Date: 01/18/99 at 11:03:48
From: Doctor Anthony
Subject: Re: Fund. Calc Theorem/Reinmann Sums
If you draw the curve and consider an element of area dA formed by
verticals through the points x and x+dx along the x axis, then by
elementary geometry (and for a very short interval dx)
dA = f(x) dx where f(x) is the ordinate at point x
So dA/dx = f(x)
Therefore f(x) is the derivative of the area A. Since by definition,
integration is the inverse of differentiation, the integral of f(x)
will be the area.
That means
F'(x) = f(x)
F(x) = INT[f(x) dx]
and on the interval [a,b] this gives
INT[f(x) dx] = F(b) - F(a)
so that F(b) - F(a) is the area under the curve between x = a and
x = b.
Riemann sums, which calculate the areas algebraically (with n ->
infinity), will therefore approach the value of the integral on the
closed interval [a,b].
The advanced calculus (analysis) is mostly in the rather long-winded
justification for algebraic manipulation of infinitesmals. You can
probably leave that to analysts and agree that the basic concept of
Riemann sums is very reasonable.
- Doctor Anthony, The Math Forum