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using the trapz function

Asked by Eliraz Nahum on 11 Oct 2018
Latest activity Commented on by Walter Roberson
on 11 Oct 2018
hey, I am trying to understand how the trapz works. I am trying to compare its results to an exact solution, but for some reason can't get the same results.
clear all
close all
clc
syms x1 y1
y1=x1^2;
y1_int=int(y1);
y1_int_f=matlabFunction(y1_int)
y_int_exact=[];
del=0.1;
x=[-5:del:5];
y=x.^2;
counter=0;
for Xval=x
counter=counter+1;
y_int_exact(counter)=y1_int_f(Xval)-y1_int_f(x(1));
y_temp=y(1:counter);
y_int_calc(counter)=trapz(y_temp);
end
plot(x,y,'b',x,y_int_calc,'--r',x,y_int_exact,'g')

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2 Answers

Answer by Eliraz Nahum on 11 Oct 2018
 Accepted Answer

according to https://uk.mathworks.com/help/matlab/ref/trapz.html I can only give the trapz function the y values of y(x). That's what I tried to do...each time giving the function a y_temp vector that contains the [y(1),y(2),...,y(counter)] values so it will find the area in each step

  1 Comment

Walter Roberson
on 11 Oct 2018
You should have read slightly further
Q = trapz(X,Y) integrates Y with respect to the coordinates or scalar spacing specified by X.
If X is a vector of coordinates, then length(X) must be equal to the size of the first dimension of Y whose size does not equal 1.
If X is a scalar spacing, then trapz(X,Y) is equivalent to X*trapz(Y).

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Answer by Walter Roberson
on 11 Oct 2018

You are not telling it what spacing to the trapz() call. Also your y_int_exact should probably be taking into account the spacing between values.

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