# Find the value from the integral equation

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Commented: Arad on 7 Feb 2023
I have the function A(e) as:
A=(cos(4*w)*cos(4*w)+1)/(1+exp(w-m0);
where the w is the variable and and m0 is the unknown value which should to be obtained by:
int(A,e,-20,20)-2=0.
I need to obtain the m0 parameter by solving the above equation.
Sargondjani on 7 Feb 2023
I'm just going to give hints:
1. make a function, or function handle of A
2. define an objective function (the equation that should be zero. You can use Matlabs "integral" function.
3. you need solve solve the objective function. You can use fsolve, or alternatively look an algorithm using Newton's method on file exhange (for example this link might work: https://uk.mathworks.com/matlabcentral/fileexchange/28227-newton-s-method)
This should get you started.

Torsten on 7 Feb 2023
fun = @(w,m0) (cos(4*w).*cos(4*w)+1)./(1+exp(w-m0));
m0sol = fsolve(@(m0)integral(@(w)fun(w,m0),-20,20)-2,1)
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.
m0sol = -18.9804
integral(@(w)fun(w,m0sol),-20,20)-2
ans = 2.6910e-11
Thank you very much.

Dyuman Joshi on 7 Feb 2023
Edited: Dyuman Joshi on 7 Feb 2023
You can solve it symbollically (This requires symbolic toolbox)
syms w m0
A=(cos(4*w)*cos(4*w)+1)/(1+exp(w-m0));
eq=int(A,w,-20,20)-2==0;
val=vpasolve(eq)
val = %Verification
syms y(x)
y(x)=(cos(4*x)*cos(4*x)+1)/(1+exp(x-val));
inty=int(y,x,-20,20)
inty = %As we can see here, we don't get the numeric value of the integral
%so we use vpa to approximate the value
vpa(inty)
ans =
2.0
Thank you very much.