9 replies to this topic

[chemiboy11]

Hi everyone,

I am currently trying to calculate the area under the curve from the graph contained in the attachment.


 for new post.doc   96.5KB   29 downloads


How do I calculate the area underthe curve from the graph? Do I have enough data?

The details are as follows:


The lower and upper limits for the REDUCEDTIME (x-axis) are 0.004 and 7.572 respectively.
The equation for y in terms of x is this:

y =-1E-08x4 + 5E-07x3 - 7E-06x2 + 4E-05x + 5E-05

waiting for replies. please help me quickly..

[breizh]

Hi ,

Let you try Simpson's method :

http://en.wikipedia.org/wiki/Simpson's_rule

Other techniques ( more less accurate) , use a planimeter or weigh the area and compare with a standard reference ( both methods) .

Note : why don't you do the math , your function is matching pretty well the graph.
At least you should compare the differerent methods


Hope this helps

Breizh

Edited by breizh, 20 August 2011 - 01:44 AM.

[Zauberberg]

As far as I remember Math classes, you need to perform an integration of the Y(X) function within given interval (X1, X2) of the independent variable. You have a polynomial function so there shouldn't be any problems.

Automated option I remember about is to use the Origin software. It calculates the area under the curve for any pre-defined function or tabular data given in the form of Y(X).

[chemiboy11]

dear Mr. Breizh, there is an error in the Simpson's rule. The content below is from wikipedia. And since my equation is of order 4 I cannot use the Simpson's rule to give reliable results. so I am still left confused.


ErrorThe error in approximating an integral by Simpson's rule is

where ξ is some number between a and b.^[3]

The error is asymptotically proportional to (b − a)⁵. However, the above derivations suggest an error proportional to (b − a)⁴. Simpson's rule gains an extra order because the points at which the integrand is evaluated are distributed symmetrically in the interval [a, b].

Note that Simpson's rule only provides exact results for any polynomial of degree three or less, since the error term involves the fourth derivative of f.

[breizh]

Hi ,

If you are not satisfied with Simpson's method , do the math it's simple !
Anyway I will encourage you to consider all the methods proposed and evaluate the gap.Think as an engineer .

Breizh

[chemiboy11]

I tried the two of the mathematical methods. the results are comparable.

the integration method gave the answer: 8.412exp-3
the Simpson's rule gave the answer: 8.715exp-3

which one is more accurate in your opinion?

[breizh]

To me your results are similar and consistent . I guess this is an intermediate calculation , let you put in perspective the final results . More important is that you learn on you can solve such kind of problem .
Good luck.
Breizh

[chemiboy11]

yes. I agree.
We have already learnt the techniques but I have never got the chance of applying the theory to practical problems. now that we are beginning to do so. I might need a little help in the start.
thanks

[kkala]

A try to clarify the matter is shown in the attached "area.xls", with following relevant clarifications.
1. You can use the trapezoidal rule (rather simplified Simpson's) to find area under the "curve" considered as a total of straight line parts. In the table of calculation (hoping to be clear enough concerning the method) the values of Y are not quite accurate, roughly read from your graph onscreen. Resulting area = 5.87E-4, to be precise when values of Y are replaced with the exact Y values that you have available.
2. Integrating f(x) from x=0.004 to x=7.572 is realized in another table (of area.xls). Resulting area = 8.73E-4 under the curve Y=f(x)=-1E-8X^4+5E-7X^3-7E-6X^2+4E-5x+5E-5.
3. To roughly check above result, shape under the curve (in the last diagram of area.xls) is more or less a trapezoid, having an area of (11E-5+6E-5)/2*7.5=6.38E-4, comparable to 8,73E-4.
4. If I do not miss something (PC is uncomfortable in the country), calculation per para 1 gives the exact area under the broken line, while calculation per para 2 gives the exact area under the curve. More precise is the result of the line representing the phenomenon more loyally. This is generally the curve, especially if carefully driven between the points (R approaching 1).
5. Some points of the curve can be improved for an apparent more precise result (reconciliation). For instance, points of x=0.326 and x=5.643 (to a lesser extent) are "suspicious" and probably need correction. Such corrections can be efficient if the experimental curve is elaborated (at least) twice to check its reproducibility. I have known of reconciliation software, but I assume it can only smooth the curve without physical insight (yet it has value).

Attached Files

•  area.xls   84.5KB   48 downloads

Edited by kkala, 23 August 2011 - 07:11 AM.

[breizh]

http://mathworld.wol...esFormulas.html

Let you consider this resource to learn about the methods and improve your calculation .

Hope this helps

Breizh