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

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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..

Accepted Answer

Walter Roberson
Walter Roberson on 28 Apr 2023
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.
  2 Comments
Walter Roberson
Walter Roberson on 28 Apr 2023
syms x y
A= [1 x^2+y^2;x-y x^2+y^2];
result = vpaintegral(vpaintegral(A, x, 0, 10), y, 0, 15)
result = 
Aninda pal
Aninda pal on 29 Apr 2023
Thank you very much sir. This is the way exactly what I was searching for.

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More Answers (2)

Torsten
Torsten on 27 Apr 2023
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 Comment
Aninda pal
Aninda pal on 28 Apr 2023
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 on 28 Apr 2023
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 Comments
Aninda pal
Aninda pal on 29 Apr 2023
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 on 29 Apr 2023
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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