# You are right, when constants are alone. I used bad terminology. The truth
is so: The integral of c*x^n is c/(n+1) * x^(n+1) for every n (except n=-1;-) )
# So, \int 2*x^2 = 2/3*x^3
# When you are integrating a function over a variable, you consider the
constant terms, (terms which doesn't have the variable in it) so:
# c = c*x^0 (x^0=1), so while integrating according to above formula: \int c
dx = c/1 * x^1 = c*x
# This is where the appended x comes from. Just for clearing the confusion.
-ugur-

On 9/7/07, Tim Mortimer <timmortimer@d2.net.au> wrote:
>
> i think i knew that about the constants et al, but was uncertain about
> making them "fit back in" to the picture in terms of getting a final
> result.... actually, I thought maybe the constants might get an x appended
> to them? (aware they "get discarded" when finding derivitives....)
>
> Thank you Ugur....
>
>
> # I think what you are saying about the perception of tempo is very
> interesting.
> # You want to integrate f(x) = (x*sqrt(1.5)/4)^2 right?
> # This is not more involved than f(x)=x^2 because constants (the
> additional
> scalars & denominators) does not affect the integration. I mean, if the
> integral of f(x) is F(x) then the integral of c*f(x) is c*F(x). In your
> case
>
> Integral[ x^2*(sqrt(1.5)/4)^2 ] = (sqrt(1.5)/4)*Integral[ x^2 ]
> = 1.5/16 Integral [ x^2 ] = 3/32 * [ 1/3 * x^3] = x^3/32
> # But I didn't understand where you get f(x) :-)
> # See you!
> -ugur guney-
>
>
>
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