Problem with integrals
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I have a integral that simplified looks like this:
quad(@(z) exp(z.^2),-2,x)
The problem here is quite obvious. You can't use quad quad a variable as a limit. So why don't I use int? Because exp(z^2) is an analytically unsolvable integral. In the task I'm trying to solve there is a second integral of the first integral and simplified it looks like this:
quad(@(x) exp(x.^2).*quad(@(z) exp(z.^2),-2,x),-2,2)
And according to my mentor there should be a way to solve this because he says that the first integration will make the other one solvable. The code above gives this error:
??? Error using ==> quad at 70
The limits of integration must be scalars.
Error in ==> @(x)exp(x.^2).*quad(@(z)exp(z.^2),-2,x)
Error in ==> quad at 77
y = f(x, varargin{:});
Error in ==> test at 54
quad(@(x) exp(x.^2).*quad(@(z) exp(z.^2),-2,x),-2,2)
So I thought I should ask you if there is a way to do what I'm trying to do (with any command, it does not have to be quad) or if its not possible.
采纳的回答
Teja Muppirala
2011-5-1
A very good question. You need to use QUADV here instead of QUAD.
quadv(@(x) exp(x.^2).*quad(@(z) exp(z.^2),-2,x),-2,2)
QUAD expects vector inputs and outputs, and so when it goes to evaluate the inner QUAD, it sends in a vector which is why you get that error.
Notice that QUAD will error on a problem like this one here below for the same reason (but QUADV will work fine):
quad(@(x) det([x 1; 1 1]),0,1)
QUADV does it without trying to send in vectors.
Just as a sanity check, compare what you get from QUADV with what you get if you did it using TRAPZ
f = @(x) exp(x.^2).*quad(@(z) exp(z.^2),-2,x);
trapz(arrayfun(f,-2:.001:2)) * 0.001
quadv(f,-2,2)
Both answers are very similar, about 541.38.
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