14. The fundamental theorem of the calculus

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1 4. The funmenl heorem of he clculus V v Figure : () Venriculr volume for subjecs wih cpciies C = 24 ml, C = 20 ml, C = 2 ml n (b) he corresponing inflow. The soli curves re ienicl o hose in Figure 3 of Lecure. From Lecure 3 we lrey know h venriculr volume ny ime equls venriculr volume ny erlier ime plus subsequen ne rechrge: V () = V () + v(x)x. () Bu from Lecure 6 we lso know h inflow equls re of chnge of volume: V () = v(). (2) So volume is he inegrl of inflow, in he sense of (); n inflow is he erivive of volume, in he sense of (2). Bu his relionship beween inegrion n iffereniion is symmeric, in he sense h lhough knowlege of V implies knowlege of v, knowlege of v oes no imply full knowlege of V. Why? In essence, becuse wo ifferen subjecs wih precisely he sme inflow rce my hve ifferen venriculr cpciies. Figure shows ( croon of) he vriion wih ime of venriculr volume C if 0 < 0.05 V () = if 0.05 < if 0.35 < if 0.4 < if for humn subjec wih venriculr cpciy V (0) = C ml for hree ifferen vlues of C. Regrless of his vlue, he unique vriion wih ime of venriculr inflow is 0 if 0 < if 0.05 < 0.35 v() = V () = 0 if 0.35 < 0.4 (4) if 0.4 < if (3)

2 (Exercise ). Clerly, herefore, knowlege of V implies knowlege of v. Bu knowlege of v oes no imply full knowlege of V, becuse lhough he meho of Lecure 3 cn be use o recover (3) from (4), he vlue of C = V (0) is uneermine. On he oher hn, knowlege of v is lmos enough o yiel knowlege of V : he only iionl informion require is he mgniue of he volume single insn, e.g., = 0. Thus v implies V excep for he iion of n rbirry consn, here C = V (0). In oher wors, he oe curves in Figure re ienicl o he soli curve, excep for rnslion in he vericl irecion (eiher up by 4 ml or own by 8 ml). The bove relionship beween v n V nmely, inflow equls re of chnge of volume, n curren volume equls iniil volume plus subsequen ne rechrge cn be reuce o single line of mhemics by wriing v() = V () V () = C + v(x) x (5) where C = V (). (5b) Furhermore, his relionship is perfecly generl: ny wo funcions f n F sisfy f() = F () F() = C + f(x)x (6) for rbirry C = F() (provie h F exiss, which we ssume). The equivlence (6) is clle he funmenl heorem of clculus; n ny funcion h ssigns o he lbel C + (6b) f(x)x, (7) where C is uneermine, is clle n ni-erivive of f. The funmenl heorem of clculus more or less sys h inegrion n iffereniion re inverse operions. There is lso n symmery, however, becuse erivives re lwys unique, wheres ni-erivives re no unique lhough ll of funcion s ni-erivives iffer only by consn; in oher wors, f is he erivive of F if n only if F is n ni-erivive of f. A proof of he heorem is skeche in he ppenix. In prcice, before pplying he funmenl heorem, we usully subsiue from one sie of (6) ino he oher. Clerly, here re wo wys in which o o so; n before consiering he firs, i is convenien o recll couple of generl poins bou iffereniion. Firs, lhough resuls concerning erivives re in principle semens bou relionship beween funcions F n F, in prcice we ofen fin i mos convenien o procee wihou explicily nming F ; so, for exmple, inse of wriing F() = ln() = F () = / (for > 0), we prefer o wrie {ln() = /. Secon, lhough erivive is in principle he limi of quoien, i is rre in prcice h we clcule limi when we wn o fin erivive: inse we employ generl resuls like he chin, prouc 2

3 n quoien rules in conjuncion wih specil rules like he one bove for he erivive of he logrihm. Likewise for ni-iffereniion. Firs, lhough resuls concerning ni-erivives re in principle semens bou relionship beween funcions f n F efine by F() = C + f(x)x, in prcice we ofen fin i mos convenien o procee wihou explicily nming F or f; so, for exmple, inse of wriing f() = / = C + f(x)x = ln(), we coul simply wrie C + x = ln(). We coul bu i woul go gins riion. If c is n rbirry consn, hen he riionl wy of sing h he funcion F x efine by F() = ln() + c is n ni-erivive of he funcion f efine by f() = / is o wrie = ln() + c. (8) Now, i hs o be si immeiely h his is lousy noion, becuse if here is one hing h he lef-hn sie of (8) shoul be inepenen of, hen i is conricing h he righ-hn sie epens on! I woul be very much beer o wrie (8) s x = ln() + c (9) x n inerpre he lef-hn sie s wy of wriing x when is compleely rbirry, x becuse x = ln() + c is jus noher wy of wriing C + x = ln(). Bu i is x x hr o buck riion, especilly one s eeply ingrine s his one. So inse we hve o lern o live wih he lousy noion, n regr (8) s simply n inverse wy of sying = {ln() + c. (0) Wih he new noion, every resul we hve ever h concerning erivives cn immeiely be rewrien s corresponing resul concerning ni-erivives. Noe, however, h i is riionl o rescle such equivlences so h he funcion on he lef-hn sie lwys, in effec, hs coefficien. For exmple, lhough {2 = 2 2 = 2 + C is perfecly correc, you re much more likely o see {2 = 2 = c (wih he rbirry consns C n c rele by C = 2c). A smple of such equivlences ppers in Tble. Noe h f(x)x is clle n inefinie inegrl, n h n inefinie inegrl is bsiclly he sme hing s n ni-erivive in precisely he sme sense h ifferenil coefficien is bsiclly he sme hing s erivive. Secon, lhough efinie inegrl is in principle he limi of sum, i is rre in prcice h we clcule limi when we wn o fin efinie inegrl; inse we use he funmenl heorem in he form F (x)x = F() F(), () which we obin from (6) by subsiuing from (6b) n he lef-hn sie of (6) ino he righ-hn sie, i.e., by subsiuing f(x) = F (x) n C = F(). Becuse () hols for Wih he rbirry consns C n c rele by c = C = ln() 3

4 {r = r r r = r+ + c (r ) r + {ln( + α) = +α = ln( + α) + c +α ( > α) {eα = αe α e α = α eα + c {sin(α) = α cos(α) cos(α) = sin(α) + c α {cos(α) = α sin(α) sin(α) = cos(α) + c α Tble : Some equivlences beween erivive n ni-erivive. In ech cse c is n rbirry consn of inegrion. ny in he omin of F, incluing he righ-hn enpoin, we hve in priculr h F (x)x = F(b) F(). (2) Wh his equion mens in prcice is h erivives re very esy o inegre. For exmple, we know h F() = ln() = F () = /; or, which of course is excly he sme hing, F(x) = ln(x) = F (x) = /x. So i follows once from (2) h x x = ( b F (x)x = F(b) F() = ln(b) ln() = ln. (3) ) (for b > > 0). Bu we hve h o nme F explicily o obin his resul, n in prcice we migh be hppier if h weren necessry. So we inrouce noher noion: we use F(x) b o enoe he jump in F(x) beween x = n x = b; i.e., we efine Now we cn re-se (2) s F(x) b = F(b) F(). (4) F (x)x = F(x) b (5) wih he vnge h we cn evlue efinie inegrls wihou ever hving o nme F explicily. For exmple, in erms of (5), (3) reuces o x x = b ) x {ln(x) x = ln(x) b = ln(b) ln() = ln(. (6) We cn similrly recover he specil resuls we obine he en of Lecure (by evluing he limi h efines he efinie inegrl); for exmple, x 2 x = { 3 x x3 x = b 3 x3 4 = 3 b3 3 3 = 3 (b3 3 ). (7)

5 No only oes using he funmenl heorem involve less work, bu lso we cn obin much more generl resul, for (if necessry, wih he help of Tble ) we hve x n x = { x n+ x = xn+ x n + n + b = bn+ n + n+ n + = bn+ n+, (8) n + provie h n ; in h cse, we hve (6). The oher wy o subsiue from one sie of (6) ino he oher is o subsiue from he righ-hn sie ino he lef-hn sie. Then we obin or f() = { C + f(x)x { f(x)x = C + { f(x)x = f() (9) becuse C = 0; inuiively, he erivive of n inegrl is he originl funcion. This form of he funmenl heorem is ofen mos useful when use in conjuncion wih he chin rule. Suppose, for exmple, h y = sin(u) + x2 x (20) n we wn o know y. Then, on using (9) wih = 0 n f(x) = + x u 2, he subsiuion = sin(u) yiels y u = y u = { + x2 x u = { + x2 x u {sin(u) = { + x2 x cos(u) = + 2 cos(u) = cos(u) + sin 2 (u).. In sum, he funmenl heorem of clculus sys h he erivive of funcion s inegrl is precisely h funcion, in he sense h { x f() = f(x) (22) x n h he inegrl of funcion s erivive is essenilly h funcion, in he sense h F (x)x = F(x) + c (23) (2) 5

6 where c is n rbirry consn. Noe h (9) n(22) re jus wo ifferen wys of sying excly he sme hing (wih x n merely juxpose); n on forge h wh (24) relly mens is x for ny x, where is n rbirry consn. F () = F(x) F() = F() x (24) Appenix: A skech of proof of he funmenl heorem of clculus y f x y x Figure 2: For Φ efine by (25), he ligher she re represens Φ() n he ol she re represens Φ( + δ); hus he rker she re represens Φ( + δ) F() The purpose of his ppenix is o show why (6) mus hol in generl (wihou cully proving i rigorously). There re wo resuls o be esblishe, nmely, h he righ-hn sie of (6) implies he lef-hn sie, n h he lef-hn sie of (6) implies he righ-hn sie. We begin wih he former, h is, we firs esblish h F() = C + f(x)x = f() = F (), which mens ssuming F() = C + f(x)x n eucing f() = F (). Accoringly, ssume h F() = C + f(x)x. I will help o efine Φ() = f(x)x, (25) which is represene by he ligher she re in Figure 2, i.e., he signe re beween he grph of f n segmen [,] of he horizonl x-xis. Noe once more he imporn poin h, becuse is n enpoin of he inervl in quesion, we cnno use o lbel he xis bu we cn use ny symbol oher hn (or y, obviously), n we hve chosen o use x. Le us now increse infiniesimlly, o + δ. Then Φ() chnges infiniesimlly o Φ( + δ), i.e., he signe re beween he grph of f n he segmen [, + δ] of he x-xis, s represene by he ol she re in Figure 2. Hence he ifference, i.e., 6

7 Φ( + δ) Φ(), is represene by he rker she re. Bu for sufficienly smll δ, his re mus fll beween h of recngle of wih δ n heigh f() n h of recngle of wih δ n heigh f( + δ) becuse, if δ is sufficienly smll, hen we cn ssume h f is eiher incresing, ecresing or consn on [, + δ]. In oher wors, for he rker she (signe) re, we hve +δ f(x)x = f()δ + o(δ). (26) As we sw momen go, however, his quniy mus lso equl Φ( + δ) Φ(). So Φ( + δ) Φ() = f()δ + o(δ), (27) implying Φ( + δ) Φ() = f() + o(δ). (28) δ δ Now king he limi s δ 0 yiels Φ () = f() + 0 = f(). We hve herefore esblishe h implying Φ() = f(x)x = Φ () = f() (29) F () = C {C + Φ() = + Φ () = 0 + f() = f() (30) Now we esblish h he lef-hn sie of (6) implies he righ-hn sie, or f() = F () = F() = C + f(x)x, which mens ssuming f() = F () n eucing F() = C + f(x)x. Accoringly, ssume h f() = F (), or which of course is excly he sme hing f(x) = F (x). Then, from he efiniion of ifferenil coefficien (Lecures 5-6), we hve F(x + δx) F(x) = f(x)δx + o(δx). (3) Recll from Lecure one of he wo noions for he efinie inegrl of f beween n b h we i no op (becuse i ws oo cumbersome): { ol signe re beween he grph of f n lim f()δ = segmen [, b] of he horizonl xis, coune δ 0 posiively bove he xis, negively below i. [,b] We cn replce b by in his equivlence, bu only if we firs replce by x on he lef-hn sie (for he reson emphsize bove): { ol signe re beween he grph of f n f(x)x = lim f(x)δx = segmen [, ] of he horizonl xis, coune (32) posiively bove he xis, negively below i. x [,] Now, becuse o(δx) 0 s δx 0, (3)-(32) imply f(x)x = lim x [,] f(x)δx = lim x [,] = lim {F(x + δx) F(x). x [,] 7 {f(x)δx + o(δx) (33)

8 In he limi s δx 0, his summion behves like he summe sequence of ifferences in Lecure 0: becuse ech F(x) is in essence F(x + δx) for he nex ifference own, he erms ll cncel in pirs excep for he very firs n ls. Th is, lim {F(x + δx) F(x) = F() F(). (34) x [,] Togeher, (33)-(34) n (6b) imply or F() = C + f(x)x, s require. Exercises f(x)x = F() C (35). Differenie (3) o obin (4) n verify h v is coninuous. Hin: Polynomils re lwys coninuous, so you nee only verify coninuiy = 0.05, = 0.35, = 0.4 n = For venriculr volume V, if b (= 0.9 in Figure ) is he perio of he cric cycle, hen mx(v, 0, b) min(v, 0, b) is clle he sroke volume. () Wh is he sroke volume in Figure? (b) Why mus i be rue h v() = 0? 0 (c) Verify (b) for v efine by (4). 3. Funcions g n G re efine on [0, 4] by 4 2 if 0 < g() = if < 3 0 if 3 4 n G() = 0 g(x) x. Using ifferen meho from he one you use for Exercise 3 of Lecure 3, fin n explici formul for G() for ll [0, 4]. 4. () Clcule { x 2x 2 (b) Show h x 3 x = ( )( + ). 2 Assume h > (hence x for ll relevn x). Hin: Use (5) wih =, b = n juicious choice of F. 5. () Clcule { x 3x 3 8

9 (b) Show h x 4 x = ( )(2 + + ). 3 3 Assume h > (hence x for ll relevn x). Hin: Use (5) wih =, b = n juicious choice of F. 6. () Clcule { x 2 x 3 x (b) Show h x(6 x) ( )(2 + 3) x =. (3 x) 2 2(3 ) Assume h < 3 (hence x < 3 for ll relevn x). Hin: Use (5) wih =, b = n juicious choice of F. Suible problems from snr clculus exs Sewr (2003): p. 402, ## 7-24 n 28-52; pp. 4-42, ## -58. Where necessry, use (9) n he chin rule. Reference Sewr, J Clculus: erly rnscenenls. Belmon, Cliforni: Brooks/Cole, 5h en. Soluions or hins for selece exercises 2. (b) V mx () V min () = = 70 ml. (b) Becuse volume is conserve. 3. G efine on [0, 4] by G() = c if 0 < c 2 if < c 3 if 3 4 is n ni-erivive of g, where he inegrion consns c, c 2 n c 3 re ifferen for ech subomin of he join. These consns re eermine by G(0) = 0 n he coninuiy of G, i.e., G( ) = G( + ), G(3 ) = G(3 + ). So c = 0, c 2 = 7 n c 2 3 = Noe h G mus be coninuous even if g is no (lhough in his cse, i is). 9