Error: "Dimension argument must be a positive integer scalar within indexing range" using trapz

x_24= [0; 4.6958; 5.6732; 14.2002; 15.4490; 16.0151;  17.9994; 21.7987; 24.1082; 29.8818; 37.4162; 41.1113; 55.6853; 59.9431]
z_24=[ 6.1876; 7.1381; 7.2346; 4.4075; 4.4321; 4.2854; 3.9659; 3.4729; 3.1695; 2.2887; 1.2414; 0.8183; -0.0147; -0.2389]

NM_18=0.82;
MHWS_18=1.5-NM_18;

z_MHWS_24=z_24-MHWS_18;

MHWS_vector_18=repmat(MHWS_18,79,1);

figure(1)
plot (x_24,z_MHWS_24,'Yellow')
hold on
plot(MHWS_vector_18,'Blue')
hold off

x=x_24;  
z=z_MHWS_24; 

xi=5.67; 
xf=14.20; 

xRange = [xi,xf];
% find logical indices for range of interest
idl = x >= xRange(1) & x <= xRange(2);
% provide just values of interest to trapz
area_selected=trapz(x(idl),z(idl))

area_total=trapz(x,z)

The error that is happening is in the last two lines of the code above (from bottom to top).

Answer 1

[Mathieu NOE]

hello Jacqueline

with the given limits xi=5.67; xf=14.20; , idl contains only one valid x value , so trapz will throw an error, because it needs at least vector of 2 scalars (and you give only one value)

I guessed that you needed to include the 14.2002 value in your x vector , so simply make the second limit a bit higher : xf=14.21;

and then it's fine, you can do the trapz computation

results :

area_selected = 43.8377

area_total = 123.3377

[Jacqueline Rigatto]

Thank you, Mathieu NOE! It helped a lot!

[Mathieu NOE]

You're welcome !