Explicit integral could not be found.
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Hello,
I am trying to get a solution for below problem, which is to evaluate double integrals of function f(x,y) where the first integral boundary is in variable y. Though the explicit integral could not be found, I expect numerical answer still exists; however I don't know how to get that. I tried double and VPA functions but they didn't work.
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syms x y
f = x^3*exp(1/x^2)*(x^(1/2)/2)^(1/2);
F = int(f,x,0,y);
ans = int(F,y,0,100); ___________________________________________________________________________
Warning: Explicit integral could not be found.
Warning: Explicit integral could not be found.
ans = int(int(x^3*exp(1/x^2)*(x^(1/2)/2)^(1/2), x = 0..y), y = 0..100)
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VPA(ans)
ans =
numeric::int(numeric::int(x^3*exp(1/x^2)*(x^(1/2)/2)^(1/2), x = 0..y), y = 0..100)
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Respuestas (2)
Babak
el 26 de Sept. de 2012
Mathematica finds the indefinite integral of f over x
Take this and evaluate it for the boundaries (i.e. x=y and x=0) and subtract the results. this will give you the definite integral from x=0 to x=y then again try to integrate the result over y.
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Matt Fig
el 26 de Sept. de 2012
Editada: Matt Fig
el 26 de Sept. de 2012
Are you sure about those limits?
Here is how one would ordinarily solve such a problem numerically. Note I use limits 1-y and 2-10:
fh = @(y) quadl(@(x) x.^3.*exp(1./x.^2).*(x.^(1/2)/2).^(1/2),1,y);
quadv(fh,2,10)
.
.
.
Let's just check the method with an easy problem. Say we want to solve:
int(int(x,0,y),1/2,1)
By simple calculus this is:
int(y^2/2,1/2,1) = 7/48
So in MATLAB
fh = @(y) quadl(@(x) x,0,y);
quadv(fh,1/2,1) % Equal to 7/48 as expected...
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