how to evaluate a double integral using the trapezoidal rule equation?

36 vues (au cours des 30 derniers jours)
Here's what I have so far
function [ I ] = myTrapz2D( f, x0, xn, y0, yn, nx, ny )
dx = (xn - x0)/nx;
dy = (yn - y0)/ny;
i = 1;
sumx = zeros(nx,1);
sumy =zeros(ny,1);
while i < nx
xi = x0 + i*dx;
sumx(i) = f(xi);
i = i+1;
end
sumx = sum(sumx);
Ix = ((dx)/2)*(f(x0)+f(xn)+(2*sumx));
fd = Ix(y);
while i < ny
yi = y0 + i*dy;
sumy(i) = fd(yi);
i = i+1;
end
sumy = sum(sumy);
I =((dy)/2)*(fd(y0)+fd(yn)+(2*sumy));
end
not sure if it's correct at all but it has to be solved using some variation of the equation for I that I used. I keep getting an error that there aren't enough input arguments. There are my input arguments: f = @(x,y) x.^2 - (2*y.^2) + (x.*y.^3); x0 = 0; xn = 2; y0 = -1; yn = 1; nx = 8; ny = 8;

Réponse acceptée

Torsten
Torsten le 16 Avr 2018
You don't need to program the trapezoidal rule in two dimensions.
Just call the trapezoidal rule in one dimension twice. In the section "Multiple Numeical Integrations" under
https://de.mathworks.com/help/matlab/ref/trapz.html
is an example with the MATALB implementation of the trapezoidal rule "trapz".
Best wishes
Torsten.
  1 commentaire
Susan Santiago
Susan Santiago le 17 Avr 2018
yes, I'm away that this can be done using trapz, however my assignment involved creating a function like the one I have shown you so, in this case, trapz does not help

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Plus de réponses (1)

Apoorv Rajput
Apoorv Rajput le 7 Oct 2021
function [ I ] = myTrapz2D( x0, xn, y0, yn, nx, ny )
syms f(x,y);
syms x;
syms y;
f(x,y)=exp(y-x);
dx = (xn - x0)/nx;
dy = (yn - y0)/ny;
i = 1;
sumx=0*x*y;
while i < nx
xi = x0 + i*dx;
sumx=sumx+ f(xi,y);
i = i+1;
end
Ix = ((dx)/2)*(f(x0,y)+f(xn,y)+(2*sumx));
syms fd(y);
fd(y) = Ix;
sumy=0*y;
i=1;
while i < ny
yi = y0 + i*dy;
sumy= sumy+fd(yi);
i = i+1;
end
I =((dy)/2)*(fd(y0)+fd(yn)+(2*sumy));
end

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