Area under Curve for Unbounded Guassian & Laurentzian Distributions

Hi - I believe (maybe mistakenly) that there are mathematical formulas for obtaining (unbounded) area under curves for Guassian & Laurentzian distributions that do NOT involve integration. I cannot seem to find those formulas. Does anyone have them or point me to a link that might.
Thank you in advance.
Sincerely,
Don

Incomplete Gamma Function
where the gamma(.5,z^2) is the incomplete gamma function;
The "Upper Incomplete Gamma Function"
(1)
For an integer
(2)
(3)
(1)
For an integer
(2)
(3)
For other
The cumulative distribution function, which gives the probability that a variate will assume a value , is then the integral of the normal distribution,
(8)
(9)
(10)
It is a bit "thrown together", I hope this helps some...

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Integral of waveform graph to calculate area under curve over time

Hello ,
Please find my attached VI and please help me for integration to get area undercut for my waveform graph.
I would like to find total Energy output from my wind power waveform graph oever one month period of time.
I believe i have to integrate to get area under curve and so i m now using integral tool in labview as attached.
As I don't have expected ans for my result and so i want to make sure the tool that i m using is correct.
Please advise me which method of integration shall i use for my graph.
Solved!
Go to Solution.
Attachments:
Wind data only.vi ‏134 KB
wind data.txt ‏1253 KB

I don't know about your calculations to get the Wattage. But as far as integration, you really should supply a dt (sample rate) in order to get a proper energy measurement. I would use your second integration method (the one with "Result" as the indicator) and give it the dt as well.
There are only two ways to tell somebody thanks: Kudos and Marked Solutions
Unofficial Forum Rules and Guidelines

Area under curve

i am having the hardest time calculating area under a closed curve using basic maths. I have tried the integrate function (by splitting the curve in two and then adding the two). I am not susre of the answer so i am using basic math functions to code the thing in. So far i have:
L1 = globusedchn
CALL FORMULACALC("CH(L1):=CH('D')*CH('F')")
CC(L1)="MULTIPLIED POINTS"
CN(L1)="MULTPTS"
R4 = 0
R1 = CHDX(1, L1)
R2 = CHDX(2, L1)
R3 = R2 - R1
R4 = R3 + R1
A = R4
FOR I = 2 TO CL(L1)
L2 = I
R5 = CHDX(L2, L1)
R6 = CHDX(L2+1, L1)
R7 = R6 - R5
R8 = R7 + A
A = R8
NEXT
CALL MSGBOXDISP(R8 & " = new area")
maybe its my data but i keep getting R8 as 0. Any pointers will be greatly appreciated. Thanks ~sn

Hi shefalika,
It looks to me like the ChnIntegrate() function is the one to use. Note that the entire area under the curve is just the last value of the integrated curve. I included your original data in the ZIP file attached below, along with a quick VBScript which loads the data, creates 2 new channels which are the integral of channels 2 and 3 vs. channel 1, respectively, and loads a REPORT layout to show the results.
The REPORT layout shows the raw data curves twice, once as a thick red line and then again as vertical blue spikes. This gives you the visual respresentation of the area under each curve being blue. I then included on each of the 2 graphs a text box which displays the last value of the integrated channel for that raw data curve. This is not a static value but an embedded @ function in the REPORT layout, specifically:
Area Under Curve = @str(ChDX(CL('Area1'), 'Area1'), 'd')@
for the 1st graph. Note that the area under the curve value is pulled explicitly from the channel named "Area1", so if you rename the integrated channel, you will need to change the channel reference in each of these 2 embedded @ formulas in the REPORT layout. The str() function formats the string result to not show any decimal points in the results (you could use 'd.ddd" instead, for instance). The ChDX(row, channel) function returns the value from the "row-th" data row of the "channel" data channel.
Also note that several items in this example are built with DIAdem 9.1 features. If you need an earlier version of the example, let me know and I can build it again in DIAdem 8.1.
Ask if you have further questions,
Brad Turpin
DIAdem Product Support Engineer
National Instruments
Attachments:
Area Under Curve.zip ‏12 KB

Regarding Area Under Curve of ROC-ODM

Which method is used by Oracle Data Miner to calculate the Area Under Curve(AUC) of ROC? either trapezoidal or other ? Please reply me

Using the trapezoid method, the numerical integration is rather direct.
See the attached vi.
You can't get the function described by the X and Y arrays. You need to know the function from start ! However, you can always fit a function to the data. Polynomial curve fitting is quite popular. Of course this is something you could have done : fit a polynomial, then integrate it. With a large number of points (such as yours), the result will not be better than the trapezoid method.
Chilly Charly (aka CC)
E-List Master - Kudos glutton - Press the yellow button on the left...
Attachments:
Area.vi ‏30 KB

Area under Curve -- ROC-- Classification

Dear All,
I had applied classification technique and plot the ROC curve, the Oracle Data miner calculates the Area under Curve of that ROC, anyone please tell me on which formula does the Area under curve is calculated....

which formula is used to calculate the area under ROC curve in oracle data miner

Calulate area under curve: Hystresis

I have a force and dispacement values in excel. I want to calulate the hystresis of this.
How can i do this through VBS?? I dont have time channel in that, How can i create it and find the hystresis?
Rsh
Solved!
Go to Solution.
Attachments:
data points.xls ‏67 KB

Hi RSH,
Ignore the above, I wasn't thinking clearly. I banged out a quick data set and tried it out, and it turns out to be much simpler-- you just need to calculated the integral channel and look at the last value in the channel, like this:
i = 1 ' GroupIndex
Call GroupDefaultSet(i)
DispCh = CNo("[" & i & "]/Displacement")
ForceCh = CNo("[" & i & "]/Force")
IntCh = CNo("[" & i & "]/Integral")
IF IntCh > 0 THEN Call ChnDel(IntCh)
Call GroupDefaultSet(i)
Call ChnIntegrate(DispCh, ForceCh, "/Integral")
IntCh = CNo("[" & "]/Integral")
Integral = ChDX(ChnLength(IntCh), IntCh)
Call ChnPropValSet(IntCh, "Integral", Integral)
The calculated area is put in the "Integral" property of the "Integral" channel in the same group as the "Displacement" and "Force" channels. In the example I'm attaching, I split the raw hysteresis curve (both forward and return) into its two components and calculated the integral of each, just to make sure that the different in their integrals really was the same as the net integral of the raw data curve-- and it was.
Brad Turpin
DIAdem Product Support Engineer
National Instruments
Attachments:
Hysteresis.zip ‏3 KB

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Is there a way to adjust brightness and contrast with curves in fcp4.5?

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you'll need to add a plug-in for this capability
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What is a quick alternative to launching an enterprise DPS app if Apple Store rejects the App? We are under a major deadline and can't wait for Apple to approve. We want to host the app elsewhere. How do we host our DPS app on our client's website? Thanks.

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How can I calculate area under a curve?

I would like to use an integral equation to obtain positive and negative areas from graphed data. For example, I have a graph representing power for a 24 hour period and want to integrate the curve to obtain daily energy.

One common way to approximate the area under a curve is to divide it into a series of trapezoids, with the area of each calculated as avg height x width. Then sum the areas.
So with this table:
x
y
trapezoid
1
0
3
2
6
11
3
16
23
4
30
39
5
48
59
6
70
83
7
96
218
You could use this formula in C2, filled down through the rest of the body cells in that column:
=IFERROR((B3+B2)÷2×(A3−A2),"")
This calculates the area of each trapezoid.
The first row is defined as a Header Row, the last row as a Footer Row. It's important not to have blank row(s) before the Footer Row. The sum formula in the Footer Row in C9 is:
=SUM(C)
Of course, the greater the number of x-y data points the better the estimate of the area under the curve.
SG

Calculating area under a curve - trapezium rule

Hi,
I'm trying to calculate the 'area under the curve' using the trapezium rule for a set of data points. i.e. I have a set of values, sampled at irregular intervals.
create table measurements (
measurement_value integer,
measurement_time timestamp
insert into measurements (measurement_value,measurement_time) values (50,TIMESTAMP'2011-01-01 00:00:00');
insert into measurements (measurement_value,measurement_time) values (40,TIMESTAMP'2011-01-01 01:00:00');
insert into measurements (measurement_value,measurement_time) values (40,TIMESTAMP'2011-01-01 03:00:00');
insert into measurements (measurement_value,measurement_time) values (50,TIMESTAMP'2011-01-01 03:30:00');
insert into measurements (measurement_value,measurement_time) values (40,TIMESTAMP'2011-01-01 05:00:00');
insert into measurements (measurement_value,measurement_time) values (40,TIMESTAMP'2011-01-01 06:00:00');
insert into measurements (measurement_value,measurement_time) values (60,TIMESTAMP'2011-01-01 07:00:00');I'd like to calculate the area of the 'curve' where the value was below a threshold of, e.g. 45.
I can use the trapezium rule and assume a straight line between subsequent measurement, i.e. for the first 2 data points, the value drops from 50 to 40 linearly over the hour, and so the area for those measurements is (30min * 5)/2. For the 2nd and 3rd data points, the area is (120min * 5), for the 3rd and 4th, the area is (15min * 5)/2. And so on. The values are not necessarily round numbers, almost any value is possible.
I've started by using 'lead' to obtain the subsequent measurement value and time, but I'm getting stuck dealing with the end points and non-trivial interpolation. Does anyone have any pointers to get me going?
Thanks,
Dan Scott

Maybe this will do it:
SELECT SUM(CASE
 WHEN curr_value > :threshold AND next_value < :threshold
 THEN -- Area of rectangle specified by threshold
 (new_measurement_time_sec-curr_measurement_time_sec)*:threshold
 + -- Area of remaining trapezoid
 ABS((next_measurement_time_sec - new_measurement_time_sec)*((next_value - :threshold)/2))
 WHEN next_value > :threshold AND curr_value < :threshold
 THEN -- Area of rectangle specified by threshold
 (next_measurement_time_sec-new_measurement_time_sec)*:threshold
 + -- Area of remaining trapezoid
 (new_measurement_time_sec - curr_measurement_time_sec)*((:threshold - curr_value)/2)
 WHEN (:threshold < curr_value AND :threshold < next_value)
 OR (:threshold > curr_value AND :threshold > next_value)
 THEN (next_measurement_time_sec-curr_measurement_time_sec)*:threshold
 ELSE -- Area of trapezoid
 ABS((next_measurement_time_sec - curr_measurement_time_sec)*((next_value - curr_value)/2))
 END) AS area
FROM (
 SELECT measurement_time
 , curr_measurement_time_sec
 , next_measurement_time_sec
 , curr_value
 , next_value
 , (:threshold - LEAD(intercept) OVER (ORDER BY measurement_time))/LEAD(slope) OVER (ORDER BY measurement_time) AS new_measurement_time_sec
 FROM ( SELECT measurement_time
 , TO_CHAR(measurement_time,'J')*24*60*60 + TO_CHAR(measurement_time,'SSSSS') AS curr_measurement_time_sec
 , TO_CHAR(LEAD(measurement_time) OVER (ORDER BY measurement_time),'J')*24*60*60 + TO_CHAR(LEAD(measurement_time) OVER (ORDER BY measurement_time),'SSSSS') AS next_measurement_time_sec
 , measurement_value AS curr_value
 , LEAD(measurement_value) OVER (ORDER BY measurement_time) AS next_value
 , REGR_SLOPE(measurement_value, TO_CHAR(measurement_time,'J')*24*60*60 + TO_CHAR(measurement_time,'SSSSS')) OVER (ORDER BY measurement_time) AS slope
 , REGR_INTERCEPT(measurement_value, TO_CHAR(measurement_time,'J')*24*60*60 + TO_CHAR(measurement_time,'SSSSS')) OVER (ORDER BY measurement_time) AS intercept
 FROM measurements
 )Results:
SQL> var threshold NUMBER;
SQL> exec :threshold := 45;
SQL> /
 AREA
1191437.2The units are the time values were converted to seconds since January 1, 4712 BC (roughly).
Edited by: Centinul on Jul 7, 2011 10:39 AM

Regression area under the curve

hello
i was trying to get the area under the curve. mathematically the
integration can b done directly from 'numeric integration' vi. but what
my problem is i cant define the time t1 to t3( as i need to set the
threshold level to eliminate noise). what i previously did was defining
the threshold level, find the 2 point of t1 and t2. but i left out the
coloured portion. using regression will it help to get t3?if yes. how
this will b done?
thanks
regards
Attachments:
graph(dt)1.PNG ‏8 KB

This has been going on for a while:
http://forums.ni.com/ni/board/message?board.id=170&message.id=226483#M226483
http://forums.ni.com/ni/board/message?board.id=MathScript&message.id=145#M145
http://forums.ni.com/ni/board/message?board.id=170&message.id=227008#M227008
If I understand this correctly, you use a threshold to find the peak, but then you want the entire area, including the "wings".
As I said before, if you pick the baseline right, your integration boundary does not need to be exact and can be significantly outside because the areas outside the peak will average to zero.
If you have a mathematical model for the peak (e.g. gaussian), you can fit it and calculate the area from the fit.
LabVIEW Champion . Do more with less code and in less time .

Area under normal curve

I need to calculate the area under a normal (gaussian) curve between two given limits. Is there some algorithm or freely usable library which I could use to calculate this? I've searched around but so far have come up with nothing.

, i.e., something relatively close to
the precision of 'double'. Why do you need it to be so precise? After all, when you're dealing with statistics there's always a level of probability. Even if you had double precision, would that make the final result more accurate?
I expect there's
probably some asymptotic approach that approaches the
desired precision in some fixed number of
iterations.
Probably so. Just out of curiosity I wanted to measure the change in area as the number of points grew by a factor of 10. here are the results:
0.9746011459589066 numPoints: 80
0.9803552459545505 numPoints: 800 area - oldArea: 0.005754099995643913
0.9808610053578404 numPoints: 8000 area - oldArea: 5.05759403289896E-4
0.9809109343345651 numPoints: 80000 area - oldArea: 4.992897672473351E-5
0.9809159208101573 numPoints: 800000 area - oldArea: 4.986475592216877E-6
0.9809164193936031 numPoints: 8000000 area - oldArea: 4.985834457515992E-7
0.980916469248969 numPoints: 80000000 area - oldArea: 4.985536594670492E-8
Basically, if you use 10 times as many points from one to the next, you get an extra decimal place of accuracy. So if you can live with 4 decimal places of accuracy, then maybe this method is ok, otherwise the computation becomes expensive.
Here is the code:
public class ATest {
 public static void main(String[] args) {
 double oldArea = -1;
 for (int k = 0; k < 6; k++) {
 double dx = 0.1 / Math.pow(10, k);
 double x1 = -4.0;
 double x2 = 4.0;
 int numPoints = (int) Math.round((x2 - x1) / dx);
 double area = 0;
 for (int i = 0; i < numPoints; i++) {
 int j = (i + 1) % numPoints;
 double xi = x1 + i * dx;
 double xj = x1 + j * dx;
 double yi = calcFunction(xi);
 double yj = calcFunction(xj);
 area += xi * yj - xj * yi;
 if (area < 0) area = -area;
 area = area / 2;
 if (oldArea != -1) {
 double dif = area - oldArea;
 System.out.println(area + " numPoints: " + numPoints + " area - oldArea: " + dif);
 else
 System.out.println(area + " numPoints: " + numPoints);
 oldArea = area;
 private static double calcFunction(double X) {
 double SIGMA = 1.0;
 double SIGMA_SQ_2 = SIGMA * SIGMA * 2;
 double C = 1.0 / Math.sqrt(Math.PI * SIGMA_SQ_2);
 double MU = 1.0;
 double xmu = X - MU;
 return C * Math.exp(-1 * xmu * xmu / SIGMA_SQ_2);
}

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Area under Curve for Unbounded Guassian & Laurentzian Distributions

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