Mathc initiation/Fichiers c : c38cc
Apparence
Installer et compiler ces fichiers dans votre répertoire de travail.
c05a.c |
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/* ---------------------------------- */
/* save as c05c.c */
/* ---------------------------------- */
#include "x_hfile.h"
#include "fc.h"
/* ---------------------------------- */
int main(void)
{
double a = -2.0;
double b = 2.0;
double n = 2*5.;
clrscrn();
printf(" Verify with a software that g(x) >= h(x) on [%.3f,%.3f].\n\n\n", a, b);
printf(" g : x-> %s\n\n", geq);
printf(" h : x-> %s\n\n", heq);
stop();
clrscrn();
printf(" Draw a typical vertical rectangle on a graph.\n\n");
printf(" Upper boundary (g) : %s \n",geq);
printf(" lower boundary (h) : %s \n\n",heq);
printf(" The length of the rectangle is : (%s) - (%s) \n",geq, heq);
printf(" The width of the rectangle is : dx \n\n");
printf(" The Area of the rectangle is : ((%s) - (%s)) * dx\n",geq, heq);
printf(" \n\n\n\n\n");
stop();
clrscrn();
printf(" If we apply \n\n\n");
printf(" (%.3f\n", b);
printf(" int( \n");
printf(" (%.3f\n\n\n", a);
printf(" to : ((%s) - (%s)) * dx\n\n", geq, heq);
printf(" We obtain a limit of sums of areas of vertical rectangles.\n\n\n");
printf(" (%.3f\n", b);
printf(" int( [(%s) - (%s)] dx = %.5f\n",
geq,heq,simpson(gminuh,a,b,n));
printf(" (%.3f\n\n\n", a);
stop();
return 0;
}
/* ---------------------------------- */
Vous trouverez ci-dessous le code Gnuplot pour dessiner les fonctions et un rectangle caractèristique. Ainsi que le code pour Octave pour vérifier le calcul.
Exemple de sortie écran :
Verify with a software that g(x) >= h(x) on [-2.000,2.000].
g : x-> x**2+1
h : x-> x-2
Press return to continue.
Copier ce code dans gnuplot :
set zeroaxis lt 8
set grid
set object rect from 1,(1-2) to 1.3,((1.3*1.3)+1)
plot [ -2.0:2.0] [-5.000:5.000] (x*x+1), (x-2)
reset
Exemple de sortie écran :
Draw a typical vertical rectangle on a graph.
Upper boundary (g) : x**2+1
lower boundary (h) : x-2
The length of the rectangle is : (x**2+1) - (x-2)
The width of the rectangle is : dx
The Area of the rectangle is : ((x**2+1) - (x-2)) * dx
Press return to continue.
Exemple de sortie écran :
If we apply
(2.000
int(
(-2.000
to : ((x**2+1) - (x-2)) * dx
We obtain a limit of sums of areas of vertical rectangles.
(2.000
int( [(x**2+1) - (x-2)] dx = 17.33333
(-2.000
Press return to continue.
Vérifier le résultat avec Octave 5.2 : I = quad (f, a, b)
>> I = quad (@(x) ((x.*x+1)-(x-2)), -2,2)
I = 17.333