Hi, all. I'm teaching second semester freshman calc (honors), using Stewart's book "Essential Calculus" (which on the whole I like), and a student asked me a question that I don't know the answer to. First, some background: Stewart defines the definite integral in terms of Riemann sums where the widths of the sub-intervals are allowed to be different, and where the sample-points within those sub-intervals are allowed to vary (i.e., they aren't required to be the left endpoints, right endpoints, or mid-points of the sub- intervals). Stewart says that a function f on the interval [a,b] is integrable with definite integral I if for all epsilon > 0 there exists a delta > 0 such that EVERY Riemann sum for f(x) on [a,b] whose mesh (the maximum of the widths of the sub-intervals) is less than delta has value within epsilon of I. The student asked, would it be equivalent to require the sub-intervals to be of equal width (still allowing the sample-points to be arbitrary within the sub-intervals)? I suspect that the answer is yes (though I also suspect that a proof would be more suitable for an upper-division course than a freshman honors course). So my first question is, can anyone come up with a really simple way to see that the two notions of integrability are the same? (Assuming that my instinct is correct and they really are the same.) And my second question is, why would it make sense pedagogically to use sub-intervals of unequal widths as part of the _definition_ of the integral, and then state a _theorem_ that says that if a function is integrable then you might as well use equal-width sub-intervals (which is what Stewart does), rather than _define_ the integral using equal-width sub-intervals, and then state a _theorem_ that says that you can use more general partitions and still get Riemann sums that converge on the correct value, as long as the mesh goes to zero? Jim Propp