MATLAB Answers

Integration of a function.

5 views (last 30 days)
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

  0 Comments

Sign in to comment.

Accepted Answer

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

  1 Comment

JACINTA ONWUKA
JACINTA ONWUKA on 16 Aug 2019
Thanks. I can now move on.

Sign in to comment.

More Answers (0)

Sign in to answer this question.


Translated by