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How can I put an integral inside a for loop when the bounds depend on the loop's variable?

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sawdiia_too
sawdiia_too on 1 Apr 2020
Commented: sawdiia_too on 2 Apr 2020
Hi, I would like to do this integration where the j of cj is the loop variable.
I tried this but when I calculate Phi_ij1(1,1) for example, Matlab says "unrecognized function or variable 'y' ". Can you help me?
for j=1:13
x_max1 = @(y,j) (-sqrt(3)/3).*y + c(j);
y_min1 = @(j) (sqrt(3)/2)*c(j);
y_max1 = @(j) c(j)*sqrt(3);
Phi_ij1 = @(i,j) integral2(B_zi(i,x,y), 0, x_max1(y,j),0,y_max1(j)) - integral2(B_zi(i,x,y),0, x_max1(y,j),y_min1(j),y_max1(j)) ;
end

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darova
darova on 1 Apr 2020
Try to specify variables integral depends on
Phi_ij1 = @(i,j) integral2(@(x,y)B_zi(i,x,y), 0, @(y)x_max1(y,j),0,y_max1(j)) - ...
integral2(@(x,y)B_zi(i,x,y), 0, @(y)x_max1(y,j),y_min1(j),y_max1(j)) ;

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Accepted Answer

darova
darova on 1 Apr 2020
Here is an example for integral3
According to this integration is done from z to x
You have only y. Looks like x=x(y) in your case
So you have to integrate by x first
Try to replace limits of integration
Phi_ij1 = @(i,j) integral2(@(x,y)B_zi(i,x,y), 0, y_max1(j)), 0, @(y)x_max1(y,j) - ...
integral2(@(x,y)B_zi(i,x,y), y_min1(j),y_max1(j), 0, @(y)x_max1(y,j)) ;

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sawdiia_too
sawdiia_too on 2 Apr 2020
In the example, the integration of x is from a scalar to a scalar while on mine the integrations are from functions to functions (one depending on y like you said x(y) and the other depending on j y(j))

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More Answers (1)

Torsten
Torsten on 2 Apr 2020
Phi_ij1 =@(i,j) integral2(@(y,x) B_zi(i,x,y),0,y_min1(j),0,@(y)x_max1(y,j));

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sawdiia_too
sawdiia_too on 2 Apr 2020
Oh that's clever. So I have to put all the limits at the same time in the same integral2? Because I don't see y_min1 in your answer. How does it work?

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