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/
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