
Re: 1.39  What exactly is the Mean value Theorem?
Posted:
Jun 19, 2014 9:51 AM


John Gabriel <thenewcalculus@gmail.com> wrote: > On Thursday, 19 June 2014 13:31:35 UTC+2, Dirk Van de moortel wrote: > >>>> If f is a differentiable function over (a,b), then >>>> {f(b)f(a)}/(ba) is the (natural) average of all the ordinates of >>>> f' over (a,b). > >>> OMG ... no. Really? That is simply hilarious. You're a total >>> fucking moron. > >> It is true. >> Try it out with a simple example such as >> f(x) = x^2 >> a = 4 >> b = 7 >> Dirk Vdm > > ( 7^2  4^2 )/ 3 = (4916)/ 3 = 33/3 = 11 > > The ordinates of f'(x) are given by 2x. The average of the ordinates > of f'(x) on the interval (4,7) is 11, as expected. > > So who is the fucking moron? YOU!
You snipped the part where I said it is true.
Wiz, the average of a function F over an interval (a,b) is defined as A(F(a,b)) = 1/(ba) Int_a^b F(x) dx Take F = f' and verify: A(f'(a,b)) = 1/(ba) Int_a^b f'(x) dx = 1/(ba) ( f(b)  f(a) ) = ( f(b)  f(a) ) / (ba) I assume you didn't notice the apostophe in f'.
Dirk Vdm

