tasks a-s

Task/Numerical-integration/C/numerical-integration-2.c

@@ -0,0 +1,71 @@
+/* test */
+double f3(double x)
+{
+  return x;
+}
+
+double f3a(double x)
+{
+  return x*x/2.0;
+}
+
+double f2(double x)
+{
+  return 1.0/x;
+}
+
+double f2a(double x)
+{
+  return log(x);
+}
+
+double f1(double x)
+{
+  return x*x*x;
+}
+
+double f1a(double x)
+{
+  return x*x*x*x/4.0;
+}
+
+typedef double (*pfunc)(double, double, double, double (*)());
+typedef double (*rfunc)(double);
+
+#define INTG(F,A,B) (F((B))-F((A)))
+
+int main()
+{
+     int i, j;
+     double ic;
+
+     pfunc f[5] = {
+       int_leftrect, int_rightrect,
+       int_midrect,  int_trapezium,
+       int_simpson
+     };
+     const char *names[5] = {
+       "leftrect", "rightrect", "midrect",
+       "trapezium", "simpson"
+     };
+     rfunc rf[] = { f1, f2, f3, f3 };
+     rfunc If[] = { f1a, f2a, f3a, f3a };
+     double ivals[] = {
+       0.0, 1.0,
+       1.0, 100.0,
+       0.0, 5000.0,
+       0.0, 6000.0
+     };
+     double approx[] = { 100.0, 1000.0, 5000000.0, 6000000.0 };
+
+     for(j=0; j < (sizeof(rf) / sizeof(rfunc)); j++)
+     {
+       for(i=0; i < 5 ; i++)
+       {
+         ic = (*f[i])(ivals[2*j], ivals[2*j+1], approx[j], rf[j]);
+         printf("%10s [ 0,1] num: %+lf, an: %lf\n",
+                names[i], ic, INTG((*If[j]), ivals[2*j], ivals[2*j+1]));
+       }
+       printf("\n");
+     }
+}