How can take an integral for a dependent expression?

Question

Jarett 2014-12-4

0
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https://ww2.mathworks.cn/matlabcentral/answers/165456-how-can-take-an-integral-for-a-dependent-expression

评论： Jarett 2014-12-9

I am trying to take the integral of an expression and am running into trouble on how to do it symbolically. I have used the trapz method and it works, however in following work this method seems to be inadequate. I have included my script for reference, ultimately I want to take the integrals denoted as Qsfun and Qcfun which are dependent on the variables xg and Vg. Thanks in advance for any insight and help.

Script

Vg = linspace(1.5e-1, 5, 3000);%Energy gap defined by us, allows optimization of where the cell has the highest efficiency.
for t = 1:length(Vg)
xg(t) = (q*Vg(t)) /(k*Ts); %Xg value operating on the energy gap value and dependent on the temp of the sun.
xgc(t) = (q*Vg(t)) /(k*Tc); %Xg value operating on the energy gap value and dependent on the temp of the cell.
x = linspace(xg(t),1000, 10000);%Size iteration of x based on values xg(t) to 1000, with 10000 interval points
for l = 1:length(x);
Qsfun(l) =(x(l)^2)/(exp(x(l))-1);
end
a = linspace(xgc(t),30*xgc(t), 30000);%Size iteration of a based on values xgc(t) to 30*xgc, with 30000 interval points
for y = 1:length(a);
Qcfun(y) =(a(y)^2)/(exp(a(y))-1);
end
Intqs(t) = trapz(x,Qsfun);
Intqc(t) = trapz(a,Qcfun);
%Iqs(t) = xg*trapz(x,Qsfun);
%Iqc(t) = xg*trapz(x,Qcfun);
%The number of quanta of frequency greater than vg incident per unit area time
Qs(t) = ((2*pi*((k*Ts)^3))/((h^3)*(c^2)))*Intqs(t);
Qc(t) = ((2*pi*((k*Tc)^3))/((h^3)*(c^2)))*Intqc(t); 
b = linspace(1e-10, 100, 3000);
for mm = 1:length(b)
Psfun(mm) = (b(mm)^3)/(exp(b(mm))-1); %Function of Plancks distribution integral
end
IntPs = trapz(b,Psfun);
u(t) = (xg(t)*Intqs(t))/IntPs;
end