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Riemann Sums and the Integral


Date: 01/18/99 at 03:34:41
From: Brian J.
Subject: Fund. Calc Theorem/Riemann Sums

Dear Dr. Math,

I have always wondered why the limit of Riemann sums just happened to 
be the integral? How is that derived? I guess a related question is 
the proof of the fundamental theorem of calculus:

     /b
    |   
    |    =    F(b) - F(a)
   /a
         
One source I had said that the proof involves advanced calculus, so it 
wasn't included, but my textbook has a proof printed. Is it that 
advanced? Thanks.


Date: 01/18/99 at 11:03:48
From: Doctor Anthony
Subject: Re: Fund. Calc Theorem/Riemann 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
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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