11 The Integral CaltechAUTHORS（11积分CaltechAUTHORS）.pdf

11 The IntegralIn this chapter we defme the integral in terms oftranSitions; i.e., by the methodofexhaustion. The reader is assumed to be familiar with the summation notationand its basic properties, as presented in most calculus texts.Piecewise Constant FunctionsIn the theory 'of differentiation, the simplest functions were the linear functionsf(x) = ax + b. We knew that the derivative ofax +b should be a, and we definedthe derivative for more general functions by comparison with the linearfunctions, using the notion of overtaking to make the comparisons.For integration theory, the comparison functions are the piecewiseconstant functions. Roughly speaking, a function f on [a, b] is piecewiseconstant if [a, b] can be broken into a fmite number of subintervals such thatfis constant on each subinterval.Definition A partition of the interval [a, b] is a sequence of numbers(to,t], ... ,tn) such thata = to k2k,--0>-----s~.Thenn m~kiAti = L hAsii=l i=lProof The idea is. to reduce the problem to the simplest case, where thesecond partition is obtained from the fIrst by the addition of a, singlepoint. Let us begin by proving the proposition for this case. Assume, then,that m = n + 1, and that (so,SI, ... ,sm) = (to,tI, ... ,tz,t*,tZ+l, ... ,tn);i.e., the s-partition is obtained from the t-partition by inserting an extrapoint t* between tz and tz+ 1. The relation between s's, ['s, j's, and k's isillustrated in Fig. 11-5. The sum obtained from the s-partition ismL h Asi= h (S1 - so) + ••• +h+l (Sl+l - sf)i=laI II I b\\•A•.,....---J '---v---'-' //toit,rt2tlt. i tl+,l·,·tn _, i tnf = k, f = k2f = k,+, f = knFig. 11·5 The s-partition is finer than the t-partition.152 CHAPTER 11: THE INTEGRALSubstituting the appropriate k's and t's for the j's and s's converts the sumtoWe can combine the two middle terms:kZ+1(t .. - tz) + kZ+1(tZ+1 - t ...) = kZ+1 (t... - tz + tZ+1 - t.)=k1+1 (t1+1 - t1)The sum is nownk1(t1 - to) + ••• + kZ":I(tZ+I - tz) + •.• + knCtn - tn-I) =~kiA.tii=1which is the sum obtained from the t-partition. (For a numerical illustra tion, see Worked Example 2.) This completes the proof for the specialcase.To handle the general case, we observe fIrst that, given two parti tio~s(to, ... ,t".) and (so, ... ,sm), we can find a partition (uo, ... ,up)which contains both of them taking all the points to, ...,tn,so"" ,sm, elim inating duplications, and putting the points in the correct order. (SeeSolved Exercise I, immediately follOwing the end of the proof.) Addingpoints to an adapted partition produces another adapted partition, since ifa function is constant on an interval, it is certainly constant on any subin terval. It follows that the u-partition is adapted to f if the s- and t-parti tions are. Now we can get from the t-partition to the u-partition by addingpoints one at a time. By the special case above, we see that the sum is un changed each time we add a point, so the sum obtained from the u-parti tion equals the sum obtained from the t-partition. In a similar way, we canget from the s-partition to the u-partition by adding one point at a time, sothe sum from the u-purtition equnls the sum from the s-partition. Since thesums from the t- and s-partitions are both equal to the sum from the u-par tition, they are equal to each other, which is what we wanted to prove.Solved Exercises1. Consider s- and t-partitions of [1,8] as follows. Let the s-partition be(1, 2, 3,4, 7,8), and let the t-partition be (1,4,5,6,8). Find the correspond ing u-partition, and show that you can get from the s- and t-partitions to theu-partition by adding one point at a time.UPPER AND LOWER SUMS AND THE DEFINITION 1532. Lel f(t) be defined byf(t) = {~-1ifO~t ... , t-n)for g and create a new function e by lowering g on the interval (to, tl)'Specifically, we let eet) = get) for all t not in (to, tl), and we puteet) = get) - p for tin (to,tl), where p is a positive constant to be chosenin such a way that the integral comes out right. Specifically, if get) = kifor t in (ti-b ti), then e(t) = k1- p for t in (to,tl), and e(t) = kj for t in(ti-l, ti), i;?:: 2. We havebi e(t)dt = (kl-p)D.tl+ k2D.t2+ ••• + knD.tn=k1D.tl+ ••• +knD.tn-pD.tlb=1g(t)dt - P D.t laSetting this equal to c and solving for p givesp~"",),,,,?.'~sf""',i7f-i""':>7""77~>'7"5.""'''>--------7)(51 (no upper sums)Lower sumsIntegralj07fffff?7f77??7ffMf",'?:.?:.~??75ff>577?7~~~~:::")':::',.,,'"» (6) (no lower sums)Upper sums~(----------------7)(7) (no upper or lower sums)Fig. 11-8 Possible configurations of lower and upper sums.Do not conclude from Fig. 11-8 that most functions are nonintegrable. Infact, cases (3), (4), (5), (6), and (7) can occur only whenfis unbounded (seeWorked Example 4). The functions for which Lfand Ufhas a gap between them(case (2» are quite "pathological" (see Solved Exercise 3). In fact, Theorem 3in the next section tells us that integrability is even more common thandifferentiability.Worked Example 4 Show that the set of upper sums for f is none。