Riemann Sums and the IntegralDate: 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/ |
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