MATLAB: Error in nested integration

integration

I want to integrate a function and program is given below:

close all; 
clear all;
fun = @(x,y)x+(x./(y.^2));
fun1= @(y) y+ integral(@(x)fun(x,y), 0,1);
c1= integral(fun1,0,1)

In the above program if we are using 'y' instead of 'y.^2', then there is no error. But if we are using 'y.^2' error occurred matrix dimension must agree. But here we are using x, y as variable hence there should be no question of matrix dimension. If, anybody can solve the problem please help me . Any suggestions regarding this will be appreciated.

Best Answer

fun1= @(y) y+ integral(@(x)fun(x,y), 0,1,'ArrayValued',true);

Note that fun is undefined at y=0. Thus c1 should come out as Inf.

Best wishes

Torsten.

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MATLAB: Error encountered for numerical integration inside for loop.

Your line

        fun = @(x)(((((besselj(0,(2*pi*q(f))/(3e8))*x*ro)-besselj(0,(2*pi*q(f))/(3e8))*x*ri)).^2)./x).*freload(x,q)));

calls freload() passing in all of q, instead of q(f)

MATLAB: Integrate multivariate fuction with respect to one variable and pass the result to another function for integration.

Notice you have:

fun= @(x, y) x.*exp(-y.*x);

which is a function of two variables. But you try

c= @(y)integral(@(x) fun, 0,1);

which does not pass any parameters to the function. You probably intended

c= @(y)integral(@(x) fun(x), 0,1);

But notice that you are not passing y, so you would need

c= @(y) integral(@(x) fun(x, y), 0,1);

Then you have

fun1= @(y) 1/2-(1/pi).*y.*c;
c1 = integral(fun1, 0,1);

this attempts to invoke c, but c is a function handle. You have to pass a value to the handle:

fun1 = @(y) 1/2-(1/pi).*y.*c(y);
c1 = integral(fun1, 0,1);

I think that should work.

Best Answer

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MATLAB: Error encountered for numerical integration inside for loop.

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