Integrating a multivariate function w.r.t. a single variable

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 1

[Teja Muppirala]

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.

[joseph erlich]

Thank you.

[ARNAB PAL]

Dear sir,

I have a question that if the function is a matrix function like,

f=@(x,y) K_2*exp(A*(x(i)-y))*B*u;

where K_2 is a (3*6) matrix,A is a (6*6) matrix ,B is (6*3) and u is (3*1) matrix. I want to find g=@(x) integral(@(y) f(x,y),0,x(i)) but when I use this it is not giving the output as a (3*1) vector?

Finally I want to find w=u-(K_2*exp(A*x(i))*X) where X is (6*1).

[Walter Roberson]

When you call integral(), it is required to return an array the same size as x, which will be a vector of varying sizes. However, there is the 'vectorvalued' option for integral(), and when set then the function will be passed scalars and can return multidimensional outputs as needed.

Answer 2

[Walter Roberson]

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.

[Adrian Martinez]

Can you please show an example of how to do this?

[Walter Roberson]

f = @(x,y) (x.^2-y.^2).*cos(x./(1+y.^2));
% Some limits of integration
a = 0;
b = 3;
syms x y
% Define g as the integral of f(x,y) dy from a to b
g(x) = int(f(x,y), y, a, b)

g(x) = 

x0 = -100;
vpasolve(g, x0)

ans = 

G = matlabFunction(g)

G = function_handle with value:
    @(x)integral(@(y)cos(x./(y.^2+1.0)).*(x.^2-y.^2),0.0,3.0)

fsolve(G, x0)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

ans = -99.0803

In some cases, the int() step would be able to calculate a closed form expression, so G will not always end up with an integral() in it.

Answer 3

[Youssef Khmou]

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]

How do you get from the "F1 and F2" to the Fx ? And where do the limits of integration over y come in?

[Youssef Khmou]

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" .

Answer 4

[Yang Zhang]

Use Symbolic Math Toolbox to your problem.