RE: average of a function

[Ted Harding]

On 09-Nov-04 Fred J. wrote:
> Hello
> 
> debian 2.6.8 i686-pc-GNU/Linux
> GNU Octave, version 2.1.60
> 
> I just started with octave, and appreciate some help
> with this simple 
> task, could not get much help from reading the FAQs
> and the manual.
> 
> OK, 
> I need to determine the average value of this function
> which I expect
> it to be -1.620993 but I am getting a different ans
> and not sure if I am
> using octave correctly.
> 
> octave> function y=x(t), y = t^2 - 5*t + 6*cos(pi*t);
> end;
> octave> quad('x', -1, 5/2)
> ans = -5.6735

If I've understood correctly, your code returns the numerical *integral*
of x(t) over the range (-1,2.5). The length of this range is 3.5.

The *average* of the function is the integral divided by the length
of the range. With your numerical answer I get

  (-5.6735)/3.5 = -1.621

which is very close to the value (-1.620993) that you expected to get.

Best wishes,
Ted.


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Date: 09-Nov-04                                       Time: 10:10:21
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