需将所有 exp(-d x) 替换为 exp(-d * x)。

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