Asked by Alexandru
on 13 Feb 2013

Hello,

I am defining a function using

**f = @(x,y) (expression in x and y)**

This definition I believe it is correct as I can call f(0,0) for example and I get the numerical value.

What I need to do next is integrate f(x,y) with respect to y between a and b and call this g(x). I need then to be able to pass g(x) to fsolve in order to compute the roots. How do I do this? I tried dblquad, but it integrates w.r.t. both variables at once. quad gives me and error as I am not sure what's the correct syntax for this.

Thanks, Alex

Answer by Teja Muppirala
on 14 Feb 2013

Accepted Answer

Define another function handle to be the integral over y. Like this:

% Just making some random 2d function f = @(x,y) (x.^2-y.^2).*cos(x./(1+y.^2));

% Some limits of integration a = 0; b = 3;

% Define g as the integral of f(x,y) dy from a to b g = @(x) integral(@(y) f(x,y) , a,b);

% Plot it, and find a zero ezplot(g) fzero(g,0)

If your version of MATLAB doesn't have the INTEGRAL function, you could use QUAD instead.

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Answer by Walter Roberson
on 13 Feb 2013

You cannot do this with numeric integration.

If you have the symbolic toolbox, then expression the function symbolically and do symbolic integration with int(). Then if you need, you can use matlabFunction to turn the symbolic result into a function handle of a numeric function.

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Answer by Youssef Khmou
on 13 Feb 2013

Hi, try this :

syms x y h=exp(-x^2-y^2) F1=int(h,x) F2=int(h,y)

Based on F1 and F2 you make function handle :

Fx=@(x) 1/2/exp(x^2)*pi^(1/2) % truncated ERF(Y)

Y=fsolve(Fx,0.1)

Walter Roberson
on 13 Feb 2013

Youssef Khmou
on 13 Feb 2013

Hi,i used the undefined integration or "primitive" , for F1 you get :

F1 =

1/2/exp(y^2)*pi^(1/2)*erf(x)

you have an analytic integration w.r.t. x, so its function of y , now manually you set a function handle Fy :

Fy=@(x) 1/2/exp(y^2)*pi^(1/2)

and then solve it with "fslove" as Alexandru said he wants to use "fslove" .

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Answer by Yang Zhang
on 15 Mar 2015

Use Symbolic Math Toolbox to your problem.

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