CHAPTER V INTEGRATION, AVERAGE BEHAVIOR A = πr 2.

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1 CHAPTER V INTEGRATION, AVERAGE BEHAVIOR A πr 2. In this hpter we will derive the formul A πr 2 for the re of irle of rdius r. As mtter of ft, we will first hve to settle on extly wht is the definition of the re of region in the plne, nd more subtle thn tht, we must deide wht kinds of regions in the plne hve res. Before we onsider the problem of re, we will develop the notion of the integrl (or verge vlue) of funtion defined on n intervl [, b], whih notion we will use lter to ompute res, one they hve been defined. The min results of this hpter inlude: () The definition of integrbility of funtion, nd the definition of the integrl of n integrble funtion, (2) The Fundmentl Theorem of Clulus (Theorem 5.9), (3) The Integrl Form of Tylor s Reminder Theorem (Theorem 5.2), (4) The Generl Binomil Theorem (Theorem 5.3), (5) The definition of the re of geometri set, (6) A πr 2 (Theorem 5.5), nd (7) The Integrl Test (Theorem 5.7). INTEGRALS OF STEP FUNCTIONS We begin by defining the integrl of ertin (but not ll) bounded, rel-vlued funtions whose domins re losed bounded intervls. Lter, we will extend the definition of integrl to ertin kinds of unbounded omplex-vlued funtions whose domins re still intervls, but whih need not be either losed or bounded. First, we rell from Chpter III the following definitions. DEFINITION. Let [, b] be losed bounded intervl of rel numbers. By prtition of [, b] we men finite set P {x 0 < x <... < x n } of n + points, where x 0 nd x n b. The n intervls {[x i, x i ]} re lled the losed subintervls of the prtition P, nd the n intervls {(x i, x i )} re lled the open subintervls or elements of P. We write P for the mximum of the numbers (lengths of the subintervls) {x i x i }, nd ll P the mesh size of the prtition P. If prtition P {x i } is ontined in nother prtition Q {y j }, i.e., eh x i equls some y j, then we sy tht Q is finer thn P. Let f be funtion on n intervl [, b], nd let P {x 0 <... < x n } be prtition of [, b]. Physiists often onsider sums of the form S P,{yi} f(y i )(x i x i ), where y i is point in the subintervl (x i, x i ). These sums (lled Riemnn sums) re pproximtions of physil quntities, nd the limit of these sums, s the mesh of the prtition beomes smller nd smller, should represent preise vlue of the physil quntity. Wht preisely is ment by the limit of suh sums is lredy subtle question, but even hving deided on wht tht definition should be, it is 2

2 22 V. INTEGRATION, AVERAGE BEHAVIOR s importnt nd diffiult to determine whether or not suh limit exists for mny (or even ny) funtions f. We pproh this question from slightly different point of view, but we will revisit Riemnn sums in the end. Agin we rell from Chpter III the following. DEFINITION. Let [, b] be losed bounded intervl in R. A rel-vlued funtion h : [, b] R is lled step funtion if there exists prtition P {x 0 < x <... < x n } of [, b] suh tht for eh i n there exists number i suh tht h(x) i for ll x (x i, x i ). REMARK. A step funtion h is onstnt on the open subintervls (or elements) of ertin prtition. Of ourse, the prtition is not unique. Indeed, if P is suh prtition, we my dd more points to it, mking lrger prtition hving more subintervls, nd the funtion h will still be onstnt on these new open subintervls. Tht is, given step funtion n be desribed using vrious distint prtitions. Also, the vlues of step funtion t the prtition points themselves is irrelevnt. We only require tht it be onstnt on the open subintervls. Exerise 5.. Let h be step funtion on [, b], nd let P {x 0 < x <... < x n } be prtition of [, b] suh tht h(x) i on the subintervl (x i, x i ) determined by P. () Prove tht the rnge of h is finite set. Wht is n upper bound on the rdinlity of this rnge? (b) Prove tht h is differentible t ll but finite number of points in [, b]. Wht is the vlue of h t suh point? () Let f be funtion on [, b]. Prove tht f is step funtion if nd only if f (x) exists nd 0 for every x (, b) exept possibly for finite number of points. (d) Wht n be sid bout the vlues of h t the endpoints {x i } of the subintervls of P? (e) Let h be step funtion on [, b], nd let j be funtion on [, b] for whih h(x) j(x) for ll x [, b] exept for one point. Show tht j is lso step funtion. (f) If k is funtion on [, b] tht grees with step funtion h exept t finite number of points, 2,..., N, show tht k is lso step funtion. Exerise 5.2. Let [, b] be fixed losed bounded intervl in R, nd let H([, b]) denote the set of ll step funtions on [, b]. () Using Prt () of Exerise 5., prove tht the set H([, b]) is vetor spe of funtions; i.e., it is losed under ddition nd slr multiplition. (b) Show tht H([, b]) is losed under multiplition; i.e., if h, h 2 H([, b]), then h h 2 H([, b]). () Show tht H([, b]) is losed under tking mximum nd minimum nd tht it ontins ll the rel-vlued onstnt funtions. (d) We ll funtion χ n inditor funtion if it equls on n intervl (, d) nd is 0 outside [, d]. To be preise, we will denote this inditor funtion by χ (,d). Prove tht every inditor funtion is step funtion, nd show lso tht every step funtion h is liner ombintion of inditor funtions: h j χ (j,d j). j

3 V. INTEGRATION, AVERAGE BEHAVIOR 23 (e) Define funtion k on [0, ] by setting k(x) 0 if x is rtionl number nd k(x) if x is n irrtionl number. Prove tht the rnge of k is finite set, but tht k is not step funtion. Our first theorem in this hpter is fundmentl onsisteny result bout the re under the grph of step funtion. Of ourse, the grph of step funtion looks like olletion of horizontl line segments, nd the region under this grph is just olletion of retngles. Atully, in this remrk, we re impliitly thinking tht the vlues { i } of the step funtion re positive. If some of these vlues re negtive, then we must re-think wht we men by the re under the grph. We first introdue the following bit of nottion. DEFINITION. Let h be step funtion on the losed intervl [, b]. Suppose P {x 0 < x <... < x n } is prtition of [, b] suh tht h(x) i on the intervl (x i, x i ). Define the weighted verge of h reltive to P to be the number S P (h) defined by S P (h) i (x i x i ). REMARK. Notie the similrity between the formul for weighted verge nd the formul for Riemnn sum. Note lso tht if the intervl is single point, i.e., b, then the only prtition P of the intervl onsists of the single point x 0, nd every weighted verge S P (h) 0. The next theorem is not surprise, lthough its proof tkes some reful thinking. It is simply the ssertion tht the weighted verges re independent of the hoie of prtition. THEOREM 5.. Let h be step funtion on the losed intervl [, b]. Suppose P {x 0 < x <... < x n } is prtition of [, b] suh tht h(x) i on the intervl (x i, x i ), nd suppose Q {y 0 < y <... < y m } is nother prtition of [, b] suh tht h(x) b j on the intervl (y j, y j ). Then the weighted verge of h reltive to P is the sme s the weighted verge of h reltive to Q. Tht is, S P (h) S Q (h). PROOF. Suppose first tht the prtition Q is obtined from the prtition P by dding one dditionl point. Then m n +, nd there exists n i 0 between nd n suh tht () for 0 i i 0 we hve y i x i. (2) x i0 < y i0+ < x i0+. (3) For i 0 < i n we hve x i y i+. In other words, y i0+ is the only point of Q tht is not point of P, nd y i0+ lies stritly between x i0 nd x i0+. Beuse h is onstnt on the intervl (x i0, x i0+) (y i0, y i0+2), it follows tht () For i i 0, i b i. (2) b i0+ b i0+2 i0+. (3) For i 0 + i n, i b i+.

4 24 V. INTEGRATION, AVERAGE BEHAVIOR So, S P (h) i (x i x i ) i 0 + i 0 + i 0 + i 0 + i (x i x i ) + i0+(x i0+ x i0 ) ii 0+2 i (x i x i ) b i (y i y i ) + i0+(y i0+2 y i0 ) ii 0+2 b i+ (y i+ y i ) b i (y i y i ) + i0+(y i0+2 y i0+ + y i0+ y i0 ) n+ ii 0+3 b i (y i y i ) b i (y i y i ) + b i0+(y i0+ y i0 ) + b i0+2(y i0+2 y i0+) m ii 0+3 m b i (y i y i ) S Q (h), b i (y i y i ) whih proves the theorem in this speil se where Q is obtined from P by dding just one more point. It follows esily now by indution tht if Q is obtined from P by dding ny finite number of dditionl points, then h is onstnt on eh of the open subintervls determined by Q, nd S Q (h) S P (h). Finlly, let Q {y 0 < y <... < y m } be n rbitrry prtition of [, b], for whih h is onstnt on eh of the open subintervls (y j, y j ) determined by Q. Define R to be the prtition of [, b] obtined by tking the union of the prtition points {x i } nd {y j }. Then R is prtition of [, b] tht is obtined by dding finite number of points to the prtition P, whene S R (h) S P (h). Likewise, R is obtined from the prtition Q by dding finite number of points, whene S R (h) S Q (h), nd this proves tht S Q (h) S P (h), s desired. DEFINITION. Let [, b] be fixed losed bounded intervl in R. We define the integrl of step funtion h on [, b], nd denote it by h, s follows: If P {x 0 < x <... < x n } is prtition of [, b], for whih h(x) i for ll x (x i, x i ),

5 V. INTEGRATION, AVERAGE BEHAVIOR 25 then h S P (h) i (x i x i ). REMARK. The integrl of step funtion h is defined to be the weighted verge of h reltive to prtition P of [, b]. Notie tht the preeding theorem is ruil in order tht this definition of h be unmbiguously defined. The integrl of step funtion should not depend on whih prtition is used. Theorem 5. sserts preisely this ft. Note lso tht if the intervl is single point, i.e., b, then the integrl of every step funtion h is 0. We use vriety of nottions for the integrl of h : h b h b h(t) dt. The following exerise provides very useful wy of desribing the integrl of step funtion. Not only does it show tht the integrl of step funtion looks like Riemnn sum, but it provides desription of the integrl tht mkes ertin lultions esier. See, for exmple, the proof of the next theorem. Exerise 5.3. Suppose h is step funtion on [, b] nd tht R {z 0 < z <... < z n } is prtition of [, b] for whih h is onstnt on eh subintervl (z i, z i ) of R. () Prove tht h S R (h) h(w i )(z i z i ), where, for eh i n, w i is ny point in (z i, z i ). (Note then tht the integrl of step funtion tkes the form of Riemnn sum.) (b) Show tht h is independent of the vlues of h t the points {z i } of the prtition R. Exerise 5.4. Let h nd h 2 be two step funtions on [, b]. () Suppose tht h (x) h 2 (x) for ll x [, b] exept for one point. Prove tht h h 2. HINT: Let P be prtition of [, b], for whih both h nd h 2 re onstnt on its open subintervls, nd for whih is one of the points of P. Now use the preeding exerise to lulte the two integrls. (b) Suppose h (x) h 2 (x) for ll but finite number of points,..., N [, b]. Prove tht h h 2. We hve used the terminology weighted verge of step funtion reltive to prtition P. The next exerise shows how the integrl of step funtion n be relted to n tul verge vlue of the funtion. Exerise 5.5. Let h be step funtion on the losed intervl [, b], nd let P {x 0 < x <... < x n } be prtition of [, b] for whih h(x) i on the intervl (x i, x i ). Let us think of the intervl [, b] s n intervl of time, nd suppose tht the funtion h ssumes the vlue i for the intervl of time between x i nd

6 26 V. INTEGRATION, AVERAGE BEHAVIOR x i. Show tht the verge vlue A(h) tken on by h throughout the entire intervl ([, b]) of time is given by h A(h) b. THEOREM 5.2. Let H([, b]) denote the vetor spe of ll step funtions on the losed intervl [, b]. Then the ssignment h h of H([, b]) into R hs the following properties: () (Linerity) H([, b]) is vetor spe. Furthermore, (h + h 2 ) h + h2, nd h h for ll h, h 2, h H([, b]), nd for ll rel numbers. (2) If h n iχ (i,d i) is liner ombintion of inditor funtions (See prt (d) of Exerise 5.2), then h n i(d i i ). (3) (Positivity) If h(x) 0 for ll x [, b], then h 0. (4) (Order-preserving) If h nd h 2 re step funtions for whih h (x) h 2 (x) for ll x [, b], then h h 2. PROOF. Tht H([, b]) is vetor spe ws proved in prt () of Exerise 5.2. Suppose P {x 0 < x <... < x n } is prtition of [, b] suh tht h (x) is onstnt for ll x (x i, x i ), nd suppose Q {y 0 < y <... < y m } is prtition of [, b] suh tht h 2 (x) is onstnt for ll x (y j, y j ). Let R {z 0 < z <... < z r } be the prtition of [, b] obtined by tking the union of the x i s nd the y j s. Then h nd h 2 re both onstnt on eh open subintervl of R, sine eh suh subintervl is ontined in some open subintervl of P nd lso is ontined in some open subintervl of Q. Therefore, h + h 2 is onstnt on eh open subintervl of R. Now, using Exerise 5.3, we hve tht (h + h 2 ) r ((h + h 2 )(w k ))(z k z k ) k r h (w k )(z k z k ) + k h + h 2. r h 2 (w k )(z k z k ) k This proves the first ssertion of prt (). Next, let P {x 0 < x <... < x n } be prtition of [, b] suh tht h(x) is onstnt on eh open subintervl of P. Then h(x) is onstnt on eh open subintervl of P, nd using Exerise 5.3 gin, we hve tht (h) h(w i )(x i x i ) h(w i )(x i x i ) h, whih ompletes the proof of the other hlf of prt ().

7 V. INTEGRATION, AVERAGE BEHAVIOR 27 To see prt (2), we need only verify tht χ (i,d i) d i i, for then prt (2) will follow from prt (). But χ (i,d i) is just step funtion determined by the four point prtition {, i, d i, b} nd vlues 0 on (, i ) nd (d i, b) nd on ( i, d i ). Therefore, we hve tht χ (i,d i) d i i. If h(x) 0 for ll x, nd P {x 0 < x <... < x n } is s bove, then h h(w i )(x i x i ) 0, nd this proves prt (3). Finlly, suppose h (x) h 2 (x) for ll x [, b]. By Exerise 5.2, we know tht the funtion h 3 h 2 h is step funtion on [, b]. Also, h 3 (x) 0 for ll x [, b]. So, by prt (3), h 3 0. Then, by prt (), 0 h 3 (h 2 h ) h 2 h, whih implies tht h h 2, s desired. Exerise 5.6. () Let h be the onstnt funtion on [, b]. Show tht h (b ). (b) Let < < d < b be rel numbers, nd let h be the step funtion on [, b] tht equls r for < x < d nd 0 otherwise. Prove tht b h(t) dt r(d ). () Let h be step funtion on [, b]. Prove tht h is step funtion, nd tht h h. HINT: Note tht h (x) h(x) h (x). Now use the preeding theorem. (d) Suppose h is step funtion on [, b] nd tht is onstnt for whih h(x) for ll x [, b]. Prove tht h (b ). INTEGRABLE FUNCTIONS We now wish to extend the definition of the integrl to wider lss of funtions. This lss will onsist of those funtions tht re uniform limits of step funtions. The requirement tht these limits be uniform is ruil. Pointwise limits of step funtions doesn t work, s we will see in Exerise 5.7 below. The initil step in rrying out this generliztion is the following. THEOREM 5.3. Let [, b] be losed bounded intervl, nd let {h n } be sequene of step funtions tht onverges uniformly to funtion f on [, b]. Then the sequene { h n } is onvergent sequene of rel numbers. PROOF. We will show tht { h n } is Cuhy sequene in R. Thus, given n ɛ > 0, hoose n N suh tht for ny n N nd ny x [, b], we hve f(x) h n (x) < ɛ 2(b ). Then, for ny m nd n both N nd ny x [, b], we hve h n (x) h m (x) h n (x) f(x) + f(x) h m (x) < ɛ b.

8 28 V. INTEGRATION, AVERAGE BEHAVIOR Therefore, h n h m (h n h m ) h n h m ɛ b ɛ, s desired. The preeding theorem provides us with perfetly good ide of how to define the integrl of funtion f tht is the uniform limit of sequene of step funtions. However, we first need to estblish nother kind of onsisteny result. THEOREM 5.4. If {h n } nd {k n } re two sequenes of step funtions on [, b], eh onverging uniformly to the sme funtion f, then lim h n lim k n. PROOF. Given ɛ > 0, hoose N so tht if n N, then h n (x) f(x) < ɛ/(2(b )) for ll x [, b], nd suh tht f(x) k n (x) < ɛ/(2(b )) for ll x [, b]. Then, h n (x) k n (x) < ɛ/(b ) for ll x [, b] if n N. So, h n k n h n k n ɛ b ɛ if n N. Tking limits gives lim h n lim k n ɛ. Sine this is true for rbitrry ɛ > 0, it follows tht lim h n lim k n, s desired. DEFINITION. Let [, b] be losed bounded intervl of rel numbers. A funtion f : [, b] R is lled integrble on [, b] if it is the uniform limit of sequene {h n } of step funtions. Let I([, b]) denote the set of ll funtions tht re integrble on [, b]. If f I([, b]), define the integrl of f, denoted f, by f lim where {h n } is some (ny) sequene of step funtions tht onverges uniformly to f on [, b]. h n, As in the se of step funtions, we use the following nottions: f b f b f(t) dt. REMARK. Note tht Theorem 5.4 is ruil in order tht this definition be unmbiguous. Indeed, we will see below tht this ritil onsisteny result is one ple where uniform limits of step funtions works while pointwise limits do not. See prts () nd (d) of Exerise 5.7. Note lso tht it follows from this definition tht f 0, beuse h 0 for ny step funtion. In ft, we will derive lmost

9 V. INTEGRATION, AVERAGE BEHAVIOR 29 everything bout the integrl of generl integrble funtion from the orresponding results bout the integrl of step funtion. No surprise. This is the essene of mthemtil nlysis, pproximtion. Exerise 5.7. Define funtion f on the losed intervl [0, ] by f(x) if x is rtionl number nd f(x) 0 if x is n irrtionl number. () Suppose h is step funtion on [0, ]. Prove tht there must exist n x [0, ] suh tht f(x) h(x) /2. HINT: Let (x i, x i ) be n intervl on whih h is onstnt. Now use the ft tht there re both rtionls nd irrtionls in this intervl. (b) Prove tht f is not the uniform limit of sequene of step funtions. Tht is, f is not n integrble funtion. () Consider the two sequenes {h n } nd {k n } of step funtions defined on the intervl [0, ] by h n χ (0,/n), nd k n nχ (0,/n). Show tht both sequenes {h n } nd {k n } onverge pointwise to the 0 funtion on [0, ]. HINT: All funtions re 0 t x 0. For x > 0, hoose N so tht /N < x. Then, for ny n N, h n (x) k n (x) 0. (d) Let h n nd k n be s in prt (). Show tht lim h n 0, but lim k n. Conlude tht the onsisteny result in Theorem 5.4 does not hold for pointwise limits of step funtions. Exerise 5.8. Define funtion f on the losed intervl [0, ] by f(x) x. () For eh positive integer n, let P n be the prtition of [0, ] given by the points {0 < /n < 2/n < 3/n <... < (n )/n < }. Define step funtion h n on [0, ] by setting h n (x) i/n if i n < x < i n, nd h n(i/n) i/n for ll 0 i n. Prove tht f(x) h n (x) < /n for ll x [0, ], nd then onlude tht f is the uniform limit of the h n s whene f I([0, ]). (b) Show tht h n () Show tht f(t) dt /2. 0 i n(n + ) n2 2n 2. The next exerise estblishes some dditionl properties of integrble funtions on n intervl [, b]. Exerise 5.9. Let [, b] be losed nd bounded intervl, nd let f be n element of I([, b]). () Show tht, for eh ɛ > 0 there exists step funtion h on [, b] suh tht f(x) h(x) < ɛ for ll x [, b]. (b) For eh positive integer n let h n be step funtion stisfying the onlusion of prt () for ɛ /n. Define k n h n /n nd l n h n + /n. Show tht k n nd l n re step funtions, tht k n (x) < f(x) < l n (x) for ll x [, b], nd tht l n (x) k n (x) l n (x) k n (x) 2/n for ll x. Hene, b (l n k n ) 2 n (b ). () Conlude from prt (b) tht, given ny ɛ > 0, there exist step funtions k nd l suh tht k(x) f(x) l(x) for whih (l(x) k(x)) < ɛ. (d) Prove tht there exists sequene {j n } of step funtions on [, b], for whih j n (x) j n+ (x) f(x) for ll x, tht onverges uniformly to f. Show lso tht there exists sequene {j n} of step funtions on [, b], for whih j n(x) j n+(x) f(x) for ll x, tht onverges uniformly to f. Tht is, if f I([, b]), then f is the

10 30 V. INTEGRATION, AVERAGE BEHAVIOR uniform limit of nonderesing sequene of step funtions nd lso is the uniform limit of noninresing sequene of step funtions. HINT: To onstrut the j n s nd j n s, use the step funtions k n nd l n of prt (b), nd rell tht the mximum nd minimum of step funtions is gin step funtion. (e) Show tht if f(x) 0 for ll x [, b], nd g is defined by g(x) f(x), then g I([, b]). HINT: Write f lim h n where h n (x) 0 for ll x nd n. Then use prt (g) of Exerise (f) (Riemnn sums gin.) Show tht, given n ɛ > 0, there exists prtition P suh tht if Q {x 0 < x <... < x n } is ny prtition finer thn P, nd {w i } re ny points for whih w i (x i, x i ), then b f(t) dt f(w i )(x i x i ) < ɛ. HINT: Let P be prtition for whih both the step funtions k nd l of prt () re onstnt on the open subintervls of P. Verify tht for ny finer prtition Q, l(w i ) f(w i ) k(w i ), nd hene l(w i )(x i x i ) i i f(w i )(x i x i ) i k(w i )(x i x i ). DEFINITION. A bounded rel-vlued funtion f on losed bounded intervl [, b] is lled Riemnn-integrble if, given ny ɛ > 0, there exist step funtions k nd l, on [, b] for whih k(x) f(x) l(x) for ll x, suh tht (l k) < ɛ. We denote the set of ll funtions on [, b] tht re Riemnn-integrble by I R ([, b]). REMARK. The notion of Riemnn-integrbility ws introdued by Riemnn in the mid nineteenth entury nd ws the first forml definition of integrbility. Sine then severl other definitions hve been given for n integrl, ulminting in the theory of Lebesgue integrtion. The definition of integrbility tht we re using in this book is slightly different nd less generl from tht of Riemnn, nd both of these re very different nd less generl from the definition given by Lebesgue in the erly twentieth entury. Prt () of Exerise 5.9 bove shows tht the funtions we re lling integrble re neessrily Riemnn-integrble. We will see in Exerise 5.0 tht there re Riemnn-integrble funtions tht re not integrble in our sense. In both ses, Riemnn s nd ours, n integrble funtion f must be trpped between two step funtions k nd l. In our definition, we must hve l(x) k(x) < ɛ for ll x [, b], while in Riemnn s definition, we only need tht l k < ɛ. The distintion is tht smll step funtion must hve smll integrl, but it isn t neessry for step funtion to be (uniformly) smll in order for it to hve smll integrl. It only hs to be smll on most of the intervl [, b]. On the other hnd, ll the definitions of integrbility on [, b] inlude mong the integrble funtions the ontinuous ones. And, ll the different definitions of integrl give the sme vlue to ontinuous funtion. The differenes then in these definitions shows up t the point of sying extly whih funtions re integrble. Perhps the most enlightening thing to sy in this onnetion is tht it is impossible to mke good definition of integrbility in suh wy tht every funtion is

11 V. INTEGRATION, AVERAGE BEHAVIOR 3 integrble. Subtle points in set theory rise in suh ttempts, nd mny fsinting nd deep mthemtil ides hve ome from them. However, we will stik with our definition, sine it is simpler thn Riemnn s nd is ompletely suffiient for our purposes. THEOREM 5.5. Let [, b] be fixed losed nd bounded intervl, nd let I([, b]) denote the set of integrble funtions on [, b]. Then: () Every element of I([, b]) is bounded funtion. Tht is, integrble funtions re neessrily bounded funtions. (2) I([, b]) is vetor spe of funtions. (3) I([, b]) is losed under multiplition; i.e., if f nd g I([, b]), then fg I([, b]). (4) Every step funtion is in I([, b]). (5) If f is ontinuous rel-vlued funtion on [, b], then f is in I([, b]). Tht is, every ontinuous rel-vlued funtion on [, b] is integrble on [, b]. PROOF. Let f I([, b]), nd write f lim h n, where {h n } is sequene of step funtions tht onverges uniformly to f. Given the positive number ɛ, hoose N so tht f(x) h N (x) < for ll x [, b]. Then f(x) h N (x) + for ll x [, b]. Beuse h N is step funtion, its rnge is finite set, so tht there exists number M for whih h N (x) M for ll x [, b]. Hene, f(x) M + for ll x [, b], nd this proves prt (). Next, let f nd g be integrble, nd write f lim h n nd g lim k n, where {h n } nd {k n } re sequenes of step funtions tht onverge uniformly to f nd g respetively. If s nd t re rel numbers, then the sequene {sh n + tk n } onverges uniformly to the funtion sf + tg. See prts () nd (d) of Exerise Therefore, sf + tg I([, b]), nd I([, b]) is vetor spe, proving prt (2). Note tht prt (3) does not follow immeditely from Exerise 3.28; the produt of uniformly onvergent sequenes my not be uniformly onvergent. To see it for this se, let f lim h n nd g lim k n be elements of I([, b]). By prt (), both f nd g re bounded, nd we write M f nd M g for numbers tht stisfy f(x) M f nd g(x) M g for ll x [, b]. Beuse the sequene {k n } onverges uniformly to g, there exists n N suh tht if n N we hve g(x) k n (x) < for ll x [, b]. This implies tht, if n N, then k n (x) M g + for ll x [, b]. Now we show tht fg is the uniform limit of the sequene h n k n. For, if n N, then f(x)g(x) h n (x)k n (x) f(x)g(x) f(x)k n (x) + f(x)k n (x) h n (x)k n (x) f(x) g(x) k n (x) + k n (x) f(x) h n (x) M f g(x) k n (x) + (M g + ) f(x) h n (x), whih implies tht fg lim(h n k n ). If h is itself step funtion, then it is obviously the uniform limit of the onstnt sequene {h}, whih implies tht h is integrble. Finlly, if f is ontinuous on [, b], it follows from Theorem 3.20 tht f is the uniform limit of sequene of step funtions, whene f I([, b]). Exerise 5.0. Let f be the funtion defined on [0, ] by f(x) sin(/x) if x 0 nd f(0) 0.

12 32 V. INTEGRATION, AVERAGE BEHAVIOR () Show tht f is ontinuous t every nonzero x nd disontinuous t 0. HINT: Observe tht, on ny intervl (0, δ), the funtion sin(/x) ttins both the vlues nd. (b) Show tht f is not integrble on [0, ]. HINT: Suppose f lim h n. Choose N so tht f(x) h N (x) < /2 for ll x [0, ]. Let P be prtition for whih h N is onstnt on its open subintervls, nd exmine the sitution for x s in the intervl (x 0, x ). () Show tht f is Riemnn-integrble on [0, ]. Conlude tht I([, b]) is proper subset of I R ([, b]). Exerise 5.. () Let f be n integrble funtion on [, b]. Suppose g is funtion for whih g(x) f(x) for ll x [, b] exept for one point. Prove tht g is integrble nd tht g f. HINT: If f lim h n, define k n (x) h n (x) for ll x nd k n () g(). Then use Exerise 5.4. (b) Agin, let f be n integrble funtion on [, b]. Suppose g is funtion for whih g(x) f(x) for ll but finite number of points,..., N [, b]. Prove tht g I([, b]), nd tht g f. () Suppose f is funtion on the losed intervl [, b], tht is uniformly ontinuous on the open intervl (, b). Prove tht f is integrble on [, b]. HINT: Just reprodue the proof to Theorem REMARK. In view of prt (b) of the preeding exerise, we see tht whether funtion f is integrble or not is totlly independent of the vlues of the funtion t fixed finite set of points. Indeed, the funtion needn t even be defined t fixed finite set of points, nd still it n be integrble. This observtion is helpful in mny instnes, e.g., in prts (d) nd (e) of Exerise 5.2. THEOREM 5.6. The ssignment f f on I([, b]) stisfies the following properties. () (Linerity) I([, b]) is vetor spe, nd (αf + βg) α f + β g for ll f, g I([, b])nd α, β R. (2) (Positivity) If f(x) 0 for ll x [, b], then f 0. (3) (Order-preserving) If f, g I([, b]) nd f(x) g(x) for ll x [, b], then f g. (4) If f I([, b]), then so is f, nd f f. (5) If f is the uniform limit of funtions f n, eh of whih is in I([, b]), then f I([, b]) nd f lim f n. (6) Let {u n } be sequene of funtions in I([, b]). Suppose tht for eh n there is number m n, for whih u n (x) m n for ll x [, b], nd suh tht the infinite series m n onverges. Then the infinite series u n onverges uniformly to n integrble funtion, nd u n u n. PROOF. Tht I([, b]) is vetor spe ws proved in prt (2) of Theorem 5.5. Let f nd g be in I([, b]), nd write f lim h n nd g lim k n, where the h n s nd the k n s re step funtions. Then αf + βg lim(αh n + βk n ), so tht, by Theorem

13 V. INTEGRATION, AVERAGE BEHAVIOR nd the definition of the integrl, we hve (αf + βg) lim (αh n + βk n ) lim(α h n + β k n ) α lim h n + β lim α f + β g, whih proves prt (). Next, if f I([, b]) stisfies f(x) 0 for ll x [, b], let {l n } be noninresing sequene of step funtions tht onverges uniformly to f. See prt (d) of Exerise 5.9. Then l n (x) f(x) 0 for ll x nd ll n. So, gin by Theorem 5.2, we hve tht f lim l n 0. This proves prt (2). Prt (3) now follows by ombining prts () nd (2) just s in the proof of Theorem 5.2. To see prt (4), let f I([, b]) be given. Write f lim h n. Then f lim h n. For f(x) h n (x) f(x) h n (x). k n Therefore, f is integrble. Also, f lim h n lim h n lim h n f. To see prt (5), let {f n } be sequene of elements of I([, b]), nd suppose tht f lim f n. For eh n, let h n be step funtion on [, b] suh tht f n (x) h n (x) < /n for ll x [, b]. Note lso tht it follows from prts (3) nd (4) tht f n Now {h n } onverges uniformly to f. For, h n < b n. f(x) h n (x) f(x) f n (x) + f n (x) h n (x) < f(x) f n (x) + n, showing tht f lim h n. Therefore, f I([, b]). Moreover, f lim h n. Finlly, f lim fn, for f f n f f h n + h n + b n. h n f n

14 34 V. INTEGRATION, AVERAGE BEHAVIOR This ompletes the proof of prt (5). Prt (6) follows diretly from prt (5) nd the Weierstrss M Test (Theorem 3.8). For, prt () of tht theorem implies tht the infinite series u n onverges uniformly, nd then u n u n follows from prt (5) of this theorem. As finl extension of our notion of integrl, we define the integrl of ertin omplex-vlued funtions. DEFINITION. Let [, b] be fixed bounded nd losed intervl. A omplexvlued funtion f u + iv is lled integrble if its rel nd imginry prts u nd v re integrble. In this se, we define b f b (u + iv) b u + i b v. THEOREM 5.7. () The set of ll integrble omplex-vlued funtions on [, b] is vetor spe over the field of omplex numbers, nd b (αf + βg) α b f + β b g for ll integrble omplex-vlued funtions f nd g nd ll omplex numbers α nd β. (2) If f is n integrble omplex-vlued funtion on [, b], then so is f, nd b f b f. PROOF. We leve the verifition of prt () to the exerise tht follows. To see prt (2), suppose tht f is integrble, nd write f u + iv. Then f u2 + v 2, so tht f is integrble by Theorem 5.5 nd prt (e) of Exerise 5.9. Now write z b f, nd write z in polr oordintes s z reiθ, where r z b f. (See Exerise 4.23.) Define funtion g by g(x) e iθ f(x) nd notie tht g f. Then b g e iθ b f r, whih is rel number. Writing g û + i v,

15 V. INTEGRATION, AVERAGE BEHAVIOR 35 we then hve tht r û + i v, implying tht v 0. So, b f r b b b g b b b b û + i û û û g f, b v s desired. Exerise 5.2. Prove prt () of the preeding theorem. HINT: Brek α, β, f, nd g into rel nd imginry prts. THE FUNDAMENTAL THEOREM OF CALCULUS We begin this setion with result tht is ertinly not surprise, but we will need it t vrious ples in lter proofs, so it s good to stte it preisely now. THEOREM 5.8. Suppose f I([, b]), nd suppose < < b. Then f I([, ]), f I([, b]), nd b f f + b f. PROOF. Suppose first tht h is step funtion on [, b], nd let P {x 0 < x <... < x n } be prtition of [, b] suh tht h(x) i on the subintervl (x i, x i ) of P. Of ourse, we my ssume without loss of generlity tht is one of the points of P, sy x k. Clerly h is step funtion on both intervls [, ] nd [, b]. Now, let Q { x 0 < x <... < x k } be the prtition of [, ] obtined by interseting P with [, ], nd let Q 2 { x k < x k+ <... < x n b} be the

16 36 V. INTEGRATION, AVERAGE BEHAVIOR prtition of [, b] obtined by interseting P with [, b]. We hve tht b h S P (h) i (x i x i ) k i (x i x i ) + S Q (h) + S Q2 (h) h + b h, ik+ i (x i x i ) whih proves the theorem for step funtions. Now, write f lim h n, where eh h n is step funtion on [, b]. Then lerly f lim h n on [, ], whih shows tht f I([, ]), nd f lim h n. Similrly, f lim h n on [, b], showing tht f I([, b]), nd Finlly, s desired. b b f lim lim( lim f lim b h n f + b h n + h n. b h n + lim b f, h n ) b h n I s time for the trumpets gin! Wht we ll the Fundmentl Theorem of Clulus ws disovered by Newton nd Leibniz more or less simultneously in the seventeenth entury, nd it is without doubt the ornerstone of ll we ll mthemtil nlysis tody. Perhps the min theoretil onsequene of this theorem is tht it provides proedure for inventing new funtions. Polynomils re rther nturl funtions, power series re simple generliztion of polynomils, nd then wht? It ll me down to thinking of funtion of vrible x s being the re beneth urve between fixed point nd the vrying point x. By now, we hve polished nd mssged these ides into reful, detiled development of the subjet, whih hs substntilly obsured the originl ingenious insights of Newton nd Leibniz. On the other hnd, our development nd proofs re omplete, while theirs were bsed hevily on their intuition. So, here it is.

17 V. INTEGRATION, AVERAGE BEHAVIOR 37 THEOREM 5.9. (Fundmentl Theorem of Clulus) Suppose f is n rbitrry element of I([, b]). Define funtion F on [, b] by F (x) x f. Then: () F is ontinuous on [, b], nd F () 0. (2) If f is ontinuous t point (, b), then F is differentible t nd F () f(). (3) Suppose tht f is ontinuous on [, b]. If G is ny ontinuous funtion on [, b] tht is differentible on (, b) nd stisfies G (x) f(x) for ll x (, b), then b f(t) dt G(b) G(). REMARK. Prt (2) of this theorem is the hert of it, the gret disovery of Newton nd Leibniz, lthough most beginning lulus students often think of prt (3) s the min sttement. Of ourse it is tht third prt tht enbles us to tully ompute integrls. PROOF. Beuse f I([, b]), we know tht f I([, x]) for every x [, b], so tht F (x) t lest is defined. Also, we know tht f is bounded; i.e., there exists n M suh tht f(t) M for ll t [, b]. Then, if x, y [, b] with x y, we hve tht F (x) F (y) x y x y x y x y f f + f f M M(x y), y x so tht F (x) F (y) M x y < ɛ if x y < δ ɛ/m. This shows tht F is (uniformly) ontinuous on [, b]. Obviously, F () f 0, nd prt () is proved. Next, suppose tht f is ontinuous t (, b), nd write L f(). Let ɛ > 0 be given. To show tht F is differentible t nd tht F () f(), we must find δ > 0 suh tht if 0 < h < δ then y f f F ( + h) F () L < ɛ. h Sine f is ontinuous t, hoose δ > 0 so tht f(t) f() < ɛ if t < δ. Now, y f

18 38 V. INTEGRATION, AVERAGE BEHAVIOR ssuming tht h > 0 for the moment, we hve tht F ( + h) F () +h f + +h f f, +h f f f nd So, if 0 < h < δ, then L +h L. h F ( + h) F () h L ɛ, +h f(t) dt h +h +h (f(t) L) dt h +h f(t) L dt h +h f(t) f() dt h +h ɛ h L h where the lst inequlity follows beuse for t [, +h], we hve tht t h < δ. A similr rgument holds if h < 0. (See the following exerise.) This proves prt (2). Suppose finlly tht G is ontinuous on [, b], differentible on (, b), nd tht G (x) f(x) for ll x (, b). Then, F G is ontinuous on [, b], differentible on (, b), nd by prt (2) (F G) (x) F (x) G (x) f(x) f(x) 0 for ll x (, b). It then follows from Exerise 4.2 tht F G is onstnt funtion C, whene, G(b) G() F (b) + C F () C F (b) nd the theorem is proved. b f(t) dt, Exerise 5.3. () Complete the proof of prt (2) of the preeding theorem; i.e., tke re of the se when h < 0. HINT: In this se, < + h <. Then, write f +h f + +h f. (b) Suppose f is ontinuous funtion on the losed intervl [, b], nd tht f exists nd is ontinuous on the open intervl (, b). Assume further tht f is integrble on the losed intervl [, b]. Prove tht f(x) f() x f for ll

19 V. INTEGRATION, AVERAGE BEHAVIOR 39 x [, b]. Be reful to understnd how this is different from the Fundmentl Theorem. () Use the Fundmentl Theorem to prove tht for x we hve nd for 0 < x < we hve ln(x) F (x) x t dt, ln(x) F (x) x t dt. HINT: Show tht these two funtions hve the sme derivtive nd gree t x. CONSEQUENCES OF THE FUNDAMENTAL THEOREM The first two theorems of this setion onstitute the bsi tehniques of integrtion tught in lulus ourse. However, the reful formultions of these stndrd methods of evluting integrls hve some subtle points, i.e., some hypotheses. Clulus students re rrely told bout these detils. THEOREM 5.0. (Integrtion by Prts Formul) Let f nd g be integrble funtions on [, b], nd s usul let F nd G denote the funtions defined by Then b F (x) x Or, relling tht f F nd g G, b f, nd G(x) x fg [F (b)g(b) F ()G()] F G [F (b)g(b) F ()G()] g. b b F g. F G. Exerise 5.4. () Prove the preeding theorem. HINT: Reple the upper limit b by vrible x, nd differentite both sides. By the wy, how do we know tht the funtions F g nd fg re integrble? (b) Suppose f nd g re integrble funtions on [, b] nd tht both f nd g re ontinuous on (, b) nd integrble on [, b]. (Of ourse f nd g re not even defined t the endpoints nd b, but they n still be integrble on [, b]. See the remrk following Exerise 5..) Prove tht b fg [f(b)g(b) f()g()] b f g.

20 40 V. INTEGRATION, AVERAGE BEHAVIOR THEOREM 5.. (Integrtion by Substitution) Let f be ontinuous funtion on [, b], nd suppose g is ontinuous, one-to-one funtion from [, d] onto [, b] suh tht g is ontinuously differentible on (, d), nd suh tht g() nd b g(d). Assume finlly tht g is integrble on [, d]. Then b f(t) dt d f(g(s))g (s) ds. PROOF. It follows from our ssumptions tht the funtion f(g(s))g (s) is ontinuous on (, b) nd integrble on [, d]. It lso follows from our ssumptions tht g mps the open intervl (, d) onto the open intervl (, b). As usul, let F denote the funtion on [, b] defined by F (x) x f(t) dt. Then, by prt (2) of the Fundmentl Theorem, F is differentible on (, b), nd F f. Then, by the hin rule, F g is ontinuous nd differentible on (, d) nd (F g) (s) F (g(s))g (s) f(g(s))g (s). So, by prt (3) of the Fundmentl Theorem, we hve tht d whih finishes the proof. f(g(s))g (s) ds d (F g) (s) ds (F g)(d) (F g)() F (g(d)) F (g()) F (b) F () b f(t) dt, Exerise 5.5. () Prove the Men Vlue Theorem for integrls: If f is ontinuous on [, b], then there exists (, b) suh tht b f(t) dt f()(b ). (b) (Uniform limits of differentible funtions. Compre with Exerise 4.26.) Suppose {f n } is sequene of ontinuous funtions on losed intervl [, b] tht onverges pointwise to funtion f. Suppose tht eh derivtive f n is ontinuous on the open intervl (, b), is integrble on the losed intervl [, b], nd tht the sequene {f n} onverges uniformly to funtion g on (, b). Prove tht f is differentible on (, b), nd f g. HINT: Let x be in (, b), nd let be in the intervl (, x). Justify the following equlities, nd use them together with the Fundmentl Theorem to mke the proof. f(x) f() lim(f n (x) f n ()) lim x f n We revisit now the Reminder Theorem of Tylor, whih we first presented in Theorem 4.9. The point is tht there is nother form of this theorem, the integrl form, nd this version is more powerful in some instnes thn the originl one, e.g., in the generl Binomil Theorem below. x g.

21 V. INTEGRATION, AVERAGE BEHAVIOR 4 THEOREM 5.2. (Integrl Form of Tylor s Reminder Theorem) Let be rel number, nd let f hve n + derivtives on ( r, + r), nd suppose tht f (n+) I([ r, + r]). Then for eh < x < + r, f(x) T n (f,) (x) x where Tf n denotes the nth Tylor polynomil for f. Similrly, for r < x <, f(x) T n (f,) (x) x f (n+) (x t)n (t) dt, n! f (n+) (x t)n (t) dt. n! Exerise 5.6. Prove the preeding theorem. HINT: Argue by indution on n, nd integrte by prts. REMARK. We return now to the generl Binomil Theorem, first studied in Theorem 4.2. The proof given there used the derivtive form of Tylor s reminder Theorem, but we were only ble to prove the Binomil Theorem for t < /2. The theorem below uses the integrl form of Tylor s Reminder Theorem in its proof, nd it gives the full binomil theorem, i.e., for ll t for whih t <. THEOREM 5.3. (Generl Binomil Theorem) Let α +bi be fixed omplex number. Then ( ) α ( + t) α t k k for ll t (, ). PROOF. For lrity, we repet some of the proof of Theorem 4.2. Given generl α +bi, onsider the funtion g : (, ) C defined by g(t) (+t) α. Observe tht the nth derivtive of g is given by g (n) (t) Then g C ((, )). For eh nonnegtive integer k define k g (k) (0)/k! k0 α(α )... (α n + ) ( + t) n α. α(α )... (α k + ) k! ( ) α, k nd set h(t) k0 kt k. The rdius of onvergene for the power series funtion h is, s ws shown in Exerise 4.3. We wish to show tht g(t) h(t) for ll < t <. Tht is, we wish to show tht g is Tylor series funtion round 0. It will suffie to show tht the sequene {S n } of prtil sums of the power series funtion h onverges to the funtion g. We note lso tht the nth prtil sum is just the nth Tylor polynomil T n g for g. Now, fix t stritly between 0 nd. The rgument for t s between nd 0 is ompletely nlogous.. Choose n ɛ > 0 for whih β ( + ɛ)t <. We let C ɛ be numbers suh tht ( α n) Cɛ ( + ɛ) n for ll nonnegtive integers n. See Exerise

22 42 V. INTEGRATION, AVERAGE BEHAVIOR 4.3. We will lso need the following estimte, whih n be esily dedued s lulus exerise (See prt (d) of Exerise 4..). For ll s between 0 nd t, we hve (t s)/(+s) t. Note lso tht, for ny s (0, t), we hve (+s) α (+s), nd this is trpped between nd ( + t). Hene, there exists number M t suh tht ( + s) α M t for ll s ( 0, t). We will need this estimte in the lultion tht follows. Then, by the integrl form of Tylor s Reminder Theorem, we hve: g(t) k t k g(t) Tg n (t) k0 t 0 t 0 t 0 t 0 g (n+) (t s)n (s) n! ds ( ) (n + ) α ( + s) α n (t s) n ds n + ( ) α ( + s) α (n + ) ( t s n + + s n ds ( ) α M t (n + )t n ds n + C ɛ M t (n + ) t 0 ( + ɛ) n+ t n ds C ɛ M t (n + )( + ɛ) n+ t n+ C ɛ M t (n + )β n+, whih tends to 0 s n goes to, beuse β <. This ompletes the proof for 0 < t <. AREA OF REGIONS IN THE PLANE It would be desirble to be ble to ssign to eh subset S of the Crtesin plne R 2 nonnegtive rel number A(S) lled its re. We would insist bsed on our intuition tht (i) if S is retngle with sides of length L nd W then the number A(S) should be LW, so tht this bstrt notion of re would generlize our intuitively fundmentl one. We would lso insist tht (ii) if S were the union of two disjoint prts, S S S 2, then A(S) should be A(S ) + A(S 2 ). (We were tught in high shool plne geometry tht the whole is the sum of its prts.) In ft, even if S were the union of n infinite number of disjoint prts, S ns n with S i S j if i j, we would insist tht (iii) A(S) n A(S n). The serh for suh definition of re for every subset of R 2 motivted muh of modern mthemtis. Whether or not suh n ssignment exists is intimtely relted to subtle questions in bsi set theory, e.g., the Axiom of Choie nd the Continuum Hypothesis. Most mthemtil nlysts ssume tht the Axiom of Choie holds, nd s result of tht ssumption, it hs been shown tht there n be no ssignment S A(S) stisfying the bove three requirements. Conversely, if one does not ssume tht the Axiom of Choie holds, then it hs lso been shown tht it is perfetly onsistent to ssume s bsi xiom tht suh n ssignment S A(S) does exist. We will not pursue these subtle points here, leving them to ourse in Set Theory or Mesure Theory. However, Here s sttement of the Axiom of Choie, nd we invite the reder to think bout how resonble it seems.

23 V. INTEGRATION, AVERAGE BEHAVIOR 43 AXIOM OF CHOICE. Let S be olletion of sets. Then there exists set A tht ontins extly one element out of eh of the sets S in S. The diffiulty mthemtiins enountered in trying to define re turned out to be involved with defining A(S) for every subset S R 2. To void this diffiulty, we will restrit our ttention here to ertin resonble subsets S. Of ourse, we ertinly wnt these sets to inlude the retngles nd ll other ommon geometri sets. DEFINITION. By (open) retngle we will men set R (, b) (, d) in R 2. Tht is, R {(x, y) : < x < b nd < y < d}. The nlogous definition of losed retngle [, b] [, d] should be ler: [, b] [, d] {(x, y) : x b, y d}. By the re of (open or losed) retngle R (, b) (, d) or [, b] [, d] we men the number A(R) (b )(d ).. The fundmentl notion behind our definition of the re of set S is this. If n open retngle R (, b) (, d) is subset of S, then the re A(S) surely should be greter thn or equl to A(R) (b )(d ). And, if S ontins the disjoint union of severl open retngles, then the re of S should be greter thn or equl to the sum of their res. We now speify preisely for whih sets we will define the re. Let [, b] be fixed losed bounded intervl in R nd let l nd u be two ontinuous rel-vlued funtions on [, b] for whih l(x) < u(x) for ll x (, b). DEFINITION. Given [, b], l, nd u s in the bove, let S be the set of ll pirs (x, y) R 2, for whih < x < b nd l(x) < y < u(x). Then S is lled n open geometri set. If we reple the < signs with signs, i.e., if S is the set of ll (x, y) suh tht x b, nd l(x) y u(x), then S is lled losed geometri set. In either se, we sy tht S is bounded on the left nd right by the vertil line segments {(, y) : l() y u()} nd {(b, y) : l(b) y u(b)}, nd it is bounded below by the grph of the funtion l nd bounded bove by the grph of the funtion u. We ll the union of these four bounding urves the boundry of S, nd denote it by C S. If the bounding funtions u nd l of geometri set S re smooth or pieewise smooth funtions, we will ll S smooth or pieewise smooth geometri set. If S is losed geometri set, we will indite the orresponding open geometri set by the symbol S 0. The symbol S 0 we hve introdued for the open geometri set orresponding to losed one is the sme symbol tht we hve used previously for the interior of set. Study the exerise tht follows to see tht the two uses of this nottion gree. Exerise 5.7. () Show tht retngles, tringles, nd irles re geometri sets. Wht in ft is the definition of irle? (b) Find some exmples of sets tht re not geometri sets. Think bout horseshoe on its side, or hert on its side. () Let f be ontinuous, nonnegtive funtion on [, b]. Show tht the region under the grph of f is geometri set. (d) Show tht the intersetion of two geometri sets is geometri set. Desribe the left, right, upper, nd lower boundries of the intersetion. Prove tht the

24 44 V. INTEGRATION, AVERAGE BEHAVIOR interior (S S 2 ) 0 of the intersetion of two geometri sets S nd S 2 oinides with the intersetion S 0 S 0 2 of their two interiors. (e) Give n exmple to show tht the union of two geometri sets need not be geometri set. (f) Show tht every losed geometri set is ompt. (g) Let S be losed geometri set. Show tht the orresponding open geometri set S 0 oinides with the interior of S, i.e., the set of ll points in the interior of S. HINT: Suppose < x < b nd l(x) < y < u(x). Begin by showing tht, beuse both l nd u re ontinuous, there must exist n ɛ > 0 nd δ > 0 suh tht < x δ < x + δ < b nd l(x) < y ɛ < y + ɛ < u(x). Now, given geometri set S (either open or losed), tht is determined by n intervl [, b] nd two bounding funtions u nd l, let P {x 0 < x <... < x n } be prtition of [, b]. For eh i n, define numbers i nd d i s follows: i sup l(x), nd d i inf u(x). x i <x<x i x i <x<x i Beuse the funtions l nd u re ontinuous, they re neessrily bounded, so tht the supremum nd infimum bove re rel numbers. For eh i n define R i to be the open retngle (x i, x i ) ( i, d i ). Of ourse, d i my be < i, in whih se the retngle R i is the empty set. In ny event, we see tht the prtition P determines finite set of (possibly empty) retngles {R i }, nd we denote the union of these retngles by the symbol C P. n (x i, x i ) ( i, d i ). The re of the retngle R i is (x i x i )(d i i ) if i < d i nd 0 otherwise. We my write in generl tht A(R i ) (x i x i ) mx((d i i ), 0). Define the number A P by A P (x i x i )(d i i ). Note tht A P is not extly the sum of the res of the retngles determined by P beuse it my hppen tht d i < i for some i s, so tht those terms in the sum would be negtive. In ny se, it is ler tht A P is less thn or equl to the sum of the res of the retngles, nd this nottion simplifies mtters lter. For ny prtition P, we hve S C P, so tht, if A(S) is to denote the re of S, we wnt to hve A(S) A(R i ) (x i x i ) mx((d i i ), 0) (x i x i )(d i i ) A P. DEFINITION. Let S be geometri set (either open or losed), bounded on the left by x, on the right by x b, below by the grph of l, nd bove by the grph of u. Define the re A(S) of S by A(S) sup A P P sup P {x 0<x <...<x n} (x i x i )(d i i ),

25 V. INTEGRATION, AVERAGE BEHAVIOR 45 where the supremum is tken over ll prtitions P of [, b], nd where the numbers i nd d i re s defined bove. Exerise 5.8. () Using the nottion of the preeding prgrphs, show tht eh retngle R i is subset of the set S nd tht R i R j if i j. It my help to drw piture of the set S nd the retngles {R i }. Cn you drw one so tht d i < i? (b) Suppose S is geometri set nd tht S 2 is nother geometri set tht is ontined in S. Prove tht A(S 2 ) A(S ). HINT: For eh prtition P, ompre the two A P s. Exerise 5.9. Let T be the tringle in the plne with verties t the three points (0, 0), (0, H), nd (B, 0). Show tht the re A(T ), s defined bove, grees with the formul A (/2)BH, where B is the bse nd H is the height. The next theorem gives the onnetion between re (geometry) nd integrtion (nlysis). In ft, this theorem is wht most lulus students think integrtion is ll bout. THEOREM 5.4. Let S be geometri set, i.e., subset of R 2 tht is determined in the bove mnner by losed bounded intervl [, b] nd two bounding funtions l nd u. Then A(S) b (u(x) l(x)) dx. PROOF. Let P {x 0 < x <... < x n } be prtition of [, b], nd let i nd d i be defined s bove. Let h be step funtion tht equls d i on the open intervl (x i, x i ), nd let k be step funtion tht equls i on the open intervl (x i, x i ). Then on eh open intervl (x i, x i ) we hve h(x) u(x) nd k(x) l(x). Complete the definitions of h nd k by defining them t the prtition points so tht h(x i ) k(x i ) for ll i. Then we hve tht h(x) k(x) u(x) l(x) for ll x [, b]. Hene, b b A P (x i x i )(d i i ) (h k) (u l). Sine this is true for every prtition P of [, b], it follows by tking the supremum over ll prtitions P tht A(S) sup A P P b (u(x) l(x)) dx, whih proves hlf of the theorem; i.e., tht A(S) b u l. To see the other inequlity, let h be ny step funtion on [, b] for whih h(x) u(x) for ll x, nd let k be ny step funtion for whih k(x) l(x) for ll x. Let P {x 0 < x <... < x n } be prtition of [, b] for whih both h nd k re onstnt on the open subintervls (x i, x i ) of P. Let, 2,..., n nd b, b 2,..., b n be the numbers suh tht h(x) i on (x i, x i ) nd k(x) b i on (x i, x i ). It follows, sine h(x) u(x) for ll x, tht i d i. Also, it follows tht b i i. Therefore, b (h k) ( i b i )(x i x i ) (x i x i )(d i i ) A P A(S).

26 46 V. INTEGRATION, AVERAGE BEHAVIOR Finlly, let {h m } be nonderesing sequene of step funtions tht onverges uniformly to u, nd let {k m } be noninresing sequene of step funtions tht onverges uniformly to l. See prt (d) of Exerise 5.9. Then b b (u l) lim (h m k m ) A(S), m whih proves the other hlf of the theorem. OK! Trumpet fnfres, plese! THEOREM 5.5. (A πr 2.) If S is irle in the plne hving rdius r, then the re A(S) of S is πr 2. PROOF. Suppose the enter of the irle S is the point (h, k). This irle is geometri set. In ft, we my desribe the irle with enter (h, k) nd rdius r s the subset S of R 2 determined by the losed bounded intervl [h r, h + r] nd the funtions u(x) k + r 2 (x h) 2 nd l(x) k r 2 (x h) 2. By the preeding theorem, we then hve tht A(S) h+r h r 2 r 2 (x h) 2 dx πr 2. We leve the verifition of the lst equlity to the following exerise. Exerise Evlute the integrl in the bove proof: h+r h r 2 r 2 (x h) 2 dx. Be reful to explin eh step by referring to theorems nd exerises in this book. It my seem like n elementry lulus exerise, but we re justifying eh step here. REMARK. There is nother formul for the re of geometri set tht is sometimes very useful. This formul gives the re in terms of double integrl. There is relly nothing new to this formul; it simply mkes use of the ft tht the number (length) u(x) l(x) n be represented s the integrl from l(x) to u(x) of the onstnt. Here s the formul: A(S) b u(x) ( l(x) dy) dx. The next theorem is result tht justifies our definition of re by verifying tht the whole is equl to the sum of its prts, something tht ny good definition of re should stisfy.