How can i find solution for un integral of bessel functions

Dear every one, i would like to find a solution for my problem in the following. Please give me some advise to do this work. Thank you in advance.

format long

% define input values

t0 = 13;

q = 700;

lamda = 1; 

rp = 0.2; 

l = 12; 

alpha = 0.6E-6;

t = 0; 

t = t+600;

r = 0.5;

rho1 = 2.7;

rho2 = 1.7; 

rho = rho1/rho2;

c1 = 0.8E-6;

c2 = 0.6E-6; 

c = c1/c2;

% Define of functions

phi = rho*c*u*besselj(0,u*0.2)+2*besselj(1,u*0.2);

psi = rho*c*u*bessely(0,u*0.2)+2*bessely(1,u*0.2);

func = (exp(-alpha*u^2*t)-1)*(bessely(0,u*r)*phi - besselj(0,u*r)*psi)/(phi^2+psi^2)/u^2;

A = 2*q/lamda/pi^2/rp/l;

t2 = t0 + A*int(func,0,100)

plot(t,t2);

And the answer of matlab:

??? Undefined function or method 'int' for input arguments of type
'double'.

Error in ==> modelcalculation at 37
t2 = t0 + A*int(func,0,100)

Answer 1

[Torsten]

Make "func" a function handle of the variable u, not a numerical value.

And use "integral" instead of "int".

See the documentation on how to use "integral".

Best wishes

Torsten.

[Tri NGUYEN]

Many thanks to Torsten, i have been re-define variables and functions as you said, it running good now.

Answer 2

[Walter Roberson]

int() is for symbolic expressions. You would use it in combination with a symbolic variable declared using "syms" or "sym". In particular, before you use "u" you would declare

syms u

and not assign a numeric value to u. (You must be have given u a numeric value somewhere along the line or your code would not have reached that point.)

[Tri NGUYEN]

Thanks to Walter Roberson.