Integration with both limits variable

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Sophie Malcolm
Sophie Malcolm el 19 de Sept. de 2020
Comentada: Bjorn Gustavsson el 20 de Sept. de 2020
Hi, I am trying to solve an integration where both limits are variable. The function itself being integrated is simple, it is just exp(-W^2). I can solve the inegration fine using this method without limits - the limits are the issue here. I think it could be because of the use of the symbolic functions to integrate? But I have tried other methods, and I cannot get it to work with these limits. I get the error:
error in sym/subsref (line 898)
R_tilde = builtin('subsref',L_tilde,Idx);
The upper limit is -phi, and the lower limit is beta-phi.
Another issue could be because the upper limit is the same as the variable being integrated. However I don't know how to get around that issue either?
I have just shown some of my code, hopefully this is enough to make sense of what I am trying to do. Any help would be greatly appreciated!
t=linspace(0,1000);
r=linspace(0,100);
D=2;
beta=r./sqrt(4*D.*t);
phi=U.*sqrt(t./D);
W=beta-phi;
% Performing integration
syms W
f=exp(-(W.^2));
a=-phi;
b=beta-phi;
int(f,W,a,b)
  3 comentarios
Sophie Malcolm
Sophie Malcolm el 20 de Sept. de 2020
Hi Bjorn,
Thank you so much for your response, I hadn't considered the use of the error function in MATLAB and will look into that now!
Thanks so much,
Sophie
Bjorn Gustavsson
Bjorn Gustavsson el 20 de Sept. de 2020
Your solution is simply:
(pi^(1/2)*(erf(b) - erf(a)))/2.

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Respuesta aceptada

Ameer Hamza
Ameer Hamza el 19 de Sept. de 2020
You are getting an error because you are passing arrays (a and b) to int() as its limit. The limits must be scalar. For example, try this
t=linspace(0,1000);
r=linspace(0,100);
D=2;
U = 3;
beta=r./sqrt(4*D.*t);
phi=U.*sqrt(t./D);
W=beta-phi;
% Performing integration
syms W
f=exp(-(W.^2));
a=-phi;
b=beta-phi;
int(f,W,a(2),b(2))
  2 comentarios
Sophie Malcolm
Sophie Malcolm el 20 de Sept. de 2020
Hi Ameer,
Thank you so much for this, I didn't realise that! That is so helpful, thanks a lot.
Sophie
Ameer Hamza
Ameer Hamza el 20 de Sept. de 2020
I am glad to be of help!

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Más respuestas (1)

Bjorn Gustavsson
Bjorn Gustavsson el 19 de Sept. de 2020
Matlab has both a numerical and a symbolic version of the error-function. That will be the answer to your problem. The numeric version is vectorized, so it can handle your inputs without problems.
HTH
  1 comentario
Sophie Malcolm
Sophie Malcolm el 20 de Sept. de 2020
Hi,
Okay that makes a lot of sense! Thank you so much for explaining. I will look into that now.
Thanks for your help,
Sophie

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