2. More Rational Fun (a) Spend two to three minutes in deep meditation on Darboux's Theorem. Pay ...

2. More Rational Fun (a) Spend two to three minutes in deep meditation on Darboux's Theorem. Pay special attention to the part in bold Darboux's Theorem on Integrability Let A C R" be bounded and let f:A-R be bounded as well. Suppose E is a bounding rectangle of A. Then f is integrable over A and f-1 iff, for every ε > 0, there is a δ > 0, such that for every partition P of E with size Pll < , if R is any point in R, for each subrectangle RE P, we have REP Recall that the size of a partition is the side of the longest edge of the subrectangle that has the longest edge in the partition (b) Let A be the set of all rational ordered pairs inside the rectangle E-[0,1 x [0, and let f:A-Rbe defined by f(x, y) 1. Then the extension f: E R is given by if x, y E Qnl0, 11 1 Explain why, for any partition P, U(f, P) = 1 and L(f,P-0. Is f integrable on A? (c) Let Pn be a uniformly spaced partition of E (so that each subrectangle in P is a square with area (2 D volume)) and for each R E P, let aR be the center point of R Compute, for all n, Σ f(xr)r(R)-1 REP (d) Let us clear the air: Is there any tension or disagreement between our conclusion in (b) and in (c)? Why or why not?