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Riemann Sums and Antiderivatives


Date: 07/11/99 at 22:43:16
From: Dan Kneezel
Subject: Riemann sums and antiderivatives

I've read the answer on Riemann Sums and the Integral:

http://mathforum.org/dr.math/problems/brian1.18.99.html   

While it was informative, I was wondering if there are any other ways 
of illustrating the link between Riemann sums and antiderivatives, 
because even though I know that the area under a curve from a to b is 
equal to F(b) - F(a) (the capital letters indicating antiderivatives), 
it seems neither to be logical nor to make sense.


Date: 07/12/99 at 08:18:12
From: Doctor Anthony
Subject: Re: Riemann sums and antiderivatives

If we work through a typical Riemann sum the general idea should be 
clear.

Consider the area under the curve y = x^2 from  x = 0 to x = a. Divide
the interval from 0 to a into n parts each of length a/n. The area
under the graph on one of these small intervals is given by

     y.(a/n) = x^2.(a/n)

and if we are at the rth interval from the origin then

       x = r(a/n)       so

     x^2 = (r.a/n)^2

So area of this element of area = (r.a/n)^2.(a/n) = r^2.(a/n)^3

We must now sum all the elements of area and allow n -> infinity.

     Total area = (a/n)^3.SUM[r^2] from 1 to n

                = (a/n)^3.(n)(n+1)(2n+1)/6

now take n^3 in denominator and put one n into each bracket in the 
numerator

                = a^3.[(1)(1+1/n)(2+1/n)]/6

and if n -> infinity this reduces to

                  a^3.[1 x 1 x 2]/6  =  a^3/3

If we took the area from x = 0 to x = b the area would be b^3/3 and so
the area between x = a and x = b (assuming a < b) is

     b^3/3 - a^3/3 = F(b) - F(a)   if F(x) = INT(a to b)[x^2.dx] 

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   

    
Associated Topics:
High School Calculus

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