Trapezoidal Estimator for the integral

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')

 Accepted Answer

David Goodmanson
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

Oh, just edited, you're right but this was just the mistake while copying. I think the problem is different thing.
After editing was done to correct the error found, what do you think to be the problem?
Total_area_Tr
Total_area_Tr =
163.13751514675
163.531023680874
163.629248948393
163.656741162872
163.661139476512
163.66176779711
163.661924876872
>> Area1_actual
Area1_actual =
163.661977236758
>> Error_Tr
Error_Tr =
-0.524462090007688
-0.130953555884616
-0.0327282883654334
-0.00523607388655023
-0.000837760245900654
-0.000209439648017451
-5.23598861263963e-05
So you have an integral that is actually not that poorly estimated, going from increments of 10 down to a fairly small number. Each estimate is a small under-estimate, not uncommon for a trapezoidal method. But are your estimates correct? That is trivial to test.
trapz(0:10:100,f1(0:10:100))
ans =
163.13751514675
So the estimate for Delta_x(1) is perfectly correct. So, I'll repeat my question. Why do you think it is wrong? I see no such evidence.
My recommendation is you accept the answer given by David. As it is now, your code now seems to work correctly.
For example, this is another script that aims to give the same estimates, but giving different results I dont understand why.
clc; clear;
f1=@(x) 1+sin(pi*x/100)-sin(pi*x/25);
Area1_actual=integral(f1,0,100);
Delta_x=[10 5 2.5 1 1/2.5 1/5 1/10];
Error_Tr_f1=zeros(length(Delta_x),1);
for i=1:length(Delta_x)
dx = Delta_x(i);
x=1:Delta_x(i):100;
%%Trapezoidal estimator
Trap_area_f1=zeros(length(Delta_x)-1,1);
for j=1:length(x)-1
Trap_area_f1(j) = dx/2*(f1(x(j))+f1(x(j+1)));
end
Error_Tr_f1(i) = sum(Trap_area_f1)-Area1_actual;
end
Hi Furkan,
see addendum to my original answer.
Do you suggest any solution for that?
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.
So, to make sure, you say that my script with this form in the question is completely true, right?

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