Trapezoidal Estimator for the integral
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FURKAN CEVAHIR
on 24 Feb 2019
Commented: FURKAN CEVAHIR
on 25 Feb 2019
I tried here to find a trapezoidal estimate of an area between different intervals(Delta_x); however, errors found are not correct numbers according to others solutions(I checked). I couldnt find my mistake.
Could you look at it?
clc; clear;
f1=@(x) 1+sin(pi*x/100)-sin(pi*x/25);
Area1_actual=integral(f1,0,100);
%%Trapezoidal estimator
c=0;
for Delta_x=[10 5 2.5 1 1/2.5 1/5 1/10];
x = 0:Delta_x:100;
for j=1:(length(x)-1)
Trap_area = Delta_x/2*(f1(x(j))+f1(x(j+1)));
mat_traparea_Tr(j,1) = Trap_area;
end
c=c+1;
Total_area_Tr(c,1) = sum(mat_traparea_Tr);
Error_Tr(c,1) = Total_area_Tr(c,1)-Area1_actual;
end
Delta_x=[10 5 2.5 1 1/2.5 1/5 1/10];
plot(Delta_x,abs(Error_Tr),'b-'); title('Trapezoidal estimator')
xlabel('DeltaX'), ylabel('Errors'); set(gca, 'XDir','reverse')
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Accepted Answer
David Goodmanson
on 24 Feb 2019
Edited: David Goodmanson
on 25 Feb 2019
Hi Furkan,
case 1: you are calculating Area2 with the trapezoidal approximation, but comparing it to Area1_actual when computing the error. Comparing it to Area2_actual gives much more satisfactory results.
case 2: Area2_actual is computed from 0 to 10, while your trapezoidal calculation is from 1 to 10.
It's true, the devil is in the details.
7 Comments
John D'Errico
on 25 Feb 2019
It aims to do so. But is incorrect in what it does. I see at least two clear errors.
First, a minor one:
Trap_area_f1=zeros(length(Delta_x)-1,1);
If you would initialize a vector, it makes sense to initialize it to the correct length. Otherwise, preallocation is just a waste of time. Better would be:
Trap_area_f1=zeros(length(x)-1,1);
Of course, the error there is a minor one, since it does not actually create a bug. Just poorly written, inefficient code.
More important?
x=1:Delta_x(i):100;
I might wonder what are the limits of integration? Do you think that might be important? ;-) (Hint.)
You can write lots of scripts that give different answers. Most randomly written scripts would give the wrong result, of course. As I showed rather carefully however, the results I showed in my comment (as generated by the code you wrote after you fixed the error found by David) were actually correct.
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