how to solve an equation where the unknown is an integration boundry for a numerical integration?

I need to solve this equation in the form for x where A is a constant and f a known function, which has to be integrated numerically.

However, when i define this function and try to solve, i get the error

"Error using integral (line 85)
A and B must be floating-point scalars."

Apparently I cannot put the variable I need to solve for in the boundry of the integral.

How can this be done?

Below is the code. I know, it is a lot of constants.

P_ep        = @(z) P_c.*(((exp(-(1/4).*(z./omega_c).^(5/12))).*((z./omega_c).^(3./2)))+(((4./C).*(2.84-0.92.*(1-cos(pi.*((sqrt(R.*z.*(z./(1.9.*omega_c)).^(0.14.*exp(23.*(S_y./E)))))./R))))).*(1-exp(-(1./25)*(z./omega_c).^(5./9))).*(z./omega_c)));  
syms        omega_m_ep
omega_m     = solve(integral(P_ep,omega_c,omega_m_ep) == (0.5 * m * V_1 ^2)- ((8/15)*E*sqrt(R)*omega_c^(5/2)),omega_m_ep);

[darova]

This happens because of solve. It's for symbolic calculations. You have complicated equation. Use fsolve to suceed

[David Welling]

I have tried the same thing using fsolve, but it still does not work. The error code appears to originate from evaluating the integral, not from the solver.

[darova]

Can you show fsolve code?

[David Welling]

P_ep        = @(z) P_c.*(((exp(-(1/4).*(z./omega_c).^(5/12))).*((z./omega_c).^(3./2)))+(((4./C).*(2.84-0.92.*(1-cos(pi.*((sqrt(R.*z.*(z./(1.9.*omega_c)).^(0.14.*exp(23.*(S_y./E)))))./R))))).*(1-exp(-(1./25)*(z./omega_c).^(5./9))).*(z./omega_c)));  % (9), with substituted in (sqrt(8/pi),10,11 and 12)

omega_m     = fsolve(integral(P_ep,omega_c,omega_m_ep) - (0.5 * m * V_1 ^2)+ ((8/15)*E*sqrt(R)*omega_c^(5/2)),0);

[David Welling]

note that all the things in these equations are constants with a known value, exept omega_m

[James Tursa]

You need to pass fslove a function handle, and that function handle then needs to call integral.

[darova]

Is it possible to see known values/constants?

Answer 1

[David Welling]

You need to pass fslove a function handle, and that function handle then needs to call integral.

This solved the problem, Thank you both