how to evaluate a double integral using the trapezoidal rule equation?
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Susan Santiago
il 15 Apr 2018
Risposto: Apoorv Rajput
il 7 Ott 2021
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;
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Risposta accettata
Torsten
il 16 Apr 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.
Più risposte (1)
Apoorv Rajput
il 7 Ott 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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