4 The Integral. 4.0 Area. (base) (height). If the units for each side of the rectangle are meters,

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1 4 The Itegrl Previous chpters delt with differetil clculus. They strted with the simple geometricl ide of the slope of tget lie to curve, developed it ito combitio of theory bout derivtives d their properties, exmied techiques for clcultig derivtives, d pplied these cocepts d techiques to rel-life situtios. This chpter begis the developmet of itegrl clculus d strts with the simple geometric ide of re ide tht will spw its ow combitio of theory, techiques d pplictios. Oe of the most importt results i mthemtics, the Fudmetl Theorem of Clculus, ppers i this chpter. It uifies differetil d itegrl clculus ito sigle grd structure. Historiclly, this uifictio mrked the begiig of moder mthemtics, d it provided importt tools for the growth d developmet of the scieces. The chpter begis with look t re, some geometric properties of res, d some pplictios. First we will exmie wys of pproximtig the res of regios such s tree leves bouded by curved edges d the res of regios bouded by grphs of fuctios. The we will develop wys to clculte the res of some of these regios exctly. Filly, we will explore the rich vriety of uses of res. 4. Are The primry purpose of this itroductory sectio is to help develop your ituitio bout res d your bility to reso usig geometric rgumets bout re. This type of resoig will pper ofte i the rest of this book d is very helpful for pplyig the ides of clculus. The bsic shpe we will use is the rectgle: the re of rectgle is (bse) (height). If the uits for ech side of the rectgle re meters, the the re will hve uits (meters) (meters) = squre meters = m. The oly other re formuls eeded for this sectio re for trigles (re = b h) d for circles (re = π r ). I dditio, we will use (d ssume to be true) three other fmilir properties of re:

2 94 the itegrl Additio Property: The totl re of regio is the sum of the res of the o-overlppig pieces tht comprise the regio: Iclusio Property: If regio B is iside regio A (see mrgi), the the re of regio B is less th or equl to the re of regio A. Loctio-Idepedece Property: The re of regio does ot deped o its loctio: Exmple. Determie the re of the regio show below left. Solutio. We c esily brek the regio ito two rectgles (show bove right), with res of 5 squre iches d squre iches respectively, so the re of the origil regio is 8 squre iches. Prctice. Determie the re of the trpezoidl regio show i the mrgi by cuttig it i two wys: () ito rectgle d trigle d (b) ito two trigles. We c use our three re properties to deduce iformtio bout res tht re difficult to clculte exctly. Let A be the regio bouded by the grph of f (x) =, the x-xis, d the verticl lies x = d x x =. Becuse the two rectgles i the mrgi figure sit iside regio A d do ot overlp ech other, the re of the rectgles, + = 5 6, is less th the re of regio A. Prctice. Build two rectgles, ech with bse uit, with boudries tht exted outside the shded regio i the mrgi figure d use their res to mke other vlid sttemet bout the re of regio A.

3 4. re 95 Prctice. Wht c you sy bout the re of regio A if we use iside d outside rectgles ech with bse uit? Exmple. The figure below right icludes drk squres, ech cetimeter o side, d lighter squres of the sme size. We c be sure tht the re of the lef below left is smller th wht umber? Solutio. The re of the lef is smller th + = 6 cm. Prctice 4. We c be sure tht the re of the lef is t lest how lrge? Fuctios c be defied i terms of res. For the costt fuctio f (t) =, defie A(x) to be the re of the rectgulr regio (top mrgi figure) bouded by the grph of f, the t-xis, d the verticl lies t t = d t = x; we c esily see tht A() = (shded regio i the secod mrgi figure). Similrly, A() = 4 d A(4) = 6. I geerl, A(x) = (bse)(height) = (x )() = x for y x. From the grph of y = A(x) (i the third mrgi figure) we c see tht A (x) = for every vlue of x >. (The fct tht A (x) = f (x) i the precedig discussio is ot coicidece, s we shll soo ler.) Prctice 5. For f (t) =, defie B(x) to be the re of the regio bouded by the grph of f, the t-xis, d verticl lies t t = d t = x (see below left). Fill i the tble below with the requested vlues of B. How re the grphs of y = A(x) d y = B(x) relted? x B(x).5 Sometimes it is useful to move regios roud. The re of prllelogrm is obvious if we move the trigulr regio from oe side of the prllelogrm to fill the regio o the other side, resultig i with rectgle (see mrgi).

4 96 the itegrl At first glce, it is difficult to estimte the totl re of the shded regios show below left: but if we slide ll of them ito sigle colum (bove right), the becomes esy to determie tht the shded re is less th the re of the eclosig rectgle = (bse)(height) = ()() =. Prctice 6. The totl re of the shded regios i the mrgi figure is less th wht umber? Some Applictios of Are Oe reso res re so useful is tht they c represet qutities other th sizes of simple geometric shpes. For exmple, if the uits of the bse of rectgle re hours d the uits of the height re miles, the the uits of the re of the rectgle re: hour (hours) ( ) miles = miles hour mesure of distce. Similrly, if the bse uits re pouds d the height uits re feet, the the re uits re foot-pouds, mesure of work. I the bottom mrgi figure, f (t) is the velocity of cr i miles per hour, d t is the time i hours. ( So the shded re will be (bse) (height) = ( hours) miles ) = 6 miles, the distce hour trveled by the cr i the hours from : p.m. util 4: p.m. Distce s Are If f (t) is the (positive) forwrd velocity of object t time t, the the re betwee the grph of f d the t-xis d the verticl lies t times t = d t = b will equl the distce tht the object hs moved forwrd betwee times d b. This re s distce cocept c mke some difficult distce problems much esier.

5 4. re 97 Exmple. A cr strts from rest (velocity = ) d stedily speeds up so tht secods lter its speed is 88 feet per secod (6 miles per hour). How fr did the cr trvel durig those secods? Solutio. We could swer the questio usig the techiques of Chpter (try this). But if stedily mes tht the velocity icreses lierly, the it is esier to use the mrgi figure d the cocept of re s distce. The re of the trigulr regio represets the distce trveled: distce = (bse)(height) = ( ( sec) 88 ft ) = 88 ft sec The cr trvels totl of 88 feet durig those secods. Prctice 7. A tri iitilly trvelig t 45 miles per hour (66 feet per secod) tkes 6 secods to decelerte to complete stop. If the tri slowed dow t stedy rte (the velocity decresed lierly), how my feet did the tri trvel before comig to stop? Prctice 8. You d fried strt off t oo d wlk i the sme directio log the sme pth t the rtes show i the mrgi figure. () Who is wlkig fster t : p.m.? Who is hed t : p.m.? (b) Who is wlkig fster t : p.m.? Who is hed t : p.m.? (c) Whe will you d your fried be together? (Aswer i words.) I the precedig Exmple d Prctice problems, fuctio represeted rte of trvel (i miles per hour, for istce) d the re represeted the totl distce trveled. For fuctios represetig other rtes, such s the productio of fctory (bicycles per dy) or the flow of wter i river (gllos per miute) or trffic over bridge (crs per miute) or the spred of disese (ewly sick people per week), the re will still represet the totl mout of somethig. Are s Totl Accumultio If f (t) represets positive rte (i uits per time itervl) t time t, the the re betwee the grph of f d the t-xis d the verticl lies t times t = d t = b will be the totl mout of {somethig} tht ccumultes betwee times d b (see mrgi). For exmple, the figure t the top of the ext pge shows the flow rte (i cubic feet per secod) of wter i the Skykomish River er the tow of Gold Br, Wshigto. The re of the shded regio represets the totl volume (cubic feet) of wter flowig pst the tow durig the moth of October:

6 98 the itegrl totl wter = re = re of rectgle + re of trigle ( ) ( ) ft ( dys) + 5 ft ( dys) sec sec ( ) ( ) = 75 ft ( dys) = 75 ft (59 sec) sec sec ft For compriso, the flow over Nigr Flls is bout. 5 ft sec. 4. Problems. () Clculte the re of the shded regio:. Clculte the re of the trpezoidl regio i the figure below left by brekig it ito trigle d rectgle. (b) Clculte the re of the shded regio:. Brek the regio show bove right ito trigle d rectgle d ( verify ) tht the totl re of h + H the trpezoid is b.

7 4. re () Clculte the sum of the rectgulr res i the regio show below left. xis. Use two well-plced trpezoids to estimte the re of this regio. 9. Let A(x) represet the re bouded by the grph of the fuctio show below, the horizotl xis, d verticl lies t t = d t = x. Evlute A(x) for x =,,, 4 d 5. (b) Wht c you sy bout the re of the shded regio show bove right? 5. () Clculte the sum of the res of the rectgles show below left.. Let B(x) represet the re bouded by the grph of the fuctio show below, the horizotl xis, d verticl lies t t = d t = x. Evlute B(x) for x =,,, 4 d 5. (b) Wht c you sy bout the re of the shded regio show bove right? 6. () Clculte the sum of the res of the trpezoids show below left.. Let C(x) represet the re bouded by the grph of the fuctio show below, the horizotl xis, d verticl lies t t = d t = x. Evlute C(x) for x =, d, d use tht iformtio to deduce formul for C(x). (b) Wht c you sy bout the re of the shded regio show bove right? 7. Cosider the regio bouded by the grph of y = + x, the positive x-xis, the positive y-xis d the lie x =. Use two well-plced rectgles to estimte the re of this regio. 8. Cosider the regio bouded by the grph of y = 9 x, the positive x-xis d the positive y-

8 the itegrl. Let A(x) represet the re bouded by the grph of the fuctio show below, the horizotl xis, d verticl lies t t = d t = x. Evlute A(x) for x =, d, d fid formul for A(x). 5. The figure below shows the velocities of two crs. From the time the brkes were pplied: () how log did it tke ech cr to stop? (b) which cr trveled frther before stoppig?. The figure below shows the velocity of cr durig -secod time frme. How fr did the cr trvel betwee t = to t = secods? 6. A speeder trvelig 45 miles per hour (i 5- mph zoe) psses stopped police cr, which immeditely tkes off fter the speeder. If the police cr speeds up stedily to 6 mph over - secod itervl d the trvels t costt 6 mph, how log d how fr will it be before the police cr ctches the speeder, who cotiued trvelig t 45 mph? (See figure below.) 4. The figure below shows the velocity of cr durig -secod time frme. How fr did the cr trvel betwee t = to t = secods? 7. Fill i the tble with the uits for re of rectgle with the give bse d height uits. bse height re miles per secod hours squre feet kilowtts houses mels secods dollrs per hour feet hours people per house mels

9 4. re 4. Prctice Aswers. () (6) + (4)() = 4 (b) ()() + (6)() = 4 ( ). outside rectgulr re = ()() + () =.5. Usig rectgles with bse = : iside re = ( ) outside re = ( ) 5 = = 7 6 =. so the re of the regio is betwee.95 d.. 4. The lef s re is lrger th the re of the drk rectgles, cm. 5. y = B(x) = x is lie with slope, so it is prllel to the lie y = A(x) = x ; see mrgi for tble. 6. Are < re of the rectgle eclosig the shifted regios = 5; see mrgi figure. 7. Drw grph of the velocity fuctio: x B(x).5 4 d the use the cocept of re s distce : distce = re of shded regio = (bse)(height) = ( (6 sec) 66 ft ) = 98 feet sec 8. () At : p.m. both re wlkig t the sme velocity. You re hed. (b) At : p.m. your fried is wlkig fster th you, but you re still hed. (The re uder your velocity curve is lrger th the re uder your fried s.) (c) You d your fried will be together o the tril whe the res (distces) uder the two velocity grphs re equl.

10 the itegrl 4. Sigm Nottio d Riem Sums Oe strtegy for clcultig the re of regio is to cut the regio ito simple shpes, clculte the re of ech simple shpe, d the dd these smller res together to get the re of the whole regio. Whe you use this pproch with my sub-regios, it will be useful to hve ottio for ddig my vlues together: the sigm (Σ) ottio. summtio sigm ottio how to red it k= k= k k the sum of k squred, from k equls to k equls 5 the sum of divided by k, from k equls to k equls j= = j the sum of to the j-th power, from j equls to j equls 5 the sum of sub, from equls to equls 7 The vrible (typiclly i, j, k, m or ) used i the summtio is clled the couter or idex vrible. The fuctio to the right of the sigm is clled the summd, while the umbers below d bove the sigm re clled the lower d upper limits of the summtio. Prctice. Write the summtio deoted by ech of the followig: () 5 k k= (b) 7 ( ) j j j= (c) 4 (m + ) m= I prctice, we frequetly use sigm ottio together with the stdrd fuctio ottio: x f (x) g(x) h(x) k= f (k + ) = f ( + ) + f ( + ) + f ( + ) 4 j= = f () + f (4) + f (5) Exmple. Use the tble to evlute f (x j ) = f (x ) + f (x ) + f (x ) + f (x 4 ) 5 5 f (k) d [5 + f (j )]. k= j= Solutio. Writig out the sum d usig the tble vlues: 5 f (k) = f () + f () + f (4) + f (5) k= = = 4

11 4. sigm ottio d riem sums while: 5 [5 + f (j )] = [5 + f ( )] + [5 + f (4 )] + [5 + f (5 )] j= = [5 + f ()] + [5 + f ()] + [5 + f ()] = [5 + ] + [5 + ] + [5 + ] which dds up to. Prctice. Use the vlues i the precedig mrgi tble to evlute: () 5 g(k) k= (b) 4 h(j) j= Exmple. For f (x) = x +, evlute (c) k= 5 [g(k) + f (k )] k= f (k). Solutio. Writig out the sum d usig the fuctio vlues: k= f (k) = f () + f () + f () + f () = ( ) ( ) ( ) ( ) = which dds up to 8. Prctice. For g(x) = 4 x, evlute g(k) d g(k + ). k= k= The summd eed ot coti the idex vrible explicitly: you c write sum from k = to k = 4 of the costt fuctio f (k) = s f (k) or 5 = = 5 = 5. Similrly: k= k= 7 = = 5 = k= Becuse the sigm ottio is simply ottio for dditio, it possesses ll of the fmilir properties of dditio. Summtio Properties: Sum of Costts: Additio: Subtrctio: Costt Multiple: k= k= k= k= C = C + C + C + + C = C ( k + b k ) = ( k b k ) = C k = C k= k= k k= k + k k= k= b k b k Problems 6 d 7 illustrte tht similr ptters for sums of products d quotiets re ot vlid.

12 4 the itegrl Sums of Ares of Rectgles I Sectio 4., we will pproximte res uder curves by buildig rectgles s high s the curve, clcultig the re of ech rectgle, d the ddig the rectgulr res together. Exmple. Evlute the sum of the rectgulr res i the mrgi figure, the write the sum usig sigm ottio. Solutio. The sum of the rectgulr res is equl to the sum of (bse) (height) for ech rectgle: () ( ) + () ( ) + () 4 ( ) = which we c rewrite s 5 k= usig sigm ottio. k Prctice 4. Evlute the sum of the rectgulr res i the mrgi figure, the write the sum usig sigm ottio. The bses of these rectgles eed ot be equl. For the rectgulr res ssocited with f (x) = x i the mrgi figure: rectgle bse height re = f () = 4 4 = 8 4 = f (4) = 6 6 = = f (5) = 5 5 = 5 so the sum of the rectgulr res is = 74. Exmple 4. Write the sum of the res of the rectgles i the mrgi figure usig sigm ottio. Solutio. The re of ech rectgle is (bse) (height): rectgle bse height re x x f (x ) (x x ) f (x ) x x f (x ) (x x ) f (x ) x x f (x ) (x x ) f (x ) The re of the k-th rectgle is (x k x k ) f (x k ), so we c express the totl re of the three rectgles s (x k x k ) f (x k ). k= Prctice 5. Write the sum of the res of the shded rectgles i the mrgi figure usig sigm ottio.

13 4. sigm ottio d riem sums 5 Are Uder Curve: Riem Sums Suppose we wt to clculte the re betwee the grph of positive fuctio f d the x-xis o the itervl [, b] (see below left). Oe method to pproximte the re ivolves buildig severl rectgles with bses o the x-xis spig the itervl [, b] d with sides tht rech up to the grph of f (bove right). We the compute the res of the rectgles d dd them up to get umber clled Riem sum of f o [, b]. The re of the regio formed by the rectgles provides pproximtio of the re we wt to compute. Exmple 5. Approximte the re show i the mrgi betwee the grph of f d the x-xis spig the itervl [, 5] by summig the res of the rectgles show i the lower mrgi figure. Solutio. The totl re is ()() + ()(5) = squre uits. I order to effectively describe this process, some ew vocbulry is helpful: prtitio of itervl d the mesh of prtitio. A prtitio P of closed itervl [, b] ito subitervls cosists of set of + poits {x =, x, x, x,..., x, x = b} listed i icresig order, so tht = x < x < x < x <... < x < x = b. (A prtitio is merely collectio of poits o the horizotl xis, urelted to the fuctio f i y wy.) The poits of the prtitio P divide [, b] ito subitervls: These itervls re [x, x ], [x, x ], [x, x ],..., [x, x ] with legths x = x x, x = x x, x = x x,..., x = x x. The poits x k of the prtitio P mrk the loctios of the verticl lies for the sides of the rectgles, d the bses of the rectgles hve

14 6 the itegrl legths x k for k =,,,...,. The mesh or orm of prtitio P is the legth of the logest of the subitervls [x k, x k ] or, equivletly, the mximum vlue of x k for k =,,,...,. For exmple, the set P = {,, 4.6, 5., 6} is prtitio of the itervl [, 6] (see mrgi) tht divides the itervl [, 6] ito four subitervls with legths x =, x =.6, x =.5 d x 4 =.9, so the mesh of this prtitio is.6, the mximum of the legths of the subitervls. (If the mesh of prtitio is smll, the the legth of ech oe of the subitervls is the sme or smller.) Prctice 6. P = {,.8, 4.8, 5., 6.5, 7, 8} is prtitio of wht itervl? How my subitervls does it crete? Wht is the mesh of the prtitio? Wht re the vlues of x d x? A fuctio, prtitio d poit chose from ech subitervl determie Riem sum. Suppose f is positive fuctio o the itervl [, b] (so tht f (x) > whe x b), P = {x =, x, x, x,..., x, x = b} is prtitio of [, b], d c k is x-vlue chose from the k-th subitervl [x k, x k ] (so x k c k x k ). The the re of the k-th rectgle is: f (c k ) (x k x k ) = f (c k ) x k Defiitio: A summtio of the form k= f (c k ) x k is clled Riem sum of f for the prtitio P d the chose poits {c, c,..., c }. This Riem sum is the totl of the res of the rectgulr regios d provides pproximtio of the re betwee the grph of f d the x-xis o the itervl [, b]. Exmple 6. Fid the Riem sum for f (x) = usig the prtitio x {, 4, 5} d the vlues c = d c = 5 (see mrgi). Solutio. The two subitervls re [, 4] d [4, 5], hece x = d x =. So the Riem sum for this prtitio is: k= f (c k ) x k = f (c ) x + f (c ) x The vlue of the Riem sum is.7. = f () + f (5) = + 5 = 7 Prctice 7. Clculte the Riem sum for f (x) = x {, 4, 5} usig the chose vlues c = d c = 4. o the prtitio

15 4. sigm ottio d riem sums 7 Prctice 8. Wht is the smllest vlue Riem sum for f (x) = x c hve usig the prtitio {, 4, 5}? (You will eed to choose vlues for c d c.) Wht is the lrgest vlue Riem sum c hve for this fuctio d prtitio? The tble below shows the output of computer progrm tht clculted Riem sums for the fuctio f (x) = with vrious umbers x of subitervls (deoted ) d differet wys of choosig the poits c k i ech subitervl. mesh c k = left edge = x k c k = rdom poit c k = right edge = x k Whe the mesh of the prtitio is smll (d the umber of subitervls,, is lrge), it ppers tht ll of the wys of choosig the c k loctios result i pproximtely the sme vlue for the Riem sum. For this decresig fuctio, usig the left edpoit of the subitervl lwys resulted i sum tht ws lrger th the re pproximted by the sum. Choosig the right edpoit resulted i vlue smller th tht re. Why? As the mesh gets smller, ll of the Riem Sums seem to be pprochig the sme vlue, pproximtely.69. (As we shll soo see, these vlues re ll pprochig l(5) ) Exmple 7. Fid the Riem sum for the fuctio f (x) = si(x) o the itervl [, π] usig the prtitio {, π 4, π }, π d the chose poits c = π 4, c = π d c = π 4. [ Solutio. The three subitervls (see mrgi) re, π ] [ π, 4 4, π ] d [ π ], π so x = π 4, x = π 4 d x = π. The Riem sum for this prtitio is: k= ( π ) f (c k ) x k = si π 4 = or pproximtely π 4 + π 4 + ( π ) 4 + si π ( π 4 + si 4 π = ( + )π 8 ) π Prctice 9. Fid the Riem sum for the fuctio d prtitio i the previous Exmple, but this time choose c =, c = π d c = π.

16 8 the itegrl Two Specil Riem Sums: Lower d Upper Sums Two prticulr Riem sums re of specil iterest becuse they represet the extreme possibilities for give prtitio. We eed f to be cotiuous i order to ssure tht it ttis its miimum d mximum vlues o y closed subitervl of the prtitio. If f is bouded but ot ecessrily cotiuous we c geerlize this defiitio by replcig f (m k ) with the gretest lower boud of ll f (x) o the itervl d f (M k ) with the lest upper boud of ll f (x) o the itervl. Defiitio: If f is positive, cotiuous fuctio o [, b] d P is prtitio of [, b], let m k be the x-vlue i the k-th subitervl so tht f (m k ) is the miimum vlue of f o tht itervl, d let M k be the x-vlue i the k-th subitervl so tht f (M k ) is the mximum vlue of f o tht subitervl. The: LS P = US P = k= k= f (m k ) x k is the lower sum of f for P f (M k ) x k is the upper sum of f for P Geometriclly, lower sum rises from buildig rectgles uder the grph of f (see first mrgi figure) d every lower sum is less th or equl to the exct re A of the regio bouded by the grph of f d the x-xis o the itervl [, b]: LS P A for every prtitio P. Likewise, upper sum rises from buildig rectgles over the grph of f (see secod mrgi figure) d every upper sum is greter th or equl to the exct re A of the regio bouded by the grph of f d the x-xis o the itervl [, b]: US P A for every prtitio P. Together, the lower d upper sums provide bouds o the size of the exct re: LS P A US P. For y c k vlue i the k-th subitervl, f (m k ) f (c k ) f (M k ), so, for y choice of the c k vlues, the Riem sum RS P = stisfies the iequlity: k= f (c k ) x k k= f (m k ) x k k= f (c k ) x k k= f (M k ) x k or, equivletly, LS P RS P US P. The lower d upper sums provide bouds o the size of ll Riem sums for give prtitio. The exct re A d every Riem sum RS P for prtitio P d y choice of poits {c k } both lie betwee the lower sum d the upper sum for P (see mrgi). Therefore, if the lower d upper sums re close together, the the re d y Riem sum for P (regrdless of how you choose the poits c k ) must lso be close together. If we kow tht the upper d lower sums for prtitio P re withi. uits of ech other, the we c be sure tht every Riem sum for prtitio P is withi. uits of the exct re A.

17 4. sigm ottio d riem sums 9 Ufortutely, fidig miimums d mximums for ech subitervl of prtitio c be time-cosumig (d tedious) tsk, so it is usully ot prcticl to determie lower d upper sums for wiggly fuctios. If f is mootoic, however, the it is esy to fid the vlues for m k d M k, d sometimes we c eve explicitly clculte the limits of the lower d upper sums. For mootoic, bouded fuctio we c gurtee tht Riem sum is withi certi distce of the exct vlue of the re it is pproximtig. Theorem: If f is positive, mootoic, bouded fuctio o [, b] the for y prtitio P d y Riem sum for f usig P, RS P A US P LS P f (b) f () (mesh of P) Proof. The Riem sum d the exct re re both betwee the upper d lower sums, so the distce betwee the Riem sum d the exct re is o bigger th the distce betwee the upper d lower sums. If f is mootoiclly icresig, we c slide the res represetig the differece of the upper d lower sums ito rectgle: Recll from Sectio. tht mootoic mes lwys icresig or lwys decresig o the itervl i questio. I words, this strig of iequlities sys tht the distce betwee y Riem sum d the re beig pproximted is o bigger th the differece betwee the upper d lower Riem sums for the sme prtitio, which i tur is o bigger th the distce betwee the vlues of the fuctio t the edpoits of the itervl times the mesh of the prtitio. See Problem 56 for the mootoiclly decresig cse. whose height equls f (b) f () d whose bse equls the mesh of P. So the totl differece of the upper d lower sums is smller th the re of tht rectgle, [ f (b) f ()] (mesh of P). 4. Problems I Problems 6, rewrite the sigm ottio s summtio d perform the idicted dditio.. ( + ) 4. = 5 si(πk) k=. 4 k. k= 5 ( + j) j= 5. 5 cos(πj) 6. j= k= k

18 the itegrl I Problems 7, rewrite ech summtio usig the sigm ottio. Do ot evlute the sums I Problems 5, use this tble: k k b k to verify the equlity for these vlues of k d b k. ( k + b k ) = k= ( k b k ) = k= k= 5 k = 5 k= k= k= k k + k k= k= For Problems 6 8, use the vlues of k d b k i the tble bove to verify the iequlity b k b k ( ) ( ) k b k = k b k k= k= k= ( ) k = k k= k= k= k b k = k= k k= b k g (j) 6. j= 4 h(k) 8. k= = f () h(). k g(k) k= 4 h(i) i= 7 g(k) h(k) k= I 6, write out ech summtio d simplify the result. These re exmples of telescopig sums [ k (k ) ]. k= 5 k= 8 k= [ k ] k + [ k + k ] [k (k ) ] k= [(k + ) k ] 4 k= 5 [x k x k ] k= I 7 4, () list the subitervls determied by the prtitio P, (b) fid the vlues of x k, (c) fid the mesh of P d (d) clculte x k. k= 7. P = {,, 4.5, 6, 7} 8. P = {,.6, 4, 4., 5, 5.5, 6} 9. P = {,,,.5, } 4. P s show below: 4. P s show below: For 9, f (x) = x, g(x) = x d h(x) = x. Evlute ech sum k= f (k). f (j). j= g(m) 4. m= k= i= f (k) f ( + i) g ( f (k)) k= 4. P s show below:

19 4. sigm ottio d riem sums 4. For x k = x k x k, verify tht: x k = legth of the itervl [, b] k= For 44 48, sketch grph of f, drw verticl lies t ech poit of the prtitio, evlute ech f (c k ) d sketch the correspodig rectgle, d clculte d dd up the res of those rectgles. 44. f (x) = x +, P = {,,, 4} () c =, c =, c = (b) c =, c =, c = f (x) = 4 x, P = {,,.5, } () c =, c =, c = (b) c =, c =.5, c = f (x) = x, P = {,, 5, } () c =, c = 4, c = 9 (b) c =, c =, c = f (x) = si(x), P = {, π 4, π, π} () c =, c = π 4, c = π (b) c = π 4, c = π, c = π 48. f (x) = x, P = {,, } () c =, c = (b) c =, c = For 49 5, sketch the fuctio d fid the smllest possible vlue d the lrgest possible vlue for Riem sum for the give fuctio d prtitio. 49. f (x) = + x () P = {,, 4, 5} (b) P = {,,, 4, 5} (c) P = {,.5,,, 4, 5} 5. f (x) = 7 x () P = {,, } (b) P = {,,, } (c) P = {,.5,,.5,, } 5. f (x) = si(x) () P = {, π, π} (b) P = {, π 4, π, π} (c) P = {, π 4, π 4, π} 5. f (x) = x x + () P = {,, } (b) P = {,,, } (c) P = {,.5,,,.5, } 5. Suppose LS P = 7.6 d US P = 7.4 for positive fuctio f d prtitio P of [, 5]. () You c be certi tht every Riem sum for the prtitio P is withi wht distce of the exct vlue of the re betwee the grph of f d the x-xis o the itervl [, 5]? (b) Wht if LS P = 7.7 d US P = 7.9? 54. Suppose you divide the itervl [, 4] ito eqully wide subitervls d clculte Riem sum for f (x) = + x by rdomly selectig poit c k i ech subitervl. () You c be certi tht the vlue of the Riem sum is withi wht distce of the exct vlue of the re betwee the grph of f d the x-xis o itervl [, 4]? (b) Wht if you use eqully wide subitervls? 55. If you divide [, 4] ito 5 eqully wide subitervls d clculte Riem sum for f (x) = + x by rdomly selectig poit c k i ech subitervl, the you c be certi tht the Riem sum is withi wht distce of the exct vlue of the re betwee f d the x-xis o the itervl [, 4]? 56. If f is mootoic decresig o [, b] d you divide [, b] ito eqully wide subitervls: the you c be certi tht the Riem sum is withi wht distce of the exct vlue of the re betwee f d the x-xis o the itervl [, b]?

20 the itegrl Summig Powers of Cosecutive Itegers The formuls below re icluded here for your referece. They will ot be used i the followig sectios, except for hdful of exercises i Sectio 4.. Formuls for some commoly ecoutered summtios c be useful for explicitly evlutig certi specil Riem sums. The summtio formul for the first positive itegers is reltively well kow, hs severl esy but clever proofs, d eve hs iterestig story behid it ( ) + = ( + ) k = k= Proof. Let S = ( ) + ( ) +, which we c lso write s S = + ( ) + ( ) Addig these two represettios of S together: S = ( ) + ( ) + + S = + ( ) + ( ) S = ( + ) + ( + ) + ( + ) + + ( + ) + ( + ) + ( + ) Krl Friedrich Guss ( ), Germ mthemtici sometimes clled the price of mthemtics. So S = ( + ) S = ( + ), the desired formul. Guss supposedly discovered this formul for himself t the ge of five whe his techer, plig to keep the clss busy for while, sked the studets to dd up the itegers from to. Guss thought few miutes, wrote his swer o his slte, d tured it i, the st smugly while his clssmtes struggled with the problem. 57. Fid the sum of the first positive itegers i two wys: () usig Guss formul, d (b) usig Guss method (from the proof). 58. Fid the sum of the first odd itegers. (Ech odd iteger c be writte i the form k for k =,,,....) 59. Fid the sum of the itegers from to. Formuls for other iteger powers of the first itegers re lso kow: k= k = k = k= k= k = + ( + ) = + + ( + )( + ) = ( ) ( + ) 4 = k 4 = = ( + )( + )( + ) k=

21 4. sigm ottio d riem sums I Problems 6 6, use the properties of summtio d the formuls for powers give bove to evlute ech sum. 6. ( + k + k ) 6. k= ( ) k k + k= 6. k (k ) k= 4. Prctice Aswers. () (b) (c). () (b) (c) 5 k = k= 7 ( ) j j = j= 4 (m + ) = m= 5 g(k) = g() + g() + g(4) + g(5) = + ( ) = 7 k= 4 h(j) = h() + h() + h() + h(4) = = j= 5 (g(k) + f (k )) = (g() + f ()) + (g(4) + f ()) + (g(5) + f (4)) k= = ( + ) + ( + ) + (5 + ) =. 4 k= g(k) = g() + g() + g(4) = = g(k + ) = g() + g() + g(4) = = k= 4. Rectgulr res = = 5 = 4 j= 5. f (x ) (x x ) + f (x ) (x x ) + f (x ) (x x ) = or k= f (x k ) (x k+ x k ) j j= f ( x j ) ( xj x j ) 6. Itervl is [, 8]; six subitervls; mesh =.; x = 4.8; x = x x = =. ( ) ( ) 7. RS = () + () 4 =.5 ( ) ( ) 8. smllest RS = () 4 + () 5 =.95 ( ) lrgest RS = ()() + () 4 =.5 9. RS = () ( π 4 ) + () ( π4 ) + () ( π ).56

22 4 the itegrl 4. The Defiite Itegrl Ech prticulr Riem sum depeds o severl thigs: the fuctio f, the itervl [, b], the prtitio P of tht itervl, d the chose vlues c k from ech subitervl of tht prtitio. Fortutely for most of the fuctios eeded for pplictios s the pproximtig rectgles get thier (d s the meshes of the prtitios P pproch d the umber of subitervls i those prtitios pproches ) the vlues of the Riem sums pproch the sme umber, idepedet of the prticulr prtitios P d the chose poits c k i the subitervls of those prtitios. This limit of the Riem sums will become the ext big topic i clculus: the defiite itegrl. Itegrls rise throughout the rest of this book d i pplictios i lmost every field tht uses mthemtics. Here we use the ottio P to me the mesh of P. The dx is differetil (see Sectio.6), the limit of the discrete qutity x i the Riem sum. Defiitio ( of the Defiite Itegrl: ) If f (c k ) x k equls fiite umber I, where ech lim P k= P is prtitio of the itervl [, b], the we sy f is itegrble o the itervl [, b] d cll the umber I the defiite itegrl of f o [, b] d write it s We red the symbol b b f (x) dx. f (x) dx s the itegrl from to b of eff of x dee x or the itegrl from to b of f (x) with respect to x. Furthermore, we cll f (x) the itegrd, the lower edpoit of itegrtio d b the upper edpoit of itegrtio. (We will sometimes lso cll d b the upper d lower limits of itegrtio.) Exmple. Describe the re betwee the grph of f (x) = x, the x-xis, d the verticl lies t x = d x = 5 s limit of Riem sums d s defiite itegrl. You my hve oticed tht we did ot precisely defie wht lim mes or how P to compute this limit. Providig defiitio turs out to be more complicted th the limits we hve ecoutered so fr, d i prctice we will rrely eed to compute such limit, so forml defiitio is left to more dvced textbooks. Solutio. Here f (x) =, = d b = 5, so: x re = lim P ( k= ) 5 x c k = k x dx which, ccordig to estimtios mde i Sectio 4., is pproximtely equl to.69. Prctice. Describe the re betwee the grph of f (x) = si(x), the x-xis, d the verticl lies t x = d x = π s limit of Riem sums d s defiite itegrl.

23 4. the defiite itegrl 5 Exmple. Usig the cocept of re, determie the vlues of: ( ) () ( + c k ) x k o the itervl [, ] (b) (c) lim P 4 k= (5 x) dx x dx Solutio. () The limit of the Riem sums represets the re betwee the grph of f (x) = + x, the x-xis, d the verticl lies t x = d x = (see mrgi); this re equls 6 squre uits. (b) The defiite itegrl represets the re betwee f (x) = 5 x, the x-xis d the verticl lies t x = d x = 4, which is trpezoid with re squre uits. (c) The defiite itegrl represets the re of the upper hlf of the circle x + y =, which hs rdius d ceter t (, ). The re of this semicircle is πr = π = π. Prctice. ( Usig the cocept of re, determie the vlues of: ) 8 () (c k ) x k o the itervl [, ] (b) 4 dx lim P k= Exmple. ( Represet ech limit of Riem sums s defiite itegrl. ) ( ) () ( + c k ) x k o [, 4] (b) ck x k o [, 9] lim P k= lim P k= Solutio. () 4 ( + x) dx (b) 9 x dx Exmple 4. Represet ech shded re i the mrgi figure s defiite itegrl. (Do ot ttempt to evlute the defiite itegrl, just trslte the picture ito symbols.) Solutio. () ( 4 x ) π dx (b) si(x) dx The vlue of defiite itegrl b f (x) dx depeds oly o the fuctio f beig itegrted d o the itervl [, b]. Replcig the vrible x tht ppers i b f (x) dx, sometimes clled dummy vrible, does ot chge the vlue of the itegrl. The followig defiite itegrls ech represet the itegrl of the fuctio f o the itervl [, b], d they re ll equl: b f (x) dx = b f (t) dt = b f (u) du = b f (w) dw

24 6 the itegrl Defiite Itegrls of Negtive Fuctios A defiite itegrl is limit of Riem sums, d you c form Riem sums usig y itegrd fuctio f, positive or egtive (or both), cotiuous or discotiuous. The defiite itegrl of itegrble fuctio will still hve geometric meig eve if the fuctio is sometimes (or lwys) egtive, d defiite itegrls of egtive fuctios lso hve meigful iterprettios i pplictios. Exmple 5. Fid the defiite itegrl of f (x) = o [, 4]. Solutio. Writig Riem sum for f (x) = o the itervl [, 4]: k= f (c k ) x k = k= ( ) x k = k= x k = () = 6 for every prtitio P d every choice of vlues for c k so: 4 dx = lim P ( ) ( ) x k = lim 6 = 6 k= P The re of the regio i i the mrgi figure is 6 uits, but becuse the regio is below the x-xis, the vlue of the itegrl is 6. the If the grph of f (x) is below the x-xis for x b ( f is egtive) b f (x) dx is times the re of the regio below the x-xis d bove the grph of f (x) betwee x = d x = b. If f (t) represets the rte of popultio chge (people per yer) for tow, the egtive vlues of f for give time itervl would idicte tht the popultio of the tow ws gettig smller, d the defiite itegrl (ow egtive umber) would represet the chge i the popultio decrese durig tht time itervl. Exmple 6. I 98 there were, ducks estig roud lke. The rte of popultio chge is show i the mrgi. Write defiite itegrl to represet the totl chge i the duck popultio betwee 98 d 99, the estimte the popultio i 99. Solutio. The totl chge i popultio is give by f (t) dt d this defiite itegrl is equl to times the re of the rectgle i the mrgi figure: so: ducks yer yers = ducks [99 popultio] = [98 popultio] + [chge from 98 to 99] = + ( ) = Approximtely, ducks were estig roud the lke i 99.

25 4. the defiite itegrl 7 If f (t) represets the velocity of cr (i miles per hour) movig i the positive directio log stright lie t time t, the egtive vlues of f idicte tht the cr ws trvellig i the egtive directio (tht is, bckwrds). The defiite itegrl of f (the itegrl will be egtive umber) represets the chge i positio of the cr durig the time itervl: how fr the cr trvelled i the egtive directio. Prctice. A bug strts t the loctio x = o the x-xis t : p.m. d wlks log the xis with the velocity show i the mrgi figure. How fr does the bug trvel betwee : p.m. d : p.m.? Where is the bug t : p.m.? Frequetly itegrd fuctio will be positive some of the time d egtive some of the time. If f represets rte of popultio icrese, the the itegrl of the positive prts of f will give the icrese i popultio d the itegrl of the egtive prts of f will give the decrese i popultio. Altogether, the itegrl of f over the etire time itervl will give the totl (et) chge i the popultio. Geometriclly, we c ow iterpret defiite itegrl s differece of res of the regio(s) betwee the grph of f d the horizotl xis: b f (x) dx = [re bove x-xis] [re below x-xis] Exmple 7. Use the mrgi figure to clculte 5 4 f (x) dx d 5 f (x) dx. Solutio. Usig the res idicted i the figure, 4 f (x) dx = 5 d f (x) dx =, while f (x) dx, 4 f (x) dx = [re bove x-xis] [re below x-xis] = [ + ] [5] = f (x) dx, f (x) dx =, where we dded the res of the regios bove the x-xis d subtrcted the re of the regio below the x-xis. Prctice 4. Use geometric resoig to evlute π si(x) dx. If f represets velocity, the itegrls o the itervls where f is positive mesure distces moved i the forwrd directio d itegrls o the itervls where f is egtive mesure distces moved i the bckwrd directio. The itegrl over the whole time itervl gives the totl (et) chge i positio: the distce moved forwrd mius the distce moved bckwrd.

26 8 the itegrl Prctice 5. A cr trvels west with the velocity show i the mrgi. () How fr does the cr trvel betwee oo d 6: p.m.? (b) At 6: p.m., where is the cr reltive to its positio t oo? Uits for the Defiite Itegrl We hve lredy see tht the re uder grph c represet qutities whose uits re ot the usul geometric uits of squre meters or squre feet. For exmple, if x mesures time i secods d f (x) gives velocity with uits feet per secod, the x hs the uits secods d f (x) x hs uits: ( ) feet (secods) = feet secod which is mesure of distce. Becuse ech Riem sum f (x) x is sum of feet d the defiite itegrl is limit of these Riem sums, the defiite itegrl hs the sme uits, feet. If the uits of f (x) re squre feet d the uits of x re feet, the b f (x) dx is umber with the uits (feet ) (feet) = feet, or cubic feet, mesure of volume. If f (x) represets force i pouds d x is distce i feet, the foot-pouds, mesure of work. I geerl, the uits for b b f (x) dx is umber with the uits f (x) dx re (uits for f (x)) (uits for x). A quick check of the uits whe computig defiite itegrl c help void errors whe solvig pplied problem. 4. Problems I Problems 4, rewrite ech limit of Riem sums s defiite itegrl. ( ). lim ( + c k ) x k o [, 4] P k=. lim P. lim P 4. lim P ( ) (c k ) x k o [, ] k= ) cos(5c k ) x k ( k= ( k= ck x k ) o [, 4] o [, 5] I Problems 5, represet the re of ech bouded regio s defiite itegrl. (Do ot ttempt to evlute the itegrl, just trslte the re ito itegrl.) 5. The regio bouded by y = x, the x-xis, d the lies x = d x = The regio bouded by y = x, the x-xis d the lie x = The regio bouded by y = x si(x), the x-xis, d the lies x = d x =.

27 4. the defiite itegrl 9 8. The shded regio show below: 6. Evlute ech itegrl usig the figure below showig the grph of f d vrious res. 9. The shded regio show below: () (d) 5 5 f (x) dx (b) f (x) dx (e) 5 7 f (x) dx f (x) dx (c) 7 5 f (x) dx 7. Evlute ech itegrl usig the figure below showig the grph of g d vrious res.. The shded regio show bove for x. I Problems 5, represet the re of ech bouded regio s defiite itegrl, the use geometry to determie the vlue of tht defiite itegrl. () g(x) dx (b) 4 g(x) dx (c) 8 4 g(x) dx. The regio bouded by y = x, the x-xis, d the lies x = d x =. (d) 8 g(x) dx (e) 8 g(x) dx. The regio bouded by y = 4 x, the x-xis d the y-xis.. The regio bouded by y = x, the x-xis d the lie x =. 8. Evlute ech itegrl usig the figure below showig the grph of h. 4. The shded regio show below left. () h(x) dx (b) 6 4 h(x) dx (c) 6 h(x) dx 5. The shded regio show bove right. (d) 4 h(x) dx (e) 4 h(x) dx

28 the itegrl For Problems 9, the figure shows your velocity (i feet per miute) log stright pth. () Sketch grph of your loctio. (b) How my feet did you wlk i 8 miutes? (c) Where, reltive to your strtig loctio, re you fter 8 miutes? 9. See figure below left.. See figure bove right. Problems 7 give the uits for x d f (x). Specify the uits of the defiite itegrl b f (x) dx.. x is time i secods ; f (x) is velocity i meters per secod. x is time i hours ; f (x) is flow rte i gllos per hour. x positio i feet ; f (x) re i squre feet 4. x is time i dys ; f (x) is temperture i degrees Celsius 5. x height i meters ; f (x) force i grms 6. x is positio i iches ; f (x) is desity i pouds per ich 7. x is time i secods ; f (x) is ( ccelertio ) i feet per secod per secod ft sec The remiig problems use the summtio formuls give t the ed of Sectio 4., s demostrted i the followig Exmple. Exmple 8. For f (x) = x, divide the itervl [, ] ito eqully wide subitervls, evlute the lower sum, d compute the limit of tht lower sum s. Solutio. The width of the itervl is b = = so ech of the subitervls should hve width x = b =. The edpoits of the k-th itervl i the prtitio re therefore (k ) d k for k =,,...,. Becuse f (x) = x is icresig o [, ] the miimum vlue of the fuctio o ech subitervl occurs t the left edpoit of the subitervl, hece we eed to choose c k = (k ). So: LS = k= = 8 ( (k ) ) k= ( ) k k + k f (c k ) x k = k= = 8 [ ( + )( + ) 6 = 8 [ 6 + ] = 8 [ k= ( + ) = 8 ] + k= k= k + (k ) k= ] = 8 [ + 6 ] As, LS 8 6 () = 8 so we c be certi tht x dx 8.

29 4. the defiite itegrl Prctice 6. Redo Exmple 6 but fid the upper Riem sum for eqully wide prtitio itervls d show tht the limit of these upper sums, s, is 8. From the previous Exmple d Prctice problem, we kow tht 8 x dx 8 so we c coclude tht x = 8. We will discover much esier method for evlutig this itegrl i Sectio For f (x) = + x, prtitio the itervl [, ] ito eqully wide subitervls of legth x =. () Compute the lower sum for this fuctio d prtitio, d clculte the limit of tht lower sum s. (b) Compute the upper sum for this fuctio d prtitio d fid the limit of the upper sum s. 9. For f (x) = x, prtitio the itervl [, ] ito eqully wide subitervls of legth x =. () Compute the lower sum for this fuctio d prtitio, d clculte the limit of tht lower sum s. (b) Compute the upper sum for this fuctio d prtitio d fid the limit of the upper sum s.. For f (x) = x, prtitio the itervl [, 9] ito subitervls by tkig x k = 9 k for k =,,...,. () Choose c k = x k for ech subitervl d compute the upper sum for this fuctio d prtitio, the clculte the limit of tht upper sum s. (b) Compute the lower sum for this fuctio d prtitio d fid the limit of the lower sum s.

30 the itegrl 4. Prctice Aswers. re = lim P. lim P 8 ( k= ( k= si (c k ) x k ) = π si(x) dx (c k ) x k ) = re of trpezoid i mrgi = 8 4 dx = re of rectgle i mrgi =. ().5 feet forwrd d.5 feet bckwrd = 5 feet totl (b) The bug eds up feet forwrd of its strtig positio t x =, so the bug s fil loctio is t x =. 4. Betwee x = d x = π, the grph of y = si(x) hs the sme re bove the x-xis s below the x-xis so the two res ccel d the defiite itegrl is : π si(x) dx =. 5. () miles west (from oo to : p.m.) plus 6 miles est (from : p.m. to 6: p.m.) yields totl trvel distce of 8 miles. (At 4: p.m. the driver is bck t the strtig positio fter drivig 4 miles: miles west d the miles est.) (b) The cr is 4 miles est of the strtig loctio. egtive of west.) 6. x = US = (Est is the =, M k = k so f (M k) = ( k ) = 4 k. The: k= = 8 k= k= 4 f (M k ) x = k k = 8 [ + + ] 6 = so the limit of these upper sums s is 8.

31 4. properties of the defiite itegrl 4. Properties of the Defiite Itegrl We hve defied defiite itegrls s limits of Riem sums, which c ofte be iterpreted s res of geometric regios. These two powerful cocepts of the defiite itegrl c help us uderstd itegrls d use them i vriety of pplictios. This sectio cotiues to emphsize this dul view of defiite itegrls d presets severl properties of defiite itegrls. We will justify these properties usig the properties of summtios d the defiitio of defiite itegrl s Riem sum, but they lso hve turl iterprettios s properties of res of regios. We will the use these properties to help uderstd fuctios tht re defied by itegrls, d lter to help clculte the vlues of defiite itegrls. Properties of the Defiite Itegrl As you red ech sttemet bout defiite itegrls, drw sketch or exmie the ccompyig figure to iterpret the property s sttemet bout res. f (x) dx = This property sys tht the defiite itegrl of fuctio over itervl cosistig of sigle poit must be. Geometriclly, we c see tht the re uder the grph of fuctio bove sigle poit should be becuse the width of poit is. I terms of Riem sums, we c t prtitio sigle poit, so isted we must defie the vlue of y defiite itegrl over o-existet itervl to be. b b f (x) dx = f (x) dx I words, this property sys tht if we reverse the limits of itegrtio, we must multiply the vlue of the defiite itegrl by. Geometriclly, if < b the the x-vlues i the first itegrl re movig bckwrds from x = b to x =, so it might seem resoble tht we should get egtive swer. I terms of Riem sums, if we move from right to left, ech x k i y prtitio P will be egtive: k= f (c k ) x k = k= f (c k ) ( x k ) = k= resultig i times the Riem sum we would use for f (c k ) x k b f (x) dx. Our defiitio of Riem sum oly llows ech x k to be positive, however, so we c simply tret this itegrl property s other defiitio.

32 4 the itegrl b k dx = k (b ) (k is y costt) Thikig geometriclly, if k > (see mrgi), the b the re of rectgle with bse b d height k, so: k dx represets Here we use the fct tht the sum of the legths of the subitervl of y prtitio of the itervl [, b] is equl to the width of [, b], which is b. b k dx = (height) (bse) = k (b ) Altertively, for y P = {x =, x, x, x,..., x, x = b} tht prtitios the itervl [, b], d every choice of poits c j from the subitervls of tht prtitio, the Riem sum is: j= f ( c j ) xj = j= k x j = k j= x j = k (b ) Becuse every Riem sum equls k (b ), the limit of those sums, s P, must lso be k (b ). b b k f (x) dx = k f (x) dx (k is y costt) I words, this property sys tht multiplyig itegrd by costt k hs the sme result s multiplyig the vlue of the defiite itegrl by tht costt. Geometriclly, multiplyig fuctio by positive costt k stretches the grph of y = f (x) by fctor of k i the verticl directio, which should multiply the re of the regio betwee tht grph d the x-xis by the sme fctor. Thikig i terms of Riem sums: j= k f ( c j ) xj = k j= f ( c j ) xj so the limit of the sum o the left over ll possible prtitios P, s P, is b o the right yields k k f (x) dx, while the correspodig limit of the sums b b f (x) dx. c f (x) dx + f (x) dx = b c f (x) dx This property is most esily uderstood (d believed) i terms of picture (see mrgi). We c lso justify this property usig Riem sums by restrictig our prtitios to iclude the poit x = b betwee x = d x = c d the splittig tht prtitio ito two sub-prtitios tht prtitio [, b] d [b, c], respectively. This property remis true, however, eve whe b c or b.

33 4. properties of the defiite itegrl 5 Properties of Defiite Itegrls of Combitios of Fuctios The ext two properties relte the vlues of itegrls of sums d differeces of fuctios to the sums d differeces of itegrls of the idividul fuctios. You will fid these properties very useful whe computig itegrls of fuctios tht ivolve the sum or differece of severl terms (such s polyomil): you c itegrte ech term d the dd or subtrct the idividul results to get the swer. Both properties hve turl iterprettios s sttemets bout res. b [ f (x) + g(x)] dx = b f (x) dx + b g(x) dx I words, this sys the itegrl of sum is the sum of the itegrls. The followig grph supplies geometricl justifictio: Usig Riem sums, we c write: [ ( ) ( )] f cj + g cj xj = j= j= f ( c j ) xj + d the tke the limit o ech side s P. g ( ) c j xj j= b [ f (x) g(x)] dx = b f (x) dx b g(x) dx I words, this sys the itegrl of differece is the differece of the itegrls. The justifictio for this differece property is quite similr to the justifictio of the sum property. (Or we c combie the sum property with the costt-multiple property, settig k =.) Prctice. Give tht the defiite itegrl 4 4 f (x) dx = 7 d tht [ f (x) g(x)] dx. If f (x) g(x) for ll x i [, b], the b 4 g(x) dx =, evlute f (x) dx b g(x) dx Geometriclly, the mrgi figure illustrtes tht if f d g re both positive d tht f (x) g(x) o the itervl [, b], the the re of regio betwee the grph of f d the x-xis is smller th the re of regio betwee the grph of g d the x-xis.

34 6 the itegrl Similr sketches for the situtios where f or g re sometimes or lwys egtive illustrte tht the property holds i other situtios s well, but we c void ll of those differet cses usig Riem sums. If we use the sme prtitio P d chose poits c j for Riem sums for f d g, the f ( c j ) g ( cj ) for ech j, so: j= f ( c j ) xj g ( ) c j xj j= Tkig the limit over ll such prtitios s the mesh of those prtitios pproches, we get b f (x) dx b g(x) dx. You my hve oticed tht we hve t clled the justifictios of these properties proofs, i prt becuse we hve t precisely defied wht lim mes, but P lso becuse of some other techicl detils left to more dvced textbooks. If m f (x) M for ll x i [, b] the m (b ) b f (x) dx M (b ) This property follows esily from the previous oe. First let g(x) = M so tht f (x) M = g(x) for ll x i [, b], hece b f (x) dx b M dx = M (b ) (usig oe of our previous properties). Likewise, tkig g(x) = m so tht f (x) m = g(x) for ll x i [, b]: b f (x) dx b m dx = m (b ) Geometriclly, this sys tht if we c trp the output vlues of fuctio o the itervl [, b] betwee two upper d lower bouds, m d M, the the vlue of the defiite itegrl must lie betwee the res of the rectgles with heights m d M. If f is cotiuous o the closed itervl [, b], the we kow tht f tkes o miimum vlue o tht itervl (cll it m) d mximum vlue (cll it M), i which cse this property just uses the lower d upper Riem sums for the simplest possible prtitio of [, b]: 5 Exmple. Determie lower d upper bouds for the vlue of with f (x) give grphiclly i the mrgi. f (x) dx

35 4. properties of the defiite itegrl 7 Solutio. If x 5, the we c estimte (from the grph) tht f (x) 9 so lower boud for 5 f (x) dx is (b ) (miimum of f o [, b]) = (4)() = 8 d upper boud is: (b ) (mximum of f o [, b]) = (4)(9) = 6 We c coclude tht 8 5 f (x) dx 6. Kowig tht the vlue of defiite itegrl is somewhere betwee 8 d 6 is ot useful for fidig its exct vlue, but the precedig estimtio property is very esy to use d provides bllprk estimte tht will help you void reportig uresoble vlue. Prctice. Determie lower boud d upper boud for the vlue of 5 f (x) dx with f s i the previous Exmple. Fuctios Defied by Itegrls If oe of the edpoits or b of the itervl [, b] chges, the the vlue of the itegrl the form x b f (t) dt typiclly chges. A defiite itegrl of f (t) dt defies fuctio of x tht possesses iterestig d useful properties. The ext exmples illustrte oe such property: the derivtive of fuctio defied by itegrl is closely relted to the itegrd, the fuctio iside the itegrl. Exmple. For the fuctio f (t) =, defie A(x) to be the re of the regio bouded by f, the t-xis, d verticl lies t t = d t = x. () Evlute A(), A(), A() d A(4). (b) Fid lgebric formul for A(x) vlid for x. (c) Clculte A (x). (d) Express A(x) s defiite itegrl. Solutio. () Referrig to the grph i the mrgi, we c see tht A() =, A() =, A() = 4 d A(4) = 6. (b) Usig the sme re ide to compute more geerl re: A(x) = re of rectgle = (bse) (height) = (x )() = x (c) A (x) = d (x ) = dx (d) x A(x) = dt

36 8 the itegrl Prctice. Aswer the questios i the previous Exmple for f (x) =. Exmple. For the fuctio f (t) = + t, defie B(x) to be the re of the regio bouded by the grph of f, the t-xis, d verticl lies t t = d t = x (see mrgi). () Evlute B(), B(), B() d B(). (b) Fid lgebric formul for B(x) vlid for x. (c) Clculte B (x). (d) Express B(x) s defiite itegrl. Solutio. () From the grph, B() =, B() =.5, B() = 4 d B() = 7.5. (b) Usig the sme re cocept: B(x) = re of trpezoid = (bse) (verge height) ( ) + ( + x) = (x) = x + x (c) B (x) = d (x + ) x dx x = + x (d) B(x) = [ + t] dt Prctice 4. Aswer the questios i the previous Exmple for f (t) = t. A curious coicidece ppered i ech of these Exmples d Prctice problems: the derivtive of the fuctio defied by the itegrl ws the sme s the itegrd, the fuctio iside the itegrl. Stted other wy, the fuctio defied by the itegrl ws tiderivtive of the fuctio iside the itegrl. I Sectio 4.4 we will see tht this coicidece is ctully property shred by ll fuctios defied by itegrl i this wy. Ad it is such importt property tht it is prt of result clled the Fudmetl Theorem of Clculus. Before we study the Fudmetl Theorem of Clculus, however, we eed to cosider existece questio: Which fuctios c be itegrted? Which Fuctios Are Itegrble? This importt questio ws filly swered i the 85s by Berhrd Riem, me tht should be fmilir to you by ow. Riem proved tht fuctio must be bdly discotiuous i order to ot be itegrble. Due to our iexct defiitio of the limit ivolved i the defiitio of the defiite itegrl, we defer proof of this theorem to more dvced textbooks. k= Theorem: Every cotiuous fuctio is itegrble. This result sys tht if f is cotiuous o the itervl [, b], the f (c k ) x k pproches the sme fiite umber, P, o mtter how we choose the prtitios P. b f (x) dx, s

37 4. properties of the defiite itegrl 9 I fct, we c geerlize this result to fuctios tht hve fiite umber of breks or jumps, s log s the fuctio is bouded: Theorem: If f is defied o itervl [, b] d bouded ( f (x) M for some umber M for ll x i [, b]) d cotiuous except t fiite umber of poits i [, b] the f is itegrble o [, b]. The fuctio f grphed i the mrgi is lwys betwee d (i fct, lwys betwee d ), so it is bouded, d it is cotiuous except t x = d x =. As log s the vlues of f () d f () re fiite umbers, their ctul vlues will ot ffect the vlue of the defiite itegrl, d we c compute the vlue of the itegrl by computig the res of the (trigulr d rectgulr) regios betwee the grph of f d the x-xis: 5 f (x) dx = Prctice 5. Evlute f (x) dx +..5 f (x) dx + 5 x dx (see mrgi). f (x) dx = = 8 The figure below depicts grphiclly the reltioships betwee differetible, cotiuous d itegrble fuctios: This sys: Every differetible fuctio is cotiuous, but there re cotiuous fuctios tht re ot differetible: simple exmple of the ltter is f (x) = x, which is cotiuous but ot differetible t x =. Every cotiuous fuctio is itegrble, but there re itegrble fuctios tht re ot cotiuous: simple exmple of the ltter situtio is the fuctio f (x) grphed i the mrgi, which is itegrble o [, 5] but discotiuous t x = d x =. Filly, s demostrted by the ext exmple, there re fuctios tht re ot itegrble.

38 the itegrl A No-itegrble Fuctio If f is cotiuous or piecewise cotiuous o [, b], the f is itegrble o [, b]. Fortutely, erly ll of the fuctios we will use throughout the rest of this book re itegrble, s re the fuctios you re likely to eed for commo pplictios. There re fuctios, however, for which the limit of the Riem sums does ot exist d hece, by defiitio, re ot itegrble. Recll the holey fuctio from Sectio.4: The fuctio h(x) = { is ot itegrble o [, ]. if x is rtiol umber if x is irrtiol umber Proof. For y prtitio P of [, ], suppose tht you, very rtiol perso, lwys choose vlues of c k tht re rtiol umbers. (Ay ope itervl o the rel-umber lie cotis rtiol umbers d irrtiol umbers, so for ech subitervl of the prtitio P you c lwys choose c k to be rtiol umber.) The h (c k ) =, so for your Riem sum: YS P = h (c k ) x k = k= k= x k = k= x k = ( ) = 6 Suppose your fried, however, lwys selects vlues of c k tht re irrtiol umbers. The h (c k ) = for ech c k, so for your fried s Riem sum: FS P = h (c k ) x k = k= k= x k = k= x k = ( ) = So the limit of your Riem sums, s the mesh of P pproches, will be 6, while the limit of your fried s sums will be. This mes tht lim P ( k= h (c k ) x k ) does ot exist (becuse there is o sigle limitig vlue of the Riem sums s P ) so h(x) is ot itegrble o [, ]. A similr rgumet shows tht h(x) is ot itegrble o y itervl of the form [, b] (where < b).

39 4. properties of the defiite itegrl 4. Problems I Problems, refer to the grph of f give below to determie the vlue of ech defiite itegrl... 5 g(x) dx. g(x) dx 4. 4 g(t) dt g(x) dx 5. 8 g(s) ds 6. 4 g(x) dx 7. g(t) dt [ + g(x)] dx 9. 6 g(u) du. 8 [t + g(t)] dt f (x) dx. f (x) dx 4. f (x) dx 6. f (t) dt 8. f (x) dx. f (x) dx. f (s) ds 4. [x + f (x)] dx 6. [ + f (x)] dx 8. f (x) dx f (x) dx f (w) dw f (x) dx f (x) dx f (x) dx f (x) dx [ + f (x)] dx [ + f (x)] dx f (x) dx [ + f (x) ] dx Problems refer to the grph of g give below. Use the grph to evlute ech itegrl. For 4, use the costt fuctios f (x) = 4 d g(x) = o the itervl [, ]. Clculte the vlue of ech itegrl d verify tht the vlue obtied i prt () is ot equl to the vlue i prt (b).. () f (x) dx g(x) dx (b) f (x) g(x) dx. (). () 4. () f (x) dx ( g(x) dx (b) [ f (x)] dx (b) f (x) dx f (x) g(x) dx ) f (x) dx (b) f (x) dx For 5 4, sketch grph of the itegrd fuctio d use it to help evlute the itegrl x dx 6. x dx 8. u du 4. [ + t ] dt [ + t ] dt [ x ] dx x dx x dx For Problems 4 46, sketch () grph of y = A(x) = x f (t) dt d (b) grph of y = A (x). 4. f (x) = x 44. f (x) = x

40 the itegrl The figure below shows the velocity of cr. Write the totl distce trveled by the cr betwee : p.m. d 4: p.m. s defiite itegrl d estimte the vlue of tht itegrl. 46. For 47 5, stte whether or ot ech fuctio is: () cotiuous o [, 4] (b) differetible o [, 4] (c) itegrble o [, 4] 47. f (x) from Problem f (x) from Problem Write the totl distce trveled by the cr i the previous problem betwee : p.m. d 6: p.m. s defiite itegrl d estimte the vlue of tht itegrl. 5. Defie g(x) = 7 for x = d g() = () Show tht the Riem sum for g(x) for y prtitio P of the itervl [, 4] is equl to 5w + 7( w), where w is the width of the subitervl tht icludes x =. (b) Compute the limit of these sums, s P (c) Compre the vlues of 4 g(x) dx d 4 7 dx. (d) Wht c you coclude bout how chgig the vlue of itegrble fuctio t sigle poit ffects the vlue of its defiite itegrl? 4. Prctice Aswers. 4 [ f (x) g(x)] dx = 7 = 4. m = d M = 6 so ()(5 ) = 4 5 f (x) dx = (6)(5 ). () A() =, A() =, A() = 6, A(4) = 9 (b) A(x) = (x )() = x (c) A (x) = (d) A(x) = 4. () B() =, B() =, B() = 4, B() = 9 (b) B(x) = (x)(x) = x (c) B (x) = x (d) A(x) = 5. (.5)() + ()() + (.)() =. x x t dt dt

41 4.4 res, itegrls d tiderivtives 4.4 Ares, Itegrls d Atiderivtives This sectio explores properties of fuctios defied s res d exmies some coectios mog res, itegrls d tiderivtives. I order to focus o these coectios d their geometric meig, ll of the fuctios i this sectio re oegtive, but i the ext sectio we will geerlize (d prove) the results for ll cotiuous fuctios. This sectio lso itroduces exmples showig how you c use the reltioships betwee res, itegrls d tiderivtives i vrious pplictios. Whe f is cotiuous, oegtive fuctio, the re fuctio A(x) = x f (t) dt represets the re of the regio bouded by the grph of f, the t-xis, d verticl lies t t = d t = x (see mrgi figure), d the derivtive of A(x) represets the rte of chge (growth) of A(x) s the verticl lie t = x moves rightwrd. Exmples d of Sectio 4. showed tht for certi fuctios f, A (x) = f (x) so tht A(x) ws tiderivtive of f (x). The ext theorem sys the result is true for every cotiuous, oegtive fuctio f. The Are Fuctio Is Atiderivtive If the so f is cotiuous, oegtive fuctio x d A(x) = f (t) dt for x ( d x ) f (t) dt = A (x) = f (x) dx A(x) is tiderivtive of f (x). This result reltig itegrls d tiderivtives is specil cse (for oegtive fuctios f ) of the first prt of the Fudmetl Theorem of Clculus (FTC ), which we will prove i Sectio 4.5. This result is importt for two resos: It sys tht lrge collectio of fuctios hve tiderivtives. It leds to esy wy to exctly evlute defiite itegrls. Exmple. Defie A(x) = x f (t) dt for the fuctio f (t) show i the mrgi. Estimte the vlues of A(x) d A (x) for x =,, 4 d 5 d use these vlues to sketch grph of y = A(x). Solutio. Dividig the regio ito squres d trigles, it is esy to see tht A() =, A() = 4.5, A(4) = 7 d A(5) = 8.5. Becuse A (x) = f (x), we kow tht A () = f () =, A () = f () =, A (4) = f (4) = d A (5) = f (5) =. A grph of y = A(x) ppers i the mrgi t the top of the ext pge.

42 4 the itegrl It is importt to recogize tht f is ot differetible t x = or x = but tht the vlues of A chge smoothly er x = d x =, d the fuctio A is differetible t those poits d t every other poit betwee x = d x = 5. Also ote tht f (4) = ( f is clerly decresig er x = 4) but tht A (4) = f (4) = is positive (the re A is growig eve though f is gettig smller). Prctice. Let B(x) be the re bouded by the horizotl xis, verticl lies t t = d t = x, d the grph of f (t) show i the mrgi. Estimte the vlues of B(x) d B (x) for x =,,, 4 d 5. Exmple. Let G(x) = d ( x ) si(t) dt. Evlute G(x) for x = π 4, π d π 4. dx Solutio. The middle mrgi figure shows A(x) = si(t) dt grphiclly. By the theorem, A (x) = si(x), so: ( π ) ( G = A π ) ( π ) = si ( π ) ( G = A π ) ( π ) = si = ( ) ( ) ( ) π G = A π π = si x =.77 =.77 The peultimte mrgi figure shows grph of y = A(x) d the bottom mrgi figure shows the grph of y = A (x) = G(x). Usig Atiderivtives to Evlute b f (x) dx Now we combie the ides of res d tiderivtives to devise techique for evlutig defiite itegrls tht is exct d ofte esy. If A(x) = A(b) = b x f (t) dt, the we kow tht A() = f (t) dt =, f (t) dt d tht A(x) is tiderivtive of f, so A (x) = f (x). We lso kow tht if F(x) is y tiderivtive of f, the F(x) d A(x) hve the sme derivtive so F(x) d A(x) re prllel fuctios d differ by costt: F(x) = A(x) + C for ll x d some costt C. As cosequece: F(b) F() = [A(b) + C] [A() + C] = A(b) A() = b f (t) dt f (t) dt = b f (t) dt This result sys tht, to evlute defiite itegrl A(b) = b f (t) dt, we c fid y tiderivtive F of f d simply evlute F(b) F().

43 4.4 res, itegrls d tiderivtives 5 This result is specil cse of the secod prt of the Fudmetl Theorem of Clculus (FTC, stted d proved i Sectio 4.5), which you will use hudreds of times over the ext severl chpters. Atiderivtives d Defiite Itegrls If the f is cotiuous, oegtive fuctio d F is y tiderivtive of f (so tht F (x) = f (x)) o [, b] b f (t) dt = F(b) F() The problem of fidig the exct vlue of defiite itegrl hs bee reduced to fidig some (y) tiderivtive F of the itegrd d the evlutig F(b) F(). Eve fidig oe tiderivtive c be difficult, so for ow we will restrict our ttetio to fuctios tht hve esy tiderivtives. Lter we will explore some methods for fidig tiderivtives of more difficult fuctios. Becuse evlutio of the form F(b) F() will occur quite ofte, we represet it symboliclly s F(x) b ] b [F(x) or Exmple. Evlute x dx i two wys: () by sketchig grph of y = x d fidig the re represeted by the defiite itegrl. (b) by fidig tiderivtive F(x) of f (x) = x d evlutig F() F(). Solutio. () A grph of y = x ppers i the mrgi; the re of the trpezoidl regio i questio hs re 4. (b) Oe tiderivtive of x is F(x) = ( x ) x (you should check for yourself tht D = x), so: F(x) = F() F() = () () = 9 = 4 which grees with the re from prt (). If someoe chose other tiderivtive of x, sy F(x) = x + 7 ( x ) (you should check for yourself tht D + 7 = x), the: [ ] [ ] F(x) = F() F() = () + 7 () + 7 = 5 = 4 No mtter which tiderivtive F we choose, F() F() = 4.. Prctice. Evlute previous Exmple. (x ) dx i the two wys specified i the

44 6 the itegrl This tiderivtive method provides extremely powerful wy to evlute some defiite itegrls, d we will use it ofte. Exmple 4. Fid the re of the regio i the first qudrt bouded by the grph of y = cos(x), the horizotl xis, d the lie x =. Solutio. The re we wt (see mrgi) is π cos(x) dx so we eed tiderivtive of f (x) = cos(x). F(x) = si(x) is oe such tiderivtive (you should check tht D (si(x)) = cos(x)), so π cos(x) dx = si(x) is the re of the regio i questio. π ( π ) = si si() = = Prctice. Fid the re of the regio bouded by the grph of y = x, the horizotl xis d the verticl lies x = d x =. Itegrls, Atiderivtives d Applictios The tiderivtive method for evlutig defiite itegrls c lso be used whe we eed to fid more geerl re, so it is ofte useful for solvig pplied problems. Exmple 5. A robot hs bee progrmmed so tht whe it strts to move, its velocity fter t secods will be t feet per secod. () How fr will the robot trvel durig its first four secods of movemet? (b) How fr will the robot trvel durig its ext four secods of movemet? (c) How log will it tke for the robot to move 79 feet from its strtig plce? Solutio. () The distce durig the first four secods will be the re uder the grph of the velocity fuctio (see mrgi figure) from t = to t = 4, re we c compute with the defiite itegrl 4 t dt. Oe tiderivtive of t is t so: 4 t dt = [t ] 4 = 4 = 64 d we c coclude tht the robot will be 64 feet wy from its strtig positio fter four secods. (b) Proceedig similrly: 8 4 t dt = [t ] 8 = = 5 64 = 448 feet

45 4.4 res, itegrls d tiderivtives 7 (c) This questio is differet from the first two. Here we kow the lower itegrtio edpoit, t =, d the totl distce, 79 feet, d eed to fid the upper itegrtio edpoit (the time whe the robot is 79 feet wy from its strtig positio). Cllig this upper edpoit T, we kow tht: T 79 = t dt = [t ] T = T = T so T = 79 = 9. The robot is 79 feet wy fter 9 secods. Prctice 4. Refer to the robot from the previous Exmple. () How fr will the robot trvel betwee t = d t = 5 secods? (b) How log will it tke for the robot to move 4 feet from its strtig plce? Exmple 6. Suppose tht t miutes fter plcig, bcteri o Petri plte the rte of growth of the bcteri popultio is 6t bcteri per miute. () How my ew bcteri re dded to the popultio durig the first seve miutes? (b) Wht is the totl popultio fter seve miutes? (c) Whe will the totl popultio rech, bcteri? Solutio. () The umber of ew bcteri is represeted by the re uder the rte-of-growth grph (see mrgi) d oe tiderivtive of 6t is t (check tht D ( t ) = 6t) so: ew bcteri = 7 6t dt = [t ] 7 = (7) () = 47 (b) [old popultio] + [ew bcteri] = + 47 = 47 bcteri. (c) Whe the totl popultio reches, bcteri, the there re = ew bcteri, hece we eed to fid the time T required for tht my ew bcteri to grow: = T 6t dt = [t ] T = (T) () = T so T = 4 T =. After miutes, the totl bcteri popultio will be + =. Prctice 5. Refer to the bcteri popultio from the previous Exmple. () How my ew bcteri will be dded to the popultio betwee t = 4 d t = 8 miutes? (b) Whe will the totl popultio rech,875 bcteri?

46 8 the itegrl 4.4 Problems I Problems 8, A(x) = x () Grph y = A(x) for x 5. f (t) dt with f (t) give. (b) Estimte the vlues of A(), A(), A() d A(4). (c) Estimte A (), A (), A () d A (4). I 9, use the give velocity of cr (i feet per secod) fter t secods to fid: () how fr the cr trvels durig the first secods. (b) how my secods it tkes the cr to trvel hlf the distce i prt (). 9. v(t) = t. v(t) = t. v(t) = 4t Problems give the velocity of cr (i feet per secod) fter t secods. () How my secods does it tke for the cr to come to stop (velocity = )? (b) How fr does the cr trvel before comig to stop? (c) How my secods does it tke the cr to trvel hlf the distce i prt (b)?. v(t) = t. v(t) = 75 t 5. f (t) = 6. f (t) = + t 7. f (t) = 6 t 8. f (t) = + t I Problems 9 8, use the Atiderivtives d Defiite Itegrls Theorem to evlute ech itegrl. 9. (). (). (). (). () 4. () 5. () 6. () 7. () 8. () 5 x dx 4x dx 6x dx x dx 4x dx 4x dx x dx (b) (b) (b) (b) (b) (b) (b) 5 dx (b) x dx (b) x dx 4x dx 6x dx x dx 4x dx 4x dx x dx (c) (c) (c) (c) (c) (c) (c) 5 dx (c) x dx [ x ] dx (b) (c) 5 x dx 4x dx 6x dx x dx 4x dx 4x dx x dx 5 dx x dx [ x ] dx 4. Fid the exct re uder hlf of oe rch of the sie curve: π si(x) dx. 5. A rtist you kow wts to mke figure cosistig of the regio betwee the curve y = x d the x-xis for x. () Where should the rtist divide the regio with verticl lie (see figure below) so tht ech piece hs the sme re? (b) Where should she divide the regio with verticl lies to get three pieces with equl res?

47 4.4 res, itegrls d tiderivtives Prctice Aswers. B() =.5, B() = 5, B() = 8.5, B(4) =, B(5) = 4.5 x B(x) = f (t) dt B (x) = d ( x ) f (t) dt = f (x) dx (by the Are Fuctio Is Atiderivtive Theorem), hece: B () = f () =, B () = f () =, B () = 4, B (4) = d B (5) =.. () (x ) dx gives the re of the trigulr regio betwee the grph of y = x d the x-xis for x : re = (bse) (height) = ()() = (b) F(x) = x x is tiderivtive of f (x) = x so:. Are = (x ) dx = F() F() = 4. () distce = x dx = x = = 8 = 7 5 [ ] [ ] = t dt = t 5 = 5 = 4 feet. (b) We kow the strtig poit is x = d the totl distce ( re uder the velocity curve) is 4 feet. We eed to fid the time T (see mrgi figure) so tht 4 feet = 4 = T hece T = 4 = 7 secods. 5. () ew bcteri = 8 4 T t dt: T t dt = t = T = T 8 6t dt = t = 64 6 = 44 bcteri. (b) We kow the totl ew popultio ( re uder the rte-ofchge grph) is 875 = 875 so: 875 = T 4 T 6t dt = t = T = T T = 65 hece T = 65 = 5 miutes.

48 4 the itegrl 4.5 The Fudmetl Theorem of Clculus This sectio cotis the most importt d most frequetly used theorem of clculus, THE Fudmetl Theorem of Clculus. Discovered idepedetly by Newto d Leibiz durig the lte 6s, it estblishes coectio betwee derivtives d itegrls, provides wy to esily clculte my defiite itegrls, d ws key step i the developmet of moder mthemtics to support the rise of sciece d techology. Clculus is oe of the most sigifict itellectul structures i the history of hum thought, d the Fudmetl Theorem of Clculus is the most importt brick i tht beutiful structure. Prior sectios hve emphsized the meig of the defiite itegrl, defied it, d beg to explore some of its pplictios d properties. I this sectio, the emphsis shifts to the Fudmetl Theorem of Clculus. You will use this theorem ofte i lter sectios. The Fudmetl Theorem hs two prts. They resemble results i the previous sectio but pply to more geerl situtios. The first prt (FTC ) sys tht every cotiuous fuctio hs tiderivtive d shows how to differetite fuctio defied s itegrl. The secod prt (FTC ) shows how to evlute the defiite itegrl of y fuctio if we kow formul for tiderivtive of tht fuctio. Prt : Atiderivtives Every cotiuous fuctio hs tiderivtive, eve fuctios with corers, such s the bsolute vlue fuctio f (x) = x, tht fil to be differetible t oe or more poits. The Fudmetl Theorem of Clculus Prt (FTC ) x If f is cotiuous d A(x) = f (t) dt the A (x) = d [ x ] f (t) dt = f (x) dx so A(x) is tiderivtive of f (x). Proof. For cotiuous fuctio f, let A(x) = defiitio of derivtive, A A(x + h) A(x) (x) = lim = lim h h h x+h x f (t) dt. f (t) dt x h f (t) dt By the Usig oe of the itegrl properties from Sectio 4., we kow tht: x+h x+h f (t) dt = f (t) dt x x f (t) dt + f (t) dt = x+h x x+h x f (t) dt f (t) dt

49 4.5 the fudmetl theorem of clculus 4 Assume for the momet tht h >. Becuse f is cotiuous o [x, x + h] we kow tht f ttis mximum d miimum o tht itervl, so there re vlues m h d M h with x < m h < x + h d x < M h < x + h so tht f (m h ) f (t) f (M h ) whe x t x + h. Hece: x+h x f (m h ) h f (m h ) f (m h ) dt x+h x x+h x x+h x f (t) dt x+h f (t) dt f (M h ) h f (t) dt f (M h h ) x f (M h ) dt Becuse x < m h < x + h, we kow lim m h + h = x; cosequetly becuse f (t) is cotiuous we lso kow tht lim f (m h) = f (x). Likewise, lim h + h + f (M h) = f (x), so the Squeezig Theorem tells us tht: lim h + x+h x f (t) dt = f (x) h Repetig this rgumet for h < is reltively strightforwrd. Exmple. Defie A(x) = x f (t) dt for f i the mrgi figure. Evlute A(x) d A (x) for x =,, d 4. Solutio. A() = f (t) dt =, A() = f (t) dt =, A() = f (t) dt = 4 d A(4) = f (t) dt =. Becuse f is cotiuous, FTC tells us tht A (x) = f (x), so A () = f () =, A () = f () =, A () = f () = d A (4) = f (4) =. Prctice. Defie A(x) = x g(t) dt for g i the mrgi figure. Evlute A(x) d A (x) for x =,, d 4. Exmple. Defie A(x) = x f (t) dt for f i the mrgi figure. For which vlue of x is A(x) mximum? For which x is the rte of chge of A mximum? Solutio. Becuse A is differetible, the oly criticl poits re where A (x) = or t edpoits. A (x) = f (x) = t x =, d A hs mximum t x =. Notice tht the vlues of A(x) icrese s x goes from to d the the vlues of A decrese. The rte of chge of A(x) is A (x) = f (x), d f (x) ppers to hve mximum t x =, so the rte of chge of A(x) is mximum whe x =. Ner x =, slight icrese i the vlue of x yields the mximum icrese i the vlue of A(x).

50 4 the itegrl Prt : Evlutig Defiite Itegrls If we kow formul for tiderivtive of fuctio, the we c compute y defiite itegrl of tht fuctio. The Fudmetl Theorem of Clculus Prt (FTC ) If the f (x) is cotiuous d F(x) is y tiderivtive of f (so tht F (x) = f (x)) b b f (x) dx = F(x) = F(b) F(). Proof. Defie A(x) = x f (t) dt. If F is tiderivtive of f, the F (x) = f (x) d by FTC we kow tht A (x) = f (x) so F (x) = A (x), hece F(x) d A(x) differ by costt: A(x) F(x) = C for ll x d some costt C. At x =, we hve C = A() F() = F() = F() so C = F() d the equtio A(x) F(x) = C becomes A(x) F(x) = F(). The A(x) = F(x) F() for ll x, so settig x = b yields A(b) = F(b) F(), hece formul we wted. b f (x) dx = F(b) F(), the We c evlute the defiite itegrl of cotiuous fuctio f by fidig tiderivtive of f (y tiderivtive of f will work) d the doig some rithmetic with this tiderivtive. FTC does ot tell us how to fid tiderivtive of f, d it does ot tell us how to fid the defiite itegrl of discotiuous fuctio. It is possible to evlute defiite itegrls of some discotiuous fuctios (s we sw i Sectio 4.) but ot by usig FTC directly. ( ) Exmple. Evlute x dx. Solutio. F(x) = x ( x is ) tiderivtive of f (x) = x (you should check tht D x x = x ), so: ( ) [ ] [ ] [ ] x dx = x x = = If your fried hd picked differet tiderivtive of x, sy G(x) = x x + 4, the her clcultios would be slightly differet : ( ) [ ] x dx = x x + 4 [ ] = + 4 but the result would be the sme. [ ] + 4 = =

51 4.5 the fudmetl theorem of clculus 4 Prctice. Evlute Exmple 4. Evlute ( ) x dx..7.5 x dx (where x = INT(x) is the lrgest iteger less th or equl to x, s i the mrgi figure). Solutio. f (x) = x is ot cotiuous t x = i the itervl [.5,.7], so we cot employ the Fudmetl Theorem of Clculus directly. We c, however, use our uderstdig of the geometric meig of defiite itegrl to compute:.7.5 x dx = (re below y = x for.5 x ) + (re below y = x for x.7) = (first bse) (first height) + (secod bse) (secod height) = (.5)() + (.7)() =.9 We could lso split the itegrl ito two pieces: x dx = x dx + x dx [ ]. [ ].7 = dx + dx = x + x = [..5] + [(.7) (.)] = =.9 usig the fct tht x = for.5 x <. d the fct tht x = for. x.7. (We lso eed to redefie the first itegrd to equl t its right edpoit d the secod itegrd to equl t its right edpoit so tht ech itegrd is cotiuous o closed itervl). Problem 5 i Sectio 4. idictes tht this redefiitio is perfectly legl. Prctice. Evlute.4. x dx. Clculus is the study of derivtives d itegrls, their meigs d their pplictios. The Fudmetl Theorem of Clculus demostrtes how differetitio d itegrtio re closely relted processes: itegrtio is relly ti-differetitio, the iverse of differetitio. Applictios: The Future Clculus is importt for my resos, but studets re usully required to study clculus becuse they will eed to pply clculus cocepts i vriety of fields. Most pplied problems i itegrl clculus require the followig steps to get from rel-life problem to umericl swer: pplied problem Riem sum defiite itegrl umber

52 44 the itegrl Step is bsolutely vitl. If we c ot trslte the ides of pplied problem ito re or Riem sum or defiite itegrl, the we c ot use itegrl clculus to solve the problem. For few specil types of pplied problems, we will be ble to move directly from the problem to itegrl, but usully it will be esier to first brek the problem ito smller pieces d to build Riem sum. Sectio 4.7 d ll of Chpter 5 focus o trsltig differet types of pplied problems ito Riem sums d defiite itegrls. Computers d clcultors re seldom of y help with Step. Step is usully esy. If we hve Riem sum k= f (c k ) x k o itervl [, b], the the limit of the sum (s ) is simply the defiite itegrl b f (x) dx. Step c be hdled i severl wys. If the fuctio f is reltively simple, we my be ble to fid tiderivtive for f (usig techiques from Sectio 4.6 d Chpter 8) d the pply FTC to get umericl swer. If the fuctio f is more complicted, the itegrl tbles or computers (Sectio 4.8) my help us fid tiderivtive for f, i which cse we c pply FTC to get umericl swer. If we cot fid tiderivtive for f, we c compute pproximte umericl swers for the defiite itegrl usig vrious pproximtio methods (Sectios 4.9 d 8.7); we typiclly employ computers to crry out the hevy-duty rithmetic. Usully y difficulties i solvig pplied problem rise i the first d third steps. There re techiques d detils to mster d uderstd, but it is lso importt to keep i mid where these techiques d detils fit ito the bigger picture. The ext Exmple illustrtes these steps for the problem of fidig volume of solid. We will explore techiques for fidig volumes of solids i greter detil i Chpter 5. Exmple 5. Fid the volume of the solid show i the mrgi for x. (Ech slice perpediculr to the xy-ple is squre.) Solutio. Step : Goig from the figure to Riem sum. If we brek the solid ito slices with cuts perpediculr to the x-xis (d the xy-ple) usig prtitio P with cuts t x, x, x,..., x (like slicig block of cheese or lof of bred), the the volume of the origil solid is equl to the sum of the volumes of the slices. The volume of the k-th slice is pproximtely equl to the volume of thi, rectgulr box: (height) (bse) (thickess) (c k + ) (c k + ) x k

53 4.5 the fudmetl theorem of clculus 45 where c k is y chose vlue betwee x k d x k. Therefore: totl volume = (volume of the k-th slice) = (c k + ) x k k= k= which is Riem sum. Step : Goig from the Riem sum to defiite itegrl. We c improve the Riem sum pproximtio of the totl volume from Step by tkig thier slices (mkig ll of the x k smller d smller) so tht the mesh of the prtitio P pproches : lim P (c k + ) x k = k= (x + ) dx = [ ] x + x + dx Step : Goig from the defiite itegrl to umericl swer. We c ow use FTC to evlute the itegrl: F(x) = x + x + x is tiderivtive of x + x + (check this by differetitig F(x)), so: [ ] [ ] x + x + dx = x + x + x [ ] [ ] = = 6 The volume of the solid shpe is exctly 6 cubic iches. Prctice 4. Fid the volume of the solid shpe i the mrgi figure for x. (Ech slice perpediculr to the xy-ple is squre.) Leibiz s Rule For Differetitig Itegrls If the edpoit of itegrl is fuctio of x rther th simply x, the we eed to use the Chi Rule together with FTC to clculte the derivtive of the itegrl. For exmple: A (x) = f (x) d dx [ A ( x )] = A (x) x = f ( x ) x We c geerlize this result by pplyig the Chi Rule to the derivtive of the itegrl: [ d g(x) ] f (t) dt = d dx dx [A (g(x))] = f (g(x)) g (x) d combie this with some itegrl properties to further exted FTC. Leibiz s Rule If f is cotiuous fuctio, A(x) = the x f (t) dt d g (x) d g (x) re both differetible fuctios [ d g ] (x) f (t) dt = f (g (x)) g dx (x) f (g (x)) g (x) g (x)

54 46 the itegrl Proof. Assume for simplicity tht f, g d g re cotiuous o (, ) d let c be y umber. The: g (x) g (x) f (t) dt = = g (x) c g (x) c c f (t) dt + f (t) dt g (x) g (x) c f (t) dt f (t) dt Now pply the precedig result. Exmple 6. If is y costt, compute the derivtives d [ 5x [ dx ] d x cos(u) du d d [ si w ] z dz. dx dw πw Solutio. Applyig Leibiz s Rule: [ d 5x ] t dt = (5x) 5 = 5x dx [ ] d x cos(u) du = cos(x ) x = x cos(x ) dx d dw [ si(w) πw ] z dz = (si(w)) cos(w) (πw) π ] t dt, The lst qutity simplifies to si (w) cos(w) π 4 w. [ Prctice 5. Compute d ] x si(t)dt. dx 4.5 Problems I Problems, () Use FTC to fid formul for A(x), differetite A(x) to obti formul for A (x), d evlute A (x) t x =, d. (b) Use FTC to evlute A (x) t x =, d.. A(x) = x t dt. A(x) = x ( + t) dt I Problems 8, compute A (), A () d A ().. A(x) = 5. A(x) = 7. A(x) = x x x t dt 4. A(x) = t dt 6. A(x) = si(t) dt 8. A(x) = x x x t dt ( t ) dt t dt I 9, A(x) = x f (t) dt, with f (t) give grphiclly. Evlute A (), A () d A ()

55 4.5 the fudmetl theorem of clculus 47 I, verify tht F(x) is tiderivtive of the itegrd d use FTC to evlute the itegrl π π π 4 e x dx, F(x) = x + 5 x dx, F(x) = x + x dx, F(x) = x [ ] x + 4x dx, F(x) = x + x x x dx, F(x) = l(x) dx, F(x) = l(x) + 4 x x dx, F(x) = l(x) dx, F(x) = l(x) + x cos(x) dx, si(x) dx, F(x) = si(x) F(x) = cos(x) x dx, F(x) = x x dx, F(x) = x x dx, F(x) = x x dx, F(x) = x x dx, F(x) = x x dx, e x dx, F(x) = x F(x) = e x x + x dx, F(x) = l( + x ) sec (x) dx, l(x) dx, F(x) = t(x) F(x) = x l(x) x x + x dx, F(x) = ( + x ) For 4 48, fid tiderivtive of the itegrd d use FTC to evlute the defiite itegrl π x dx 5. [ ] x + 4x dx 7. π 4 si(x) dx e 5 x dx 4. x dx x dx x dx x dx x dx 4. e x dx π x + x dx e x dx 47. (x ) dx π 6 sec (x) dx si(x) l(x) dx I 49 54, fid the re of the shded regio

56 48 the itegrl 55. Give tht A (x) = t(x), fid D (A(x)), D ( A(x ) ) d D (A(si(x))). 56. Give tht B (x) = sec(x), fid D (B(x)), D ( B(x ) ) d D (B(si(x))) I 57 68, pply Leibiz s Rule. d dx d dx [ 5x ] + t dt 58. [ si(x) ] + t dt 6. [ ] d x + t dt dx [ +x d dx ( ) ] t + 5 dt d dx d dx d dx [ x [ 9 x [ π 7x ( ) ] t + dt ( ) ] t + dt ] cos(t) dt [ d π ] cos(t) dt dx [ π d dx ] t(7t) dt [ d π dx x [ x d dx d dx d dy x [ l(x) ] cos(t) dt t(t) dt ] ] 5t cos(t) dt [ ] y t(θ) dθ 4.5 Prctice Aswers. A() =, A() =.5, A() =, A(4) =.5; A (x) = f gx) so A () = g() =, A () = g() =, A () =, A (4) =.. F(x) = x x is tiderivtive of f (x) = x so: [ ] [ ] [ ] [ ] x dx = x x = = 4 F(x) = x x + 7 is other tiderivtive of f (x) = x so: [ ] [ ] [ ] [ ] x dx = x x + 7 = = 4 No mtter which tiderivtive of f (x) = x you use, the vlue [ ] of the defiite itegrl x dx is 4.. Becuse f (x) = x is ot cotiuous o [.,.4] we cot use the Fudmetl Theorem of Clculus. Isted, we c thik of the defiite itegrl s re (see mrgi figure) d compute:.4. x dx =.9 4. First brek the solid ito slices d pproximte the volume of the k-th slice by ( c k ) x k where c k is y poit i the k-th subitervl. Next dd up these pproximte volumes to get Riem Sum: ( c k ) x k k=

57 4.5 the fudmetl theorem of clculus d the tke the limit of these Riem sums s the mesh of the prtitios pproches (d, where is the umber of subitervls i the prtitio): lim P [ k= ( c k ) x k ] = = = = ( x) dx ( 9 6x + x ) dx [ 9x x + ] x [ ] [ + ] = 6 [ ] d x si(t) dt = si(x d [ ) x ] = x si(x ) dx dx

58 5 the itegrl 4.6 Fidig Atiderivtives I order to use the secod prt of the Fudmetl Theorem of Clculus, we eed tiderivtive of the itegrd, but sometimes it is ot esy to fid oe. This sectio collects some of the iformtio we lredy kow bout geerl properties of tiderivtives d bout tiderivtives of prticulr fuctios. It shows how to use this iformtio to fid tiderivtives of more complicted fuctios d itroduces chge of vrible techique to mke tht job esier. Idefiite Itegrls d Atiderivtives Atiderivtives rise so ofte tht there is specil ottio to idicte the tiderivtive of fuctio: If you ve bee woderig why we clled b f (t) dt defiite itegrl, ow you kow. A defiite itegrl hs specific upper d lower limits, while idefiite itegrl does ot. f (x) dx, red s the idefiite itegrl of f or s the tiderivtives of f, represets the collectio (or fmily) of ll fuctios whose derivtives re f. If F is tiderivtive of f, the y member of the fmily f (x) dx hs the form F(x) + C for some costt C. We write f (x) dx = F(x) + C, where C represets rbitrry costt. There re o smll fmilies i the world of tiderivtives: if f hs oe tiderivtive F, the f hs ifiite umber of tiderivtives d ech hs the form F(x) + C, which mes there re my wys to write prticulr idefiite itegrl d some of them my look very differet. You c check tht F(x) = si (x), G(x) = cos (x) d H(x) = si (x) + cos (x) ll hve the sme derivtive, f (x) = si(x) cos(x), so the idefiite itegrl of si(x) cos(x), si(x) cos(x) dx, c be writte i severl wys: si (x) + C or cos (x) + K or si (x) + cos (x) + C. Prctice. Verify tht t(x) sec (x) dx = t (x) + C d tht t(x) sec (x) dx = sec (x) + K. Properties of Atiderivtives (Idefiite Itegrls) These sum, differece d costtmultiple properties follow directly from correspodig properties for derivtives. If f d g re itegrble fuctios, the [ f (x) ± g(x)] dx = f (x) dx ± g(x) dx k f (x) dx = k f (x) dx

59 4.6 fidig tiderivtives 5 Although we kow geerl rules for derivtives of products d quotiets, ufortutely there re o esy geerl ptters for tiderivtives of products d quotiets we will oly be ble to dd oe more geerl property to this list (i Sectio 8.). We lredy kow tiderivtives for severl importt fuctios. Costt Fuctios: k dx = kx + C Powers of x: x p dx = xp+ p + + C if p =, dx = l x + C x Expoetil Fuctios: e x dx = e x + C Trig Fuctios: cos(x) dx = si(x) + C, si(x) dx = cos(x) + C sec (x) dx = t(x) + C, csc (x) dx = cot(x) + C sec(x) t(x) dx = sec(x) + C, csc(x) cot(x) dx = csc(x) + C Our list of tiderivtives of prticulr fuctios will grow i comig chpters d will evetully iclude tiderivtives of dditiol trigoometric fuctios, the iverse trigoometric fuctios, logrithms, rtiol fuctios d more. (See Appedix I.) All of these tiderivtives c be verified by differetitig. For x dx you my be woderig bout the presece of the bsolute vlue sigs i the tiderivtive. If x >, you c check tht: D (l( x )) = D (l(x)) = x If x <, the you c check tht: D (l( x )) = D (l( x)) = x = x Whe computig defiite itegrl of the form b x dx, either d b will both be positive or both be egtive, becuse the itegrd is ot defied t x =, so x = cot be icluded i the itervl of itegrtio. Atiderivtives of More Complicted Fuctios Atiderivtives re very sesitive to smll chges i the itegrd, so we should be very creful. Exmple. We kow D (si(x)) = cos(x), so cos(x) dx = si(x) + C. Fid: () cos(x + ) dx (b) cos(5x 7) dx (c) cos(x ) dx Fortutely, tiderivtive c lwys be checked by differetitig, so eve though we my ot fid the correct tiderivtive, we should be ble to determie whether or ot tiderivtive cdidte is ctully tiderivtive. Solutio. () Becuse si(x) is tiderivtive of cos(x), it is resoble to hope tht si(x + ) will be tiderivtive of cos(x + ). Ufortutely, we see tht D (si(x + )) = cos(x + ), exctly twice the result we wt. Let s try gi by modifyig our guess to be hlf the origil guess: ( ) D si(x + ) = cos(x + ) = cos(x + ) which is wht we wt, so cos(x + ) dx = si(x + ) + C. (b) D (si(5x ( 7)) = cos(5x ) 7) 5, so dividig the origil guess by 5 we get D 5 si(5x 7) = 5 cos(5x 7) 5 = cos(5x 7) d coclude tht cos(5x 7) dx = si(5x 7) + C. 5 (c) D ( si(x ) ) = cos(x ) x. It ws esy eough i prts () d (b) to modify our guesses to elimite the costts d 5, but here the x is much hrder to elimite:

60 5 the itegrl ( ( D x si x )) ( si(x = ) ) D x = x D ( si(x ) ) si(x ) D(x) (x) ) = (x) cos(x ) si(x ) (x) = cos(x ) si(x ) x = cos(x ) Our guess did ot check out we re stuck. Advced mthemticl techiques beyod the scope of this text c show tht cos(x ) does ot hve elemetry tiderivtive composed of polyomils, roots, trigoometric fuctios, expoetil fuctios or their iverses. The vlue of defiite itegrl of cos(x ) could still be pproximted s ccurtely s eeded by usig Riem sums or oe of the umericl techiques i Sectios 4.9 d 8.7, but o mtter how hrd we try, we cot fid cocise formul for tiderivtive of cos(x ) i order to use the Fudmetl Theorem of Clculus. Eve simple-lookig itegrd c be very difficult. At this poit, there is o quick wy to tell the differece betwee esy idefiite itegrl d difficult or impossible oe. Gettig the Costts Right The previous exmple illustrted oe techique for fidig tiderivtives: guess the form of the swer, differetite your guess d the modify your origil guess so its derivtive is exctly wht you wt it to be. Exmple. Kowig tht sec (x) dx = t(x) + C d x + C, fid () sec (x + 7) dx (b) dx. 5x + x dx = Solutio. () If we guess swer of t(x + 7) d the differetite it, we get D (t(x + 7)) = sec (x + 7) D(x + 7) = sec (x + 7), which is three times wht we wt. If we divide our origil guess by d try gi, we hve: ( ) D t(x + 7) = D (t(x + 7)) = sec (x + 7) = sec (x + 7) so sec (x + 7) dx = t(x + 7) + C. (b) If we guess 5x + d the differetite it, we get: D ( (5x + ) ) = (5x + ) D(5x + ) = 5 (5x + )

61 4.6 fidig tiderivtives 5 which is five times wht we wt. Dividig our guess by 5 d differetitig, we hve: ( ) D 5 (5x + ) = 5 (5x + ) 5 = 5x + so 5x + dx = 5 5x + + C. Prctice. Fid sec (7x) dx d x + 8 dx. The guess d check method is very effective techique if you c mke good first guess, oe tht misses the desired result oly by costt multiple. I tht situtio, just divide the first guess by the uwted costt multiple. If the derivtive of your guess misses by somethig other th costt multiple, the more drstic modifictios re eeded. Sometimes the ext techique c help. Mkig Ptters More Obvious: Chgig the Vrible Successful itegrtio is mostly mtter of recogizig ptters. The chge of vrible techique c mke some uderlyig ptters of itegrl esier to recogize. Essetilly, the techique ivolves rewritig itegrl tht is origilly i terms of oe vrible, sy x, i terms of other vrible, sy u, with the hope tht it will be esier to fid tiderivtive of the ew itegrd. For exmple, we c rewrite cos(5x + ) dx by settig u = 5x +. The cos(5x + ) becomes cos(u) but we must lso covert the dx i the origil itegrl. We kow tht du = 5, so rewritig this lst dx expressio i differetil ottio, we get du = 5 dx; isoltig dx yields dx = 5 du so: cos(5x + ) dx = cos(u) 5 du = cos(u) du 5 This ew itegrl is esier: cos(u) du = 5 5 si(u) + C but our origil problem ws i terms of x d our swer is i terms of u, so we must resubstitute usig the reltioship u = 5x + : 5 si(u) + C = si(5x + ) + C 5 We c ow coclude tht: cos(5x + ) dx = As si(5x + ) + C 5 We first discussed differetil ottio i Sectio.8; lthough you my ot hve used them much i differetil clculus, you will ow use them extesively. lwys, you c check this result by differetitig.

62 54 the itegrl We c summrize the steps of this chge of vrible (or usubstitutio ) method s: Ofte u is set equl to some iterior prt of the origil itegrd fuctio. set ew vrible, sy u, equl to some fuctio of the origil vrible x clculte the differetil du i terms of x d dx rewrite the origil itegrl i terms of u d du itegrte the ew itegrl to get swer i terms of u resubstitute for u to get result i terms of the origil vrible x Exmple. Mke the suggested chge of vrible, rewrite ech itegrl i terms of u d du, d evlute the itegrl. () cos(x) e si(x) dx with u = si(x) (b) x dx with u = 5 + x 5 + x Solutio. () u = si(x) du = cos(x) dx d e si(x) = e u : cos(x)e si(x) dx = e u du = e u + C = e si(x) + C (b) u = 5 + x du = x dx, so: x 5 + x dx = u du = l u + C = l 5 + x + C Becuse 5 + x >, we c lso write the swer s l ( 5 + x ). I ech exmple, the chge of vrible did ot fid the tiderivtive, but it did mke the ptter of the itegrd more obvious, which i tur mde it esier to determie tiderivtive. Prctice. Mke the suggested chge of vrible, rewrite ech itegrl i terms of u d du d evlute the itegrl. () (7x + 5) dx with u = 7x + 5 (b) ( ) x si x dx with u = x The previous exmples hve supplied suggested substitutio, but i the future you will eed to decide wht u should equl. Ufortutely there re o rules tht gurtee your choice will led to esier itegrl sometimes you will eed to resort to tril d error util you fid prticulr u-substitutio tht works for your itegrd. There is, however, rule of thumb tht frequetly results i esier itegrls. Eve though the followig suggestio comes with o gurtees, it is ofte worth tryig.

63 4.6 fidig tiderivtives 55 A Rule of Thumb for Chgig the Vrible If prt of the itegrd cosists of compositio of fuctios, f (g(x)), try settig u = g(x), the ier fuctio. The key to becomig skilled t selectig good u d correctly mkig the substitutio is prctice. If prt of the itegrd is beig rised to power, try settig u equl to the prt beig rised to the power. For exmple, if the itegrd icludes ( + si(x)) 5, try u = + si(x). If prt of the itegrd ivolves trigoometric (or expoetil or logrithmic) fuctio of other fuctio, try settig u equl to the iside fuctio: if the itegrd icludes the fuctio si ( + x ), try u = + x. Exmple 4. Select u for ech itegrd d rewrite the ssocited itegrl i terms of u d du. () cos(x) + si(x) dx (b) 5e x + e x dx (c) e x si (e x ) dx Solutio. () If u = + si(x), du = cos(x) dx du = cos(x) dx so the itegrl becomes u du. (b) With u = + e x du = e x dx, 5 the itegrl becomes u du. (c) With u = ex du = e x dx, the itegrl becomes si(u) du. Chgig Vribles with Defiite Itegrls If we eed to chge vribles i defiite itegrl, we hve two choices: First work out the correspodig idefiite itegrl d the use tht tiderivtive d FTC to evlute the defiite itegrl. Chge vribles i the defiite itegrl, which requires chgig the limits of itegrtio from x limits to u limits. For the secod optio, if the origil itegrl hd edpoits x = d x = b, d we mke the substitutio u = g(x) du = g (x) dx, the the ew itegrl will hve edpoits u = g() d u = g(b): x=b x= Exmple 5. Evlute f (g(x)) g (x) dx = (x ) 4 dx. u=g(b) u=g() f (u) du Solutio. Usig the first optio with u = x du = dx du = dx, the correspodig idefiite itegrl becomes: (x ) 4 dx = u 4 du = 5 u5 + C = 5 (x )5 + C

64 56 the itegrl Both optios require roughly the sme mout of work d computtio. I prctice you should choose the optio tht seems esiest for you d poses the lest risk of error. We ow use this result to evlute the origil defiite itegrl: [ ] [ ] [ ] (x ) 4 dx = (x )5 = ( )5 = 5 5 = 5 = 5 For the secod optio, we mke the sme substitutio u = x du = dx while lso computig x = u = = d x = u = = : x= x= (x ) 4 dx = u= u= u4 du = We rrive t the sme swer either wy. 5 u5 = 5 5 ( )5 5 = 5 Prctice 4. If the origil itegrls i Exmple 4 hd edpoits () x = to x = π (b) x = to x = or (c) x = to x = l(), the the ew itegrls should hve wht edpoits? Specil Trsformtios for si (x) dx d cos (x) dx cos(x) = si (x) cos(x) = cos (x) si(x) = si(x) cos(x) The itegrls of si (x) d cos (x) rise ofte, d we c fid their tiderivtives with the help of some trigoometric idetities. Solvig the first idetity i the mrgi for si (x), we get: si (x) = cos(x) d solvig the secod idetity for cos (x), we get: cos (x) = + cos(x) Itegrtig the first of these ew idetities yields: [ si (x) dx = ] cos(x) dx = x 4 si(x) + C Usig the idetity si(x) = si(x) cos(x), we c lso write: si (x) dx = x si(x) cos(x) + C I prctice, it s esier to remember the ew trig idetities d use them to work out these tiderivtives, rther th memorizig the tiderivtives directly. Similrly, usig cos (x) = + cos(x) yields: cos (x) dx = x + 4 si(x) + C = x + si(x) cos(x) + C

65 4.6 fidig tiderivtives Problems For Problems 4, put f (x) = x d g(x) = x to verify the iequlity. ( ) ( ). f (x) g(x) dx = f (x) dx g(x) dx f (x) dx = g(x) f (x) dx g(x) dx ( f (x) g(x) dx = f (x) dx = g(x) 4 4 f (x) dx g(x) dx ) ( ) f (x) dx g(x) dx For 5 4, use the suggested u to fid du d rewrite the itegrl i terms of u d du. The fid tiderivtive i terms of u d, filly, rewrite your swer i terms of x. 5. cos(x) dx, u = x 6. si(7x) dx, u = 7x 7. e x si( + e x ) dx, u = + e x 8. e 5x dx, u = 5x 9. cos(x) sec (si(x)) dx, u = si(x) cos(x) si(x) dx, u = si(x) 5 dx, u = + x + x ( x 5 + x ) 7 dx, u = 5 + x ( x si + x ) dx, u = + x e x dx, u = + ex + ex For 5 6, use the chge-of-vrible techique to evlute the idefiite itegrl cos(4x) dx 6. x ( 5 + x 4) dx 8. x dx. + x e x dx ( x si x ) dx si(x) cos(x) dx l(x) x dx. ( + x) 7 dx 4. e x sec (e x ) t (e x ) dx x cos ( x ) dx I 7 4, evlute the itegrl. π cos(x) dx 8. e x si ( + e x ) dx. π x + x dx si(l(x)) dx x cos(4x) dx e 5x dx ( x + x ) 5 ( dx. x 4 x 5 ) dx 5 l() + x dx 4. 5 x x dx 6. + x dx si (5x) dx [ (x)] si dx 4. e x + e x dx + x dx + x dx cos (x) dx [ ] e x + si (x) dx 4. Fid the re uder oe rch of the grph of y = si (x). 44. Evlute π si (x) dx. I 45 5, expd the itegrd d the fid tiderivtive. ( ) ( 45. x + dx 46. x + 5) dx 47. (e x + ) ( dx 48. x + x ) dx 49. (x + )(x + 5) dx 5. (7 + si(x)) dx ( 5. e x e x + e x) dx 5. ( + si(x)) si(x) dx 5. x ( ) x + x dx

66 58 the itegrl I 54 64, divide, the fid tiderivtive. 54. x + dx 55. x x x + dx 56. x x + dx 57. x x + dx 58. x x + 5 x x + 8 dx 59. x x 6. x x + x + dx 6. x x dx 6. e x + e x x + 4 e x dx 6. dx x dx 64. x + dx x The defiite itegrls i 65 7 ivolve res ssocited with prts of circles; use your kowledge of circles d their res to evlute them. (Suggestio: Sketch grph of the itegrd fuctio.) x dx x dx 68. [ + ] x dx 7. 4 x dx 6 x dx [ ] x dx 4.6 Prctice Aswers. D ( t (x) + C ) = t (x) D (t(x)) = t(x) sec (x) D ( sec (x) + C ) = sec (x) D (sec(x)) = sec(x) sec(x) t(x). We kow D (t(x)) = sec (x), so it is resoble to try t(7x): D (t(7x)) = sec (7x) D(7x) = 7 sec (7x), result seve times the result we wt, so divide the origil guess by 7 d try gi: ( ) D 7 t(7x) = 7 sec (7x) 7 = sec (7x) so sec (7x) dx = t(7x) + C. 7 ( ) D (x + 8) = (x + 8) D(x + 8) = (x + 8) so multiply our origil guess by : ( ) D (x + 8) hece = (x + 8) D(x + 8) = x + 8 dx = x C. x + 8. () u = 7x + 5 du = 7 dx dx = 7 du so: (7x + 5) dx = u 7 du = 7 4 u4 + C = 8 (7x + 5)4 + C (b) u = x du = x dx so si(x ) x dx becomes: ( ) si(u) du = cos(u) + C = cos x + C 4. () u = + si(x) so x = u = + si( ) = d x = π u = + si(π) =. (This itegrl is ow esy; why?) (b) u = + e x so x = u = + e = d x = u = + e (c) u = e x so x = u = e = d x = l() u = e l() =

67 4.7 first pplictios of defiite itegrls First Applictios of Defiite Itegrls The developmet of clculus by Newto d Leibiz ws vitl step i the dvcemet of pure mthemtics, but Newto lso dvced the scieces d pplied mthemtics. Not oly did he discover theoreticl results, he immeditely used those results to swer importt questios bout grvity d motio. The success of these pplictios of mthemtics to the physicl scieces helped estblish wht we ow tke for grted: mthemtics c d should be used to swer questios bout the world. Newto pplied mthemtics to the outstdig problems of his dy, problems primrily i the field of physics. Durig the iterveig -plus yers, thousds upo thousds of people hve cotiued these theoreticl d pplied trditios, usig mthemtics to help develop our uderstdig of the physicl d biologicl scieces, s well s the behviorl scieces d ecoomics. Mthemtics is still used to swer ew questios i physics d egieerig, but it is lso importt for modelig ecologicl processes, for uderstdig the behvior of DNA, for determiig how the bri works, d eve for devisig ficil strtegies. The mthemtics you re lerig ow c help you become prt of this trditio, d you might eve use it to dd to our uderstdig of the world. It is importt to uderstd the specil pplictios of itegrtio we will study i cse you eed to use those prticulr pplictios. But it is lso importt tht you uderstd the process of buildig models with itegrls so you c pply tht process to other situtios i vriety of fields of study. Coceptully, covertig pplied problem to Riem sum is the most vluble step. Typiclly, it is lso the most chllegig. Are betwee Two Curves We hve lredy used itegrls to fid the re betwee the grph of fuctio d the horizotl xis. We c lso use itegrls to fid the re betwee the grphs of two fuctios. If f (x) g(x) for ll x i [, b], the we c pproximte the re betwee the grphs of f d g by prtitioig the itervl [, b] d formig Riem sum (see mrgi). The height of ech rectgle is f (c k ) g(c k ) so the re of the k-th rectgle is: (height) (bse) = [ f (c k ) g(c k )] x k d pproximtio of the totl re is give by which is Riem sum. [ f (c k ) g(c k )] x k k=

68 6 the itegrl The limit of this Riem sum, s the mesh of the prtitios pproches, is defiite itegrl: b [ f (x) g(x)] dx We will sometimes use rrow to idicte the limit of the Riem sum s the mesh of the prtitios pproches, writig: [ f (c k ) g(c k )] x k k= b [ f (x) g(x)] dx If the T(x) B(x) for x b the re of the regio bouded by the grphs of the top fuctio T(x), the bottom fuctio B(x), d the lies x = d x = b is give by: b [T(x) B(x)] dx Exmple. Fid the re bouded betwee the grphs of f (x) = x d g(x) = for x 4. Solutio. It is cler from the mrgi figure tht the re betwee f d g is.5 squre uits. Usig the itegrtio procedure bove, we eed to idetify top fuctio d bottom fuctio. For x, g(x) = x = f (x) so the re of the left-hd trigle is give by the itegrl: [ x] dx = [ x ] [ x = 9 9 ] [ ] = For the itervl x 4, g(x) = x = f (x) so the re of the right-hd trigle is give by the itegrl: 4 [x ] dx = [ ] 4 [ ] 9 x x = [8 ] 9 = Addig these two res, we get +.5 =.5. Grphig the regio i questio to determie which fuctio is o top d which is o bottom is ofte crucil to gettig the right swer to problem ivolvig the re betwee two curves. If we hd midlessly itegrted i the previous Exmple without cosultig grph: 4 [ x] dx = [ x ] 4 [ x = [ 8] ] = we would hve rrived t icorrect swer. Prctice. Use itegrls d the grphs of f (x) = + x d g(x) = x to determie the re betwee the grphs of f d g for x.

69 4.7 first pplictios of defiite itegrls 6 Exmple. Objects A d B strt from the sme loctio t the sme time d trvel log the sme pth with respective velocities v A (t) = t + d v B (t) = t 4t + meters per secod (see mrgi). How fr hed is A fter secods? After 5 secods? Solutio. From the grph, it ppers tht v A (t) v B (t), t lest for t, but for the secod questio we eed to kow whether this holds for t 5 s well. Settig v A (t) = v B (t) to see where the grphs itersect: t + = t 4t + t 5t = t = or t = 5 Checkig tht v A () = 4 > = v B () (or referrig to the grph), we c coclude tht v A (t) v B (t) o the itervl [, 5]. Becuse v A (t) v B (t), the re betwee the grphs of v A d v B over itervl [, x] represets the distce betwee the objects fter x secods. After three secods, the distce prt is: [v A (t) v B (t)] dt = = [ ] (t + ) (t 4t + ) dt = [ 5 t t ] = [ ] 45 9 [ ] = 7 or.5 meters. After five secods, the distce prt is [ 5t t ] dt 5 [v A (t) v B (t)] dt = or pproximtely.8 meters. [ 5 t ] 5 t = 5 6 If f (x) g(x) o itervl [, b] (s illustrted i the mrgi figure), we could hve used simpler geometric rgumet tht the re betwee the grphs of f d g is just the re below the grph of f mius the re below the grph of g: b b b f (x) dx g(x) dx = [ f (x) g(x)] dx which grees with our previous result. We took differet pproch t the begiig of this sectio, however, becuse it provides ice (yet simple) exmple of trsltig geometric or physicl problem ito Riem sum d the ito defiite itegrl. Exmple. Fid the re of the shded regio i the mrgi figure. Solutio. These re the sme two fuctios from our previous Exmple; i our previous solutio we observed tht t + t 4t + for t 5, d it is strightforwrd to check tht t + t 4t + for t 5 (d, i prticulr, for 5 t 7).

70 6 the itegrl We therefore eed to split our problem ito two pieces d subtrct the bottom fuctio from the top fuctio o ech itervl. The re of the left regio is: 5 [ ] [ 5 (t + ) (t 4t + ) dt = t ] 5 t = 5 6 (s worked out i the previous exmple), while the re of the regio o the right is: 7 5 [ ] [ (t 4t + ) (t + ) dt = t 5 ] 7 t = 8 5 so the totl re is = 67 =.5. Averge Vlue of Fuctio We compute the verge (or me vlue) of umbers,,,..., by ddig them up d dividig by : A br bove qutity typiclly idictes the me of tht quitity. verge = = k k= but computig the verge vlue of fuctio requires itegrl. To estimte the verge vlue of f o the itervl [, b], we c prtitio [, b] ito eqully log subitervls of legth x = b, the choose vlue c k i ech subitervl, d fid the verge of the fuctio vlues f (c k ) t those poits: f = verge of f f (c ) + f (c ) + + f (c ) = k= f (c k ) While this lst term resembles Riem sum, it does ot hve the form f (c k ) x k, becuse b = x =. But multiplyig d dividig by b yields: k= f (c k ) = k= f (c k ) b b = b k= This lst (Riem) sum coverges to defiite itegrl: f (c k ) b b k= f (c k ) b = b k= f (c k ) x b f (x) dx b s the umber of subitervls gets lrger d the mesh, x = b, pproches.

71 4.7 first pplictios of defiite itegrls 6 Defiitio: Averge (Me) Vlue of Fuctio The verge vlue of itegrble fuctio f o [, b] is b f (x) dx b The verge vlue of positive fuctio hs ice geometric iterprettio. Imgie tht the re uder f (see mrgi) represets liquid trpped bove by the grph of f d o the other sides by the x-xis d the lies x = d x = b. If we remove the lid (the grph of f ), the liquid would settle ito the shpe of rectgle with the sme re s the regio uder the grph of f. If the height of this rectgle is H, the the re of the rectgle is H (b ), so: H (b ) = b f (x) dx H = b f (x) dx b The verge vlue of positive fuctio f is the height H of the rectgle whose re is the sme s the re uder f. Exmple 4. Fid the verge vlue of si(x) o the itervl [, π]. Solutio. Usig our defiitio, the verge vlue is: π si(x) dx = [ ] π cos(x) π π = π [() ( )] = π.666 A rectgle with height.64 o the itervl [, π] ecloses the π sme re s oe rch of the sie curve. If the itervl i the previous Exmple hd bee [, π], the verge vlue would be. (Why?) Prctice. Durig ie-hour work dy, the productio rte t time t hours ws r(t) = 5 + t crs per hour. Fid the verge hourly productio rte. Fuctio verges, ivolvig mes s well s more complicted techiques, re used to smooth dt so tht uderlyig ptters become more obvious d to remove high frequecy oise from sigls. I these situtios, the vlue of the origil fuctio f t poit is replced by some verge of f over itervl icludig tht poit. If f is the grph of rther jgged dt (see mrgi), the the -yer verge of f is the itegrl: g(x) = x+5 x 5 f (t) dt verge of f over timesp of five yers o either side of x.

72 64 the itegrl The figure below shows the grphs of mothly verge (rther oisy dt) of surfce-temperture dt, ul verge (still rther jgged ) d five-yer verge ( much smoother fuctio): This movig verge of oisy dt is frequetly used with dt such s wether iformtio d stock prices. Typiclly this movig verge fuctio revels ptter much more clerly th the origil dt. Work I physics, the mout of work doe o object is defied s the force pplied to the object times the displcemet of the object (the distce the object is moved while the force is pplied). Or, more succictly: work = (force) (displcemet) If you lift three-poud book two feet, the the force is pouds (the weight of the book), d the displcemet is feet, so you hve doe ( pouds) ( feet) = 6 foot-pouds of work. Whe the pplied force d the displcemet re both costts, clcultig work is simply mtter of multiplictio. Prctice. How much work is doe liftig -poud object from the groud to the top of -foot buildig? If either the force or the displcemet vries, however, we eed to use itegrtio. Exmple 5. How much work is doe liftig -poud object from the groud to the top of -foot buildig usig cble tht weighs pouds per foot?

73 4.7 first pplictios of defiite itegrls 65 Solutio. This is more chllegig situtio. We kow the work eeded to move the object is ()() = foot-pouds, but oce we strt pullig the cble oto the roof, we eed to do less d less work to pull the remiig prt of the cble. Let s prtitio the cble ito smll icremets so the displcemet of ech smll piece of the cble is roughly costt. If we brek the cble ito smll pieces, ech piece hs legth x =, so its weight (the force required to move it) is: ( ( x ft) lbs ) = x lbs ft If this smll piece of cble is iitilly c k feet bove the groud, the its displcemet is c k feet, so the work doe o this smll piece is ( c k ) x ft-lbs d the totl work doe o the etire cble is (pproximtely): ( c k ) x k= ( x) dx Oce gi we hve formed Riem sum, which coverges to defiite itegrl s we chop the cble ito smller d smller pieces. This itegrl represets the work eeded to lift the cble to the roof: ( x) dx = (6 x) dx = 6x x = [8 9] [ ] = 9 ft-lbs so the totl work required to lift the object d the cble to the roof is + 9 = ft-lbs. Prctice 4. Suppose the buildig i Exmple 5 is 5 feet tll d the cble weighs pouds per foot. () Compute the work doe risig the object from the groud to height of feet. (b) From height of feet to height of feet. The situtio i the previous Exmple d Prctice problems is but oe of my tht rise whe computig work. We will exmie others i Sectio 5.4. Summry The re, verge d work pplictios i this sectio merely itroduce few of the my pplictios of defiite itegrls. They illustrte the ptter of movig from pplied problem to Riem sum, to defiite itegrl d, filly, to umericl swer. We will explore my more pplictios i Chpter 5.

74 66 the itegrl 4.7 Problems I Problems 4, use the vlues i the tble below to estimte the idicted res. x f(x) g(x) h(x) Estimte the re betwee f d g for x 4.. Estimte the re betwee f d g for x 6.. Estimte the re betwee f d h for x Estimte the re betwee g d h for x Estimte the re of the isld i the figure below. 4. f (x) = (x ), g(x) = x +, x 5. f (x) = e x, g(x) = x, x 6. f (x) = cos(x), g(x) = si(x), x π 4 7. f (x) =, g(x) = x, x 8. f (x) =, g(x) = 4 x, x I Problems 9, use the vlues of f i the tble t the begiig of the pge to estimte the verge vlue of f o the idicted itervl. 9. [.5, 4.5]. [.5, 6.5]. [.5,.5]. [.5, 6.5] I 6, fid the verge vlue of the fuctio whose grph ppers below o the give itervl.. [, ] 4. [, 4] 5. [, 6] 6. [4, 6] 6. Estimte the re of the isld i figure bove if the distces betwee the lies is 5 feet isted of 4 feet. I Problems 7 8, sketch grph of ech fuctio d fid the re betwee the grphs of f d g for x i the give itervl. 7. f (x) = x +, g(x) =, x 8. f (x) = x +, g(x) = + x, x 9. f (x) = x, g(x) = x, x. f (x) = 4 x, g(x) = x +, x. f (x) =, g(x) = x, x e x. f (x) = x, g(x) = x, x 4. f (x) = x +, g(x) = cos(x), x π 4 I Problems 7, fid the verge vlue of the give fuctio o the idicted itervl. 7. f (x) = x +, x 4 8. f (x) = x, x 9. f (x) = x, x. f (x) = x, x 4. f (x) = si(x), x π. f (x) = cos(x), x π. Clculte the verge vlue of f (x) = x o the itervl [, C] for C =, 9, 8,. Wht is the ptter? 4. Clculte the verge vlue of f (x) = x o the itervl [, C] for C =,, 8,. Wht is the ptter?

75 4.7 first pplictios of defiite itegrls The figure below shows the umber of telephoe clls per miute t lrge compy. Estimte the verge umber of clls per miute: () from 8:.m. to 5: p.m. (b) from 9:.m. to : p.m. 6. The figure below shows the velocity of cr durig five-hour trip. () Estimte how fr the cr trveled. (b) At wht costt velocity should you drive i order to trvel the sme distce i five hours? 7. () How much work is doe liftig -poud bucket from the groud to the top of - foot buildig with cble tht weighs three pouds per foot? (b) How much work is doe liftig the sme bucket from the groud to height of 5 feet with the sme cble? 8. () How much work is doe liftig 6-poud chir from the groud to the top of -foot buildig with cble tht weighs poud per foot? (b) How much work is doe liftig the sme chir from the groud to height of 5 feet with the sme cble? 9. () How much work is doe liftig -poud clculus book from the groud to the top of -foot buildig with cble tht weighs pouds per foot? (b) From the groud to height of feet? (c) From height of feet to height of feet? 4. How much work is doe liftig 8-poud ijured child to the top of -foot hole usig stretcher weighig 4 pouds d cble tht weighs poud per foot? 4. How much work is doe liftig 6-poud ijured child to the top of 5-foot hole usig stretcher weighig pouds d cble tht weighs poud per foot? 4. How much work is doe liftig -poud ijured dult to the top of -foot hole usig stretcher weighig pouds d cble tht weighs poud per foot?

76 68 the itegrl 4.7 Prctice Aswers. Referrig to grph (see mrgi figure) d usig geometry: A = ()() = d B = (4)() = 4 so the totl re is + 4 = 5. Referrig to grph of the fuctios d usig itegrls: A = [( x) ( + x)] dx = = [x x ] = [ ] [ ] = [ x] dx B = [( + x) ( x)] dx = [x ] dx [ ] = x x = [9 6] [ )] = 4 which lso results i totl re of + 4 = 5.. Usig the verge vlue formul: 9 9 [ 5 + ] t dt = 9 = 9 9 [ ] 5 + t dt = [ 5t + ] 9 9 t ] [( ) ( + ) so the verge hourly productio rte is 7 crs per hour. = = 7. (force) (displcemet) = ( pouds) ( feet) = foot-pouds 4. () The work required to move the object distce of feet is ( pouds) ( feet) = foot-pouds. The work required to move the top feet of the cble oto the roof is: ( x) dx = [x ] x = [ 5] [] = 5 ft-lbs d the force required to move the remiig 4 feet of cble is: ( (4 ft) lbs ) ( ft) = ft-lbs ft so the totl work required is = 45 footpouds. (b) The work required to move the object distce of feet is gi ( pouds) ( feet) = foot-pouds. The work required to move the top feet of the cble oto the roof is gi 5 footpouds, d the force required to move the remiig feet of cble is: ( ( ft) lbs ) ( ft) = 9 ft-lbs ft so the totl work required is = 5 foot-pouds.

77 4.8 usig tbles (d techology) to fid tiderivtives Usig Tbles (d Techology) to Fid Atiderivtives Appedix I shows ptters for my tiderivtives some of which you should lredy kow bsed o your work i this chpter. My referece books d Web sites coti fr more th the oes listed i the ppedix. A tble of itegrls helps you while you re lerig clculus d serves s referece lter whe you re usig clculus. Thik of itegrl tble s dictiory: somethig to use whe you eed to spell chllegig word or eed the meig of ew word. It would be difficult to write report if you hd to look up the spellig of every word, d it will be difficult to ler d use clculus if you hve to look up every tiderivtive. Tbles of tiderivtives re limited by ecessity d ofte tke loger to use th fidig tiderivtive from scrtch, but they c lso be very vluble d useful. This sectio shows how to trsform some itegrls ito forms foud i Appedix I d how to use recursio formuls foud i itegrl tbles. The first Exmples d Prctice problems illustrte some of the techiques used to chge itegrl ito stdrd form. Exmple. Use Appedix I to fid 9 + x dx. Solutio. The itegrd is rtiol fuctio, d the first etry you see listed i the Rtiol Fuctios sectio of Appedix I should be: + x dx = ( x ) rct + C This resembles the ptter we eed, so replcig the with we hve: 9 + x dx = + x dx = ( x ) rct + C These techiques re useful whether tht stdrd form resides i tble or i your hed. Appedix I (like some other itegrl tbles) omits the +C rbitrry costt for cociseess, but you eed to remember to iclude it whe usig the results of the tble to fid idefiite itegrl. You c (d should) check this swer by differetitig. Prctice. Use Appedix I to fid 5 + x dx d 5 x dx. Notice tht smll chge i the form of the itegrd (from + to here) c led to very differet result. Exmple. Use Appedix I to fid 5 + x dx. Solutio. The itegrd is gi rtiol fuctio, d the geerl form is the sme s i the previous Exmple: 5 + x dx = but here we eeded to put = 5. ( 5) + x dx = ( ) x rct 5 + C 5 The costt i the deomitor of this itegrd ws ot perfect squre, but the process is exctly the sme eve if the result looks bit messier due to the presece of the rdicl. Prctice. Use Appedix I to fid 7 + x dx d 7 x dx.

78 7 the itegrl We ofte eed to perform some lgebric mipultios to chge itegrd ito oe tht exctly mtches ptter i the tble. Exmple. Use Appedix I to fid 9 + 4x dx. Solutio. The itegrd is gi rtiol fuctio, d the geerl form resembles the oe used i the previous Exmples, but here we hve 4x where we oly see x i the tble ptter. To get the itegrd i the form we wt, we c fctor 4 out of the deomitor: 9 + 4x dx = 4 ( x ) dx = 4 ( ) = 4 rct x ( ) + x dx + C = 6 rct ( x Aother pproch ivolves chge of vrible. First write: 9 + 4x dx = + (x) dx ) + C We hve x where we would like to see stdloe vrible. To get tht ptter, put u = x du = dx dx = du: + (x) dx = + u du = ( u ) rct + C = 6 rct ( x ) + C which yields the sme result s our previous method. Prctice. Use Appedix I to fid 5 + 9x dx d Sometimes chge of vrible is bsolutely ecessry. e x Exmple 4. Use Appedix I to fid dx. 9 + ex 5 9x dx. Solutio. Here the itegrd is ot rtiol fuctio, but we c trsform it ito oe by usig the substitutio u = e x du = e x dx so tht u = (e x ) = e x : e x 9 + e x dx = + (e x ) ex dx = + u du = ( u ) rct + C = ( e x ) rct + C If you do t see the exct ptter you eed i itegrl tble, try substitutio first. cos(x) cos(x) Prctice 4. Evlute 5 + si dx d (x) 5 si (x) dx. How should you recogize whether lgebr or chge of vrible is eeded? Experiece d prctice, prctice, prctice.

79 4.8 usig tbles (d techology) to fid tiderivtives 7 Usig Recursio Formuls A recursio formul gives oe tiderivtive i terms of other tiderivtive. Usully the ew tiderivtive is somehow simpler th the origil oe. For exmple, the first recursio formul for trigoometric fuctio listed i Appedix I sttes: si (x) dx = si (x) cos(x) + si (x) dx This formul would llow us to write si 8 (x) dx, for istce, i terms of si 6 (x) dx, which should (theoreticlly, t lest) be esier to compute th the origil itegrl. Exmple 5. Use recursio formul to evlute si 4 (x) dx. We will develop this formul from scrtch i Problem 5 of Sectio 8.. For ow, you c check tht it works by comprig the derivtive of your swer to the origil itegrd for itegrtio problem tht uses this or y other recursio formul. Solutio. Applyig the formul give i the discussio bove: si 4 (x) dx = 4 si (x) cos(x) + si (x) dx 4 This ew itegrl is oe we lredy kow how to evlute: [ si (x) dx = ] cos(x) dx = x 4 si(x) + K Puttig this together with the result of the recursio formul, we get: si 4 (x) dx = 4 si (x) cos(x) + [ 4 x ] 4 si(x) + C = 4 si (x) cos(x) + 8 x 6 si(x) + C We could hve used Appedix I to fid si (x) dx isted or eve pplied the recursio formul secod time to rewrite si (x) dx i terms of si (x) dx = dx. We could hve icluded the +K here but the the result t the ext stge would hve icluded the costt terms + 4 K + C which is lso rbitrry costt. Prctice 5. Use Appedix I to evlute cos 4 (x) dx d cos 4 (7x) dx. Usig Techology My Web sites (such s Wolfrm Alph, computer progrms (wxmxim is good free oe) d clcultors (such s the TI-89 or TI-Nspire CAS) feture computer lgebr systems tht c fid tiderivtives of wide vriety of fuctios. For exmple, typig itegrl si^4(x) ito Wolfrm Alph yields: Although techology c help us fid tiderivtive d evlute defiite itegrl, i pplictio problem you still eed to set up the Riem sum tht leds to the defiite itegrl. which (pplyig some trig ideities) grees with our result bove.

80 7 the itegrl 4.8 Problems I Problems 48, use ptters d recursio formuls from the itegrl tble i Appedix I s ecessry (log with y other tiderivtives d itegrtio techiques you lredy kow) to evlute ech itegrl x dx. 4 x dx 5. + x dx 8. 4 x dx. [ ] x dx. x x dx [ ] 9 x dx 6. cos(x) + 5 x dx [ x dx 9. e x + 7 ] + x dx 5 x dx. 4 x dx x dx 4. sec(x + 5) dx x dx x dx x dx x sec(x + 7) dx 9. l(x + ) dx. l(x ) dx. x l(5x + 7) dx. e x l (e x ) dx. cos(x) l (si(x)) dx 4. x 9 dx x dx x dx 9. + x dx. 9 + x dx 7. [ x + ] 5 + x dx. [ e x + 7 ] + x dx. 6 + x dx 9 x dx 5 x dx x dx x dx x dx 7. 6 l(x + ) dx 8. x l(5x + 7) dx 9. π cos(x) l ( + si(x)) dx x dx 4. si (x) dx x dx 4. cos (x) dx x dx cos 5 (x) dx 46. sec 5 (x) dx 47. x cos(x) dx 48. x si 5 (x) dx

81 4.8 usig tbles (d techology) to fid tiderivtives Before doig y clcultios, predict which you expect to be lrger: the verge vlue of si(x) o [, π] the verge vlue of si (x) o [, π] The clculte ech verge to see if your predictio ws correct. 5. Fid the re of the regio bouded by the grph of f (x) = l(x), the x-xis d the lies x = d x = C whe C = e,, d. 5. Fid the verge vlue of f (x) = l(x) o the itervl x C whe C = e,,,. 5. Before doig y clcultios, predict which of the followig itegrls you expect to be the lrgest, the evlute ech itegrl. () e x dx (c) (b) x e x dx xe x dx 5. Before doig y clcultios, predict which of the followig itegrls you expect to be the lrgest, the evlute ech itegrl. () e x dx (c) (b) x e x dx xe x dx 54. Before doig y clcultios, predict which of the followig itegrls you expect to be the lrgest, the evlute ech itegrl. () (b) (c) π π π si(x) dx x si(x) dx x si(x) dx C 55. Evlute dx for C =,, d. + x Before doig the clcultio, estimte the vlue of the itegrl whe C = Prctice Aswers. The itegrl dx resembles the ptter from Exmple : 5 + x 5 + x dx = 5 + x dx = ( x ) 5 rct + C 5 The itegrd i dx is lso rtiol fuctio, but we 5 x eed differet ptter from Appedix I (see mrgi) with = 5: 5 x dx = 5 x dx = l x + 5 x 5 + C. The itegrl For 7 + x dx = dx mtches the ptter i Exmple : 7 + x ( 7) + x dx = ( ) x rct 7 + C 7 7 x dx we eed the ptter i the mrgi with = 7: 7 x dx = ( 7) x dx = 7 l x + 7 x 7 + C x dx = l x + x

82 74 the itegrl u = x du = dx dx = du u = si(x) du = cos(x) dx u = 7x du = 7 dx dx = 7 du cos 4 (7x) dx = cos 4 (u) du 7. For the itegrl 5 + 9x dx = 5 + 9x dx we c fctor 9 from the deomitor: 9 ( x ) dx = 9 ( 5 ) dx + x ( ) + C = ( ) x 5 rct + C 5 = 9 rct 5 d proceed s before. We could proceed similrly for x 5 5 9x dx or we could substitute u = x (see mrgi): 5 9x dx = 5 (x) dx = 5 u du = 5 l u + 5 u 5 + C = l x + 5 x 5 + C 4. For cos(x) 5 + si dx, first use the substitutio i the mrgi: (x) cos(x) 5 + si (x) dx = 5 + u du followed by the result of the first prt of Prctice : 5 + u du = ( u ) 5 rct + C = ( ) si(x) 5 5 rct + C 5 cos(x) For 5 si dx use the sme substitutio, followed by the (x) result from the secod prt of Prctice : 5 u du = l u + 5 u 5 + C = l si(x) + 5 si(x) 5 + C 5. For cos 4 (x) dx we eed the recursio formul: cos (x) dx = cos (x) si(x) + cos (x) dx with = 4: cos 4 (x) dx = 4 cos (x) si(x) + cos (x) dx 4 = 4 cos (x) si(x) + [ 4 + ] cos(x) dx = 4 cos (x) si(x) + [ 4 x + ] 4 si(x) + C For cos 4 (7x) dx, first use substitutio (see mrgi) d the the result of the previous itegrtio: cos 4 (7x) dx = 8 cos (7x) si(7x) + [ 8 (7x) + ] 4 si(4x) + C

83 4.9 pproximtig defiite itegrls Approximtig Defiite Itegrls The Fudmetl Theorem of Clculus tells how to clculte the exct vlue of defiite itegrl if the itegrd is cotiuous d if we c fid formul for tiderivtive of the itegrd. I prctice, however, we my eed to compute the defiite itegrl of fuctio for which we oly hve tble vlues or grph or of fuctio tht does ot hve elemetry tiderivtive. This sectio presets severl techiques for gettig pproximte umericl vlues for defiite itegrls without usig tiderivtives. Mthemticlly, exct swers re preferble d stisfyig, but for most pplictios umericl swer ccurte to severl digits is just s useful. The ides behid these methods re geometric d rther simple, but usig the methods to get good pproximtios typiclly requires lots of rithmetic, somethig clcultors d computers re very good (d quick) t doig. The Geerl Approch The methods i this sectio pproximte the defiite itegrl of fuctio f by prtitioig the itervl of itegrtio d buildig esy fuctio with vlues close to those of f o ech itervl, the evlutig the defiite itegrls of the esy fuctios exctly. If the esy fuctios re close to f, the the sum of the defiite itegrls of the esy fuctios should be close to the defiite itegrl of f. The Left, Right d Midpoit Rules pproximte f with horizotl lies o ech prtitio itervl so the esy fuctios re costt fuctios, d the pproximtig regios re rectgles (see top mrgi figure). The Trpezoidl Rule pproximtes f with slted lies, so the esy fuctios re lier d the pproximtig regios re trpezoids (see middle mrgi figure). Filly, Simpso s Rule pproximtes f with prbols, so the esy fuctios re qudrtic polyomils (see bottom mrgi figure). The Left d Right pproximtio rules re simply Riem sums with the poit c k i the k-th subitervl chose to be the left or right edpoit of tht subitervl. They typiclly require lrge umber of computtios to get eve mediocre pproximtios to the defiite itegrl of f d re seldom used i prctice. Alog with the Midpoit Rule (which chooses ech c k to be the midpoit of the k-th subitervl), they re discussed er the ed of the Problems for this sectio. All of these methods prtitio the itervl [, b] ito subitervls of equl width, so ech subitervl hs legth h = x k = b. The poits of the prtitio re x =, x = + h, x = + h, x = + h, d so o. The k-th poit i the prtitio is give by the formul x k = + k h d the lst (-th) poit is thus: ( ) b x = + h = + = + b = b

84 76 the itegrl Approximtig Defiite Itegrl Usig Trpezoids See Problem 9. If the grph of f is curved, the slted lies typiclly come closer to the grph of f th horizotl oes do. These slted lies led to trpezoidl pproximtig regios. The re of trpezoid is (bse) (verge height) so the re of the first trpezoid i the mrgi figure is: ( ) y + y ( x) Similrly, the res of the ext few trpezoids re: ( ) ( ) ( ) y + y ( x) y + y, ( x) y + y, ( x) 4 d so o, with the re of the lst regio beig ( ) y + y ( x) The sum of these trpezoidl res is: ( ) ( ) ( ) y + y T = ( x) y + y + ( x) y + y + ( x) + ( ) y + y. + ( x) ( ) x = [(y + y ) + (y + y ) + (y + y ) + + (y + y )] ( ) h = [y + y + y + y + + y + y ] ( ) h = [ f (x ) + f (x ) + f (x ) + f (x ) + + f (x ) + f (x )] Ech f (x k ) vlue, except the first (k = ) d the lst (k = ), is the right-edpoit height of oe trpezoid d the left-edpoit height of the ext, so it shows up i the clcultio for two trpezoids d is multiplied by i the formul for the trpezoidl pproximtio. Trpezoidl Approximtio Rule If the f is itegrble o [, b] d [, b] is prtitioed ito subitervls of width h = b the Trpezoidl pproximtio of b f (x) dx is: T = h [ f (x ) + f (x ) + f (x ) + f (x ) + + f (x ) + f (x )]

85 4.9 pproximtig defiite itegrls 77 Exmple. Compute T 4, the Trpezoidl pproximtio of for = 4, with the vlues of f i the mrgi tble. Solutio. The step size is h = b = 4 = so: T 4 = h [ f (x ) + f (x ) + f (x ) + f (x ) + f (x 4 )] =.5 so we c sy tht f (x) dx [4. + (.4) + (.8) + (.6) + (.)] = (.5)(7) = 6.75 f (x) dx Let s see how well the Trpezoidl Rule pproximtes itegrl whose vlue we c compute exctly: x f (x) x dx = x = [7 ] = Exmple. Clculte T 4 for x dx. Solutio. The step size is h = b = 4 = so: T 4 = h [ f (x ) + f (x ) + f (x ) + f (x ) + f (x 4 )] =.5 [(.) + (.5) + (.) + (.5) + (.) ] = (.5) [ + (.5) + (4) + (6.5) + 9] = 8.75 which is withi. of the exct swer. Lrger vlues for give better pproximtios: T = 8.67 d T = Prctice. O summer dy, the level of the pod show i the mrgi fell. feet becuse of evportio. Use the Trpezoidl Rule to pproximte the surfce re of the pod d the estimte how much wter evported. Approximtig Defiite Itegrl Usig Prbols If the grph of f is curved, the slted lies from the Trpezoidl Rule my ot fit the grph of f s closely s we would like, requirig lrge umber of subitervls to chieve good pproximtio of the defiite itegrl. Curves typiclly fit the grph of f better th stright lies i such situtios, d the esiest olier curves we kow re prbols. Just s we eed two poits to determie equtio of lie, we will eed three poits to determie equtio of prbol. This prbolic method is kow s Simpso s Rule, med fter British mthemtici d ivetor Thoms Simpso (7 76); Germs cll it Kepler sche Fssregel, fter Johes Kepler, who developed it yers before Simpso.

86 78 the itegrl This result is ot obvious; see Problem for the ecessry lgebr. Cllig these poits (x, y ), (x, y ) d (x, y ), the re uder prbolic regio with evely spced x k vlues (see mrgi) is: [ ] y + 4y ( x) + y = x 6 [y + 4y + y ] Tkig the subitervls i pirs, the res of the ext few prbolic regios re: x [y + 4y + y 4 ], x [y 4 + 4y 5 + y 6 ], d so o, with the re of the lst pir of regios beig: x [y + 4y + y ] so the sum of ll prbolic res (see mrgi) is: x [y 6 + 4y 7 + y 8 ] S = x [y + 4y + y ] + x [y + 4y + y 4 ] + + x [y + 4y + y ]. ( ) h = [(y + 4y + y + y + 4y + y 4 + y + 4y + y ] ( ) h = [y + 4y + y + 4y + y y + 4y + y )] ( ) h = [ f (x ) + 4 f (x ) + f (x ) + 4 f (x ) + f (x 4 ) +. + f (x ) + 4 f (x ) + f (x )] I order to use pirs of subitervls, the umber of subitervls must be eve. Notice tht the coefficiet ptter for the re uder sigle prbolic regio is 4, but whe we put severl prbols ext to ech other, they shre some edges d the ptter becomes with the shred edges gettig couted twice. Prbolic Approximtio Rule (Simpso s Rule) If the f is itegrble o [, b] d [, b] is prtitioed ito subitervls of legth h = b the Prbolic pproximtio of b f (x) dx is: S = h [ f (x ) + 4 f (x ) + f (x ) + 4 f (x ) +... x f (x) f (x ) + 4 f (x ) + f (x )] Exmple. Clculte S 4, the Simpso s Rule pproximtio of for the fuctio f with vlues i the mrgi tble. f (x) dx

87 4.9 pproximtig defiite itegrls 79 Solutio. The step size is h = b = 4 =, so: S 4 = h [ f (x ) + 4 f (x ) + f (x ) + 4 f (x ) + f (x 4 )] = [4. + 4(.4) + (.8) + 4(.6) + (.)] = 6 (4) = 4 6 or pproximtely 6.8. Exmple 4. Clculte S 4 for x dx. Solutio. As i the previous Exmples, h = b =.5 d x =, x =.5, x =, x =.5 d x 4 =. S 4 = h [ f (x ) + 4 f (x ) + f (x ) + 4 f (x ) + f (x 4 )] = [ f () + 4 f (.5) + f () + 4 f (.5) + f ()] ( ) [ ( = + 4.5) ( + ) ( + 4.5) + ( )] 6 ( ) [ = + 4( ) + (4) + 4(4 ( ) [ ) + 8] = 8 + ] 6 6 or pproximtely The exct vlue of the itegrl is: [ x x dx = l() ] = 8 l() l() = 6 l() Lrger vlues of give better pproximtios: S = d S = Prctice. Use Simpso s Rule to estimte the surfce re of the pod i the mrgi figure. Which Method Is Best? The most difficult d time-cosumig prt of these pproximtios, whether doe by hd or by computer, is the evlutio of the fuctio t the x k vlues. For subitervls, ll of the methods require bout the sme umber of fuctio evlutios. The tble o the ext pge 5 illustrtes how closely ech method pproximtes dx = l(5) x usig severl vlues of. The results i the tble lso show how quickly the ctul error shriks s the vlue of icreses: just doublig from 4 to 8 cuts the ctul error of the Simpso s Rule pproximtio of this defiite itegrl by fctor of 9 good rewrd for our extr work.

88 8 the itegrl The error bouds i the third colum re discussed below. Notice tht for ech vlue of, the Simpso s Rule pproximtio S hs the smllest error, d tht the error for the Midpoit Rule pproximtio M (discussed i the Problems) is roughly hlf the error for the Trpezoidl Rule T. L d R deote the Left d Right pproximtios, respectively. method pproximtio error boud ctul error T S L R M T S L R M T S L R M How Good Are the Approximtios? The pproximtio rules re vluble by themselves, but they re prticulrly useful becuse we c fid error boud formuls tht gurtee how close these pproximtios come to the exct vlues of the defiite itegrl. It is useful to kow tht the vlue of itegrl is bout.7, but we c hve more cofidece i our pproximtio if we kow tht vlue is withi. of.7. The we c decide if our pproximtio is good eough for the job t hd or if we eed to improve it. We c lso solve the formuls for the error bouds provided below to determie how my subitervls we eed to gurtee tht our pproximtio is withi some specified distce of the exct swer. There is o reso to use subitervls if 8 will give the eeded ccurcy. Ufortutely, the formuls for the error bouds require iformtio bout the derivtives of the itegrds, so we cot use these error boud formuls for the pproximtios of itegrls of fuctios defied oly by tbles or grphs or of cotiuous (hece itegrble) fuctios tht fil to hve cotiuous derivtives. The error boud formul for the Trpezoidl Rule pproximtio give t the top of the ext pge is just gurtee : the ctul error is gurteed to be o lrger th the error boud. I fct, the ctul error is usully much smller th the error boud (compre the error bouds with the ctul error for T 4, T 8 d T i the tble bove to see this priciple i ctio).

89 4.9 pproximtig defiite itegrls 8 The word error does ot idicte mistke, it simply mes the devitio or distce of the pproximte swers from the exct swer. Error Boud for Trpezoidl Approximtio If f is cotiuous o [, b] d f (x) B the the error of the T pproximtio of b f (x) dx stisfies: b error = f (x) dx T (b ) B While it s possible to prove this error boud formul usig mthemtics you ve lredy lered, the proof is highly techicl d sheds little or o isight ito the workigs of the Trpezoidl Rule, so we (like the uthors of most clculus books) hve omitted it. Exmple 5. You c be certi tht the T pproximtio of is withi wht distce of the exct vlue of the itegrl? si(x ) dx Solutio. We kow tht b =, = d f (x) = si ( x ), so f (x) = 4x si ( x ) + cos ( x ) is cotiuous o [, ]. We ow eed upper boud for f (x). If f (x) is differetible (it is here) the we could use the techiques of Chpter to fid its mximum vlue o [, ] but tht would require fidig third derivtive of f, s well s some chllegig lgebr. Usig the trigle iequlity d the fcts tht si(θ) d cos(θ), we c coclude: f (x) ( = 4x si x ) ( + cos x ) 4 + = 8 so we could tke B = 8. We c do bit better, however, by cosultig grph of f (x) o [, ] (see mrgi); it ppers cler from the grph tht f (x), so we tke B = isted. Usig these vlues for, b, d B i the error boud formul: error = si(x ) dx T = 88 = 5 <.74 so we c be certi tht our T pproximtio of the defiite itegrl is withi.74 of the exct vlue: T.74 si(x ) dx T +.74 Computig T = , we c be certi tht the vlue of the itegrl si(x ) dx is somewhere betwee.7 d.87. Prctice. Fid error boud for the T pproximtio of 5 x dx. Exmple 6. How lrge must be to be certi tht T is withi. of si(x ) dx? Prctice your differetitio skills by verifyig this. Notice tht ( boud for) the error depeds o three thigs: the size of the itervl of itegrtio (the bigger the itervl, the bigger the potetil error); the umber of subitervls i the prtitio (the more subitervls, the smller the potetil error); d the size of the secod derivtive of the itegrd. We ve lredy see tht the secod derivtive of fuctio is relted to the cocvity of its grph lter o we will ler tht the secod derivtive helps mesure the curvture of the grph of f ; it should mke sese tht the more curvy fuctio is, the less effective lier pproximtio techique would be.

90 8 the itegrl Solutio. Here we kow the llowble error of. d we must fid. From Exmple 5 we kow tht b = d B =, so we wt the error boud to be less th the llowble error of.: As ofte hppes, T 86 is eve closer th. to the exct vlue of the itegrl: ( T 86 si x ) dx. <. > > > 85.6 Becuse must be iteger, we c tke = 86. Computig T , we c be certi tht the exct vlue of the itegrl is betwee.865 d Prctice 4. Determie how lrge must be i order to esure tht T 5 is withi. of x dx. Error Boud for Simpso s Prbolic Approximtio If f (4) is cotiuous o [, b] d f (4) (x) B 4 the the error of the S pproximtio of b f (x) dx stisfies: b error = f (x) dx S (b )5 B Exmple 7. Fid error boud for the S pproximtio of si(x ) dx. Solutio. We hve b =, = d f (x) = si(x ), so f (4) (x) = (6x 4 ) si(x ) 48x cos(x ) is cotiuous o [, ]. From grph of f (4) (x) o [, ] (see mrgi), we c estimte tht B 4 = 65, so error = si(x ) dx S = 58 8 <. d we c be certi tht our S pproximtio of si(x ) dx is withi. of the exct vlue: S. si(x ) dx S +. Computig S =.85765, we re certi tht the exct vlue of si(x ) dx is betwee.8765 d Notice tht we chieved much rrower gurtee usig S compred to usig T to pproximte the sme itegrl. Exmple 8. Determie how lrge must be to esure tht S is withi. of the exct vlue of si(x ) dx?

91 4.9 pproximtig defiite itegrls 8 Solutio. We wt the error boud to be less th. d eed to fid. We kow tht b = d B 4 = < > 4 88 > > = 88 Becuse must be eve iteger, we c tke = 4 d be certi tht S 4 is withi. of Altertive Methods si(x ) dx. As we hve come to expect, S 4 is eve closer th. to the exct vlue of the itegrl; usig dvced methods, we c show tht: si(x ) dx S 4.5 I Sectio 8.7 d i Chpter, you will ler how to pproximte fuctio f over etire itervl [, b] usig sigle polyomil p(x) of degree ; you c the pproximte b f (x) dx with b p(x) dx, which is reltively esy to compute. Oe dvtge of this method is tht (oce we hve foud p(x)), we oly eed to evlute other polyomil (P(x) where P (x) = p(x)) t two vlues (P() d P(b)) to compute b p(x) dx b f (x) dx d we c get better pproximtios by icresig d usig polyomils of higher d higher degree; usig the Trpezoidl Rule or Simpso s Rule requires us to evlute f (x) t + poits. A disdvtge of this pproch is tht our origil f (x) must hve cotiuous derivtives, which is ot lwys the cse, d we eed to be ble to compute those derivtives t sigle poit. Most textbooks o Numericl Alysis offer more sophisticted techiques for pproximtig defiite itegrls. Usig Techology If you hve writte eve the most bsic computer code, you should be ble to write progrm to compute y Trpezoidl Rule or Simpso s Rule pproximtio you wt (ccurte up to the flotig-poit limittios of the mchie ruig your code). If you hve grphig clcultor, it likely hs oe or more umericl itegrtio utilities (see the mrgi for TI-8 output). The Web site Wolfrm Alph ( c pproximte defiite itegrls to y desired ccurcy; typig itegrl si(x^) from x= to x= yields: Wolfrm Alph c lso be used to quickly pply Simpso s Rule: use Simpso s rule si(x^) from to with itervls yields pproximtio of.848 for si(x ) dx.

92 84 the itegrl 4.9 Problems. Use the vlues i the tble below left to pproximte 6 f (x) dx by clcultig T 4 d S 4.. Use the vlues i the tble below left to pproximte 6 f (x) dx by clcultig T 8 d S 8. x f (x) x g(x) Use the vlues i the tble bove right to pproximte g(x) dx by clcultig T 8 d S Use the vlues i the tble bove right to pproximte g(x) dx by clcultig T 4 d S π x dx 8. x dx. si(x) dx. 6 [ x] dx x dx x dx. Estimte the re of the piece of ld locted betwee the river d the rod i the figure below. 4. Estimte the re of the isld i the figure below. For Problems 5, clculte () T 4, (b) S 4 d (c) the exct vlue of the itegrl π x dx 6. x dx 8. si(x) dx. 6 [ x] dx x dx x dx For Problems 6, clculte () T 6 d (b) S x dx. x dx 4 x dx 4. e x dx si(x) x dx 6. + si(x) dx For 7, clculte () the error boud for T 4, (b) the error boud for S 4, (c) the vlue of so tht the error boud for T is less th., d (d) the vlue of so tht the error boud for S is less th.. 5. Estimte the volume of wter i the reservoir show below if the verge depth is feet.

93 4.9 pproximtig defiite itegrls The tble below left shows the speedometer redigs (i feet per miute) for cr t oe-miute itervls. Estimte how fr the cr trveled () durig the first 5 miutes of the trip d (b) durig the first miutes of the trip. t v(t) t v(t) t v(t) t v(t) The tble bove right shows the speed (i feet per miute) of jogger t oe-miute itervls. Estimte how fr the jogger r durig her workout. 8. Use the error-boud formul for Simpso s Rule to show tht the prbolic pproximtio gives the exct vlue of b f (x) dx if f (x) = Ax + Bx + Cx + D is polyomil of degree or less. 9. A trpezoidl regio with bse b d heights h d h (ssume h = h ) c be cut ito rectgle with bse b d height h d trigle with bse b d height h h (see figure t right). Show tht the sum of the re of [ the rectgle ] h + h d the re of the trigle is b.. Let f (m) be the miimum vlue of f o the itervl [x, x ], f (M) be the mximum vlue of f o [x, x ], d h = x x. Show tht: [ ] f (x ) + f (x h f (m) b h f (M) d use this result to show tht the trpezoidl pproximtio is betwee the lower d upper Riem sums for f. Becuse the limit (s h ) of these Riem sums is b f (x) dx, coclude tht the limit of the trpezoidl sums must equl b f (x) dx.. Let f (m) be the miimum vlue of f o the itervl [x, x ], f (M) the mximum of f o [x, x ] d h = x x = x x. Show tht the vlue [ ] f (x ) + 4 f (x h ) + f (x ) 6 is betwee h f (m) d h f (M) d use this result to show tht the prbolic pproximtio f (x) dx is betwee the lower d upper Rie- of b m sums for f. Coclude tht the limit of the prbolic sums must equl b f (x) dx.. This problem guides you through the steps to show tht the re uder prbolic regio (see mrgi) with evely spced x k vlues (which, for the purposes of this problem we will cll x = m h, x = m d x = m + h) is: h [ f (x ) + 4 f (x ) + f (x )] = h [y + 4y + y ] () For f (x) = Ax + Bx + C, verify tht: m+h m h f (x) dx = A x + B x + Cx m+h m h = Am h + Ah + Bmh + Ch

94 86 the itegrl (b) Expd ech of the polyomils: y = f (m h) = A(m h) + B(m h) + C y = f (m) = Am + Bm + C y = f (m + h) = A(m + h) + B(m + h) + C d use the results to verify tht: [ ] h f (m h) + 4 f (m) + f (m + h) [y + 4y + y ] = h 6 = Am h + Ah + Bmh + Ch (c) Compre the results of prts () d (b) to coclude tht for y qudrtic fuctio f (x) = Ax + Bx + C: m+h m h f (x) dx = h [y + 4y + y ] Left-Edpoit, Right-Edpoit d Midpoit Rules The rectgulr pproximtio methods pproximte itegrd with horizotl lies, so tht the pproximtig regios re rectgles d the sum of the res of these rectgulr regios is Riem sum. The Left- d Right-Edpoit Rules re esy to uderstd d use, but they typiclly require very lrge umber of subitervls to esure good pproximtios of defiite itegrl. The Midpoit Rule uses the vlue of the itegrd t the midpoit of ech subitervl: if these midpoit vlues of f re vilble (for exmple, whe f is give by formul) the the Midpoit Rule is ofte more efficiet th the Trpezoidl rule. The rectgulr pproximtio rules re: L = h [ f (x ) + f (x ) + f (x ) + + f (x )] R = h [ f (x ) + f (x ) + f (x ) + + f (x )] M = h [ f (c) + f (c + h) + f (c + h) + + f (c + ( )h)] where c = x + h so tht the poits c, c + h, c + h, etc. re the midpoits of the subitervls. The error bouds for these methods re: error for L or R (b ) B error for M (b ) B 4 where B f (x) o [, b] d B f (x) o [, b]. Notice tht the error boud for M is hlf the error boud of T, the trpezoidl pproximtio.

95 4.9 pproximtig defiite itegrls 87 For Problems 8, clculte () L 4, (b) R 4, (c) M 4 d (d) the exct vlue of the itegrl π x dx 4. x dx 6. si(x) dx 8. 6 [ x] dx x dx x dx 9. Show tht the Trpezoidl pproximtio is the verge of the Left- d Right-Edpoit pproximtios: T = (L + R ). The itegrls i Problems 4 4 will rise i pplictios from Chpter 5. Use techology to pproximte ech itegrl by pplyig Simpso s Rule with = d = 4 to pproximte their vlues. (Is S 4 very differet from S?) π π π e x dx + 4x dx + cos (x) dx 6 si (t) + 9 cos (t) dt 4.9 Prctice Aswers. Usig the Trpezoidl Rule to pproximte the pod s surfce re: T 5 ft [( ) ft] = 9 ft so the volume is (surfce re)(depth) ( 9 ft ) (. ft) = 9 ft.. Usig Simpso s Rule to pproximte the pod s surfce re: S 5 ft [( ) ft] 4 ft. b =, = d f (x) = x f (x) = x f (x) = x, so o the itervl [, 5]: f (x) = We c therefore tke B = 4, so: 4. We wt: so solvig for : = 4 x error (b ) B 4 () = error (b ) B 4 = 7 48 < > > 7 = 5 > Usig = 4 will work. We c be certi tht T 4 is withi. of the exct vlue of the itegrl. (We cot gurtee tht T is withi. of the exct vlue of the itegrl, but it probbly is.)