Evaluating double integral with unknown variable in
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Hi,
I am trying to evaluate a double integral of x,y with an unkown constant c. The result for c should resemble a binomial distribution. I am trying to solve this using integral2 as follows:
fun = @(x,y)(((c-y+x)^50)*((1-c+y-x)^3280)*(y^50)*((1-y)^130)*((x^16)*((1-x)^3308)))
xmax = @(y)(1-c+y)
xmin = @(y)(y-c)
q = integral2(fun,0,1,xmin,xmax)
But I get a list of errors:
Unrecognized function or variable 'c'.
Error in stupidstupid>@(y)(y-c) (line 3)
xmin = @(y)(y-c)
Error in integral2Calc>integral2t/tensor (line 189)
bottom = YMIN(x);
Error in integral2Calc>integral2t (line 55)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 9)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 106)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Error in stupidstupid (line 4)
q = integral2(fun,0,1,xmin,xmax)
I think this has to do with the fact that I have not define the variable c, but I do now know how to define it. Any help would be greatly appreciated. Thanks in Advance!
3 Comments
VBBV
on 10 Oct 2020
Edited: Walter Roberson
on 10 Oct 2020
Use syms to define it as symbolic variable
syms c
Tron
on 10 Oct 2020
Walter Roberson
on 10 Oct 2020
integral() and integral2() require that the function returns a numeric value. They are adaptive numeric integrations and need the numeric values to figure out whether they need to refine further.
Answers (1)
Walter Roberson
on 10 Oct 2020
0 votes
you can never use integral() or intrgral2() when you have undefined variables. You have to use symbolic integration and hope that matlab can solve it.
3 Comments
Tron
on 10 Oct 2020
Walter Roberson
on 10 Oct 2020
It is not necessarily impossible to integrate symbolically, but symbolic integration will probably require expanding out the (1-x)^3308 and (1-c+y-x)^3280 and so on, and doing all of the polynomial multiplications, and finally doing the integration.
Those are really large powers... how did they originate ?
Walter Roberson
on 10 Oct 2020
My tests suggest that the final result would be a polynomial of degree 16+3308 = 3324. where those numbers come from ((x^16)*((1-x)^3308))
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