integration with limits using an array

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James Keiser
James Keiser 2022-10-6
评论： the cyclist 2022-10-6
I am trying to integrate an equation and it works when I do not have limits but when I add them it does not work.
u_wake is a 1x56 array, rho is a constant and 0.07 and 0.13 are the limit boundaries.
This works when I use this:
Drag = rho*trapz(u_wake.*(21.15-u_wake));
But not when I do this:
Drag = rho*trapz(u_wake.*(21.15-u_wake),0.07,0.13);
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James Keiser
James Keiser 2022-10-6
编辑：James Keiser 2022-10-6
I cannot upload the data because it is 56 files and I am only allowed to upload 10. The data is 56 numbers all below 21.15 .
The error code is:
Error in wake_scan_sample (line 35)
Drag = rho*trapz(u_wake.*(21.15-u_wake),0.07,0.13);
the cyclist
the cyclist 2022-10-6
It's unclear to me why you can't just upload u_wake, which is just a 1x56 vector.
Also, that is not the complete error message, which I am guessing looks something more like ...
Error using matlab.internal.math.getdimarg
Dimension argument must be a positive integer scalar within indexing range.
Error in trapz>getDimArg (line 90)
dim = matlab.internal.math.getdimarg(dim);
Error in trapz (line 36)
dim = min(ndims(y)+1, getDimArg(dim));
Error in wake_scan_sample (line 35)
Drag = rho*trapz(u_wake.*(21.15-u_wake),0.07,0.13);

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回答（1 个）

the cyclist
the cyclist 2022-10-6
Looking at the documentation for trapz, it doesn't seem like the syntax you are trying to use is valid.
  2 个评论
the cyclist
the cyclist 2022-10-6
Ran in:
I assume that u_wake is the "y" value of the integral. Are 0.07 and 0.13 the endpoints of the "x" value? Are the values of u_wake at evenly spaced points along x? Then
Drag = rho*trapz(linspace(0.07,0.13,length(u_wake)),u_wake.*(21.15-u_wake));
should calculate what you want. For example,
% I just made up some u_wake and rho data
u_wake = sort(rand(1,56));
rho = 1;
Drag = rho*trapz(linspace(0.07,0.13,length(u_wake)),u_wake.*(21.15-u_wake))
Drag = 0.5618

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