Chapter 6. The Integral. 6.1 Measuring Work. Human Work

Transcription

1 Chpter 6 The Integrl There re mny contexts work, energy, re, volume, distnce trvelled, nd profit nd loss re just few where the quntity in which we re interested is product of known quntities. For exmple, the electricl energy needed to burn three 1 wtt light bulbs for t hours is 3 t wtt-hours. In this exmple, though, the clcultion becomes more complicted if lights re turned off nd on during the time intervl t. We fce the sme compliction in ny context in which one of the fctors in product vries. To describe such product we will introduce the integrl. As you will see, the integrl itself cn be viewed s vrible quntity. By nlyzing the rte t which tht quntity chnges, we will find tht every integrl cn be expressed s the solution to prticulr differentil eqution. We will thus be ble to use ll our tools for solving differentil equtions to determine integrls. 6.1 Mesuring Work Humn Work Let s mesure the work done by the stff of n office tht processes ctlog orders. Suppose typicl worker in the office cn process 1 orders n hour. Then we would expect 6 people to process 6 orders n hour; in two hours, they could process 12 orders. Processing ctlog orders 1 orders per hour person 6 persons 2 hours = 12 orders. 337

2 338 CHAPTER 6. THE INTEGRAL Notice tht stff of 4 people working 3 hours could process the sme number of orders: 1 orders per hour person 4 persons 3 hours = 12 orders. Humn work is mesured s product Productivity rte Mowing lwns It is nturl to sy tht 6 persons working two hours do the sme mount of work s 4 persons working three hours. This suggests tht we use the product number of workers elpsed time to mesure humn work. In these terms, it tkes 12 person-hours of humn work to process 12 orders. Another nme tht hs been used in the pst for this unit of work is the mn-hour. If the tsk is lrge, work cn even be mesured in mnmonths or mn-yers. The term we will use most of the time is stffhour. Mesuring the work in terms of person-hours or stff-hours my seem little strnge t first fter ll, typicl mnger of our ctlog order office would be most interested in the number of orders processed; tht is, the production of the office. Notice, however, tht we cn re-phrse the rte t which orders re processed s 1 orders per stff-hour. This is sometimes clled the productivity rte. The productivity rte llows us to trnslte humn work into production: production = productivity rte humn work orders 12 orders = 1 12 stff-hours. stff-hour As this eqution shows, production vries linerly with work nd the productivity rte serves s multiplier (see our discussion of the multiplier on pges 31 33). If we modify the productivity rtein suitble wy, we cn use this eqution for other kinds of jobs. For exmple, we cn use it to predict how much mowing lwn mowing crew will do. Suppose the productivity rte is.7 cres per stff-hour. Then we expect tht stff of S working for H hours cn mow cres.7 SH stff-hours =.7 SH cres stff-hour of lwn ltogether.

3 6.1. MEASURING WORK 339 Production is mesured differently in different jobs s orders processed, or cres mowed, or houses pinted. However, in ll these jobs humn work is mesured in the sme wy, s stff-hours, which gives us common unit tht cn be trnslted from one job to nother. A stff of S working stedily for H hours does SH stff-hours of work. Suppose, though, the stffing level S is not constnt, s in the grph below. Cn we still find the totl mount of work done? Stff-hours provides common mesure of work in different jobs Non-constnt stffing stff S hours hours hours time hours The bsic formul works only when the stffing level is constnt. But stffing is constnt over certin time intervls. Thus, to find the totl mount of work done, we should simply use the bsic formul on ech of those intervls, nd then dd up the individul contributions. These clcultions re done in the following tble. The totl work is 42.5 stff-hours. So if the productivity rte is 1 orders per stff-hour, 425 orders cn be processed. The work done is sum of products Accumulted work 2 stff 1.5 hours = 3. stff-hours = = 12. totl = 42.5 stff-hours The lst clcultion tells us how much work got done over n entire dy. Wht cn we tell n office mnger who wnts to know how work is progressing during the dy? At the beginning of the dy, only two people re working, so fter the first T hours (where T 1.5) work done up to time T = 2 stff T hours = 2 T stff-hours.

4 34 CHAPTER 6. THE INTEGRAL Work ccumultes t rte equl to the number of stff S is the derivtive of W, so... Even before we consider wht hppens fter 1.5 hours, this expression clls our ttention to the fct tht ccumulted work is function let s denote it W(T). According to the formul, for the first 1.5 hours W(T) is liner function whose multiplier is W = 2 stff-hours. hour This multiplier is the rte t which work is being ccumulted. It is lso the slope of the grph of W(T) over the intervl T 1.5. With this insight, we cn determine the rest of the grph of W(T). Wht must W(T) look like on the next time intervl 1.5 T 7? Here 5 members of stff re working, so work is ccumulting t the rte of 5 stff-hours per hour. Therefore, on this intervl the grph of W is stright line segment whose slope is 5 stff-hours per hour. On the third intervl, the grph is nother stright line segment whose slope is 3 stff-hours per hour. The complete grph of W(T) is shown below. As the grphs show, the slope of the ccumulted work function W(T) is the height of the stffing function S(T). In other words, S is the derivtive of W: W (T) = S(T). stff S T hours stff-hours W 4 3 (11, 42.5) 3 (7, 3.5) T 2 (1.5, 3) hours The ccumulted work function W(T)

5 6.1. MEASURING WORK 341 Notice tht the units for W nd for S re comptible: the units for W re stff-hours per hour, which we cn think of s stff, the units for S. We cn describe the reltion between S nd W nother wy. At the moment, we hve explined S in terms of W. However, since we strted with S, it is relly more pproprite to reverse the roles, nd explin W in terms of S. Chpter 4.5 gives us the lnguge to do this: W is n ntiderivtive of S. In other words, y = W(T) is solution to the differentil eqution dy dt = S(T). As we find ccumultion functions in other contexts, this reltion will give us crucil informtion. Before leving this exmple we note some specil fetures of S nd W. The stffing function S is sid to be piecewise constnt, or step function. The grphs illustrte the generl fct tht the derivtive of piecewise liner function (W, in this cse) is piecewise constnt.... W is n ntiderivtive of S The derivtive of piecewise liner function Summry The exmple of humn work illustrtes the key ides we will meet, gin nd gin, in different contexts in this chpter. Essentilly, we hve two functions W(t) nd S(t) nd two different wys of expressing the reltion between them: On the one hnd, while on the other hnd, W(t) is n ccumultion function for S(t), S(t) is the derivtive of W(t). Exploring the fr-reching implictions of functions connected by such twofold reltionship will occupy the rest of this chpter.

6 342 CHAPTER 6. THE INTEGRAL The nlogy between electricl energy nd humn work Electricl Energy Just s humns do work, so does electricity. A power compny chrges customers for the work done by the electricity it supplies, nd it mesures tht work in wy tht is strictly nlogous to the wy we mesure humn work. The work done by electricity is usully referred to s (electricl) energy. For exmple, suppose we illuminte two light bulbs one rted t 1 wtts, the other t 6 wtts. It will tke the sme mount of electricl energy to burn the 1-wtt bulb for 3 hours s it will to burn the 6-wtt bulb for 5 hours. Both will use 3 wtt-hours of electricity. The power of the light bulb mesured in wtts is nlogous to the number of stff working (nd, in fct, workers hve sometimes been clled mnpower). The time the bulb burns is nlogous to the time the stff work. Finlly, the product energy = power elpsed time for electricity is nlogous to the product work = number of stff elpsed time for humn effort. Electric power is mesured in wtts, in kilowtts (= 1, wtts), nd in megwtts (= 1,, wtts). Electric energy is mesured in wtt-hours, in kilowtt-hours (bbrevited kwh ) nd in megwtt-hours (bbrevited mwh ). Since n individul electricl pplince hs power demnd of bout one kilowtt, kwh re suitble units to use for describing the energy consumption of house, while mwh re more nturl for whole town. Suppose the power demnd of town over 24 hour period is described by the following grph: power 9 megwtts 6 3 time hours

7 6.1. MEASURING WORK 343 Since this grph decribes power, its verticl height over ny point t on the time xis tells us the totl wttge of the light bulbs, dishwshers, computers, etc. tht re turned on in the town t tht instnt. This demnd fluctutes between 3 nd 9 megwtts, roughly. The problem is to determine the totl mount of energy used in dy how mny megwtt-hours re there in this grph? Although the eqution Power is nlogous to stffing level energy = power elpsed time, gives the bsic reltion between energy nd power, we cn t use it directly becuse the power demnd isn t constnt. The stffing function S(t) we considered erlier wsn t constnt, either, but we were still ble to compute stff-hours becuse S(t) ws piecewise constnt. This suggests tht we should replce the power grph by piecewise constnt grph tht pproximtes it. Here is one such pproximtion: A piecewise constnt pproximtion power megwtts time hours As you cn see, the step function hs five steps, so our pproximtion to the totl energy consumption of the town will be sum of five individul products: energy = 1447 mwh. This vlue is only n estimte, though. How cn we get better estimte? The nswer is cler: strt with step function tht pproximtes the power grph more closely. In principle, we cn get s good n pproximtion s we might desire this wy. We re limited only by the precision of the power grph itself. As our pproximtion to the power grph improves, so does the ccurcy of the clcultion tht estimtes energy consumption. Better estimtes

8 344 CHAPTER 6. THE INTEGRAL In summry, we determine the energy consumption of the town by sequence of successive pproximtions. The steps in the sequence re listed in the box below. 1. Approximte the power demnd by step function. 2. Estimte energy consumption from this pproximtion. 3. Improve the energy estimte by choosing new step function tht follows power demnd more closely. Accumulted energy consumption Energy ccumultion Energy is being consumed stedily over the entire dy; cn we determine how much energy hs been used through the first T hours of the dy? We ll denote this quntity E(T) nd cll it the energy ccumultion function. For exmple, we lredy hve the estimte E(24) = 1447 mwh; cn we estimte E(3) or E(17.6)? Once gin, the erlier exmple of humn effort cn guide us. We sw tht work ccumultes t rte equl to the number of stff present: W (T) = S(T). Since S(T) ws piecewise constnt, this rte eqution llowed us to determine W(T) s piecewise liner function. We clim tht there is n nlogous reltion between ccumulted energy consumption nd power demnd nmely E (T) = p(t). Estimting E (T) Unlike S(T), the function p(t) is not piecewise constnt. Therefore, the rgument we used to show tht W (T) = S(T) will not work here. We need nother rgument. To explin why the differentil eqution E (T) = p(t) should be true, we will strt by nlyzing E (T) E T E(T + T) E(T) =. T Assume we hve mde T so smll tht, to the level of precision we require, the pproximtion E/ T grees with E (T). The numertor E is, by

9 6.1. MEASURING WORK 345 definition, the totl energy used up to time T + T, minus the totl energy used up to time T. This is just the energy used during the time intervl T tht runs from time T to time T + T: E = energy used between times T nd T + T. Since the elpsed time T is smll, the power demnd should be nerly constnt, so we cn get good estimte for energy consumption from the bsic eqution energy used = power elpsed time. In prticulr, if we represent the power by p(t), which is the power demnd t the beginning of the time period from T to T + T, then we hve E p(t) T. Using this vlue in our pproximtion for the derivtive E (T), we get E (T) E T p(t) T T = p(t). Tht is, E (T) p(t), nd the pproximtion becomes more nd more exct s the time intervl T shrinks to. Thus, E E (T) = lim T T = p(t). Here is nother wy to rrive t the sme conclusion. Our strting point is the bsic formul E p(t) T, which holds over smll time intervl T. This formul tells us how E responds to smll chnges in T. But tht is exctly wht the microscope eqution tells us: E E (T) T. Since these equtions give the sme informtion, their multipliers must be the sme: p(t) = E (T). In words, the differentil eqution E = p sys tht power is the rte t which energy is consumed. In purely mthemticl terms: A second wy to see tht E = p

10 346 CHAPTER 6. THE INTEGRAL The energy ccumultion function y = E(t) is solution to the differentil eqution dy/dt = p(t). In fct, y = E(t) is the solution to the initil vlue problem dy dt = p(t) y() =. We cn use ll the methods described in chpter 4.5 to solve this problem. The reltion we hve explored between power nd energy cn be found in n nlogous form in mny other contexts, s we will see in the next two sections. In section 4 we will turn bck to ccumultion functions nd investigte them s solutions to differentil equtions. Then, in chpter 11, we will look t some specil methods for solving the prticulr differentil equtions tht rise in ccumultion problems. Exercises Humn work 1. House-pinting is job tht cn be done by severl people working simultneously, so we cn mesure the mount of work done in stff-hours. Consider house-pinting business run by some students. Becuse of clss schedules, different numbers of students will be pinting t different times of the dy. Let S(T) be the number of stff present t time T, mesured in hours from 8 m, nd suppose tht during n 8-hour work dy, we hve 3, T < 2, S(T) = 2, 2 T < 4.5, 4, 4.5 T 8. ) Drw the grph of the step function defined here, nd compute the totl number of stff hours. b) Drw the grph tht shows how stff-hours ccumulte on this job. This is the grph of the ccumulted work function W(T). (Compre the grphs of stff nd stff-hours on pge 34.) c) Determine the derivtive W (T). Is W (T) = S(T)?

11 6.1. MEASURING WORK Suppose tht there is house-pinting job to be done, nd by pst experience the students know tht four of them could finish it in 6 hours. But for the first 3.5 hours, only two students cn show up, nd fter tht, five will be vilble. ) How long will the whole job tke? [Answer: 6.9 hours.] b) Drw grph of the stffing function for this problem. Mrk on the grph the time tht the job is finished. c) Drw the grph of the ccumulted work function W(T). d) Determine the derivtive W (T). Is W (T) = S(T)? Averge stffing. Suppose job cn be done in three hours when 6 people work the first hour nd 9 work during the lst two hours. Then the job tkes 24 stff-hours of work, nd the verge stffing is verge stffing = 24 stff-hours 3 hours = 8 stff. This mens tht constnt stffing level of 8 persons cn ccomplish the job in the sme time tht the given vrible stffing level did. Note tht the verge stffing level (8 persons) is not the verge of the two numbers 9 nd 6! 3. Wht is the verge stffing of the jobs considered in exercises 1 nd 2, bove? 4. ) Drw the grph tht shows how work would ccumulte in the job described in exercise 1 if the work-force ws kept t the verge stffing level insted of the vrying level described in the exercise. Compre this grph to the grph you drew in exercise 1 b. b) Wht is the derivtive W (T) of the work ccumultion function whose grph you drew in prt ()? 5. Wht is the verge stffing for the job described by the grph on pge 339? Electricl energy 6. On Mondy evening, 15 wtt spce heter is left on from 7 until 11 pm. How mny kilowtt-hours of electricity does it consume?

12 348 CHAPTER 6. THE INTEGRAL 7. ) Tht sme heter lso hs settings for 5 nd 1 wtts. Suppose tht on Tuesdy we put it on the 1 wtt setting from 6 to 8 pm, then switch to 15 wtts from 8 till 11 pm, nd then on the 5 wtt setting through the night until 8 m, Wednesdy. How much energy is consumed (in kwh)? b) Sketch the grphs of power demnd p(t) nd ccumulted energy consumption E(T) for the spce heter from Tuesdy evening to Wednesdy morning. Determine whether E (T) = p(t) in this cse. c) The verge power demnd of the spce heter is defined by: verge power demnd = energy consumption. elpsed time If energy consumption is mesured in kilowtt-hours, nd time in hours, then we cn mesure verge power demnd in kilowtts the sme s power itself. (Notice the similrity with verge stffing.) Wht is the verge power demnd from Tuesdy evening to Wednesdy morning? If the heter could be set t this verge power level, how would the energy consumption compre to the ctul energy consumption you determined in prt ()? 8. The grphs on pges 342 nd 343 describe the power demnd of town over 24-hour period. Give n estimte of the verge power demnd of the town during tht period. Explin wht you did to produce your estimte. [Answer: 6.29 megwtts is one estimte.] Work s force distnce The effort it tkes to move n object is lso clled work. Since it tkes twice s much effort to move the object twice s fr, or to move nother object tht is twice s hevy, we cn see tht the work done in moving n object is proportionl to both the force pplied nd to the distnce moved. The simplest wy to express this fct is to define work = force distnce. For exmple, to lift weight of 2 pounds stright up it tkes 2 pounds of force. If the verticl distnce is 3 feet then 2 pounds 3 feet = 6 foot-pounds

13 6.1. MEASURING WORK 349 of work is done. Thus, once gin the quntity we re interested in hs the form of product. The foot-pound is one of the stndrd units for mesuring work. 9. Suppose trctor pulls loded wgon over rod whose steepness vries. If the first 15 feet of rod re reltively level nd the trctor hs to exert only 2 pounds of force while the next 4 feet re inclined nd the trctor hs to exert 55 pounds of force, how much work does the trctor do ltogether? 1. A motor on lrge ship is lifting 2 pound nchor tht is lredy out of the wter t the end of 3 foot chin. The chin weighs 4 pounds per foot. As the motor lifts the nchor, the prt of the chin tht is hnging gets shorter nd shorter, thereby reducing the weight the motor must lift. ) Wht is the combined weight of nchor nd hnging chin when the nchor hs been lifted x feet bove its initil position? b) Divide the 3-foot distnce tht the nchor must move into 3 equl intervls of 1 feet ech. Estimte how much work the motor does lifting the nchor nd chin over ech 1-foot intervl by multiplying the combined weight t the bottom of the intervl by the 1-foot height. Wht is your estimte for the totl work done by the motor in rising the nchor nd chin 3 feet? c) Repet ll the steps of prt (b), but this time use 3 equl intervls of 1 foot ech. Is your new estimte of the work done lrger or smller thn your estimte in prt (b)? Which estimte is likely to be more ccurte? On wht do you bse your judgment? d) If you ignore the weight of the chin entirely, wht is your estimte of the work done? How much extr work do you therefore estimte the motor must do to rise the hevy chin long with the nchor?

14 35 CHAPTER 6. THE INTEGRAL 6.2 Riemnn Sums In the lst section we estimted energy consumption in town by replcing the power function p(t) by step function. Let s puse to describe tht process in somewht more generl terms tht we cn dpt to other contexts. The power grph, the pproximting step function, nd the energy estimte re shown below. power 9 megwtts t 1 t 2 t 3 t 4 t 5 t 1 t 2 t 3 t 4 t time hours energy = 1447 mwh. Smpling the power function The height of the first step is 28.5 megwtts. This is the ctul power level t the time t 1 indicted on the grph. Tht is, p(t 1 ) = 28.5 megwtts. We found power level of 28.5 megwtts by smpling the power function t the time t 1. The height of the first step could hve been different if we hd smpled the power function t different time. In generl, if we smple the power function p(t) t the time t 1 in the intervl t 1, then we would estimte the energy used during tht time to be energy p(t 1 ) t 1 mwh. Notice tht t 1 is not in the middle, or t either end, of the first intervl. It is simply time when the power demnd is representtive of wht s hppening over the entire intervl. Furthermore, t 1 is not even unique; there is nother smpling time (ner t = 5 hours) when the power level is gin 28.5 megwtts. We cn describe wht hppens in the other time intervls the sme wy. If we smple the k-th intervl t the point t k, then the height of the k-th power step will be p(t k ) nd our estimte for the energy used during tht

15 6.2. RIEMANN SUMS 351 time will be energy p(t k ) t k mwh. We now hve generl wy to construct n pproximtion for the power function nd n estimte for the energy consumed over 24-hour period. It involves these steps. A procedure for pproximting power level nd energy use 1. Choose ny number n of subintervls, nd let them hve rbitrry positive widths t 1, t 2,..., t n, subject only to the condition t t n = 24 hours. 2. Smple the k-th subintervl t ny point t k, nd let p(t k ) represent the power level over this subintervl. 3. Estimte the energy used over the 24 hours by the sum energy p(t 1 ) t 1 + p(t 2 ) t p(t n ) t n mwh. The expression on the right is clled Riemnn sum for the power function p(t) on the intervl t 24 hours. The work of Bernhrd Riemnn ( ) hs hd profound influence on contemporry mthemticins nd physicists. His revolutionry ides bout the geometry of spce, for exmple, re the bsis for Einstein s theory of generl reltivity. The enormous rnge of choices in this process mens there re innumerble wys to construct Riemnn sum for p(t). However, we re not relly interested in rbitrry Riemnns sums. On the contrry, we wnt to build Riemnn sums tht will give us good estimtes for energy consumption. Therefore, we will choose ech subintervl t k so smll tht the power demnd over tht subintervl differs only very little from the smpled vlue p(t k ). A Riemnn sum constructed with these choices will then differ only very little from the totl energy used during the 24-hour time intervl. Essentilly, we use Riemnn sum to resolve dilemm. We know the bsic formul energy = power time works when power is constnt, but in generl power isn t constnt tht s the dilemm. We resolve the dilemm by using insted sum of terms of the form power time. With this sum we get n estimte for the energy. Choices tht led to good estimtes The dilemm

16 352 CHAPTER 6. THE INTEGRAL In this section we will explore some other problems tht present the sme dilemm. In ech cse we will strt with bsic formul tht involves product of two constnt fctors, nd we will need to dpt the formul to the sitution where one of the fctors vries. The solution will be to construct Riemnn sum of such products, producing n estimte for the quntity we were fter in the first plce. As we work through ech of these problems, you should puse to compre it to the problem of energy consumption. Estimting velocity nd distnce Clculting Distnce Trvelled It is esy to tell how fr cr hs trvelled by reding its odometer. The problem is more complicted for ship, prticulrly siling ship in the dys before electronic nvigtion ws common. The crew lwys hd instruments tht could mesure or t lest estimte the velocity of the ship t ny time. Then, during ny time intervl in which the ship s velocity is constnt, the distnce trvelled is given by the fmilir formul distnce = velocity elpsed time. Smpling the velocity function If the velocity is not constnt, then this formul does not work. The remedy is to brek up the long time period into severl short ones. Suppose their lengths re t 1, t 2,..., t n. By ssumption, the velocity is function of time t; let s denote it v(t). At some time t k during ech time period t k mesure the velocity: v k = v(t k ). Then the Riemnn sum v(t 1 ) t 1 + v(t 2 ) t v(t n ) t n is n estimte for the totl distnce trvelled. For exmple, suppose the velocity is mesured five times during 15 hour trip once every three hours s shown in the tble below. Then the bsic formul distnce = velocity elpsed time. gives us n estimte for the distnce trvelled during ech three-hour period, nd the sum of these distnces is n estimte of the totl distnce trvelled during the fifteen hours. These clcultions pper in the right-hnd column of the tble. (Note tht the first mesurement is used to clculte the distnce trvelled between hours nd 3, while the lst mesurement, tken 12 hours fter the strt, is used to clculte the distnce trvelled between hours 12 nd 15.)

17 6.2. RIEMANN SUMS 353 smpling time elpsed time velocity distnce trvelled (hours) (hours) (miles/hour) (miles) = = = = = Thus we estimte the ship hs trvelled miles during the fifteen hours. The estimted distnce The number 61.65, obtined by dding the numbers in the right-most column, is Riemnn sum for is Riemnn sum for the velocity function. the velocity function Consider the specific choices tht we mde to construct this Riemnn sum: t 1 = t 2 = t 3 = t 4 = t 5 = 3 t 1 =, t 2 = 3, t 3 = 6, t 4 = 9, t 5 = 12. These choices differ from the choices we mde in the energy exmple in two notble wys. First, ll the subintervls here re the sme size. This Intervls nd smpling is becuse it is nturl to tke velocity redings t regulr time intervls. times re chosen in systemtic wy By contrst, in the energy exmple the subintervls were of different widths. Those widths were chosen in order to mke piecewise constont function tht followed the power demnd grph closely. Second, ll the smpling times lie t the beginning of the subintervls in which they pper. Agin, this is nturl nd convenient for velocity mesurements. In the energy exmple, the smpling times were chosen with n eye to the power grph. Even though we cn mke rbitrry choices in constructing Riemnn sum, we will do it systemticlly whenever possible. This mens choosing subintervls of equl size nd smpling points t the sme plce within ech intervl. Let s turn bck to our estimte for the totl distnce. Since the velocity of the ship could hve fluctuted significntly during ech of the three-hour periods we used, our estimte is rther rough. To improve the estimte we could mesure the velocity more frequently for exmple, every 15 minutes. If we did, the Riemnn sum would hve 6 terms (four distnces per hour for 15 hours). The individul terms in the sum would ll be much smller, though, becuse they would be estimtes for the distnce trvelled in 15 Improving the distnce estimte

18 354 CHAPTER 6. THE INTEGRAL minutes insted of in 3 hours. For instnce, the first of the 6 terms would be 1.4 miles.25 hours =.35 miles. hour Of course it my not mke prcticl sense to do such precise clcultion. Other fctors, such s wter currents or the inccurcy of the velocity mesurements themselves, my keep us from getting good estimte for the distnce. Essentilly, the Riemnn sum is only model for the distnce covered by ship. Clculting Ares The re of rectngle is just the product of its length nd its width. How cn we mesure the re of region tht hs n irregulr boundry, like the one t the left? We would like to use the bsic formul re = length width. However, since the region doesn t hve stright sides, there is nothing we cn cll length or width to work with. We cn begin to del with this problem by breking up the region into smller regions tht do hve stright sides with, t most, only one curved side. This cn be done mny different wys. The lower figure shows one possibility. The sum of the res of ll the little regions will be the re we re looking for. Although we hven t yet solved the originl problem, we hve t lest reduced it to nother problem tht looks simpler nd my be esier to solve. Let s now work on the reduced problem for the shded region.

19 6.2. RIEMANN SUMS 355 Here is the shded region, turned so tht it sits flt on one of its stright sides. We would like to clculte its re using the formul width height, but this formul pplies only to rectngles. We cn, however, pproximte the region by collection of rectngles, s shown t the right. The formul does pply to the individul rectngles nd the sum of their res will pproximte the re of the whole region. y To get the re of rectngle, we must mesure its width nd height. Their heights vry with the height of the curved top of the shded region. To f(x) describe tht height in systemtic wy, we hve plced the shded region in coordinte plne so tht it sits on the x-xis. The other two stright sides lie on the verticl lines x = nd x = b. x b x The curved side defines the grph of function y = f(x). Therefore, t ech point x, the verticl height from the xis to the curve is f(x). By introducing coordinte plne we gin ccess to mthemticl tools such s the lnguge of functions to describe the vrious res. The k-th rectngle hs been singled out on the left, below. We let x k denote the width of its bse. By smpling the function f t properly chosen point x k in the bse, we get the height f(x k ) of the rectngle. Its re is therefore f(x k ) x k. If we do the sme thing for ll n rectngles shown on the right, we cn write their totl re s f(x 1 ) x 1 + f(x 2 ) x f(x n ) x n. Clculting the res of the rectngles y y f(x k ) x k x k b x b x

20 356 CHAPTER 6. THE INTEGRAL The re estimte is Riemnn sum Notice tht our estimte for the re hs the form of Riemnn sum for the height function f(x) over the intervl x b. To get better estimte, we should use nrrower rectngles, nd more of them. In other words, we should construct nother Riemnn sum in which the number of terms, n, is lrger nd the width x k of every subintervl is smller. Putting it yet nother wy, we should smple the height more often. Consider wht hppens if we pply this procedure to region whose re we know lredy. The semicircle of rdius r = 1 hs n re of πr 2 /2 = π/2 = The semicircle is the grph of the function f(x) = 1 x 2, which lies over the intervl 1 x 1. To get the figure on the left, we smpled the height f(x) t 2 evenly spced points, strting with x = 1. In the better pproximtion on the right, we incresed the number of smple points to 5. The vlues of the shded res were clculted with the progrm RIEMANN, which we will develop lter in this section. Note tht with 5 rectngles the Riemnn sum is within.5 of π/2, the exct vlue of the re. y y -1 1 x -1 1 x shded re = shded re = Clculting Lengths It is to be expected tht products nd ultimtely, Riemnn sums will be involved in clculting res. It is more surprising to find tht we cn use them to clculte lengths, too. In fct, when we re working in coordinte plne, using product to describe the length of stright line is even quite nturl.

21 6.2. RIEMANN SUMS 357 To see how this cn hppen, consider line segment in the x, y-plne tht hs known slope m. If we lso know the horizontl seprtion between the two ends, we cn find the length of the segment. Suppose the horizontl seprtion is x nd m the verticl seprtion y. Then the length of the segment is x x2 + y 2 The length of stright line by the Pythgoren theorem (see pge 9). Since y = m x, we cn rewrite this s x2 + (m x) 2 = x 1 + m 2. In other words, if line hs slope m nd it is x units wide, then its length is the product 1 + m2 x. Suppose the line is curved, insted of stright. Cn we describe its length the sme wy? We ll ssume tht the curve is the grph y = g(x). The compliction is tht the slope m = g (x) now vries with x. If g (x) doesn t vry too much over n intervl of length x, then the curve is nerly stright. Pick point x in tht intervl nd smple the slope g (x ) x g (x ) there; we expect the length of the curve to be pproximtely x 1 + (g (x )) 2 x. The length of curved line As the figure shows, this is the exct length of the stright line segment tht lies over the sme intervl x nd is tngent to the curve t the point x = x. If the slope g (x) vries pprecibly over the intervl, we should subdivide the intervl into smll pieces x 1, x 2,..., x n, over which the curve is nerly stright. Then, if we smple the slope t the point x k in the k-th subintervl, the sum 1 + (g (x 1 )) 2 x (g (x n )) 2 x n will give us n estimte for the totl length of the curve.

22 358 CHAPTER 6. THE INTEGRAL The length of curve is estimted by Riemnn sum Once gin, we find n expression tht hs the form of Riemnn sum. There is, however, new ingredient worth noting. The estimte is Riemnn sum not for the originl function g(x) but for nother function f(x) = 1 + (g (x)) 2 tht we constructed using g. The importnt thing is tht the length is estimted by Riemnn sum for some function. y y π x π x 4-segment length = segment length = The figure bove shows two estimtes for the length of the grph of y = sin x between nd π. In ech cse, we used subintervls of equl length nd we smpled the slope t the left end of ech subintervl. As you cn see, the four segments pproximte the grph of y = sin x only very roughly. When we increse the number of segments to 2, on the right, the pproximtion to the shpe of the grph becomes quite good. Notice tht the grph itself is not shown on the right; only the 2 segments. To clculte the two lengths, we constructed Riemnn sums for the function f(x) = 1 + cos 2 x. We used the fct tht the derivtive of g(x) = sin x is g (x) = cosx, nd we did the clcultions using the progrm RIEMANN. By using the progrm with still smller subintervls you cn show tht the exct length = Thus, the 2-segment estimte is lredy ccurte to four deciml plces. We hve lredy constructed estimtes for the length of curve, in chpter 2 (pges 89 91). Those estimtes were sums, too, but they were not Riemnn sums. The terms hd the form x2 + y 2 ; they were not products of the form 1 + m 2 x. The sums in chpter 2 my seem more strightforwrd. However, we re developing Riemnn sums s powerful generl tool for deling with mny different questions. By expressing lengths s Riemnn sums we gin ccess to tht power.

23 6.2. RIEMANN SUMS 359 Definition The Riemnn sums tht pper in the clcultion of power, distnce, nd length re instnces of generl mthemticl object tht cn be constructed for ny function whtsoever. We puse now to describe tht construction prt from ny prticulr context. In wht follows it will be convenient for us to write n intervl of the form x b more compctly s [, b]. Nottion: [, b] Definition. Suppose the function f(x) is defined for x in the intervl [, b]. Then Riemnn sum for f(x) on [, b] is n expression of the form f(x 1 ) x 1 + f(x 2 ) x f(x n ) x n. The intervl [, b] hs been divided into n subintervls whose lengths re x 1,..., x n, respectively, nd for ech k from 1 to n, x k is some point in the k-th subintervl. Notice tht once the function nd the intervl hve been specified, Riemnn sum is determined by the following dt: Dt for Riemnn sum A decomposition of the originl intervl into subintervls (which determines the lengths of the subintervls). A smpling point chosen from ech subintervl (which determines vlue of the function on ech subintervl). A Riemnn sum for f(x) is sum of products of vlues of x nd vlues of y = f(x). If x nd y hve units, then so does the Riemnn sum; its units re the units for x times the units for y. When Riemnn sum rises in prticulr context, the nottion my look different from wht ppers in the definition just given: the vrible might not be x, nd the function might not be f(x). For exmple, the energy pproximtion we considered t the beginning of the section is Riemnn sum for the power demnd function p(t) on [, 24]. The length pproximtion for the grph y = sin x is Riemnn sum for the function 1 + cos 2 x on [, π]. It is importnt to note tht, from mthemticl point of view, Riemnn sum is just number. It s the context tht provides the mening: Riemnn sums for power demnd tht vries over time pproximte totl Units A Riemnn sum is just number

24 36 CHAPTER 6. THE INTEGRAL energy consumption; Riemnn sums for velocity tht vries over time pproximte totl distnce; nd Riemnn sums for height tht vries over distnce pproximte totl re. To illustrte the generlity of Riemnn sum, nd to stress tht it is just number rrived t through rbitrry choices, let s work through n exmple without context. Consider the function f(x) = 1 + x 3 on [1, 3]. The dt We will brek up the full intervl [1, 3] into three subintervls [1, 1.6], [1.6, 2.3] nd [2.3, 3]. Thus x 1 =.6 x 2 = x 3 =.7. Next we ll pick point in ech subintervl, sy x 1 = 1.3, x 2 = 2 nd x 3 = 2.8. Here is the dt lid out on the x-xis. x 1 =.6 x 2 =.7 x 2 = x 3 With this dt we get the following Riemnn sum for 1 + x 3 on [1, 3]: f(x 1 ) x 1 + f(x 2 ) x 2 + f(x 3 ) x 3 = = Systemtic dt In this cse, the choice of the subintervls, s well s the choice of the point x k in ech subintervl, ws hphzrd. Different dt would produce different vlue for the Riemnn sum. Keep in mind tht n individul Riemnn sum is not especilly significnt. Ultimtely, we re interested in seeing wht hppens when we reclculte Riemnn sums with smller nd smller subintervls. For tht reson, it is helpful to do the clcultions systemticlly. Clculting Riemnn sum lgorithmiclly. As we hve seen with our contextul problems, the dt for Riemnn sum is not usully chosen in hphzrd fshion. In fct, when deling with functions given by formuls, such s the function f(x) = 1 x 2 whose grph is semicircle, it pys to be systemtic. We use subintervls of equl size nd pick the sme point

25 6.2. RIEMANN SUMS 361 from ech subintervl (e.g., lwys pick the midpoint or lwys pick the left endpoint). The benefit of systemtic choices is tht we cn write down the computtions involved in Riemnn sum in simple lgorithmic form tht cn be crried out on computer. Let s illustrte how this strtegy pplies to the function 1 + x 3 on [1, 3]. Since the whole intervl is 3 1 = 2 units long, if we construct n subintervls of equl length x, then x = 2/n. For every k = 1,..., n, we choose the smpling point x k to be the left endpoint of the k-th subintervl. Here is picture of the dt: x x x x... x 1 x 2 x 3 x n 3 x x x 1 + (n 1) x In this systemtic pproch, the spce between one smpling point nd the next is x, the sme s the width of subintervl. This puts the k-th smpling point t x = 1 + (k 1) x. In the following tble, we dd up the terms in Riemnn sum S for f(x) = 1 + x 3 on the intervl [1, 3]. We used n = 4 subintervls nd lwys smpled f t the left endpoint. Ech row shows the following: 1. the current smpling point; 2. the vlue of f t tht point; 3. the current term S = f x in the sum; 4. the ccumulted vlue of S. left current current ccumulted endpoint 1 + x 3 S S The Riemnn sum S ppers s the finl vlue in the fourth column. The progrm RIEMANN, below, will generte the lst two columns in the bove tble. The sttement x = on the sixth line determines the position

26 362 CHAPTER 6. THE INTEGRAL of the first smpling point. Within the FOR NEXT loop, the sttement x = x + deltx moves the smpling point to its next position. Progrm: RIEMANN Left endpoint Riemnn sums DEF fnf (x) = SQR(1 + x ^ 3) = 1 b = 3 numberofsteps = 4 deltx = (b - ) / numberofsteps x = ccumultion = FOR k = 1 TO numberofsteps delts = fnf(x) * deltx ccumultion = ccumultion + delts x = x + deltx PRINT delts, ccumultion NEXT k By modifying RIEMANN, you cn clculte Riemnn sums for other smpling points nd for other functions. For exmple, to smple t midpoints, you must strt t the midpoint of the first subintervl. Since the subintervl is x units wide, its midpoint is x/2 units from the left endpoint, which is x =. Thus, to hve the progrm generte midpoint sums, just chnge the sttement tht initilizes x (line 6) to x = + deltx / 2. Summtion Nottion Becuse Riemnn sums rise frequently nd becuse they re unwieldy to write out in full, we now introduce method clled summtion nottion tht llows us to write them more compctly. To see how it works, look first t the sum Using summtion nottion, we cn express this s 5 k=1 k 2.

27 6.2. RIEMANN SUMS 363 For somewht more bstrct exmple, consider the sum n, which we cn express s n k. k=1 We use the cpitl letter sigm from the Greek lphbet to denote sum. For this reson, summtion nottion is sometimes referred to s sigm nottion. You should regrd s n instruction telling you to sum the numbers of the indicted form s the index k runs through the integers, strting t the integer displyed below the nd ending t the integer displyed bove it. Notice tht chnging the index k to some other letter hs no effect on the sum. For exmple, Sigm nottion 2 k=1 2 k 3 = j 3, since both expressions give the sum of the cubes of the first twenty positive integers. Other spects of summtion nottion will be covered in the exercises. Summtion nottion llows us to write the Riemnn sum more efficiently s j=1 f(x 1 ) x f(x n ) x n n f(x k ) x k. k=1 Be sure not to get tied into one prticulr wy of using these symbols. For exmple, you should instntly recognize m t i g(t i ) i=1 s Riemnn sum. In wht follows we will commonly use summtion nottion when working with Riemnn sums. The importnt thing to remember is tht summtion nottion is only shorthnd to express Riemnn sum in more compct form.

28 364 CHAPTER 6. THE INTEGRAL Exercises Mking pproximtions 1. Estimte the verge velocity of the ship whose motion is described on pge 352. The voyge lsts 15 hours. 2. The im of this question is to determine how much electricl energy ws consumed in house over 24-hour period, when the power demnd p ws mesured t different times to hve these vlues: time (24-hour clock) 1:3 5: 8: 9:3 11: 15: 18:3 2: 22:3 23: power (wtts) Notice tht the time intervl is from t = hours to t = 24 hours, but the power demnd ws not smpled t either of those times. ) Set up n estimte for the energy consumption in the form of Riemnn sum p(t 1 ) t p(t n ) t n for the power function p(t). To do this, you must identify explicitly the vlue of n, the smpling times t k, nd the time intervls t k tht you used in constructing your estimte. [Note: the smpling times come from the tble, but there is wide ltitude in how you choose the subintervls t k.] b) Wht is the estimted energy consumption, using your choice of dt? There is no single correct nswer to this question. Your estimte depends on the choices you mde in setting up the Riemnn sum. c) Plot the dt given in the tble in prt () on (t, p)-coordinte plne. Then drw on the sme coordinte plne the step function tht represents your estimte of the power function p(t). The width of the k-th step should be the time intervl t k tht you specified in prt (); is it?

29 6.2. RIEMANN SUMS 365 d) Estimte the verge power demnd in the house during the 24-hour period. Wste production. A colony of living yest cells in vt of fermenting grpe juice produces wste products minly lcohol nd crbon dioxide s it consumes the sugr in the grpe juice. It is resonble to expect tht nother yest colony, twice s lrge s this one, would produce twice s much wste over the sme time period. Moreover, since wste ccumultes over time, if we double the time period we would expect our colony to produce twice s much wste. These observtions suggest tht wste production is proportionl to both the size of the colony nd the mount of time tht psses. If P is the size of the colony, in grms, nd t is short time intervl, then we cn express wste production W s function of P nd t: W = k P t grms. If t is mesured in hours, then the multiplier k hs to be mesured in units of grms of wste per hour per grm of yest. The preceding formul is useful only over time intervl t in which the popultion size P does not vry significntly. If the time intervl is lrge, nd the popultion size cn be expressed s function P(t) of the time t, then we cn estimte wste production by breking up the whole time intervl into succession of smller intervls t 1, t 2,..., t n nd forming Riemnn sum k P(t 1 ) t k P(t n ) t n W grms. The time t k must lie within the time intervl t k, nd P(t k ) must be good pproximtion to the popultion size P(t) throughout tht time intervl. 3. Suppose the colony strts with 3 grms of yest (i.e., t time t = hours) nd it grows exponentilly ccording to the formul P(t) = 3 e.2t. If the wste production constnt k is.1 grms per hour per grm of yest, estimte how much wste is produced in the first four hours. Use Riemnn sum with four hour-long time intervls nd mesure the popultion size of the yest in the middle of ech intervl tht is, on the hlf-hour.

30 366 CHAPTER 6. THE INTEGRAL Using RIEMANN 4. ) Clculte left endpoint Riemnn sums for the function 1 + x 3 on the intervl [1, 3] using 4, 4, 4, nd 4 eqully-spced subintervls. How mny ddeciml points in this sequence hve stbilized? b) The left endpoint Riemnn sums for 1 + x 3 on the intervl [1, 3] seem to be pproching limit s the number of subintervls increses without bound. Give the numericl vlue of tht limit, ccurte to four deciml plces. c) Clculte left endpoint Riemnn sums for the function 1 + x 3 on the intervl [3, 7]. Construct sequence of Riemnn sums using more nd more subintervls, until you cn determine the limiting vlue of these sums, ccurte to three deciml plces. Wht is tht limit? d) Clculte left endpoint Riemnn sums for the function 1 + x 3 on the intervl [1, 7] in order to determine the limiting vlue of the sums to three deciml plce ccurcy. Wht is tht vlue? How re the limiting vlues in prts (b), (c), nd (d) relted? How re the corresponding intervls relted? 5. Modify RIEMANN so it will clculte Riemnn sum by smpling the given function t the midpoint of ech subintervl, insted of the left endpoint. Describe exctly how you chnged the progrm to do this. 6. ) Clculte midpoint Riemnn sums for the function 1 + x 3 on the intervl [1, 3] using 4, 4, 4, nd 4 eqully-spced subintervls. How mny deciml points in this sequence hve stbilized? b) Roughly how mny subintervls re needed to mke the midpoint Riemnn sums for 1 + x 3 on the intervl [1, 3] stbilize out to the first four digits? Wht is the stble vlue? Compre this to the limiting vlue you found erlier for left endpoint Riemnn sums. Is one vlue lrger thn the other; could they be the sme? c) Comment on the reltive efficiency of midpoint Riemnn sums versus left endpoint Riemnn sums (t lest for the function 1 + x 3 on the intervl [1, 3]). To get the sme level of ccurcy, n efficient clcultion will tke fewer steps thn n inefficient one. 7. ) Modify RIEMANN to clculte right endpoint Riemnn sums, nd use it to clculte right endpoint Riemnn sums for the function 1 + x 3 on

31 6.2. RIEMANN SUMS 367 the intervl [1, 3] using 4, 4, 4, nd 4 eqully-spced subintervls. How mny digits in this sequence hve stbilized? b) Comment on the efficiency of right endpoint Riemnn sums s compred to left endpoint nd to midpoint Riemnn sums t lest s fr s the function 1 + x 3 is concerned. 8. Clculte left endpoint Riemnn sums for the function f(x) = 1 x 2 on the intervl [ 1, 1]. Use 2 nd 5 eqully-spced subintervls. Compre your vlues with the estimtes for the re of semicircle given on pge ) Clculte left endpoint Riemnn sums for the function f(x) = 1 + cos 2 x on the intervl [, π]. Use 4 nd 2 eqully-spced subintervls. Compre your vlues with the estimtes for the length of the grph of y = sin x between nd π, given on pge 358. b) Wht is the limiting vlue of the Riemnn sums, s the number of subintervls becomes infinite? Find the limit to 11 deciml plces ccurcy. 1. Clculte left endpoint Riemnn sums for the function f(x) = cos(x 2 ) on the intervl [, 4], using 1, 1, nd 1 eqully-spced subintervls. [Answer: With 1 eqully-spced intervls, the left endpoint Riemnn sum hs the vlue ] 11. Clculte left endpoint Riemnn sums for the function f(x) = cosx on the intervl [2, 3], 1 + x2 using 1, 1, nd 1 eqully-spced subintervls. The Riemnn sums re ll negtive; why? (A suggestion: sketch the grph of f. Wht does tht tell you bout the signs of the terms in Riemnn sum for f?)

32 368 CHAPTER 6. THE INTEGRAL 12. ) Clculte midpoint Riemnn sums for the function H(z) = z 3 on the intervl [ 2, 2], using 1, 1, nd 1 eqully-spced subintervls. The Riemnn sums re ll zero; why? (On some computers nd clcultors, you my find tht there will be nonzero digit in the fourteenth or fifteenth deciml plce this is due to round-off error.) b) Repet prt () using left endpoint Riemnn sums. Are the results still zero? Cn you explin the difference, if ny, between these two results? Volume s Riemnn sum If you slice rectngulr prllelepiped (e.g., brick or shoebox) prllel to fce, the re A of cross-section does not vry. The sme is true for cylinder (e.g., cn of spinch or coin). For ny solid tht hs constnt cross-section (e.g., the object on the right, below), its volume is just the product of its cross-sectionl re with its thickness. x x x A A A volume = re of cross-section thickness = A x Most solids don t hve such regulr shpe. They re more like the one shown below. If you tke cross-sectionl slices perpendiculr to some fixed line (which will become our x-xis), the slices will not generlly hve regulr shpe. They my be roughly ovl, s shown below, but they will generlly vry in re. Suppose the re of the cross-section x inches long the xis is A(x) squre inches. Becuse A(x) vries with x, you cnnot clculte

33 6.2. RIEMANN SUMS 369 the volume of this solid using the simple formul bove. However, you cn estimte the volume s Riemnn sum for A. A(x) x x x 1 x 2 x 1 x 2 x The procedure should now be fmilir to you. Subdivide the x-xis into segments of length x 1, x 2,..., x n inches, respectively. The solid piece tht lies over the first segment hs thickness of x 1 inches. If you slice this piece t point x 1 inches long the x-xis, the re of the slice is A(x 1 ) squre inches, nd the volume of the piece is pproximtely A(x 1 ) x 1 cubic inches. The second piece is x 2 inches thick. If you slice it x 2 inches long the x-xis, the slice hs n re of A(x 2 ) squre inches, so the second piece hs n pproximte volume of A(x 2 ) x 2 cubic inches. If you continue in this wy nd dd up the n volumes, you get n estimte for the totl volume tht hs the form of Riemnn sum for the re function A(x): volume A(x 1 ) x 1 + A(x 2 ) x A(x n ) x n cubic inches. One plce where this pproch cn be used is in medicl dignosis. The X-ry technique known s CAT scn provides sequence of precisely-spced cross-sectionl views of ptient. From these views much informtion bout the stte of the ptient s internl orgns cn be gined without invsive surgery. In prticulr, the volume of specific piece of tissue cn be estimted, s Riemnn sum, from the res of individul slices nd the spcing between them. The next exercise gives n exmple. 13. A CAT scn of humn liver shows us X-ry slices spced 2 centimeters prt. If the res of the slices re 72, 145, 139, 127, 111, 89, 63, nd 22 squre centimeters, estimte the volume of the liver. 14. The volume of sphere whose rdius is r is exctly V = 4πr 3 /3.