Integration of a function.

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JACINTA ONWUKA
JACINTA ONWUKA le 16 Août 2019
Commenté : JACINTA ONWUKA le 16 Août 2019
I'm trying to integrate a function but is giving be this error below
Incorrect dimensions for matrix multiplication.
Check that the number of columns in the first matrix matches the number
of rows in the second matrix. To perform elementwise multiplication, use '.*'.
I cant identify to set the dimention right but is still giving error. Please help me.
L=1;
T=100;
r=0.03;
I1=0.1;
p=0.1;
epsilon=0.3;
rho=1000;
beta= 0.1;
Cr=3000;
RhoP=0;
Mycp = 0:10:100;
n = zeros(numel(Mycp),1 );
n2 = zeros(numel(Mycp),1 );
n3 = zeros(numel(Mycp),1 );
Jcp = zeros(numel(Mycp),1 );
for i = 1:numel(Mycp )
MycpCurrent=Mycp(i);
delta = 1-MycpCurrent/100;
tau = (1/(beta*(L+delta*p)))*log((L*(I1+delta*p))/(delta*p*(L-I1 )));
t05 =(1/(beta*(L+delta*p)))*log((L*(0.05*L+delta*p))/(delta*p*(L-0.05*L )));
I2= @(t)(L*delta*p*(exp (beta*(L+delta*p)*t)-1)) ./ (L + delta*p* exp(beta*(L+delta*p)*t ));
I3= @(t)(L*(I1+delta*p)*exp((epsilon*beta)*(L+delta*p)*(t-tau))-...
delta*p*(L -I1))./(L-I1+(I1+delta*p)*exp(epsilon*beta*(L+delta*p)*(t-tau)));
fun = @(t,MycpCurrent) MycpCurrent*L*exp(-r*t);
fun2=@(t) rho*I2(t)*I3(t).*exp(-r*t)+((Cr*L*exp(-r*tau)));
fun3=@(t)RhoP*I2(t)*I3(t).*exp(-r*t);
n(i) = integral(@(t)fun(t,MycpCurrent),0,100, 'ArrayValued',1);
n2(i)= integral(fun2,t05,tau); %this is giving me error.
n3(i)= integral(fun3,tau,100);
end

Réponse acceptée

infinity
infinity le 16 Août 2019
Hello,
You just need to remove "." in the formulation of I2, I3, fun2 and fun3. The code will run withour errors.
  1 commentaire
JACINTA ONWUKA
JACINTA ONWUKA le 16 Août 2019
Thanks. I can now move on.

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