Integral Error

[Winzer]

1. The problem statement, all variables and given/known data
I had to do a curve fit on some data and got an equation to the form:


2. Relevant equations
F(t) = a_0 + a_1 t + a_2 t^2


3. The attempt at a solution
Each parameter has an associated uncertainty.
I need to integrate F(t) over a range to get I. How do I find the the uncertainty in delta I with the error for each parameter.

[tiny-tim]

Hi Winzer! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
Each parameter has an associated uncertainty.
I need to integrate F(t) over a range to get I. How do I find the the uncertainty in delta I with the error for each parameter.

I'm not sure what you're asking …

the integral is a0t + a1t2/2 + a2t3/3 + constant …

what is the problem with the uncertainty in that?

[Gib Z]

The op means that he has 3 values for t, experimentally measured so they have some uncertainty. So they could be A, with an error of +/- a, B with an error of +/- b, and C with an error of +/- c, where A,B and C are the measured values, and a,b and c are the uncertainties in measuring them.

To get the error range for I, you need to set up all 3 quadratic equations each fitted to
A+a, B+b, C+c, another to A,B,C and another to A-a, B-b, C-c.

Once you have those 3 equations, the integral for the first one is the upper limit for I, the integral for the second one is the estimate you will be using, and the integral for the 3rd one is the lower limit for I (assuming the parabola is concave up, otherwise the first integral is the lower limit and third one is upper).

[Winzer]

Thanks Gib Z. For some reason I thought there was a quadrature way of doing it.

[Gib Z]

Thanks Gib Z. For some reason I thought there was a quadrature way of doing it.

Ahh actually there is, and it might have been quicker ! =[ Using the Newton-Cotes formula for n=3 (Simpson's Rule) takes away the need to find parabolas fitting points and does the integration automatically! The formula is

\frac{b-a}{6} (f_0 + 4 f_1 + f_2)

with error term of -\frac{(b-a)^5}{2880}\,f^{(4)}(\xi) but these are parabolas anyway so the error term is zero. Hence we can easily see that if \Delta f_n is the maximum uncertainty in measurement, the maximum uncertainty in the integral is :

\pm \frac{b-a}{6} ( \Delta f_0 + 4\Delta f_1 + \Delta f_2 ).

Somewhat an easier process to carry out!