A REA AND INTEGRALS. 1 Sigma Notation 2. 2 Area 7. 3 The Definite Integral 16

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1 ALTERNATIVE TREATMENT: A REA AND INTEGRALS Sigm Nottio Are 7 The Defiite Itegrl Reprited with permissio from Clculus: Erl Trscedetls, Third Editio Jmes Stewrt 995 Brooks/Cole Pulishig Comp, A divisio of Itertiol Thomso Pulishig Co.

2 SECTION SIGMA NOTATION SECTION Sigm Nottio I fidig res d evlutig itegrls we ofte ecouter sums with m terms. A coveiet w of writig such sums uses the Greek letter (cpitl sigm, correspodig to our letter S) d is clled sigm ottio. This tells us to ed with i=. This tells us to dd. This tells us to strt with i=m. µ i im Defiitio If m, m,..., re rel umers d m d re itegers such tht m, the i m m m im With fuctio ottio, Defiitio c e writte s f i f m f m f m f f im im Thus, the smol idictes summtio i which the letter i (clled the ide of summtio) tkes o the vlues m, m,...,. Other letters c lso e used s the ide of summtio. EXAMPLE () () (c) (d) (e) (f) 4 i 4 i 4 5 i 5 j 4 5 j k k i i EXAMPLE Write the sum i sigm ottio. SOLUTION There is o uique w of writig sum i sigm ottio. We could write i i or or j j k k The followig theorem gives three simple rules for workig with sigm ottio.

3 SECTION SIGMA NOTATION () (c) Theorem c i c i im im If c is costt (tht is, it does ot deped o i), the i i i i im im im () i i i i im im im Proof To see wh these rules re true, ll we hve to do is write oth sides i epded form. Rule () is just the distriutive propert of rel umers: c m c m c c m m Rule () follows from the ssocitive d commuttive properties: m m m m m m m m Rule (c) is proved similrl. EXAMPLE Fid SOLUTION. terms PRINCIPLE OF MATHEMATICAL INDUCTION Let S e sttemet ivolvig the positive iteger. Suppose tht. S is true.. If S k is true, the S k is true. The S is true for ll positive itegers. EXAMPLE 4 Prove the formul for the sum of the first positive itegers: i SOLUTION This formul c e proved mthemticl iductio (see pge 59) or the followig method used the Germ mthemtici Krl Friedrich Guss ( ) whe he ws te ers old. Write the sum S twice, oce i the usul order d oce i reverse order: S S Addig ll colums verticll, we get S O the right side there re terms, ech of which is, so S or S EXAMPLE 5 Prove the formul for the sum of the squres of the first positive itegers: i

4 4 SECTION SIGMA NOTATION SOLUTION Let S e the desired sum. We strt with the telescopig sum (or collpsig sum): Most terms ccel i pirs. i i 4 O the other hd, usig Theorem d Emples d 4, we hve i i i i i i S S 5 Thus we hve S 5 Solvig this equtio for S, we oti S or S See pges 59 d for more thorough discussio of mthemticl iductio. SOLUTION Let e the give formul. S S. is true ecuse S k. Assume tht is true; tht is, The k kk k k k k So S k is true. k k k 7k B the Priciple of Mthemticl Iductio, is true for ll. kk k k k k k k k S k kk k We list the results of Emples, 4, d 5 together with similr result for cues d fourth powers (see Eercises 7 4) s Theorem. These formuls re eeded for fidig res i the et sectio.

5 SECTION SIGMA NOTATION 5 () Theorem Let c e costt d positive iteger. The () c c (c) (e) i i (d) (f) i i 4 The tpe of clcultio i Emple 7 rises i the et sectio whe we compute res. EXAMPLE Evlute i4i. SOLUTION Usig Theorems d, we hve EXAMPLE 7 Fid lim. SOLUTION i4i 4i i 4 i i l i lim i l l 4 l l l l i i 4

6 SECTION SIGMA NOTATION Eercises Write the sum i epded form.. si k. k j. Write the sum i sigm ottio.. 4. s s4 s5 s s i i4 k i j Fid the vlue of the sum. 8 i4. i. j j i 7. i i i.. i i 4.. i i i i. 7. Prove formul () of Theorem. i i4 8 k k5 j j 4 i f i i ii i 8 cos k k 4 i i 5i i ii i k k k 8. Prove formul (e) of Theorem usig mthemticl iductio. 9. Prove formul (e) of Theorem usig method similr to tht of Emple 5, Solutio [strt with i 4 i Prove formul (e) of Theorem usig the followig method pulished Au Bekr Mohmmed i Alhusi Alkrchi i out A.D.. The figure shows squre ABCD i which sides AB d AD hve ee divided ito segmets of legths,,,...,. Thus the side of the squre hs legth so the re is. But the re is lso the sum of the res of the gomos G, G,..., G show i the figure. Show tht the re of G is i i d coclude tht formul (e) is true. 4. Evlute ech telescopig sum. () (c) () (d) 4. Prove the geerlized trigle iequlit 4 4 Fid ech limit i 4 i 4 i 99 i i lim l lim l D 5 4 i i lim i l lim l i. G G A G B i 5 i G i i 5 i 5 i 47. Prove the formul for the sum of fiite geometric series with first term d commo rtio r : r r r r r r... G C

7 SECTION AREA Evlute. 49. Evlute i i. 5. Evlute m j i j 5. Fid the umer such tht i () Use the product formul for si cos (see 8 i Appedi D) to show tht si cos i si(i ) si(i ) () Use the idetit i prt () d telescopig sums to prove the formul cos i si( ) si si where is ot iteger multiple of cos i si cos si. Deduce tht 5. Use the method of Eercise 5 to prove the formul si i si si si where is ot iteger multiple of. SECTION Are We egi ttemptig to solve the re prolem: Fid the re of the regio S tht lies uder the curve f from to. This mes tht S, illustrted i Figure, is ouded the grph of cotiuous fuctio f [where f ], the verticl lies d, d the -is. =ƒ FIGURE S=s(, ), ƒd = S = I trig to solve the re prolem we hve to sk ourselves: Wht is the meig of the word re? This questio is es to swer for regios with stright sides. For rectgle, the re is defied s the product of the legth d the width. The re of trigle is hlf the se times the height. The re of polgo is foud dividig it ito trigles (s i Figure ) d ddig the res of the trigles. A A w h A A l FIGURE A=lw A= h A=A +A +A +A However, it is t so es to fid the re of regio with curved sides. We ll hve ituitive ide of wht the re of regio is. But prt of the re prolem is to mke this ituitive ide precise givig ect defiitio of re. Recll tht i defiig tget we first pproimted the slope of the tget lie slopes of sect lies d the we took the limit of these pproimtios. We pursue sim-

8 8 SECTION AREA ilr ide for res. We first pproimte the regio S polgos d the we tke the limit of the res of these polgos. The followig emple illustrtes the procedure. EXAMPLE Let s tr to fid the re uder the prol regio S illustrted i Figure ). from to (the prolic (, ) = FIGURE Oe method of pproimtig the desired re is to divide the itervl, ito suitervls of equl legth d cosider the rectgles whose ses re these suitervls d whose heights re the vlues of the fuctio t the right-hd edpoits of these suitervls. Figure 4 shows the pproimtio of the prolic regio four, eight, d rectgles (, ) (, ) (, ) FIGURE 4 4 () 4 8 () (c) Let S e the sum of the res of the rectgles i Figure 4(c). Ech rectgle hs width d the heights re the vlues of the fuctio f t the poits,,,..., ; tht is, the heights re,,,...,. Thus S i Usig the formul for the sum of the squres of the first itegers [Formul.(d)], we c write S For istce, the sum of the res of the four shded rectgles i Figure 4() is S

9 SECTION AREA 9 S d the sum of the res of the eight rectgles i FIgure 4() is The results of similr clcultios re show i the tle i the mrgi. It looks s if is ecomig closer to s icreses. I fct S S lim S l l l l From Figure 4 it ppers tht, s icreses, S ecomes etter d etter pproimtio to the re of the prolic segmet. Therefore we defie the re A to e the limit of the sums of the res of the pproimtig rectgles, tht is, A l S I pplig the ide of Emple to the more geerl regio S of Figure, we hve o eed to use rectgles of equl width. We strt sudividig the itervl, ito smller suitervls choosig prtitio poits,,,..., so tht The the suitervls re,,,,,,...,, This sudivisio is clled the prtitio of, d we deote it P. We use the ottio i for the legth of the ith suitervl, i. Thus i i This legth of the logest suitervl is deoted P d is clled the orm of P. Thus P m,,..., Figure 5 illustrtes oe possile prtitio of,. Î Î Î Î i Î FIGURE 5 =... i- i... - = B drwig the lies,,,...,, we use the prtitio P to divide the regio S ito strips S, S,..., S s i Figure. Net we pproimte these strips S i rectgles R i. To do this we choose umer * i i ech suitervl, i d costruct rectgle R i with se i d height f* i s i Figure 7.

10 SECTION AREA =ƒ Î i f( i *) S S S S i S R R R R i R... i- i... - i- i - FIGURE FIGURE 7 * * * i * * Ech poit * i c e where i its suitervl t the right edpoit (s i Emple ) or t the left edpoit or somewhere etwee the edpoits. The re of the ith rectgle R i is A i f * i i The rectgles R,..., R form polgol pproimtio to the regio S. Wht we thik of ituitivel s the re of S is pproimted the sum of the res of these rectgles, which is A i f * i i f * f * Figure 8 shows this pproimtio for prtitios with, 4, 8, d. * * () = () =4 (c) =8 (d) = FIGURE 8 Notice tht this pproimtio ppers to ecome etter d etter s the strips ecome thier d thier, tht is, s P l. Therefore we defie the re A of the regio S s the limitig vlue (if it eists) of the res of the pproimtig polgos, tht is, the limit of the sum () of the res of the pproimtig rectgles. I smols: A Pl f * i i The precedig discussio d the digrms i Figures 7 d 8 show tht the defiitio of re i () correspods to our ituitive feelig of wht re ought to e. The limit i () m or m ot eist. It c e show tht if f is cotiuous, the this limit does eist; tht is, the regio hs re. [The precise meig of the limit i Defiitio is tht for ever there is correspodig umer such tht A f * i i wheever P I other words, the re c e pproimted sum of res of rectgles to withi ritrr degree of ccurc ( ) tkig the orm of the prtitio sufficietl smll.

11 SECTION AREA EXAMPLE () If the itervl, is divided ito suitervls the prtitio P d the set of prtitio poits is,.,.,.,.5,, fid P. () If f 4 5 d * i is chose to e the left edpoit of the ith suitervl, fid the sum of the res of the pproimtig rectgles. (c) Sketch the pproimtig rectgles. SOLUTION () We re give tht,,.,., 4, 5.5, d, so FIGURE (See Figure 9.) Therefore P m.,.,.4,.4,.5,.5 () Sice * i, the sum of the res of the pproimtig rectgles is, (), = -4+5 f * i i f i f f. f. f. 4 f 5 FIGURE f.5 (c) The grph of f d the pproimtig rectgles re sketched i Figure. EXAMPLE Fid the re uder the prol from to. SOLUTION Sice f is cotiuous, the limit () tht defies the re must eist for ll possile prtitios P of the itervl, s log s P l. To simplif thigs let us tke the prtitio P tht divides, ito suitervls of equl legth. (This is clled regulr prtitio.) The the prtitio poits re,, 4,..., i i,..., d i so the orm of P is P m i The poit * i c e chose to e where i the ith suitervl. For the ske of defiiteess, let us choose it to e the right-hd edpoit: i * i i

12 SECTION AREA Sice P, the coditio P l is equivlet to l. So the defiitio of re () ecomes A P l f* i i i l i l l i l 8 l 8 i l f i 8i 4 ( Theorem.) ( Theorem.) 4 4 The sum i this clcultio is represeted the res of the shded rectgles i Figure. Notice tht i this cse, with our choice of * i s the right-hd edpoit d = R =5.8 = R Å4.85 =5 R =4.747 FIGURE Right sums = L =4.8 = L Å4.548 =5 L =4.587 FIGURE Left sums

13 SECTION AREA sice f is icresig, f * i is the mimum vlue of f o, i, so the sum R of the res of the pproimtig rectgles is lws greter th the ect re A 4. We could just s well hve chose * i to e the left-hd edpoit, tht is, * i i. The f * i is the miimum vlue of f o, i, so the sum L of the res of the pproimtig rectgles i Figure is lws less th A. The clcultio with this choice is s follows: A P l f* i i i l l l 8 l 8 l 4 f i 8 i i l i i i Notice tht we hve otied the sme swer with the differet choice of * i. I fct, we would oti the sme swer if * i ws chose to e the midpoit of, i (see Eercise ) or ideed other poit of this itervl. EXAMPLE 4 Fid the re uder the cosie curve from to, where. SOLUTION As i the first prt of Emple, we choose regulr prtitio P so tht P d we choose * i to e the right-hd edpoit of the ith suitervl: * i i i Sice P l s l, the re uder the cosie curve from to is A P l f* i i i l cosi i l cosi To evlute this limit we use the formul of Eercise 5 i Sectio : cos i si cos si

14 4 SECTION AREA with. The Equtio ecomes si cos 4 A l si Now cos cos l cos s l sice cosie is cotiuous. Lettig t d usig Theorem.5., we hve lim l si t l t si t lim t l t si t Puttig these limits i Equtio 4, we oti A si cos si I prticulr, tkig, we hve proved tht the re uder the cosie curve from to is si (see Figure ). =cos re= FIGURE π NOTE The re clcultios i Emple d 4 re ot es. We will see i Sectio 5., however, tht the Fudmetl Theorem of Clculus gives much esier method for computig these res. Eercises 8 You re give fuctio f, itervl, prtitio poits, d descriptio of * i withi the ith suitervl. () Fid P. () Fid the sum of the res of the pproimtig rectgles, s give i (). (c) Sketch the grph of f d the pproimtig rectgles.. f,, 4,,,,, 4, * i left edpoit. f,, 4,,,,, 4, * i right edpoit. f,, 4,,,,, 4, * i midpoit 4. f,, 4,,.5,,, 4, * i left edpoit 5. f,,,,.5,,.5,.,.5,, * i right edpoit. f,,,,.5,.,.5,, *.5, *, *.5, * 4 7. f si,,,, 4,, 4,, *, *, *, * f 4 cos,,,,, 4,,, * i left edpoit

15 SECTION AREA 5 9. () Sketch grph of the regio tht lies uder the prol from to d use it to mke rough visul estimte of the re of the regio. () Fid epressio for R, the sum of the res of the pproimtig rectgles, tkig * i i () to e the right edpoit d usig suitervls of equl legth. (c) Fid the umericl vlues of the pproimtig res R for,, d 4. (d) Fid the ect re of the regio. ;. () Use grphig device to sketch grph of the regio tht lies uder the curve 4 from to d use it to mke rough visul estimte of the re of the regio. () Fid epressio for R, the sum of the res of the pproimtig rectgles, tkig * i i () to e the right edpoit d usig suitervls of equl legth. (c) Fid the umericl vlues of the pproimtig res R for,, d. (d) Fid the ect re of the regio.. Fid the re from Emple tkig * i to e the midpoit of, i. Illustrte the pproimtig rectgles with sketch.. Fid the re uder the curve from to usig suitervls of equl legth d tkig * i i () to e the () left edpoit, () right edpoit, d (c) midpoit of the ith suitervl. I ech cse, sketch the pproimtig rectgles. 8 Use () to fid the re uder the give curve from to. Use equl suitervls d tke * i to e the right edpoit of the ith suitervl. Sketch the regio..,, 5 4.,, ,,.,, 7.,, 8. 4,, CAS CAS 9 If ou hve progrmmle clcultor (or computer), it is possile to evlute the epressio () for the sum of res of pproimtig rectgles, eve for lrge vlues of, usig loopig. (O TI use the Is commd, o Csio use Isz, o HP or i BASIC use FOR-NEXT loop.) Compute the sum of the res of pproimtig rectgles usig equl suitervls d right edpoits for,, d 5. The guess the vlue of the ect re. 9. The regio uder si from to. The regio uder from to. Some computer lger sstems hve commds tht will drw pproimtig rectgles d evlute the sums of their res, t lest if * i is left or right edpoit. (For istce, i Mple use lefto, righto, leftsum, d rightsum.) () If f s, 4, fid the left d right sums for,, d 5. () Illustrte grphig the rectgles i prt (). (c) Show tht the ect re uder f lies etwee 4. d () If f sisi,, use the commds discussed i Eercise to fid the left d right sums for,, d 5. () Illustrte grphig the rectgles i prt (). (c) Show tht the ect re uder f lies etwee.87 d.9. 4 Determie regio whose re is equl to the give limit. Do ot evlute the limit. i. lim 4. l t 4 4 lim l i 5. Fid the re uder the curve si from to. [Hit: Use equl suitervls d right edpoits, d use Eercise 5 i Sectio.]. () Let A e the re of polgo with equl sides iscried i circle with rdius r. B dividig the polgo ito cogruet trigles with cetrl gle, show tht A r si. () Show tht lim l A r. [Hit: Use Equtio.5..]

16 SECTION THE DEFINITE INTEGRAL SECTION The Defiite Itegrl We sw i the preceedig sectio tht limit of the form A P l f* i i i rises whe we compute re. It turs out tht this sme tpe of limit occurs i wide vriet of situtios eve whe f is ot ecessril positive fuctio. I Chpters 5 d 8 we will see tht limits of the form () lso rise i fidig legths of curves, volumes of solids, res of surfces, ceters of mss, fluid pressure, d work, s well s other qutities. We therefore give this tpe of limit specil me d ottio. Defiitio of Defiite Itegrl If f is fuctio defied o closed itervl,, let P e prtitio of, with prtitio poits,,...,, where Choose poits * i i, i d let i i d P m i. The the defiite itegrl of f from to is f d P l f* i i i if this limit eists. If the limit does eist, the itervl,. f is clled itegrle o the NOTE The smol ws itroduced Leiiz d is clled itegrl sig. It is elogted S d ws chose ecuse itegrl is limit of sums. I the ottio f d, f is clled the itegrd d d re clled the limits of itegrtio; is the lower limit d is the upper limit. The smol d hs o meig itself; f d is ll oe smol. The procedure of clcultig itegrl is clled itegrtio. NOTE The defiite itegrl f d is umer; it does ot deped o. I fct, we could use letter i plce of without chgig the vlue of the itegrl: f d f t dt f r dr NOTE The sum f * i i FIGURE tht occurs i Defiitio is clled Riem sum fter the Germ mthemtici Berhrd Riem (8 8). The defiite itegrl is sometimes clled the Riem itegrl. If f hppes to e positive, the the Riem sum c e iterpreted s sum of res of pproimtig rectgles [Compre () with (.).] If f tkes o oth positive d egtive vlues, s i Figure, the the Riem sum is the sum of the res of the rectgles tht lie ove the -is d the egtives of the res of the rectgles tht lie elow the -is (the res of the gold rectgles mius the res of the lue rectgles). EXAMPLE Let f 5 d cosider the prtitio P of the itervl, mes of the set of prtitio poits,.5,,.,.,. I this emple

17 SECTION THE DEFINITE INTEGRAL 7,, 5, d,.5,,., 4., d 5. the legths of the suitervls re Thus the orm of the prtitio P is Suppose we choose *.8, *., *., * 4, d * 5.7. The the correspodig Riem sum is 5 f * i i f.8 f. f. f 4 f P m.5,.5,.7,.5,.8.8 FIGURE Notice tht, i this emple, f is ot positive fuctio d so the Riem sum does ot represet sum of res of rectgles. But it does represet the sum of the res of the gold rectgles (ove the -is) mius the sum of the res of the lue rectgles (elow the is) i Figure. NOTE 4 A itegrl eed ot represet re. But for positive fuctios, itegrl c e iterpreted s re. I fct, comprig Defiitio with the defiitio of re (.), we see the followig: For the specil cse where f, f d the re uder the grph of f from to I geerl, defiite itegrl c e iterpreted s differece of res: f d A A + _ + where A is the re of the regio ove the -is d elow the grph of f d A is the re of the regio elow the -is d ove the grph of f. (This seems resole from compriso of Figures d.) FIGURE NOTE 5 The follows: precise meig of the limit tht defies the itegrl i Defiitio is s f d I mes tht for ever there is correspodig umer such tht I f * i i for ll prtitios P of, with P d for ll possile choices of * i i, i.

18 8 SECTION THE DEFINITE INTEGRAL Berhrd Riem received his Ph.D. uder the directio of the legedr Guss t the Uiversit of Göttige d remied there to tech. Guss, who ws ot i the hit of prisig other mthemticis, spoke of Riem s cretive, ctive, trul mthemticl mid d gloriousl fertile origilit. The defiitio () of itegrl tht we use is due to Riem. He lso mde mjor cotriutios to the theor of fuctios of comple vrile, mthemticl phsics, umer theor, d the foudtios of geometr. Riem s rod cocept of spce d geometr tured out to e the right settig, 5 ers lter, for Eistei s geerl reltivit theor. Riem s helth ws poor throughout his life, d he died of tuerculosis t the ge of 9. This mes tht defiite itegrl c e pproimted to withi desired degree of ccurc Riem sum. NOTE I Defiitio we re delig with fuctio f defied o itervl,, so we re implicitl ssumig tht. But for some purposes it is useful to eted the defiitio of to the cse where f d or s follows: EXAMPLE Epress If, the f d. f d If, the f d. lim P l * i * i si * i i i s itegrl o the itervl,. SOLUTION Comprig the give limit with the limit i Defiitio, we see tht the will e ideticl if we choose f si We re give tht d. Therefore, Defiitio, we hve lim P l i * i * i si * i i si d EXAMPLE Evlute the followig itegrls iterpretig ech i terms of res. () s d () d = œ - or + = SOLUTION () Sice f s, we c iterpret this itegrl s the re uder the curve s from to. But, sice, we get, which shows tht the grph of f is the qurter-circle with rdius i Figure 4. Therefore s d 4 4 FIGURE 4 (I Sectio 8. we will e le to prove tht the re of circle of rdius r is r evlutig the itegrl r sr d usig the techiques of Chpter 8.) () The grph of is the lie with slope show i Figure 5. We compute the itegrl s the differece of the res of the two trigles: d A A.5 (, ) =- A A FIGURE 5 _

19 SECTION THE DEFINITE INTEGRAL 9 The itegrls i Emple were simple to evlute ecuse we were le to epress them i terms of res of simple regios, ut ot ll itegrls re tht es. I fct, the itegrls of some fuctios do t eve eist. So the questio rises: Which fuctios re itegrle? A prtil swer is give the followig theorem, which is proved i courses o dvced clculus. 4 Theorem If f is either cotiuous or mootoic o,, the f is itegrle o, ; tht is, the defiite itegrl f d eists. FIGURE Discotiuous itegrle fuctio FIGURE 7 Noitegrle fuctio If f is discotiuous t some poits i,, the f d might eist or it might ot eist (see Eercises 7 d 7). If f hs ol fiite umer of discotiuities d these re ll jump discotiuities, the f is clled piecewise cotiuous d it turs out tht f is itegrle. (See Figure.) It c e show tht if f is itegrle o,, the f must e ouded fuctio o, ; tht is, there eists umer M such tht f M for ll i,. Geometricll, this mes tht the grph of f lies etwee the horizotl lies M d M. I prticulr, if f hs ifiite discotiuit t some poit i,, the f is ot ouded d is therefore ot itegrle. (See Eercise 7 d Figure 7.) If f is itegrle o,, the the Riem sums () must pproch f d s P l o mtter how the prtitios P re chose d o mtter how the poits * i re chose i, i. Therefore, if it is kow eforehd tht f is itegrle o, ( for istce, if it is kow tht f is cotiuous or mootoic), the i clcultig the vlue of itegrl we re free to choose prtitios P d poits * i i w we like s log s P l. For purposes of clcultio, it is ofte coveiet to tke P to e regulr prtitio; tht is, ll the suitervls hve the sme legth. The d,,,..., i i If we choose * i to e the right edpoit of the ith suitervl, the i * i i i Sice P, we hve P l s l, so Defiitio gives f d P l f* i i l i f i Sice does ot deped o i, Theorem. llows us to tke it i frot of the sigm sig, d we hve the followig formul for clcultig itegrls. 5 Theorem If f is itegrle o,, the f d l f i

20 SECTION THE DEFINITE INTEGRAL EXAMPLE 4 Evlute 5 d. SOLUTION Here we hve f 5,, d. Sice f is cotiuous, we kow it is itegrle d so Theorem 5 gives 5 d l l 8 l l f i lim l i i i i FIGURE 8 This itegrl cot e iterpreted s re ecuse f tkes o oth positive d egtive vlues. But it c e iterpreted s the differece of res A A, where A d re show i Figure 8. Figure 9 illustrtes the clcultio showig the positive d egtive terms i the right Riem sum R for 4. The vlues i the tle show the Riem sums pprochig the ect vlue of the itegrl,.5, s l. A R FIGURE 9 A much simpler method for evlutig the itegrl i Emple 4 will e give i Sectio 5.4 fter we hve proved the Fudmetl Theorem of Clculus. The Midpoit Rule We ofte choose the smple poit * i to e the right edpoit of the ith suitervl ecuse it is coveiet for computig the limit. But if the purpose is to fid pproimtio to itegrl, it is usull etter to choose * i to e the midpoit of the itervl, which we deote i. A Riem sum is pproimtio to itegrl, ut if we use midpoits d regulr prtitio we get the followig pproimtio: Midpoit Rule f d f i f f where d i i midpoit of, i

21 SECTION THE DEFINITE INTEGRAL = EXAMPLE 5 Use the Midpoit Rule with 5 to pproimte. d SOLUTION The prtitio poits re,.,.4,.,.8, d., so the midpoits of the five itervls re.,.,.5,.7, d.9. The width of the itervls is 5 5, so the Midpoit Rule gives d f. f. f.5 f.7 f FIGURE Sice f for, the itegrl represets re d the pproimtio give the Midpoit Rule is the sum of the res of the rectgles show i Figure. At the momet we do t kow how ccurte the pproimtio i Emple 5 is, ut i Sectio 8.7 we will ler method for estimtig the error ivolved i usig the Midpoit Rule. At tht time we will discuss other methods for pproimtig defiite itegrls. If we ppl the Midpoit Rule to the itegrl i Emple 4, we get the picture i Figure. The pproimtio M 4.5 is much closer to the true vlue.5 th the right edpoit pproimtio, R show i Figure 9. Properties of the Defiite Itegrl FIGURE M 4.5 We ow develop some sic properties of itegrls tht will help us to evlute itegrls i simple mer. Properties of the Itegrl Suppose tht ll of the followig itegrls eist. The. cd c, where c is costt. f t d f d t d. cf d c f d, where c is costt f t d f d t d f d f d f d The proof of Propert is requested i Eercise 5. This propert ss tht the itegrl of costt fuctio f c is the costt times the legth of the itervl. If c d, this is to e epected ecuse c is the re of the shded rectgle i Figure. c =c FIGURE j c d=c(-) re=c(-)

22 SECTION THE DEFINITE INTEGRAL Proof of Propert Sice f t d eists, we c compute it usig regulr prtitio d choosig * i to e the right edpoit of the ith suitervl, tht is, * i i. Usig the fct tht the limit of sum is the sum of the limits, we hve f t d l f i t i l f i t i l f i l t i f d t d ( Theorem.) FIGURE =ƒ c Propert ss tht the itegrl of sum is the sum of the itegrls. Propert c e proved i similr mer (see Eercise 5) d ss tht the itegrl of costt times fuctio is the costt times the itegrl of the fuctio. I other words, costt (ut ol costt) c e tke i frot of itegrl sig. Propert 4 is proved writig f t f td usig Properties d with c. Propert 5 is somewht more complicted d is proved t the ed of this sectio, ut for the cse where f d c, it c e see from the geometric iterprettio i Figure. For positive fuctios f, f d is the totl re uder f from to, which is the sum of (the re from to c ) d c f d f d (the re from c c to ). EXAMPLE Use the properties of itegrls d the results d cos d (from Eercise i this sectio d Emple 4 i Sectio ) to evlute the followig itegrls. () cos d () 5 4 d SOLUTION () Usiig Properties d of itegrls, we get () Sice cos d d if if cos d 8 we use Propert 5 to split the itegrl t : 5 d d 5 d 4 4 d 5 4 d 4 d 5 d Notice tht Properties 5 re true whether,, or. The followig properties, however, re true ol if.

23 SECTION THE DEFINITE INTEGRAL Order Properties of the Itegrl Suppose the followig itegrls eist d.. If f for, the f d. 7. If f t for, the f d t d. 8. If m f M for, the m f d M 9. f d f d M m FIGURE 4 =ƒ If f, the f d represets the re uder the grph of f, so the geometric iterprettio of Propert is simpl tht res re positive. But the propert c e proved from the defiitio of itegrl (Eercise ). Propert 7 ss tht igger fuctio hs igger itegrl. It follows from Properties d 4 ecuse f t. Propert 8 is illustrted Figure 4 for the cse where f. If f is cotiuous we could tke m d M to e the solute miimum d mimum vlues of f o the itervl,. I this cse Propert 8 ss tht the re uder the grph of f is greter th the re of the rectgle with height m d less th the re of the rectgle with height M. Proof of Propert 8 Sice m f M, Propert 7 gives md f d Md Usig Propert to evlute the itegrls o the left- d right-hd sides, we oti m f d M The proof of Propert 9 is left s Eercise 7. 4 EXAMPLE 7 Use Propert 8 to estimte the vlue of s d. =œ 4 FIGURE 5 SOLUTION Sice f s is icresig fuctio, its solute miimum o, 4 is m f d its solute mimum o, 4 is M f 4 s4. Thus Propert 8 gives or The result of Emple 7 is illustrted i Figure 5. The re uder s from to 4 is greter th the re of the lower rectgle d less th the re of the lrge rectgle. EXAMPLE 8 Show tht s d s d 4 4 s d SOLUTION The miimum vlue of f s o, 4 is m f s, sice f is icresig. Thus Propert 8 gives 4 s d s4 s 4.4

24 4 SECTION THE DEFINITE INTEGRAL This result is ot good eough, so isted we use Propert 7. Notice tht Sice for, we hve s for 4. Thus, Propert 7, 4? s s s d 4 d [Here we hve used the fct tht d from Eercise.] Proof of Propert 5 We first ssume tht c. Sice we re ssumig tht f d eists, we c compute it s limit of Riem sums usig ol prtitios P tht iclude c s oe of the prtitio poits. If P is such prtitio, let P e the correspodig prtitio of, c determied those prtitio poits of P tht lie i, c. Similrl, P will deote the correspodig prtitio of c,. Note tht P P d P P. Thus, if P l, it follows tht P d P l l. If i i is the set of prtitio poits for P d k m, where k is the umer of suitervls i, cd m is the umer of suitervls i c,, the i i k is the set of prtitio poits for P. If we write t j for the prtitio poits to the right of c kj, the t i j m is the set of prtitio poits for. Thus we hve P k k Choosig * d lettig t i i j t j t j, we compute f d s follows: f P l f i i P l k f i i P l k f i i m j P l k f i i P l m f t j t j j c f d f t dt Now suppose tht c. B wht we hve lred proved, we hve c c t t m ik f i i f t j t j f f d f d c Therefore f c f d f d c f d f d (See Note.) The proofs re similr for the remiig four orderigs of,, d c. c

25 SECTION THE DEFINITE INTEGRAL 5 Eercises You re give fuctio f, itervl, prtitio poits Riem sums for the fuctio f s o the itervl tht defie prtitio P, d poits * i i the ith suitervl., with. Epli wh these estimtes show tht () Fid P. () Fid the Riem sum ().. f 7,, 5,,.,.,., 4., 5, * i midpoit.85 s d.85. f,,,,.,.,,.8,.,, Deduce tht the pproimtio usig the Midpoit Rule with * i midpoit i Eercise is ccurte to two deciml plces.. f,,,,.4,,,.8,.4,, 5 Use Theorem 5 to evlute the itegrl. * i right edpoit 4. f,,,,.5,,.7,.4,, 5. c d. 7 d * i left edpoit 5. f,,,,.5,,.5,, *, *.4, *., * 4. f si,,,,,,,,, *.5, *.5, *.5, * 4.5, * d d. 5 d 5 4 d 7. The grph of fuctio f is give. Estimte f d usig four equl suitervls with () right edpoits, () left edpoits, d (c) midpoits. f 8. Prove tht d.. Prove tht d. 8 Evlute the itegrl iterpretig it i terms of res.. d 4. s4 d 5. ( s9 ) d. d CAS 8. The tle gives the vlues of fuctio otied from eperimet. Use them to estimte f d usig three equl suitervls with () right edpoits, () left edpoits, d (c) midpoits. If the fuctio is kow to e decresig fuctio, c ou s whether our estimtes re less th or greter th the ect vlue of the itegrl? 9 Use the Midpoit Rule with the give vlue of to pproimte the itegrl. Roud the swer to four deciml plces d, f s d,.. If ou hve CAS tht evlutes midpoit pproimtios d grphs the correspodig rectgles (use middlesum d middleo commds i Mple), check the swer to Eercise d illustrte with grph. The repet with d. 4. With progrmmle clcultor or computer (see the istructios for Eercise 9 i Sectio ), compute the left d right 4 d, 4 7 t d, Epress the limit s defiite itegrl o the give itervl. 9. lim * i 5* i i,. lim s* i i,. lim cos i i, t i. lim i, 5 Epress the limit s defiite itegrl.. lim [Hit: Cosider f 4.] ( ) d Pl Pl Pl Pl i 4 l 5 lim l i i lim l i, 4 5,, 4, 5 d

26 SECTION THE DEFINITE INTEGRAL. Evlute cos d. 7. Give tht s d 8, wht is st dt? () Fid pproimtio to the itegrl d usig Riem sum with right edpoits d 8. () Drw digrm like Figure to illustrte the pproimtio i prt (). (c) Evlute 4 d. (d) Iterpret the itegrl i prt (c) s differece of res d illustrte with digrm like Figure Use the properties of itegrls to evlute ech itegrl. You m ssume from Sectio tht d ou m use the results of Eercises d. 9. s d d 4. 5 cos 4 d if 4. where f f d if Write the give sum or differece s sigle itegrl i the form f d Use the properties of itegrls to verif the iequlit without evlutig the itegrls f d f d f d f d 5 f d f d 7 f d s5 d s d d 4 8 d si d 5. s d s f d f d f d f d f d f d 5 si d 4 si d 4 4 cos d si d Use Propert 8 to estimte the vlue of the itegrl d 57. d s 4 d. 4 Use properties of itegrls, together with Eercises d, to prove the iequlit Prove Propert of itegrls.. Prove Propert of itegrls. 7. Prove Propert 9 of itegrls. [Hit:.] 8. Suppose tht f is cotiuous o, d f for ll i,. Prove tht f d. [Hit: Use the Etreme Vlue Theorem d Propert 8.] 9. Which of the followig fuctios re itegrle o the itervl,? () f si () f sec if (c) f if (d) 7. Let f si d f d s 4 d 5 s d.5 si d 8 cos d f f f f if if f t if if s d 4 si d 4 () Show tht f is ot cotiuous o,. () Show tht f is uouded o,. (c) Show tht f d does ot eist, tht is, f is ot itegrle o,. [Hit: Show tht the first term i the Riem sum, f*, c e mde ritrril lrge.] 4 cos d

27 SECTION THE DEFINITE INTEGRAL A7 7. Let Show tht f is ouded ut ot itegrle o,. [Hit: Show tht, o mtter how smll P is, some Riem sums re wheres others re equl to.] 7. Evlute d usig prtitio of, poits of geometric progressio:,,,...,,..., i i. Tke * i i d use the formul i Eercise 47 i Sectio for the sum of geometric series. if is rtiol f if is irrtiol 7. Fid d. Hit: Use regulr prtitio ut choose * to i e the geometric me of d i ( * i s i ) d use the idetit mm m m ; 74. () Drw the grph of the fuctio f cos i the viewig rectgle,,. () If we defie ew fuctio t t, the cost dt t is the re uder the grph of f from to [util f ecomes egtive, t which poit t ecomes differece of res.] Use the grph of f from prt () to estimte the vlue of t whe,.,.4,.,... up to. At wht vlue of does t strt to decrese? (c) Use the iformtio from prt () to sketch rough grph of t. (d) Sketch more ccurte grph of t usig our clcultor or computer to estimte t., t.4,.... (Use the itegrtio commd, if ville, or the Midpoit Rule.) (e) Use our grph of t from prt (d) to sketch the grph of t usig the iterprettio of t s the slope of tget lie. How does the grph of t compre with the grph of f?