Integration of a function.

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JACINTA ONWUKA
JACINTA ONWUKA il 16 Ago 2019
Commentato: JACINTA ONWUKA il 16 Ago 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

Risposta accettata

infinity
infinity il 16 Ago 2019
Hello,
You just need to remove "." in the formulation of I2, I3, fun2 and fun3. The code will run withour errors.

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