using trapz in a for loop

I am trying to use trapz in a for loop to compute an integral with multiple bounds.

This is the code I'm trying to use:

clear;clc
mu=1.837e-5; %dynamic viscosity, Pa*s
nu=1.552e-5; %kinetmatic viscosity, m^2/s
input=readmatrix('ProjectPartCTestInput.txt');
s=input(:,1);
ue=input(:,2);
n=100;
ue6=ue.^6;
for i=1:length(n)
    int=trapz(s(1:i),ue(1:i));
    %delta22(i)=(0.45*nu*int)/ue6; %delta2^2
    %delta2(i)=sqrt(delta22); %delta2
end

And this is the error I keep getting:

Error using permute

ORDER contains an invalid permutation index.

Error in trapz (line 51)

y = permute(y,perm);

Error in Untitled (line 14)

int=trapz(s(1:i),ue(1:i));

Any suggestions?

Answer 1

[Star Strider]

The problem:

for i=1:length(n)

This returns 1 since ‘n’ is a scalar.

You probably intend:

for i=1:n

however that is still incorrect.

This:

int=trapz(s(1:n),ue(1:n));

without the loop will likely work.

If you want to partition ‘s’ and ‘ue’ into 100-element columns, consider using reshape and iterate to integrate over the columns.

.

Answer 2

[Voss]

When you call trapz with two arguments and the second argument is a scalar (like what happens in your loop when i == 1), then trapz treats the second argument as the dimension argument (which is usually the third input argument). In your case, presumably ue(1) is a value that's not a valid dimension, such as 0, and that causes the error.

To avoid this error, explicitly state the dimension argument (in this case it's 1 since s and ue are column vectors):

int=trapz(s(1:i),ue(1:i),1);
%                       ^^ explicitly include dimension argument

Answer 3

[Walter Roberson]

 int=trapz(s(1:i),ue(1:i));

when i is 1, that would be trapz(scalar,scalar) . But when you trapz() and the second parameter is a scalar, the second parameter is interpreted as being a dimension index.

That is, the syntax that the help summarizes as

trapz(Y,DIM)

has priority over the syntax summarized over

trapz(X,Y)

in the case that Y is a scalar.

Note: trapz() with a scalar Y always gives a result of 0. trapz() requires at least two points to give a non-zero result.