Function handle, integral respect.

I have function u(x)=integral exp(5*i*abs(x-y))*(-1-2*sinh(y))*exp(i*5*y)dy y=0..2,when x=0..2,and I try to find her

    for x=0:0.01:2
    N=N+1;
    syms f(x,y)
    f(x,y)=exp(5*i*abs(x-y))*(-1-2*sinh(y))*exp(i*5*y);
    f2=integral(f,0,2);
    u(N)=integral(f2,0,2);
    end

What wrong?

[Adam]

I don't know. What is wrong? Are you getting an error? Incorrect result?

[Opera Era]

I have error: First input argument must be a function handle.

[Opera Era]

Modified to this:

 syms x y 
    f=int(exp(5*i*abs(x-y))*(-1-2*sinh(y))*exp(i*5*y),y);
    f2=vpa(subs(f,y,2))-vpa(subs(f,y,0));
    u(N)=vpa(subs(f,x,k));

but get error

[Steven Lord]

And the full text of the error message ( EVERYTHING in red ) that you received from your modified code is ... ?

Answer 1

[Steven Lord]

There are two main functions for integrating a function.

Use the int function from Symbolic Math Toolbox to integrate a symbolic expression.

Use the integral function from MATLAB to numerically integrate a function handle.

Trying to use int to integrate a function handle won't work.

Trying to use integral to integrate a symbolic expression won't work.

You have a symbolic expression, so int is the right tool to use.