Negative Value when using Trapz
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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 个评论
Star Strider
2019-10-3
‘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.
Jana Stucke
2019-10-3
采纳的回答
Star Strider
2019-10-3
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
4 个评论
Ian Hunter
2019-12-20
thanks, this was very helpful!
Star Strider
2019-12-20
My pleasure!
Andrea Mira
2021-3-24
In case the "function" has a part where dx decreases and another where dx increases. Could fliplr be used?
Or would it be necessary to calculate the area in two parts? The part where dx decreases applying fliplr and the part where dx increases without fliplr?
(I'm interested in calculating the red area between the "scatter function" and the blue horizontal line)
Thanks in advance
Star Strider
2021-3-24
Andrea Mira — If I understand correctly what you want to do, I would simply calculate (integrate) those two red areas separately and then add their absolute values if you want to get the total area. Otherwise, they would subtract from each other, producing an area that would be essentially (within calculation error) 0.
更多回答（2 个）
Guillaume
2019-10-3
Can someone tell me as to why this is the case
Because your x vector is decreasing, so 
is negative for each trapeze
is negative for each trapezea = trapz(fliplr(x), fliplr(y))
to use increasing x and matching y.
Steven Lord
2019-10-3
编辑：Steven Lord
2019-10-3
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.]
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