3 6 6 CHAPTER 5 INTEGRALS CAS 5. Fid the ect re uder the cosie curve cos from to, where. (Use computer lger sstem oth to evlute the sum d compute the it.) I prticulr, wht is the re if? 6. () 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 l A r. [Hit: Use Equtio 3.3..] 5. THE DEFINITE INTEGRAL We sw i Sectio 5. tht it of the form l f * i f * f * f * l rises whe we compute re. We lso sw tht it rises whe we tr to fid the distce trveled oject. It turs out tht this sme tpe of it occurs i wide vriet of situtios eve whe f is ot ecessril positive fuctio. I Chpters 6 d 8 we will see tht its of the form () lso rise i fidig legths of curves, volumes of solids, ceters of mss, force due to wter pressure, d work, s well s other qutities. We therefore give this tpe of it specil me d ottio. DEFINITION OF A DEFINITE INTEGRAL If f is fuctio defied for, we divide the itervl, ito suitervls of equl width. We let,,,..., ( ) e the edpoits of these suitervls d we let *, *,..., * e smple poits i these suitervls, so * i lies i the ith suitervl i, i. The the defiite itegrl of f from to is f d l f * i provided tht this it eists. If it does eist, we s tht f is itegrle o,. The precise meig of the it tht defies the itegrl is s follows: For ever umer there is iteger N such tht for ever iteger N d for ever choice of * i i i, i. NOTE The smol ws itroduced Leiiz d is clled itegrl sig. It is elogted S d ws chose ecuse itegrl is it of sums. I the ottio f d, f is clled the itegrd d d re clled the its of itegrtio; is the lower it d is the upper it. For ow, the smol d hs o meig itself; f d is ll oe smol. The d simpl idictes tht the idepedet vrile is. The procedure of clcultig itegrl is clled itegrtio. f d f * i
SECTION 5. THE DEFINITE INTEGRAL 3 6 7 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 3 The sum f * i RIEANN 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 39. tht occurs i Defiitio is clled Riem sum fter the Germ mthemtici Berhrd Riem (86 866). So Defiitio ss tht the defiite itegrl of itegrle fuctio c e pproimted to withi desired degree of ccurc Riem sum. We kow tht if f hppes to e positive, the the Riem sum c e iterpreted s sum of res of pproimtig rectgles (see Figure ). B comprig Defiitio with the defiitio of re i Sectio 5., we see tht the defiite itegrl f d c e iterpreted s the re uder the curve f from to. (See Figure.) Î =ƒ i * FIGURE 3 µ f( i *) Î is pproimtio to the et re =ƒ =ƒ FIGURE 4 j ƒ d is the et re FIGURE If ƒ, the Riem sum µ f( i *) Î is the sum of res of rectgles. If f tkes o oth positive d egtive vlues, s i Figure 3, 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). Whe we tke the it of such Riem sums, we get the situtio illustrted i Figure 4. A defiite itegrl c e iterpreted s et re, tht is, 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. FIGURE If ƒ, the itegrl j ƒ d is the re uder the curve =ƒ from to. NOTE 4 Although we hve defied f d dividig, ito suitervls of equl width, there re situtios i which it is dvtgeous to work with suitervls of uequl width. For istce, i Eercise 4 i Sectio 5. NASA provided velocit dt t times tht were ot equll spced, ut we were still le to estimte the distce trveled. Ad there re methods for umericl itegrtio tht tke dvtge of uequl suitervls.
3 6 8 CHAPTER 5 INTEGRALS If the suitervl widths re,,...,, we hve to esure tht ll these widths pproch i the itig process. This hppes if the lrgest width, m i, pproches. So i this cse the defiitio of defiite itegrl ecomes f d m i l f * i i NOTE 5 We hve defied the defiite itegrl for iegrle fuctio, ut ot ll fuctios re itegrle (see Eercises 67 68). The followig theorem shows tht the most commol occurrig fuctios re i fct itegrle. It is proved i more dvced courses. 3 THEORE If f is cotiuous o,, or if f hs ol fiite umer of jump discotiuities, the f is itegrle o, ; tht is, the defiite itegrl f d eists. If f is itegrle o,, the the it i Defiitio eists d gives the sme vlue o mtter how we choose the smple poits * i. To simplif the clcultio of the itegrl we ofte tke the smple poits to e right edpoits. The * i i d the defiitio of itegrl simplifies s follows. 4 THEORE If f is itegrle o,, the f d l f i where d i i EXAPLE Epress s itegrl o the itervl,. l 3 i i si i SOLUTION Comprig the give it with the it i Theorem 4, we see tht the will e ideticl if we choose f 3 si. We re give tht d. Therefore, Theorem 4, we hve l 3 i i si i 3 si d Lter, whe we ppl the defiite itegrl to phsicl situtios, it will e importt to recogize its of sums s itegrls, s we did i Emple. Whe Leiiz chose the ottio for itegrl, he chose the igrediets s remiders of the itig process. I geerl, whe we write l we replce, * i, d d. f * i f d
SECTION 5. THE DEFINITE INTEGRAL 3 6 9 EVALUATING INTEGRALS Whe we use it to evlute defiite itegrl, we eed to kow how to work with sums. The followig three equtios give formuls for sums of powers of positive itegers. Equtio 5 m e fmilir to ou from course i lger. Equtios 6 d 7 were discussed i Sectio 5. d re proved i Appedi E. 5 6 7 i i 6 i 3 The remiig formuls re simple rules for workig with sigm ottio: N Formuls 8 re proved writig out ech side i epded form. The left side of Equtio 9 is c c c The right side is c These re equl the distriutive propert. The other formuls re discussed i Appedi E. 8 9 c c c i c i i i i i i i i i EXAPLE () Evlute the Riem sum for f 3 6 tkig the smple poits to e right edpoits d, 3, d 6. 3 () Evlute 3 6 d. SOLUTION () With 6 the itervl width is 3 6 d the right edpoits re.5,., 3.5, 4., 5.5, d 6 3.. So the Riem sum is R 6 6 f i f.5 f. f.5 f. f.5 f 3..875 5 5.65 4.65 9 3.9375
3 7 CHAPTER 5 INTEGRALS 5 = -6 FIGURE 5 3 Notice tht 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 5. () With suitervls we hve 3 Thus, 3, 6, 3 9, d, i geerl, i 3i. Sice we re usig right edpoits, we c use Theorem 4: N I the sum, is costt (ulike i), so we c move 3 i frot of the sig. 5 = -6 A 3 A 3 3 6 d l f i l 3 l 3i 3 6 3i 3 l 7 l l 8 4 l 8 4 8 4 8 4 7 7 4 3 i 3 8 i i 3 54 i 54 7 6.75 f 3i 3 (Equtio 9 with c 3 ) (Equtios d 9) (Equtios 7 d 5) FIGURE 6 3 j ( -6) d=a -A =_6.75 This itegrl c t 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 6. Figure 7 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, 6.75, s l. A R 5 = -6 3 4 6.3998 6.63 5 6.79 6.7365 5 6.7473 FIGURE 7 R Å_6.39 9 8
SECTION 5. THE DEFINITE INTEGRAL 3 7 A much simpler method for evlutig the itegrl i Emple will e give i Sectio 5.3. N Becuse f e is positive, the itegrl i Emple 3 represets the re show i Figure 8. = EXAPLE 3 () Set up epressio for 3 e d s it of sums. () Use computer lger sstem to evlute the epressio. SOLUTION () Here we hve f e,, 3, d So,, 4, 3 6, d 3 i i FIGURE 8 From Theorem 4, we get 3 e d l f i l f i l e i () If we sk computer lger sstem to evlute the sum d simplif, we oti N A computer lger sstem is le to fid eplicit epressio for this sum ecuse it is geometric series. The it could e foud usig l Hospitl s Rule. e i e 3 e e Now we sk the computer lger sstem to evlute the it: 3 e d l e 3 e e 3 e e We will ler much esier method for the evlutio of itegrls i the et sectio. V EXAPLE 4 Evlute the followig itegrls iterpretig ech i terms of res. () s d () 3 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 9. Therefore s d 4 4 FIGURE 9 (I Sectio 7.3 we will e le to prove tht the re of circle of rdius r is r.)
3 7 CHAPTER 5 INTEGRALS () The grph of is the lie with slope show i Figure. We compute the itegrl s the differece of the res of the two trigles: 3 d A A.5 (3, ) =- A A 3 FIGURE _ THE IDPOINT RULE We ofte choose the smple poit * i to e the right edpoit of the ith suitervl ecuse it is coveiet for computig the it. 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 we get the followig pproimtio. TEC odule 5. /7.7 shows how the idpoit Rule estimtes improve s icreses. IDPOINT RULE f d f i f f where d i i i midpoit of i, i = EXAPLE 5 Use the idpoit Rule with to pproimte V 5. d SOLUTION The edpoits of the five suitervls re,.,.4,.6,.8, d., so the midpoits re.,.3,.5,.7, d.9. The width of the suitervls is 5 5, so the idpoit Rule gives d f. f.3 f.5 f.7 f.9 5..3.5.7.9.6998 FIGURE Sice f for, the itegrl represets re, d the pproimtio give the idpoit Rule is the sum of the res of the rectgles show i Figure.
SECTION 5. THE DEFINITE INTEGRAL 3 73 At the momet we do t kow how ccurte the pproimtio i Emple 5 is, ut i Sectio 7.7 we will ler method for estimtig the error ivolved i usig the idpoit Rule. At tht time we will discuss other methods for pproimtig defiite itegrls. If we ppl the idpoit Rule to the itegrl i Emple, we get the picture i Figure. The pproimtio 4 6.7563 is much closer to the true vlue 6.75 th the right edpoit pproimtio, R 4 6.3998, show i Figure 7. TEC I Visul 5. ou c compre left, right, d midpoit pproimtios to the itegrl i Emple for differet vlues of. 5 = -6 3 FIGURE Å_6.7563 PROPERTIES OF THE DEFINITE INTEGRAL Whe we defied the defiite itegrl f d, we implicitl ssumed tht. But the defiitio s it of Riem sums mkes sese eve if. Notice tht if we reverse d, the chges from to. Therefore f d f d If, the d so f d We ow develop some sic properties of itegrls tht will help us to evlute itegrls i simple mer. We ssume tht f d t re cotiuous fuctios. c =c PROPERTIES OF THE INTEGRAL. c d c, where c is costt. f t d f d t d 3. cf d c f d, where c is costt re=c(-) 4. f t d f d t d FIGURE 3 j c d=c(-) 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 3.
3 7 4 CHAPTER 5 INTEGRALS g f+g FIGURE 4 j [ƒ+ ] d= j ƒ d+j d N Propert 3 seems ituitivel resole ecuse we kow tht multiplig fuctio positive umer c stretches or shriks its grph verticll fctor of c. So it stretches or shriks ech pproimtig rectgle fctor c d therefore it hs the effect of multiplig the re c. f Propert ss tht the itegrl of sum is the sum of the itegrls. For positive fuctios it ss tht the re uder f t is the re uder f plus the re uder t. Figure 4 helps us uderstd wh this is true: I view of how grphicl dditio works, the correspodig verticl lie segmets hve equl height. I geerl, Propert follows from Theorem 4 d the fct tht the it of sum is the sum of the its: f t d l f i t i l f i t i l f i l t i f d t d Propert 3 c e proved i similr mer 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 3 with c. EXAPLE 6 Use the properties of itegrls to evlute 4 3 d. SOL UTION Usig Properties d 3 of itegrls, we hve We kow from Propert tht 4 3 d 4 d 3 d 4 d 3 d d we foud i Emple i Sectio 5. tht 4 d 4 4 d 3. So 4 3 d 4 d 3 d 4 3 3 5 The et propert tells us how to comie itegrls of the sme fuctio over djcet itervls: =ƒ 5. c f d f d f d c FIGURE 5 c This is ot es to prove i geerl, ut for the cse where f d c Propert 5 c e see from the geometric iterprettio i Figure 5: The re uder f from to c plus the re from c to is equl to the totl re from to.
SECTION 5. THE DEFINITE INTEGRAL 3 7 5 V EXAPLE 7 If it is kow tht f d 7 d f d, fid f d. SOL UTION B Propert 5, we hve 8 8 8 f d f d f d 8 so 8 f d f d 8 f d 7 5 Properties 5 re true whether,, or. The followig properties, i which we compre sizes of fuctios d sizes of itegrls, re true ol if. COPARISON PROPERTIES OF THE INTEGRAL 6. If f for, the f d. 7. If f t for, the f d t d. 8. If m f for, the m f d m =ƒ If f, the f d represets the re uder the grph of f, so the geometric iterprettio of Propert 6 is simpl tht res re positive. But the propert c e proved from the defiitio of itegrl (Eercise 64). Propert 7 ss tht igger fuctio hs igger itegrl. It follows from Properties 6 d 4 ecuse f t. Propert 8 is illustrted Figure 6 for the cse where f. If f is cotiuous we could tke m d 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. FIGURE 6 PROOF OF PROPERTY 8 Sice m f, Propert 7 gives m d f d d Usig Propert to evlute the itegrls o the left d right sides, we oti m f d Propert 8 is useful whe ll we wt is rough estimte of the size of itegrl without goig to the other of usig the idpoit Rule. EXAPLE 8 Use Propert 8 to estimte e d. SOL UTION Becuse f e is decresig fuctio o,, its solute mimum vlue is f d its solute miimum vlue is m f e. Thus,
3 7 6 CHAPTER 5 INTEGRALS Propert 8, = =e or e e d e e d =/e Sice e.3679, we c write.367 e d FIGURE 7 The result of Emple 8 is illustrted i Figure 7. The itegrl is greter th the re of the lower rectgle d less th the re of the squre. 5. EXERCISES. Evlute the Riem sum for f 3, 4, with si suitervls, tkig the smple poits to e left edpoits. Epli, with the id of digrm, wht the Riem sum represets. 6. The grph of t is show. Estimte t d with si suitervls usig () right edpoits, () left edpoits, d 3 (c) midpoits. 3. If f, 3, evlute the Riem sum with 6, tkig the smple poits to e right edpoits. Wht does the Riem sum represet? Illustrte with digrm. g 3. If f e,, fid the Riem sum with 4 correct to si deciml plces, tkig the smple poits to e midpoits. Wht does the Riem sum represet? Illustrte with digrm. 4. () Fid the Riem sum for f si, 3, with si terms, tkig the smple poits to e right edpoits. (Give our swer correct to si deciml plces.) Epli wht the Riem sum represets with the id of sketch. () Repet prt () with midpoits s smple poits. 5. The grph of fuctio f is give. Estimte f d usig four suitervls with () right edpoits, () left edpoits, d (c) midpoits. f 8 7. A tle of vlues of icresig fuctio f is show. Use the tle to fid lower d upper estimtes for f d. 5 5 5 5 f 4 37 5 6 5 36 8. The tle gives the vlues of fuctio otied from eperimet. Use them to estimte 9 f d usig three equl 3 suitervls with () right edpoits, () left edpoits, d (c) midpoits. If the fuctio is kow to e icresig fuctio, c ou s whether our estimtes re less th or greter th the ect vlue of the itegrl? 3 4 5 6 7 8 9 f 3.4..6.3.9.4.8
SECTION 5. THE DEFINITE INTEGRAL 3 7 7 CAS 9 Use the idpoit Rule with the give vlue of to pproimte the itegrl. Roud the swer to four deciml plces. 9.. si d, 5.. cos 4 d, 3. If ou hve CAS tht evlutes midpoit pproimtios d grphs the correspodig rectgles (use middlesum d middleo commds i ple), check the swer to Eercise d illustrte with grph. The repet with d. 4. With progrmmle clcultor or computer (see the istructios for Eercise 7 i Sectio 5.), compute the left d right Riem sums for the fuctio f si o the itervl, with. Epli wh these estimtes show tht Deduce tht the pproimtio usig the idpoit Rule with 5 i Eercise is ccurte to two deciml plces. 5. Use clcultor or computer to mke tle of vlues of right Riem sums R for the itegrl si d with 5,, 5, d. Wht vlue do these umers pper to e pprochig? 6. Use clcultor or computer to mke tle of vlues of left d right Riem sums L d R for the itegrl with 5,, 5, d. Betwee wht two e d umers must the vlue of the itegrl lie? C ou mke similr sttemet for the itegrl? Epli. e d 7 Epress the it s defiite itegrl o the give itervl. 7. 8. 9. s 3 d, 4 l i l i, l s* i * i,. 4 3 i * 6 i * 5, l 5 Use the form of the defiitio of the itegrl give i Theorem 4 to evlute the itegrl. 5.36 si d.35 cos i l,, i. 3 d., 6, 8], 4 4 5 e d, 4 5 d CAS 6. () Fid pproimtio to the itegrl 3 d usig Riem sum with right edpoits d 8. () Drw digrm like Figure 3 to illustrte the pproimtio i prt (). (c) Use Theorem 4 to evlute 4 3 d. (d) Iterpret the itegrl i prt (c) s differece of res d illustrte with digrm like Figure 4. 7. Prove tht d. 8. Prove tht d 3 3. 3 9 3 Epress the itegrl s it of Riem sums. Do ot evlute the it. 9. 6 3. d 5 3 3 Epress the itegrl s it of sums. The evlute, usig computer lger sstem to fid oth the sum d the it. 3. si 5 d 3. 33. The grph of f is show. Evlute ech itegrl iterpretig it i terms of res. () f d (c) 7 f d 5 34. The grph of t cosists of two stright lies d semicircle. Use it to evlute ech itegrl. () (d) () t d () t d (c) 4 6 =ƒ 4 6 8 = 4 6 d 5 f d 9 f d 4 l d 7 t d 3. d 4. 5 3 d 4 7 5. 3 d
3 7 8 CHAPTER 5 INTEGRALS 35 4 Evlute the itegrl iterpretig it i terms of res. 35. ( d 36. 3 s4 d 54. s 4 4 cos d s3 6 4 37. ( s9 ) d 3 38. 39. 4. d 4. Evlute si cos 4 d. 3 3 d 5 d 55 6 Use Propert 8 to estimte the vlue of the itegrl. 4 55. s d 56. 3 57. t d 58. 4 59. e d 6. 3 d 3 3 d si d 4. Give tht 3s 4 d 5s5 8, wht is 3usu 4 du? 43. I Emple i Sectio 5. we showed tht d. 3 Use this fct d the properties of itegrls to evlute 5 6 d. 44. Use the properties of itegrls d the result of Emple 3 to evlute 3 e d. 45. Use the result of Emple 3 to evlute e d. 46. Use the result of Eercise 7 d the fct tht cos d (from Eercise 5 i Sectio 5.), together with the properties of itegrls, to evlute cos 5 d. Write s sigle itegrl i the form f d: 48. If 5 f d d 5 f d 3.6, fid 4 f d. 4 49. If d 9 f d 37 9 t d 6, fid 9 f 3t d. 5. Fid f d if 5. Suppose f hs solute miimum vlue m d solute mimum vlue. Betwee wht two vlues must f d lie? Which propert of itegrls llows ou to mke our coclusio? 5 54 Use the properties of itegrls to verif the iequlit without evlutig the itegrls. 5. 53. 5 s d 47. f d 5 f d f d 3 for 3 f for 3 s d s d s 3 6 6 Use properties of itegrls, together with Eercises 7 d 8, to prove the iequlit. 6. s 4 d 6 6. 3 63. Prove Propert 3 of itegrls. 64. Prove Propert 6 of itegrls. 65. If f is cotiuous o,, show tht [Hit:.] 66. Use the result of Eercise 65 to show tht 67. Let f if is rtiol umer d f if is irrtiol umer. Show tht f is ot itegrle o,. 68. Let f d f if. Show tht f is ot itegrle o,. [Hit: Show tht the first term i the Riem sum, f i *, c e mde ritrril lrge.] 69 7 Epress the it s defiite itegrl. 69. [Hit: Cosider f 4.] 7. 3 i 4 l 5 l i f d f d f f f f si d f d 7. Fid d. Hit: Choose i * to e the geometric me of i d i (tht is, i * s i i ) d use the idetit m m m m si d 8
SECTION 5.3 THE FUNDAENTAL THEORE OF CALCULUS 3 7 9 D I S C O V E R Y P R O J E C T AREA FUNCTIONS. () Drw the lie t d use geometr to fid the re uder this lie, ove the t-is, d etwee the verticl lies t d t 3. () If, let A e the re of the regio tht lies uder the lie t etwee t d t. Sketch this regio d use geometr to fid epressio for A. (c) Differetite the re fuctio A. Wht do ou otice?. () If, let A A represets the re of regio. Sketch tht regio. () Use the result of Eercise 8 i Sectio 5. to fid epressio for A. (c) Fid A. Wht do ou otice? (d) If d h is smll positive umer, the A h A represets the re of regio. Descrie d sketch the regio. (e) Drw rectgle tht pproimtes the regio i prt (d). B comprig the res of these two regios, show tht A h h A (f) Use prt (e) to give ituitive epltio for the result of prt (c). ; 3. () Drw the grph of the fuctio f cos i the viewig rectgle,.5,.5. () If we defie ew fuctio t t cos t dt the t is the re uder the grph of f from to [util f ecomes egtive, t which poit t ecomes differece of res]. Use prt () to determie the vlue of t which t strts to decrese. [Ulike the itegrl i Prolem, it is impossile to evlute the itegrl defiig t to oti eplicit epressio for t.] (c) Use the itegrtio commd o our clcultor or computer to estimte t., t.4, t.6,..., t.8, t. The use these vlues to sketch grph of t. (d) Use our grph of t from prt (c) 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? 4. Suppose f is cotiuous fuctio o the itervl, d we defie ew fuctio t the equtio t f t dt Bsed o our results i Prolems 3, cojecture epressio for t. t dt 5.3 THE FUNDAENTAL THEORE OF CALCULUS The Fudmetl Theorem of Clculus is ppropritel med ecuse it estlishes coectio etwee the two rches of clculus: differetil clculus d itegrl clculus. Differetil clculus rose from the tget prolem, wheres itegrl clculus rose from seemigl urelted prolem, the re prolem. Newto s metor t Cmridge,