MATLAB Answers

0

Negative Value when using Trapz

Asked by Jana Stucke on 3 Oct 2019
Latest activity Edited by Steven Lord
on 3 Oct 2019
Hi guys,
I try to calculate the area under this curve using trapz. However, it returns a negative value. Can someone tell me as to why this is the case when my x and y-values are positive?
x=[1 0.938524445788592 0.928012855054005 0.869986463167799 0.866618294101049 0.851905469533143 0.816509718296436 0.804756601303802 0.773481908667312 0.743487036373908 0.721555011244502 0.692238382577883 0.660395319804889 0.622278234454403 0.600185408288678 0.582390124224061 0.534500435615996 0.496551977223480 0.460628844607043 0.403312845618717 0.396635208896749 0.369880255480953 0.330164761722580 0.320181673106196 0.266016313621435 0.232051898082808 0.207117082563950 0.160899350279211 0.149854984446954 0.0908664503933046 0.0762867242364327 0.00582165604699889 0];
y=[0.4503 0.9715 1.0442 1.1506 1.1598 1.2079 1.3224 1.3278 1.3576 1.2198 1.0836 0.8967 0.6814 0.5081 0.4139 0.3949 0.3297 0.3276 0.3335 0.3500 0.3516 0.3560 0.3627 0.3634 0.3651 0.3640 0.3629 0.3594 0.3587 0.3522 0.3511 0.3488 0.3486]
a= trapz(x,y)

  2 Comments

Can someone tell me as to why this is the case when my x and y-values are positive?
Not without your code and data.
edited

Sign in to comment.

3 Answers

Answer by Star Strider
on 3 Oct 2019
 Accepted Answer

The problem is easiest to see with:
dx = diff(x)
All the ‘dx’ results are negative because ‘x’ is goinmg from highest-to-lowest.
To get a positive result:
a = trapz(fliplr(x),fliplr(y))
produces:
a =
0.60535

  0 Comments

Sign in to comment.


Answer by Steven Lord
on 3 Oct 2019
Edited by Steven Lord
on 3 Oct 2019

Your x vector is sorted descending.
>> issorted(x, 'ascend')
ans =
logical
0
>> issorted(x, 'descend')
ans =
logical
1
In essence, you're integrating the function represented by the y data from x = 1 to x = 0, not from x = 0 to x = 1. If you flip your x vector so you're integrating from x = 0 to x = 1 (essentially swapping the limits of integration) the area will be positive.
>> trapz(flip(x), flip(y))
[edited: I had forgotten to flip y until I saw Guillaume's answer.]

  0 Comments

Sign in to comment.


Answer by Guillaume
on 3 Oct 2019

Can someone tell me as to why this is the case
Because your x vector is decreasing, so is negative for each trapeze
a = trapz(fliplr(x), fliplr(y))
to use increasing x and matching y.

  0 Comments

Sign in to comment.