Definite integral with parameter

Hi, I have the following equation to integrate:

c = (5.*x.^2-Fa.*x)./(((a.*x+q).*W-((a.*x+q)-2.*t).*(W-2.*t)).*(0.5.*(a.*x+q)-0.5.*q)+(W.*(a.*x+q).^3)./12-((W-2.*t).*(a.*x+q-2.*t).^3)./12);

I need to integrate in terms of x and I don't know the value of Fa (but it is a real value). I tried using int or integral alone or combined with vpa but Matlab gives as a result the same equation without integrating. I tried integrating it as a function but still no luck.

The integration is between 0 and 2300 and the parameters are:

a = 0.043478;

q = 100; % Ha

W = 40;

t = 4;

Answer 1

[Alan Stevens]

You could try something like this:

a = 0.043478;
q = 100; % Ha
W = 40;
t = 4;
c = @(x,Fa )(5.*x.^2-Fa.*x)./(((a.*x+q).*W-((a.*x+q)-2.*t).*(W-2.*t)).*(0.5.*(a.*x+q)-0.5.*q)...
    +(W.*(a.*x+q).^3)./12-((W-2.*t).*(a.*x+q-2.*t).^3)./12) - Fa; % Subtract Fa for use in integral
lo = 0; hi = 2300;
Fa = 1; % Set rhe value of Fa when you know it
Q = integral(@(x)c(x,Fa),lo,hi);
disp(Q)

   1.6580e+03

[Edoardo Marelli]

Thank you for your answer, I just realised I didn't explain it well.

I need to find Fa equating the integral to 0. So I would need to integrate first, and then equate the integrated function to 0 to get the Fa.

[Alan Stevens]

Like this then

a = 0.043478;
q = 100; % Ha
W = 40;
t = 4;
c = @(x,Fa )(5.*x.^2-Fa.*x)./(((a.*x+q).*W-((a.*x+q)-2.*t).*(W-2.*t)).*(0.5.*(a.*x+q)-0.5.*q)...
    +(W.*(a.*x+q).^3)./12-((W-2.*t).*(a.*x+q-2.*t).^3)./12) - Fa; % Subtract Fa for use in integral
lo = 0; hi = 2300;
Q = @(Fa) integral(@(x)c(x,Fa),lo,hi);
Fa0 = 1; % Initial guess
Fa = fzero(Q,Fa0);
disp(Fa)

    1.7207