I want to integrate a symbolic matrix numerically. How can I do it??

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clear all
clc
syms x y
A= [1 x^2+y^2;x-y x^2+y^2];
I want to numerically integrate all elements of A for x limits (0-10) and ylimits (0-15) .
Thanks for your time and help..

采纳的回答

Walter Roberson
Walter Roberson 2023-4-28
Use nested vpaintegral() calls. The inner call will execute first and figure out that it cannot integrate because there is an additional free variable, so it will return a vpaintegral() data form. The outer vpaintegral will recognize the vpaintegral data form from the inner and will be able to proceed with the 2d integration.

更多回答(2 个)

Torsten
Torsten 2023-4-27
syms x y
A = [1 x^2+y^2;x-y x^2+y^2];
IntA = int(int(A,x,0,10),y,0,15)
IntA = 
  1 个评论
Aninda pal
Aninda pal 2023-4-28
I want to do this integration numerically.. I have a larger system of equations where in matrix formation each element are functions of different variables. some cases arise where the value to integrate is constant. "int" operation is taking very long for my case. Thank you anyways..

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Paul
Paul 2023-4-28
I don't think there's a function for numerically integrating an arrayvalued function of two variables. Here's a loop approach to integrating each element of A individually using integral2
syms x y
A = [1 x^2+y^2;x-y x^2+y^2];
for ii = 1:2,
for jj = 1:2
Afun = matlabFunction(A(ii,jj),'vars',{'x' 'y'});
% add 0*x to ensure proper dimension when integrating A(1,1)
result(ii,jj) = integral2(@(x,y) Afun(x,y)+0*x,0,10,0,15);
end
end
format short e
result
result = 2×2
1.0e+00 * 1.5000e+02 1.6250e+04 -3.7500e+02 1.6250e+04
  2 个评论
Aninda pal
Aninda pal 2023-4-29
Thank you sir for your kind contribution, however for larger matrices using the loop is very much time taking. That is why I am searching for a another way.
Paul
Paul 2023-4-29
I'd be surprised if vpaintegral is faster than integral2 for the same tolerances. Let's try it.
syms x y
A = [1 x^2+y^2;x-y x^2+y^2];
tic
for kk = 1:100
result = nan(2,2);
for ii = 1:2,
for jj = 1:2
Afun = matlabFunction(A(ii,jj),'vars',{'x' 'y'});
% add 0*x to ensure proper dimension when integrating A(1,1)
result(ii,jj) = integral2(@(x,y) Afun(x,y)+0*x,0,10,0,15);
end
end
end
toc
Elapsed time is 4.600858 seconds.
tic
for kk = 1:100
result = vpaintegral(vpaintegral(A, x, 0, 10), y, 0, 15);
end
toc
Elapsed time is 20.426502 seconds.

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