Jim asked about the integrability of 2 x sin(1/x) - cos(1/x) if x is not zero, f(x) = 0 if x is zero. The function is Riemann integrable on [0,1]. Consider the intervals from 0 to \eps, and from \eps to 1. By uniform continuity, there is some \delta such that on the larger interval, any partitions of mesh \delta, on a subinterval of \eps to 1, with any associated Riemann sums, will yield values within \eps of one another. Since making \delta even smaller won't hurt, be sure that \delta < \eps. Then on any single interval, of mesh \delta, straddling the division point, the value of the integrand cannot vary by more than, say, 6; so throwing that into a Riemann sum cannot alter the value by more than 6\delta < 6\eps. (I'm taking 3 as an absolute upper bound for the integrand on the entire interval.) Finally, for any partition whatever entirely contained in the left interval, the value of a Riemann sum cannot vary by more than 6 \eps. Thus for any \eps, there is a mesh size \delta such that all Riemann sums are within 13\eps of one another. QED -- Fred Kochman -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of James Propp Sent: Thursday, February 07, 2008 11:11 PM To: math-fun@mailman.xmission.com Subject: [math-fun] Re: another calculus question The sort of function I'm worried about is 2 x sin(1/x) - cos(1/x) if x is not zero, f(x) = 0 if x is zero. This function has an antiderivative, namely x^2 sin(1/x) if x is not zero, F(x) = 0 if x is zero, but is f(x) Riemann integrable? (Note that it isn't continuous.) Jim _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun