Negative Value when using Trapz

72 次查看（过去 30 天）
Jana Stucke
Jana Stucke 2019-10-3
评论： Star Strider 2021-3-24
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
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.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Star Strider
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
显示 2更早的评论
Andrea Mira
Andrea Mira 2021-3-24
Excuse me @Star StriderStar Strider, could I ask you a doubt about your comment?
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
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.

请先登录，再进行评论。

更多回答（2 个）

Guillaume
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
a = trapz(fliplr(x), fliplr(y))
to use increasing x and matching y.
Steven Lord
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.]

类别

在 Help Center 和 File Exchange 中查找有关 Numerical Integration and Differentiation 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by