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NUMERICAL INTEGRATION USING SIMPSONS

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This is the program to evalute intergration using Simpsons. There is an error in line 11.
lambda=1:0.1:2;
gamma1=0.4;
zeta=0.3;
a=0;%lower limit
b=1;%lower limit
n=10;%number of sub-intervals
f1=@(x)((1-x).^((gamma1./zeta)-1)).*(exp(-(lambda.*x)./zeta));
h=(b-a)/n;
for k=1:1:n
x(k)=a+k*h;
y(k)=f1(x(k));
end
so=0;se=0;
for k=1:1:n-1
if rem(k,2)==1
so=so+y(k);%sum of odd terms
else
se=se+y(k); %sum of even terms
end
end
SOL=h/3*(f1(a)+f1(b)+4*so+2*se);

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Accepted Answer

Alan Stevens
Alan Stevens on 24 Oct 2020
The problem is that you have many values of lambda. To deal with this use a loop to do the integration for each value of lambda (this requires f1 to have an extra input parameter):
lambda=1:0.1:2;
gamma1=0.4;
zeta=0.3;
a=0;%lower limit
b=1;%lower limit
n=10;%number of sub-intervals
f1=@(x,p)((1-x).^((gamma1./zeta)-1)).*(exp(-(lambda(p).*x)./zeta));
h=(b-a)/n;
for p = 1:numel(lambda)
for k=1:n
x(k)=a+k*h;
y(k)=f1(x(k),p);
end
so=0;se=0;
for k=1:n-1
if rem(k,2)==1
so=so+y(k);%sum of odd terms
else
se=se+y(k); %sum of even terms
end
end
SOL(p)=h/3*(f1(a,p)+f1(b,p)+4*so+2*se);
end

More Answers (1)

VBBV
VBBV on 24 Oct 2020
In the for loop use
% if true
% code
%end
y(k) = f1(k);
Use a space btw @(x) ((1-x) ...) for the anonymous function.

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