Integration of a function.

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JACINTA ONWUKA
JACINTA ONWUKA 2019 年 8 月 16 日
コメント済み: JACINTA ONWUKA 2019 年 8 月 16 日
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

採用された回答

infinity
infinity 2019 年 8 月 16 日
Hello,
You just need to remove "." in the formulation of I2, I3, fun2 and fun3. The code will run withour errors.
  1 件のコメント
JACINTA ONWUKA
JACINTA ONWUKA 2019 年 8 月 16 日
Thanks. I can now move on.

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