Re: The definite integral.

From: William Hughes (wpihughes_at_hotmail.com)
Date: 03/16/05 

Date: 15 Mar 2005 19:13:47 -0800

David C. Ullrich wrote:

     <snip>

> You've said a few times that one can prove ftc without mvt.
> How do you do that? Without using mvt at _any_ point in the
> argument, I mean. (Like, the fact that f' = 0 implies f is
> constant uses mvt, at least the obvious/standard proof does.)

I am pretty sure I have seen an elegant proof of this, but
I don't have a reference. Perhaps I dreamt it.

However, one can always cheat.

We only need mvt to show f' = 0 implies f constant.

So let f'= 0 on (a,b). Assume wlog that f(0) = 0.
Assume that f is not constant. Then, there is some
c in (a,b) with f(c) <> 0. Draw the line from (a,0) to
(c,f(c)). Note that the slope of this line is not 0.
Consider the case where the slope is positive.
Assume that there is a point d in (a,c) with f(x)
below the line. But now f(x) must cross
the line from below, or be below the line at c.
Either leads to a contradiction.
Assume therfore that f(x) lies above the line.
By continuity of f, there must be a point d in (a,c)
such that f(d) < f(c)/2. Draw a new line from
(d,f(d)) to (c,f(c)). This line also has positive
slope and f(x) must start out below it ( f'(d) = 0 ).
But this means that f(x) must cross the second line from
below, or be below the second line at c. Both
lead to contradictions.
The case for negative slope is similar.

Not only is this as ugly as sin, but it has the same
"feel" as the usual proof of the mvt. Still I can
claim honour is satisfied.

             <snip>

                             - William Hughes

Relevant Pages

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