Plotting graph for integration with parameters

I would like to draw a graph . I can't get any graph with my code. Interval of Fn is 0.05 and want it to be plotted on x axis.

T = 0.08; B = 0.1; L = 1; S = 0.188;

syms u

for Fn = 0.1 : 0.05 : 0.7

K = @(u) 32./pi.*B.^2.*(T./L).^2./S./Fn.^4.*((sin(cosh(u)./(2.*Fn.^2))./(cosh(u)./(2.*Fn.^2)).^2 - cos(cosh(u)./(2.*Fn.^2))./(cosh(u)./(2.*Fn.^2))).*((1-2./(T./L.*(cosh(u)./Fn).^2).^2)./(T./L.*(cosh(u)./Fn).^2)+2./(T./L.*(cosh(u)./Fn).^2).^2.*exp(-T./L.*(cosh(u)./Fn).^2).^2.*(1+1./(T./L.*(cosh(u)./Fn).^2))).*cosh(u)).^2;

Q = int(K,u,0,10);

end

hold on

for Fn = 0.1 : 0.05 : 0.7

plot(Fn,Q)

end

hold off

It's my first time making a code so there would be lots of errors.

Q converges before inf. So doesn,t need to be integrated to infinity.

K would be like below as make it simpler.

T = 0.08; B = 0.1; L = 1; S = 0.188;

a = @(u) cosh(u)/(2*Fn^2);

b = @(u) T/L*(cosh(u)/Fn)^2;

X=@(u,a) sin(a)/a.^2-cos(a)/a;

Z=@(u,b) 1/b*(1-2/b.^2)+2/b.^2*(1+1/b)*exp(-b);

K=@(u,X,Z,a,b) 32/pi*B.^2/S*(T/L).^2/Fn.^4*(X*Z*cosh(u)).^2

Answer 1

[Anilcan Taner]

Sadly I have no idea about your code but I guess you should create empty Q variable with lets say ones code or zeros code. By that I mean that your Q variable might not be saving along the way in the for loop. Q=ones(your_size_in_row,your_size_in_column)

Same follows for Fn too, create Fn in variable form than try to plot(Fn,Q).

Hope it helps. I am newbie like you too.