HOW TO USE INTEGRALS E L Ldy (December 21, 1998) Consider the following set of formuls from high-school geometry nd physics: Are = Width Length Are of Rectngle Distnce = Velocity Time Distnce Trveled by Moving Object Volume = Bse Are Height Volume of Cylinder Work = Force Displcement Work Done by Constnt Force Force = Pressure Are Force resulting from constnt pressure Mss = Density Volume Mss of solid with constnt density There s common simple pttern (structure) to ll these formuls, nmely A = BC The first four of these formuls cn be interpreted grphiclly Nmely, in the second formul, for instnce, if one constructs rectngle, where the verticl side corresponds to velocity V nd the horizontl side to time T, then the re of this rectngle will represent the distnce trveled by n object moving t velocity V for time T 10 cm/sec Distnce = 250 cm 25 sec (Notice tht since horizontl units in this picture represent seconds, nd verticl units represent cm/sec, the units for the re of the rectngle should be s shown sec cm sec =cm,
2 Likewise, the fourth formul could be represented grphiclly by rectngle where the horizontl side represents distnce (displcement) nd the verticl side represents force Once gin, the re of the rectngle represents the work done by the moving force 6 lb Work =66ft-lb 11 ft A force of 6 pounds is pplied to n object which moves for distnce of 11 feet (The lst two formuls could lso be represented by rectngles, but in this cse the horizontl side of the rectngle would be one-dimensionl representtive re or volume In these cses, it s more useful to represent the eqution by three or four-dimensionl figure rther thn by rectngle) The Need for Integrls All six of the formuls bove re simplistic In the rel world, objects don t trvel t constnt speed They speed up, slow down, even stop for while to rest or get fuel To hve relly useful formul relting velocity nd distnce, we need to consider the possibility tht velocity v is not constnt but insted is function of time t Likewise, insted of the third formul, giving the volume of cylinder, it would be more useful to consider solid where the horizontl cross section chnges s the height chnges, i e the re of the cross section t height h is function A(h) ofh And in the remining exmples, it is importnt to del with situtions where force, pressure, or density re vrible And even in cse of the first formul, we relize tht not ll plne regions re rectngles, so it would be useful to hve formul for the re of region where the width is vrible function of the horizontl position A rectngle with deformed top
3 r In this cone, the horizontl cross section t height h is circle with rdius r =8 2h, where h rnges from 0 t the bottom to h =4 t the top The volume of the cone turns out to be given by the formul Vol = 4 0 π(8 2h)2 dh More Relistic Problems Consider now simple, but slightly more relistic, time-velocity problem: The velocity of n object between time t =3 secnd t=6 secisgivenbythe formul v(t) = 1 2t 1 (mesured in units of cm/sec) How fr does the object trvel during this time period? Wht this problem is more or less sking is: How does one multiply velocity by elpsed time in cse when velocity is ctully chnging s time progresses? We hve seen bove tht re cn be geometric wy of representing multipliction Now represent the bove problem by grphing the function v = 1 2 t 1 v cm/sec 2 1 v = 1 2 t 1 3 6 t sec It turns out tht the distnce trveled by the object between time t =3 ndtime t=6,s
4 described in the problem, will be equl to the re over tht prt of the t-xis between t =3 nd t= 6 nd under the grph of the function v =2t 1, s indicted by the shded region in the grph Note tht in the grph, the horizontl units represent seconds nd the verticl units re mesured in cm/sec Logiclly, then, the units for re should be cm sec = cm, sec s required Likewise, consider force-displcement-work problem A verticl force on n object moving horizontlly is given by the formul F (x) =5 x (where x is mesured in feet nd F is mesured in pounds) Find the work done by this force while the object moves between the points x =2 ftnd x=4 ft Anlogously to the velocity-time problem, we cn multiply the vrible force in this problem by the displcement, by mesuring the re under the grph of the function F (x) = 5 x between the points x =2 nd x= 4, s indicted on the grph below (The units for work in this cse re ft-lbs) F lbs 3 2 1 2 3 4 F = 5 x x ft Similrly, if one constructs the grph of the cross-sectionl re of solid, where the horizontl coordinte h represents height nd the verticl coordinte represents the re A(h) of the horizontl cross-section of the solid t tht height, then the volume of tht solid will be given by the re between this curve nd tht prt of the h-xis between the bottom height (h strt )ndtheheightt the top (h end ) (In this cse, no prt of the curve will be below the h-xis, since the cross-sectionl re is never negtive)
5 A cm 2 3 π 2 π π A = π(2 1 2 x)2 1 2 3 4 h cm The re of the horizontl cross section of cone t height h equls π(2 1 2 h)2 cm 2 The volume of the cone will equl the re under the curve A = π(2 1 2 h)2 for h between 0 nd 4 In the lnguge of clculus, the six simplistic high-school formuls t the beginning of these notes re replced by formuls given by integrls A = D = V = W = xend x strt w(x) dx tend t strt v(t) dt hend h strt A(h) dh xend F = M = x strt F (x) dx p(x, y) dx dy Ω T ρ(x, y, z) dx dy dz There re mny other problems where this sme ide pplies In order to be ble to tell whether it pplies to given sitution, we re going to consider below the resons why it is true
6 In physics books, concepts such s work re often simply defined by formuls in the form of integrls: Work = b F (x) dx This netly sidesteps the need to prove tht these formuls re correct However since such definitions re, for most students, extremely non-intuitive, one might wonder to wht extent they re rbitrry Could one lso get stisfctory theory by using some completely different definition for work? In fct, from the principles to be given below, one cn see tht even before one knows how to formlly define the concept of work, the formul Work = b F (x) dx cn be seen to be n essentil consequence of simple xioms bout the reltionship between force nd work tht lmost everyone will ccept s self-evident Nmely, it follows from the fct tht the reltionship between force nd work is (using words tht will be defined below) cumultive nd incresing Wht Mkes Reltionship Expressible by n Integrl? The purpose of these notes is to give two simple principles which will enble one in most cses to recognize when mthemticl reltionship cn be expressed in terms of n integrl nd to be ble to set up the correct integrl with fir mount of confidence Before explining these principles, it will be useful to note severl exmples of formuls in physics nd other sciences where the bsic pttern A = BC is vlid even without simplistic ssumptions nd does not generlize to formul given by n integrl Nmely, consider Ohm s Lw relting voltge, current nd resistnce; Newton s Second Lw of Motion (F = m); Hook s Lw for the force exerted by stretched (or compressed) spring; nd the Boyle-Chrles Lw relting volume, temperture, nd pressure for n idel gs
7 Some Formuls Which Never Generlize to Integrls E = IR F = m W = EI T = kpv F = kx (Ohm s Lw for voltge, current nd resistnce) (Newton s Second Lw of Motion) (Electricl power is the product of current & voltge) (The Boyle-Chrles Lw for temperture, pressure, nd volume) (Hook s Lw for springs) Betweenness One of the most importnt things tht mkes the formuls Are = Height Length Distnce = Velocity Time Volume = Bse Are Height Work = Force Displcement different from formuls such s Ohm s Lw nd Hook s Lw is tht these four formuls ll hve the property tht the second fctor on the right ctully represents the mount by which certin vrible chnges For instnce, the second formul could be better written s Distnce = Velocity Elpsed Time or Distnce = Velocity Time Intervl Likewise, in the formul for the volume of cylinder, the fctor clled height is ctully the difference in height between the top nd the bottom of the cylinder And in the fourth formul, for work, displcement is the mesure of chnge in position And the sme thing is even true of length in the formul for the re of rectngle, i e length is the difference between the position of the right edge of the rectngle nd the position of the left edge If we now use the vrible x to represent horizontl position, y to represent verticl position, nd t to represent time, then the four bove formuls could be expressed s, using rther obvious notion, A =(x end x strt )W D =(t end t strt )V V =(y end y strt )A W =(x end x strt )F
8 By contrst, in Ohm s Lw E = IR, the voltge E is determined purely by the size of the resistnce nd current t given instnt, nd is not influenced by ny other vlues tht resistnce nd current might hve tken in the pst, so tht formul given by n integrl would not be pproprite If we try to imgine generliztion of Ohm s Lw written in the form of n integrl, E = Voltge =?? R(i) di, (where i represents current nd R(i) resistnce), there is no resonble choice for wht vlues to put t the top nd the bottom of the integrl sign, since the current I cnnot nturlly thought of s the mount of chnge mde by some vrible Furthermore, the nottion R(i) is inpproprite since resistnce is not normlly function of current One might sy tht in those situtions where high-school formuls generlize to integrls there is notion of betweenness A moving object trvels between strting time nd n ending time, nd to know how fst it trvels we need to know the velocity t ll the instnts between these two times Likewise, solid body hs horizontl cross-sections t ll heights between some strting height h 0 nd some ending height h 1, nd we cn compute the volume of the solid if we know the re of these cross sections t ll the heights between these two As rule of thumb, ny time one of the two fctors on the right-hnd side of formul A = BC represents time or distnce, one cn suspect tht the formul corresponding to the generl sitution will be given by n integrl (Note tht Hook s Lw for springs, mentioned bove, F = kx, isone exception Even if imgines sitution where the spring constnt k would be function of x, Hook s Lw would still not be given by n integrl, becuse the force exerted by the spring would still depend only on the length x to which the spring hs been stretched (or compressed), nd not by nything tht hppens t points in between the spring s rest point nd x) Cumultive Reltionships I think tht it is pretty fir to sy tht if one thinks tht mthemticl reltionship might hve some expression in the form of n integrl, then there will in fct exist n integrl expressing tht reltionship The problem then becomes to find the correct integrl formul, which cn be more difficult A more sophisticted criterion for the existence of n integrl expressing given reltionship involves the notion of cumultive reltionship An exmple will mke the ide cler The reltionship between velocity nd distnce is cumultive in tht if one considers time t 2 in between two other times t 1 nd t 3, then the distnce n object trvels between times t 1 nd t 3 cn be obtined by dding together the distnce trveled between time t 1 nd time t 2 plus the distnce trveled between t 2 nd t 3
9 SIDEBAR: Wht Is n Integrl? In beginning clculus courses, the integrl is introduced by discussing the problem of finding the re under curve Dividing the re into tiny verticl strips, one rrives t the concept of Riemnn sum (or some vrition on this ide) A theorem is then proved stting tht under resonble conditions such Riemnn sum will converge to limit s the width of the rectngles goes to zero The integrl is then defined to be the limit gurnteed to exist by this theorem This seems to sy tht the wy to find the re under curve involves monstrosity tht pprently no one could ever compute in prctice There s point here tht mthemticins tke for grnted, but students re often not explicitly told Nmely, it doesn t mtter if the definition of the integrl is completely imprcticl, becuse the definition of concept doesn t hve to be something one ctully uses except to prove few theorems In mthemtics, it s not importnt wht things re Wht s importnt is how they behve the rules they obey The definition of concept is simply wy of getting your foot in the door It gives you firm foundtion to develop the rules which the concept obeys nd which re the things tht everybody relly uses in prctice (In mny cses, such s with the integrl, wht definition relly is is n existence theorem) In prctice, integrls cn be computed by nti-derivtives, so tht the Riemnn sums re for the most prt irrelevnt The point of view of these notes is to encourge students to think of the integrl of function s the re under its curve between the two prescribed points (with the dded proviso tht re below the x-xis should be considered negtive) (The grph of function of two vribles is of course surfce, nd double integrl is equl to the volume under tht surfce Unfortuntely, triple integrls re difficult to visulize in n nlogous wy) As mthemticin, however, I feel compelled to point out tht defining the integrl s the re under grph ultimtely doesn t simplify things t ll This is becuse giving rigorous mthemticl definition of re is not ny esier (nor much more difficult) thn developing the integrl on the bsis of Riemnn sums Secondly, there exist functions so unruly tht their grphs re extremely disconnected nd don t even look like coherent curves, so tht it doesn t mke ny sense to tlk bout the re under these grphs (The ide is little like tht of frctl curve But insted of rdiclly chnging direction infinitely often like frctl, these curves hve discontinuous breks infinitely often) Fortuntely, however, beginning clculus students (nd most people who use clculus s tool) don t hve to del with such functions nd cn mnge quite well by depending on their intuitive ides bout re
10 Time Time Time = t 1 = t 2 = t 3 Distnce Likewise, the reltionship between pressure nd force is cumultive If function p(x, y) describes pressure pplied to certin plnr region, nd if one considers two non-overlpping (two-dimensionl) pieces of tht region, then the force exerted by tht pressure on the union of the two pieces will be the sum of the forces exerted on ech piece One rough, informl, non-technicl definition of the integrl is tht b f(x) dx gives the cumultive effect of the function f(x) when pplied to ll the vlues of x between nd b For instnce, the cumultive effect when velocity function v(t) is pplied t ll the moments of time t between t = nd t = b will be the distnce n object whose velocity is given by tht function would trvel As prcticl mtter, lmost ny time scientific reltionship between quntity nd the vlues of function over n intervl hs the property of betweenness, then tht reltionship will be cumultive A reltionship which is not cumultive would be roughly comprble to rel-life situtions such s ir trvel, where the time to trvel between New York nd Los Angeles would not be the sum of the time to trvel from New York to Chicgo nd the time from Chicgo to Los Angeles (ssuming the first flight ws non-stop) I cn now present the min point of these notes, nmely two rules of thumb for expressing mthemticl reltionship in the formul This is s fr in the rticle s mny people will need to red Two Rules of Thumb (1) In generl, lmost ny time quntity is determined by the vlues of function over n intervl in cumultive wy, one cn be firly certin tht the reltionship between the quntity nd the function be expressed s n some sort of integrl (2) In most cses, if it is known tht the reltionship between quntity nd function is expressible by some integrl, nd if suggested integrl formul for this reltionship yields the correct nswers in cses when the function is constnt, then the formul will be correct The first rule of these rules of thumb is more universlly vlid thn the second To enble students to understnd when the second rule of thumb will pply, I will first discuss the resons why it works, nd the bsic ssumptions involved After this, I will discuss one exmple of sitution where the second principle fils: nmely, the formul for the length of curve
11 Why The Second Rule of Thumb Works To understnd why n integrl formul which gives the correct nswer for constnt functions is usully the correct formul in generl, let s go bck to the cnonicl exmple of velocity nd distnce The min signicnce of the fct tht the reltionship between distnce nd velocity is cumultive is tht if we cn brek the time intervl from t 0 to t 1 up into pieces, nd if the formul we re trying to prove is correct on ech piece, then the formul must be correct for the whole intervl (since the distnce corresponding to the whole intervl is the sum of the distnces on the seprte pieces) Nowweknowthtifv(t) is constnt function, then the distnce trveled by n object whose velocity is v(t) will be given by the re under the grph of v(t) And we lso know tht the reltionship between velocity nd v distnce is cumultive It follows tht the cm/sec equlity between distnce nd the re under the grph of the velocity function v(t) =2 2 will vlid for ny function which is Distnce mde up of number of pieces where ech 1 =6cm piece is constnt function In the trde, t function of this sort is clled step function However if we fill in the re 1 2 3 4 5 sec underneth ech horizontl piece, wht step function relly looks like is br grph The fct tht distnce corresponds to re in the cse of constnt functions An object trvels with constnt velocity v(t) =2 cm/sec from time t =1 to t =4 The distnce trveled is the sme s the re under the grph mens tht the re in ech bnd of the br grph corresponds to the distnce n object would trvel in tht little piece of time if its velocity were given by the height of the br grph (step function) t tht point Adding ll the pieces together, we see tht the re comprised by this br grph is equl to the distnce n object would trvel if its velocity were given by the step function (br grph) Now let s go bck to the question we rised erlier: Suppose tht n object is trveling between time t =2 nd t= 5 nd tht its velocity t time t is, sy, v(t) =t/2 How cn we multiply velocity by elpsed time in cse like this when velocity chnges s time progresses? We climed bove tht the nswer to this conundrum cn be obtined by mesuring the re under the grph of the velocity function between the strting time nd ending time As wy seeing why this is true, we imgine temporrily tht insted of incresing continuously, the velocity ctully chnges by mking very lrge number of extremely tiny quntum jumps By mking the time intervl between jumps smll enough, we cn get something tht pproximtes tht
12 A Velocity Determined by Step Function cm/sec 2 1 2 3 4 5 sec An object trvels from time t =2 until t= 5 strting t velocity of 1 cm/sec Every hlf second, the velocity increses by 25 cm/sec, nd is constnt in between jumps (Thus, for instnce, during the finl hlf second, between time t =45nd t= 5, the object is trveling t velocity of 225 cm/sec It thus trvels distnce of 5 225 = 1125 cm during this finl hlf second) The distnce trveled by the object cn be esily computed s 5+625 + 75 + 8725 + 10+1125 = 4875 cm which is numericlly the sme s the re comprising the br grph the ctul velocity function v(t) (or in fct ny velocity function tht occurs in physics books nd clculus books; ny continuous function, for instnce) extremely closely In other words, the given velocity function cn be pproximted extremely closely by step function For instnce, here is step function tht looks firly close to the curve v(t) =t/2 between t =2 nd t=5
13 cm/sec 2 1 t 2 3 4 5 sec A crude step-function pproximtion to the grph of v = 1 2 t with few of the verticl lines of the corresponding br grph In this pproximtion, the length of the horizontl lines, usully denoted by t, is 125 Despite the fct tht those with poor vision (especilly those with stigmtism) my hve difficulty in distinguishing the individul horizontl lines tht mke up this step function, by mthemticl stndrds, this pproximtion is quite crude If we use t which is one-qurter this size, we get the following grph, we is strting to fll within the rnge where the eye (nd the lser printer) re unble to distinguish it from the line v = t/2 And yet in terms of the sheer mthemtics setting side the questions of vision nd drwing we cn do much much better cm/sec 2 1 t 2 3 4 5 sec A step-function pproximtion to the grph of v = 1 2t with t = 03125 If we mke t ny smller, then we fll below the level of resolution tht most lser printers cn esily del with And yet, t lest, in principle, we could mke the pproximte fr better We could construct step function where t is smller thn the eye cn distinguish, or, for tht mtter, smller thn the dimeter of n electron At tht point, for ll prcticl purposes there would be no difference between the step function nd the function we strted with Now t ech step, we compute the distnce trveled by multiplying the velocity t tht step by the width of the step Adding these ll together, we see tht the totl distnce trveled equls the re under the step function As we mke the steps smller nd smller, we get closer nd closer to the
14 distnce ctully trveled But we re lso getting closer nd closer to the re under the grph of the originl velocity function Cse closed Cse closed? Hm At the very lest, it would be worthwhile to spell out the resoning here more crefully And when we do tht, it will turn out tht there re couple of loose ends tht need to be tied up But essentilly, except for the fine points, this is the resoning tht shows tht the re under the grph of velocity function equls the distnce trveled by n object whose velocity is described by tht function In fct, lot of students my not wnt to red ny further But let s look t the fine points Pssge to the Limit The fct tht we cn find the distnce trveled by n object whose velocity is described by step function by mesuring the re under the grph of tht function, plus the fct tht there lwys exists step function which pproximtes given function to within n ccurcy so gret tht neither the humn eye nor electronic microscopes cn distinguish the two seems to indicte tht the Distnce = Are principle is true to within very high degree of ccurcy Let s consider, in fct, the question of exctly how much ccurcy we cn clim By tking t =10 7, we cn construct step function tht pproximtes the function v(t) =t/2 (witht rnging from 2 to 5) to within n ccurcy better thn 6 deciml plces It would seem resonble to conclude, then, tht the Distnce = Are principle is true for the velocity function v(t) = t/2 tlest to within n ccurcy of 6 deciml plces (Perhps this resoning is not quite s creful s it ought to be, but it s not off by much We ll show lter how to get quite precise estimte of the error involved) But we could just s well tke t =10 16, nd thus chieve n ccurcy of 15 deciml plces Or by choosing step function with still smller steps, we could chieve n ccurcy of 100 deciml plces If we now consider ll possible step function pproximtions to given function, we cn see tht the Distnce = Are principle is true up to ny conceivble degree of ccurcy In other words, the principle is just plin true, period This resoning is completely different from wht one sees nywhere in pre-clculus mthemtics nd it is the very hert of wht mkes clculus different from high school lgebr It goes bck to wht Archimedes clled the Method of Exhustion Nmely, in clculus one uses the ide tht by tking sequence of closer nd closer pproximtions one cn finlly rrive t limit which is exct, even though none of the pproximtions themselves re exct Stting the resoning bove in more conventionl mthemticl lnguge: s we consider ll possible step functions pproximting the velocity function, the re under these step functions converges to the re under the velocity function, nd the distnce corresponding to these step
functions converges to the distnces trveled by n object whose velocity is given by the originl function But for the step functions, the re nd the distnce trveled re the sme Therefore they converge to the sme limit, so the re under the originl velocity function nd the distnce trveled by the object re the sme If we use n rrow to indicte convergence to given function s we tke step functions where the width of the steps become smller nd smller, we cn cn grphiclly show this resoning by the following digrm: 15 Are of step function Are under grph of given function Distnce corresponding Distnce corresponding to step function to given velocity function The Step-function Approximtion Principle The preceding resoning, which we hve given in terms of the reltionship between velocity nd distnce, pplies just s well to the reltionship between force nd work, between cross-sectionl re nd volume, nd to ll the other mthemticl reltionships we hve considered In fct, this resoning seems to completely estblish the Second Rule of Thumb: if formul given by n integrl yields the correct nswer for constnt functions, then it is the correct formul Unfortuntely, however, the resoning given is flwed, becuse it depends on hidden ssumption Theredoinfctexistfew cumultive reltionships where the Second Rule of Thumb is not vlid The most common of these is the clcultion of the length of the grph of function y = f(x) The formul Length = b dx gives the correct nswer for the length of the curve y = f(x) between x = nd x = b in the cse when f(x) is constnt function (since in this cse Length = b ), nd yet is not correct in ny other cse (Notice tht the function f(x) is not even prt of the integrl) 3 2 1 b The grph of the constnt function f(x) =2 The length of the grph between x = nd x = b is b
16 Step Function Approximtions for the Motion of Flling Object Stted in prcticl terms, the fct tht the reltionship between the velocity of n object nd the distnce it trvels stisfies the Step Function Approximtion Principle sys tht if one only knows the velocity t some finite (but very lrge) number of time points nd computes distnce by mking the ssumption tht the velocity in between these time points is constnt, then by using enough different time points one cn get n rbitrrily good pproximtion to the true distnce the object trvels Let s try this out for the cse of flling object The velocity function of flling object is v(t) = 32tft/sec, (where t is mesured in seconds) We ll see wht hppens when we pproximte this by step function To strt with, let s ssume tht we re given the velocity of the object t intervls of 1 second nd mke the pproximting ssumption tht the velocity is constnt in between these time points Thus, for instnce, we might ssume tht during the first tenth of second the object s velocity is 0 (This is obviously incorrect, but we re using it s n pproximting ssumption) During the next tenth of second, we tke the object s velocity s 32 ft/sec, nd the corresponding distnce is 32 1 =3 ft Adding up the corresponding distnces, we get vlue of 0+32 + 64 + 96 + 128 + +896 + 928 = 1392 ft for the distnce the object flls during 3 seconds Since the true vlue is 16 t 2 =16 9 = 144, we see tht our pproximtion is considerbly on the low side Of course we didn t hve to tke the lowest possible velocity during ech time intervl s the vlue of the step function during tht intervl It would lso hve been resonble to hve ssumed tht the velocity is 32 ft/sec during the first tenth of second, 64 ft/sec during the second one, nd 96 ft/sec during the third 1 second This would give n pproximtion of 32 + 64 + 96 + +928 + 96 = 1488 ft for the totl distnce trveled, which is s much too lrge s the originl pproximtion ws too smll We cn notice, though, tht lthough neither of these two pproximtions is very good, the two pproximtions do brcket the true vlue of the distnce the object flls To improve the ccurcy, we might consider, for instnce, step function where the width of ech step is 01 sec (Thus we would be using 300 time points s the bsis for our pproximtion) For low-end pproximtion, we would ssume tht the object ws trveling t velocity of 0 during the first hundredth of second, velocity of 32 ft/sec during the second 01 second, etc For high-end pproximtion, we would ssume velocity of 32 ft/sec for the first 01 sec, 64 ft/sec for the second 01 sec, etc Without doing the rithmetic, let s note tht it is cler tht once gin the low-end pproximtion nd the high-end one will brcket the true vlue for the distnce fllen Furthermore, notice tht during the finl 01 sec, the lower step function uses vlue of 299 32 = 9568 ft/sec nd the higher one uses vlue of 3 32 = 96 ft/sec Thus the discrepncy between the two pproximting velocities used during the finl tenth of second is 32 ft/sec Furthermore, this is the biggest discrepncy for ny of the time intervls Thus we cn sy tht over ech time intervl, the difference between the higher step function nd the lower is t most 32 ft/sec This mens tht the difference between the distnce computed over the 3 second intervl on the bsis of the higher step function nd tht computed on the bsis of the lower will be smller (considerbly smller, in fct) then 3 32 = 96 ft Since these two pproximtions brcket the true vlue, we cn thus conclude tht the error in these pproximtions is smller thn 96 ft At this point, we re pproching n ccurcy tht might be stisfctory for mny engineering purposes But beyond this, we cn see from this logic tht by tking time intervl of, sy, 0001 sec, we would get n error smller thn 0096 ft, nd in fct, by using sufficiently short time intervls one could get ny desired degree of ccurcy
17 SIDEBAR: The Men Vlue Property In going through the clcultion for flling body, we defined two step functions For one of these, we mde the vlue of the step function t point t between t i nd t i+1 equl equl to the vlue v(t) tkes t the beginning of this intervl This choice ws obviously too smll For the second step function, we chose the vlue the v(t) tkes t the right end of the intervl, which ws clerly too lrge It might hve occurred to the reder tht it would hve been more intelligent to hve chosen the vlue tht v(t) tkes in the middle of the intervl Or perhps one could choose the verge of the vlues t the two ends The question of how to estimte n integrl most efficiently by using step functions is essentilly the topic of numericl integrtion nd is not relly the concern here It is interesting to notice, though, tht in the cse of the velocity function v(t) =32t, if one chooses the vlue tht v(t) tkes in the middle of ech intervl, then the nswer obtined is exctly correct, regrdless of the size of t This is essentilly n ccident More precisely, it is true for ny liner function More generlly, though, given ny resonble (for instnce, continuous) function v(t) defined between points t = nd t = b, then for ny t, even lrge one, there exists some step function pproximtion v 1 (t) tov(t) such tht the pproximtion to b v(t) dt obtined by using v 1(t) will be exctly correct In other words, if one divides the intervl [, b] up into sequence of n points t 0 =, t 1, t 2,, t n, where the distnce between ech pir of points is some pre-ssigned t, then it is possible to find numbers C 1, C 2, C 3,etc,suchthtechC i lies somewhere in between the smllest nd the lrgest vlue tht v(t) tkes on the intervl [t i 1,t i ], nd so tht n 1 C i t = b v(t)dt (In fct, if the originl velocity function v(t) is continuous, then one cn choose C i = v( t i )forsome t i with t i 1 t i t i ) This is becuse the reltionship between velocity nd distnce hs the Men Vlue Property Nmely, if n object trvels from time t 0 to time t 1 ccording to continuous velocity function v(t), then there exists time t in between t 0 nd t 1 such tht the distnce the object trvels equls (t 1 t 0 ) v( t ) Restted, this sys tht there is some moment in between t 0 nd t 1 when the velocity of the object is the sme s the verge velocity over the whole intervl The reson for this is esy to see Consider ll the possible vlues which cn be obtined in the form (t 1 t 0 )v(t), where t lies somewhere between t 0 nd t 1 Some of these vlues re clerly less thn the ctul distnce the object trvels For instnce, if we choose time t when the velocity tkes its minimum vlue, then (t 1 t 0 )v(t) gives vlue which is too low On the other hnd, for some t, (t 1 t 0 )v(t) is too lrge (for instnce when v(t) tkes its mximum vlue) (If cr trvels for n hour t speeds which re lwys between 30 mph nd 50 mph, then the distnce trveled will be greter thn 30 miles but less thn 50 miles We re ssuming here tht v(t) is not constnt) But since (t 1 t 0 )v(t) is continuous function of t, s it vries between vlues tht re too low nd vlues tht re too high, somewhere the must be time t where the vlue of (t 1 t 0 )v(t) is exctly correct A cumultive reltionship between function f(x) nd quntity Q will lwys hve the Men Vlue Property whenever (1) it is n incresing reltionship; nd (2) Q =(x 1 x 0 )f whenever f(x) isconstntfnd is pplied between x 0 nd x 1
18 Furthermore, by being sufficiently devious one cn sbotge the Second Rule of Thumb even in cses where it ought to work For instnce, the formul Distnce = t1 t 0 v(t)+8v (t)dt gives the correct nswer for constnt velocity functions, since if v(t) is constnt then the derivtive v (t) is 0, nd yet it is wrong in lmost ll other cses For this reson, it s good ide to understnd the hidden ssumption underlying our proof of the Second Rule of Thumb, even though when pplied to the exmples we hve been considering this ssumption is so nturl tht most clculus books tke it for grnted without even mentioning it The Hidden Assumption In the resoning bove we hve tken it for grnted tht if step function is n extremely good pproximtion to the ctul velocity function describing the motion of n object, then the distnce clculted by using this step function will be very close to the ctul distnce trveled by the object For convenience, I will cll this principle the Step Function Approximtion Principle The Step-function Approximtion Principle, which pplies not just to the reltionship between velocity nd distnce, but lso to tht between cross-sectionl re nd volume, force nd work, pressure nd force, s well s to mny other cumultive reltionships, is the missing piece we need in order to conclude from the resoning given previously the formul given by n integrl will be correct provided tht it gives the correct nswer for constnt functions This principle is in fct vlid for most cumultive reltionships Unfortuntely, though, there re few exceptions The length of curve is n exmple where the Step-function Approximtion Principle is not vlid Even if one chooses step function which is extremely close to given non-function, the length of the step function will not be close to the length of the given function Consequently, s seen bove, one cn t find the correct integrl formul for the length of the grph of function by looking for formul which gives the correct nswer for constnt functions The Step-function Approximtion Principle is n cid test for formuls given by integrls If the quntity in question cnnot be pproximted to within n rbitrry degree of ccurcy by replcing the function in question by step functions, then one cnnot find correct integrl formul to express this reltionship merely by choosing one which gives the correct nswer for constnt functions In most clculus books, formuls for n ppliction of integrtion re developed by first constructing pproximtions for the quntity in question by using step functions nd then tking the limit s the size of the steps goes to zero This is wht one is doing in the disk method for finding the volume of solid of revolution, for instnce To think of solid of revolution s being
19 SIDEBAR: The Volume of Solid of Revolution We hve mentioned before tht the volume of solid cn computed s the integrl of its horizontl or verticl cross-sections This cn be justified by the Step-function Approximtion Principle Consider in prticulr the cse of solid whose horizontl cross sections re circles If the rdius r(h) of the cross-section t height h is step-function, this would sy tht the solid consists of stck of disks h h r = r(h) r r = r(h) r We ssume tht the horizontl rdius of the solid of revolution t height h is determined by step-function r(h) (Note tht the independent vrible here is verticl, so tht the grph is turned 90 from the expected orienttion) The solid looks like stck of disks Ech disk hs cross-sectionl re of πr(h) 2, nd thus hs volume πr(h) 2 h This formul cn lso be written s π h2 h 1 r(h) 2 dh, since by ssumption r(h) is constnt between h 1 nd h 2 The formul H Volume = π r(h) 2 dh 0 for the volume of solid of revolution is correct under the ssumption tht the rdius r(h) is step-function of h, since it is correct for the cse of disk (i e cylinder) nd the volume of the whole is the sum of the volumes of the disks Since it seems intuitively cler tht s we mke the width of the steps smller nd smller, the resulting solid cn be mde to pproch ny desired solid of revolution rbitrrily closely nd tht the volumes will lso converge to within ny desired degree of ccurcy, we see tht the Step-function Approximtion Principle pplies nd so the formul H Volume = π f(h) 2 dh 0 is vlid for ny solid of revolution round the verticl xis, where f(h) denotes the rdius of the horizontl cross-section t height h
20 pproximted by bunch of disks is simply to think of the originl function f(x) which ws revolved round the x-xis s being replced by step function (See sidebr) However this is completely unnecessry You don t need to ctully use step functions in order to set n integrl up Step functions re needed for the proof, not the ctul clcultion You simply need to find formul tht gives the right nswer for constnt functions Wht is crucil, though, is to know tht one could in principle get n rbitrrily good pproximtion if one did pproximte the given function by step function This crucil step, however, is the one tht most clculus books give very little ttention to The conventionl tretment of pplictions of integrtion in most clculus books often ssumes without justifiction tht the Step-function Approximtion Principle will pply to the prticulr ppliction under considertion ( As the thickness of the disks goes pproches 0, the corresponding volume will pproch the volume of the given solid of revolution ) For most pplictions, this is highly plusible Furthermore, in trying to justify this ssumption more rigorously, one runs into the problem tht there s the sme difficulty in defining concepts such s work, volume, nd the like precisely tht there is in defining the concept of re rigorously In fct, in most physics books these concepts re simply defined by formuls in the form of integrls Work, for instnce, is defined to be the integrl of force with respect to distnce Stbility Wht is t issue in deciding whether the Step-function Approximtion Principle pplies in given sitution is not relly bout step functions t ll Rther, borrowing word from some other prts of mthemtics (nd perhps not using it quite correctly), the issue is one of stbility The reltionship between velocity nd distnce is stble, mening tht if one chnges the velocity function of n object by very smll mount (or imgines two objects whose velocity functions re very close to ech other), then the distnce trveled will not be very different Likewise the reltionship between the cross-sectionl res of solid nd its volume is stble: if the solid is chnged in such wy tht the cross-sectionl res re only slightly different, then the volume will lso chnge very little The notion of stbility rectifies the flw in my erlier proof of the Second Rule of Thumb With this flw remedied, this becomes no longer rule of thumb but theorem Suppose tht one is looking for formul for vrible quntity Q [for instnce, work] tht is determined by the vlues of function f(x) [such s force] for x between x = nd x = b Suppose tht the reltionship between the function f(x) nd the quntity Q is cumultive nd is stble in the sense tht if one mkes very smll chnge to the function f(x) then the resulting chnge in Q will lso be smll In this cse, the formul for Q is given by n integrl Furthermore, if resonble integrl formul gives the correct result in the cse of constnt functions, then this formul is in fct correct
21 SIDEBAR: Resonble Integrl Formuls The phrse resonble integrl formul occurs bove becuse, s n exmple further on will show, by being sufficiently dibolicl, one cn indeed contrive exceptions to the principle bove: nmely formuls which give the correct nswers for constnt functions nd yet fil for other functions One will hve resonble integrl formul if the expression one is integrting is obtined from the bsic function in question by pplying some continuous function of two vribles to it nd x Assuming tht f(x) is the bsic function involved, the following re exmples of legitimte integrnds when pplying the Step-function Approximtion Principle b f(x) 2 +4f(x)dx b xdx f(x) 2 +1 b x 2 e f(x) dx (As will be indicted below, the most common wy to go wrong is to use n integrnd tht involves f (x) orf (x), etc) A More Complicted Exmple: A Volume of Revolution Not ll importnt formuls given by integrls, hve the simple form Q = b f(x) dx Asn exmple of how the principle bove cn be pplied in more complicted sitution, consider the clssic problem of determining the volume of revolution resulting from revolving curve y = f(x) round the y-xis (The sitution is much simpler if one revolves the curve round the x-xis) Now if the function f(x) isconstnth, then the volume of revolution is cylinder with rdius b nd height H, nd its volume is known to be πb 2 H We wnt to see how to use this to derive the integrl formul for the volume when f(x) is not constnt For purposes of explntion, insted of merely considering cylinder, it is essentil to consider the volume between two concentric cylinders, which looks like cylinder with hole If the rdius of the inside cylinder is nd the outside rdius is b, then the volume in between is obtined by simply subtrcting the volume of the inside cylinder (the hole ) from tht of the cylinder s whole This gives Volume = πb 2 H π 2 H = π(b 2 2 )H
22 y y =8 2x x This cone cn be seen s the solid resulting from revolving the line y =8 2x round the y-xis As previously discussed, the volume cn be seen s determined by its horizontl cross-sections, whose re t height y is π(y 8) 2 /4, giving formul Volume = π 8 0 (y 8) 2 dy 4 But the volume cn lso be seen s determined by cylindricl verticl cross-sections (indicted by the dshed verticl lines), whose re t distnce x from the origin re given by 2πx(8 2x) This suggests formul Volume =2π 4 0 x(8 2x) dx It is not esy, though, to see how to verify the correctness of this formul We now wnt to replce this by n integrl formul, where the constnt height H is replced by function f(x) This is bit perplexing, though, becuse it s hrd to see wht the fctor (b 2 2 ) should become in the integrl formul Simply tking Volume = b πf(x) dx (?) is clerly not going to work, becuse when f(x) =H this gives the incorrect nswer Volume = πbh πh = π(b )H To try nd remedy this by writing Volume = π b doesn t even give well-formed integrl The formul Volume = π b f(x)(dx) 2 (?) f(x) d(x 2 ) (?) seems eqully nonsensicl (Actully, this lst one cn be justified theoreticlly, nd if interpreted in the right wy is ctully correct But, for beginners t lest, it just looks too flky) To find the correct formul, slightly rewrite the formul for the cse f(x) = H ( constnt): V = πh(b 2 2 )=πhx 2 b This wy of writing it mke it esy to see tht if we wnt n integrl formul V = b dx tht will produce this result, we simply need n integrnd tht will produce πhx 2 s its nti-derivtive when x=
H is constnt In other words, we need n nti-nti-derivtive for x 2 But n nti-nti-derivtive is simply derivtive, nd the derivtive of x 2 is 2x So to produce the correct nswer when f(x) =H, constnt, we should integrte 2πxH Thus V= b 2πxf(x) dx should be the desired formul In fct in the cse, if f(x) =H ( constnt), we get Volume = b = πh 2πxH dx b 2xdx= =πh(b 2 2 ), which is the correct nswer Since the formul yields the correct nswer when f(x) isconstnth, it is the correct formul in generl 23 A Reltionship Tht Is Not Stble As n exmple of mthemticl reltionship where one does not hve stbility, nd where the Step-function Approximtion Principle does not pply, consider the length of the grph of function y = f(x) between two points x = nd x = b The length of this curve surely hs cumultive reltionship to the function, since if c is vlue of x between nd b, then the length of the entire curve cn be obtined by dding together the length of tht portion between nd c plus the length of the portion between c nd b Therefore it is lmost certin tht the formul for the length will be given by n integrl c b The length of the curve for x between nd b is the sum of the length of tht portion between x = nd x = c plus the length between c nd b However the length of curve is not stbly relted to the function determining the curve One cn chnge the function in such wy tht t no point is the chnge very lrge, nd yet the chnge in length is enormous One cn, for instnce, wlk stright down street in such wy tht one s pth is stright line Or one could wlk down the sme street, but this time crossing from one side to nother every few feet The new criss-crossing pth would never be tht fr wy from the originl stright-line pth, but the distnce one wlks would be enormously longer (This ide occurs in the theory of frctls One cn strt by tking reltively nice curve, nd then chnge it by dding little bumps ll long it One cn then chnge it still more by dding little bumps long the little bumps, nd then dd still more bumps to those bumps Eventully one reches
24 SIDEBAR: The Leibnitz Approch Newton nd Leibnitz rgued fiercely s to which hd the right explntion of clculus, lthough in truth, neither ws completely correct Leibnitz s wy of explining things seems obviously crzy, nd indeed is crzy, s befits Germn philosopher whose min clim to fme is not his mthemtics but the crzy philosophicl ide tht the world consists of something clled monds (More precisely, mond is n entire world in itself, centered round one individul Every person in the world lives in his own mond Well, never mind) But if you cn get pst the fct tht it s crzy, Leibnitz s wy of looking t clculus is ctully quite nice nd gives relible results Furthermore, t lest his explntion is consistent with the nottion we ctully use for integrls I m going to suggest n explntion which is slightly different thn Leibnitz s, but is still crzy Nmely, suppose tht insted of the intervl between x = nd x = b being continuous, it is ctully mde up of huge number of extremely smll quntum pieces We write dx for the length of ech quntum piece (It s s if dx is the distnce from one tom of the number line to the next In fct, of course, the mthemticl number line, unlike lines in the physicl world, does not hve toms) Wht b f(x) dx then mens, ccording to this explntion, is tht we let x rnge over the huge but finite number of points between x = nd x = b, nd t ech of those points we compute f(x) nd multiply it by dx Then we dd up ll these vlues for f(x)dx Despite the wy tht this explntion is wrong nd even crzy, it does produce relible nswers, nd in my experience it s the wy most people think who ctully use clculus s tool When pplied to the volume of revolution exmple, this wy of thinking leds us to think of the volume s being mde up by gluing together n incredible number of incredibly thin concentric sheets (Anyone who s ever done ppier mché will understnd the ide) But these sheets hve width dx much thinner thn sheet of pper The totl volume of the solid will then be equl to the sum of the incredibly smll volumes of ll these ultr-thin sheets Now, t given distnce x from the xis of the solid (i e the y-xis), the sheet of pper (s it were) will hve length of 2πx nd height of f(x), nd therefore n re of 2πxf(x) Since the thickness is dx, the volume of the sheet is 2πxf(x) dx (This seems like completely vlid explntion, but creful clcultion will show tht it s not It doesn t tke into ccount the fct tht the sheet is curved nd the fct tht the top edge of the sheet is beveled to mtch the slope of the grph of y = f(x) However becuse of the ultr-thinness of the sheet, the error involved is fr smller thn dx; so smll tht it drops below the quntum level nd thus disppers This is the relly crzy prt of the explntion) Leibnitz s symbol is ctully n elongted S (but don t write it tht wy, unless you wnt everyone to know wht totl dork you re!) So b 2πxf(x) dx mens (ccording to this crzy explntion) tht we dd up ll these tiny little volumes This will give the totl volume of the solid Mthemticins generlly don t pprove of this explntion becuse it doesn t mke sense, nd insted put into clculus books rigorous clcultions tht re so tedious tht very few students re willing to go through them My suggestion is to use the Leibnitz pproch to come up with the formul in the first plce, nd then, if you hve ny doubts bout its correctness, use the principles I ve been explining here: in lmost ll cses, ll you hve to do is to check tht the formul you cme up with gives the correct nswer for constnt functions
25 the points where the chnges one is mking to the curve become so smll tht the eye cn t even detect them, nd yet if one tkes this process to the limit one gets curve which is infinitely long) When f(x) isconstnt function, then the grph of f(x) is horizontl line, nd its length is simply b Thus the formul Length = b dx (?) A stright-line pth nd two zig-zg pths between the sme two points The zig-zg pths re 26 timesslongsthe stright-line pth This rtio depends only on gives the correct nswer for the the ngle of the zig-zgs, not on their height length of the grph of constnt Consequently, we could mke the zig-zg pth function Nonetheless, this so close to the stright-line pth tht the eye could not distinguish them, nd yet the formul does not give the correct zig-zg pth would be much longer result for functions which re not constnt In fct, it lwys give vlues which re too smll (usully much too smll) for functions which re not constnt For instnce, if one considers the function f(x) =2x, then its grph is stright line with slope of 2, nd the length of this line between the points x =0 nd x= 1 is esily seen to be 5, s contrsted with the vlue 1 produced by the integrl formul bove This pprent prdox occurs becuse one does not get rbitrrily good pproximtions to the length of curve by replcing tht curve by step function In fct, the length of the grph (1,2) of step function between points x = nd x = b is lwys b Mking the jumps in the step function very smll does not ffect its length t ll (It is little strnge to even tlk bout the length of step function, since the grph hs breks in it However if we gree tht length is cumultive, then it is esy to see tht the length of step function hs to be the sum of the lengths of the horizontl pieces The verticl jumps do not contribute to the length) Since telling whether the Step Function Approximtion Principle pplies cn conceivbly sometimes be difficult judgement to mke, it s good to hve one tht s even esier to use