Integrating without using the symbolic toolbox
13 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Opariuc Andrei
le 14 Déc 2020
Commenté : Opariuc Andrei
le 14 Déc 2020
syms x
f=log(45.*sin(x.^2) + cos(x)); % initial function
g=sqrt(1+diff(f).^2); % requirement step 1
h=int(g,0,pi/2) ;% requirement final step
instead of int i want to use " integral " and solve numerically ,if i remove the syms and introduce @(x) then i won't get a displayed result
and the result of final step while using the syms method is :
int(((sin(x) - 90*x*cos(x^2))^2/(45*sin(x^2) + cos(x))^2 + 1)^(1/2), x, 0, pi/2)
x belongs (0,pi/2)
How can i get rid of syms ?
0 commentaires
Réponse acceptée
John D'Errico
le 14 Déc 2020
Modifié(e) : John D'Errico
le 14 Déc 2020
This is an arc length integral. Apparently you wish to compute the length of that curve in the plane, with x going from 0 to pi/2. As a numerical integral, this is easy.
syms x
f=log(45.*sin(x.^2) + cos(x)); % initial function
fplot(f,[0,pi/2])
g=sqrt(1+diff(f).^2); % requirement step 1
gfun = matlabFunction(g);
First, it alway makes sense to plot everything, before you just throw it into any general tool.
fplot(g,[0,pi/2])
I don't epect that arc length integral has an exact solution. I might also be slightly worried about what seems to be a singularity at 0, though an integration tool should survive that. You can use vpaintegral, if you wish to stay in the symbolic domain.
vpaintegral(g,[0,pi/2])
Or you can use integral.
integral(gfun,0,pi/2)
Which seems to agree.
And, since I have a tool on the file exchange that can compute the arclength of a curve, I might try this:
ffun = matlabFunction(f);
xint = linspace(0,pi/2);
arclength(xint,ffun(xint),'spline')
ans =
4.7383
which also seems to agree pretty well.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Calculus dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!