2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).


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1 AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following techniques or comintions thereof: the Power Rule for Antiderivtives recognition usustitution integrtion y prts sustitution with Pythgoren identities trigonometric sustitution prtil frctions nd other methods s specified in clss Approimte integrls with left right midpoint nd trpezoidl Riemnn sums (not new mteril) nd Simpson s Rule Given n error ound formul for ny of the ove pproimtions use it to find necessry numers of suintervls etc.; you need not memorize these formule ut e sure to understnd wht ech prt mens. The formuls re given on pges 557 nd 56. Understnd the two principl types of improper integrls nd how to determine whether ech converges; if n integrl converges now how to evlute it in terms of its (if possile) nd how to determine the error if n pproimtion is mde to the improper integrl with n infinite it. Find the rc length of eplicitlydefined curves (oth y in terms of nd in terms of y) using the formuls for endpoints nd s = + d d or d s = + dy. dy Understnd where these formuls cme from (pges ). Find the surfce re of surfce of revolution: if the surfce is given y function rotted out the is its re is S= πyds where ds is defined y rc length s ove or if the function is rotted out the yis its re is S= π ds. Agin understnd where these formuls cme from (pges 59 59). Prctice Prolems These prolems my e done with clcultor ecept where noted otherwise. The originl test of course required tht you show ll clcultions. The rte t which wter flows out of pipe in gllons per hour is given y differentile function R of time t. The tle elow shows the rtes s mesured every hours for hour period. hr t [ ] ( ) R t gl hr Use midpoint Riemnn sum with suintervls of equl length to pproimte R() t dt. Using correct hr units eplin the mening of your nswer in terms of wter flow. hr Use Simpson s Rule with 8 suintervls of equl length to pproimte R() t dt. c Use the Trpezoid Rule with sudivisions of equl length to pproimte R() t dt. Find the rc length long the grph of y 8 you my use your clcultor to evlute the integrl. hr hr hr = from ( y ) = ( ) to ( ) ( ) hr y = 88. Show the complete setup; Find the re of the surfce of revolution tht is generted y revolving the curve is. Show the complete setup; you my use your clcultor to evlute the integrl. y = e for out the y
2 + sin Let I e defined y I = d. Show using nonclcultor wor tht I converges. + sin Find vlue of for which I d hs n error of less thn.. Justify your nswer using nonclcultor wor. 5 5 Let I e defined y I = ln d. Estimte midpoint sum for I using suintervls of equl length. Determine the numer n of suintervls of equl length needed so tht the error ound for midpoint sum M n given y given intervl.) I K( ) M is less thn or equl to.. (Let K e n upper ound on f ( ) n n on the 6 Let f e the function defined y f ( ) = ln on nd the is. Find the volume of the solid generted y revolving R out the yis or show tht it is infinite. Justify your nswer. d 7 Evlute showing nonclcultor wor. 8 If f ( ) is positive nd f ( ) > on the intervl nd let R e the region etween the grph of y = f ( ) which of the following pproimtions to ( ) must e too lrge? I the right rectngle sum II the midpoint sum III the trpezoidl sum I only II only c I nd II only d III only e II nd III only e d? + + e + e + e c e + e + e + e e e e e e + e + e + e 9 If equl sudivisions of [ ] re used wht is the trpezoidl pproimtion of e e e d ( ) e ( ) f d The epression L + is Riemnn sum pproimtion for d d 5 c d 5 d d 5 5 e 5 d The length of curve from = to = is given y + 9. of the following could e n eqution for this curve? y = + y = 5 + c d y = 6 e y = If the curve contins the point ( ) y = which
3 Answers 58.6 gl 57. gl c 55 gl see solutions = ln.59 5 n = 6 π 6 V = 7 ln5 8 d 9 e Solutions Firstly ny integrl of rte gives totl mount. Therefore this integrl gives n pproimtion to the totl mount of wter tht flowed out of the pipe from t = hr to t = hr. To find the pproimtion to the integrl note tht ech suintervl hs length 6 hr ecuse the intervl s totl length is hr nd we re dividing it into hr prts; then = 6 hr. Therefore our suintervls if they re to e of equl length re t ( hr6 hr ) t ( 6 hr hr ) t ( hr8 hr ) nd t ( 8 hr hr ). Then ecuse we re using the midpoint rule our smple points re t = hr t = 9 hr t = 5 hr nd t = hr. So the suintervl midpoint pproimtion to the R t dt 6 hr. gl hr +. gl hr +. gl hr +. gl hr = 58.6 gl. hr integrl is () ( )( ) hr hr This time our suintervls hve length hr ecuse = hr. The smple points re ll the given vlues of t. 8 A formul for Simpson s Rule is given on the top of pge 56; using this the pproimtion to the integrl is hr R() t dt ( hr )( R( ) + R( ) + R( 6 ) + R( 9 ) + R( ) + R( 5 ) + R( 8 ) + R( ) + R( )) hr = hr = 57. gl. ( )( ) c We re c to suintervls of length 6 hr s in prt. However the smple points re now the endpoints of the suintervls with ll ut the first nd lst ones counted twice ech nd the totl divided y two (verging the two ses of ech trpezoid). Thus the trpezoidl pproimtion to the integrl in question is given y hr R t dt 6 R + R 6 + R + R 8 + R = ( ) () ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) hr = 55 gl. dy First solve the eqution for y eplicitly in terms of : y =. Therefore d the rc length formul from to is s = + d d we hve 6 s8 = + d 9 unplesnt to evlute y hnd ut ny CAS will gldly tell you tht it s equl to 5 nd you cn lso 7 pproimte it with This is est done in terms of where = ln y nd. y = The its will e e e = [ e ] = y the Power Rule. Since 8. This is thoroughly. Then the re is given e y S= π ln y + dy. y The ntiderivtive of this is not n elementry function ut cn e pproimted; CAS gives n pproimtion to the integrl s 7.55.
4 + sin for [ ) nd f ( ) d cn e shown to converge. The numertor of the originl integrnd + sin rnges 5 + sin in vlue from to 5 since sin rnges from to ; therefore lwys nd it is firly convenient to show tht d converges: it is equl to d. This integrl simplifies to sin = + = which is finite. Therefore d converges. + sin + sin + sin We cn divide d into two prts: d d + in which the second term represents the error of the pproimtion. Thus we need to find such tht.. + sin d However n + sin ntiderivtive of cnnot e epressed in terms of elementry functions so we insted consider 5 5 d. which is vlid ecuse the ltter is lwys greter mening the error in pproimting d will lwys e greter thn the error in pproimting the integrl in which we re ctully interested. Since d = = + = we re fced with the rther simple ts of finding such 5 tht.. Now ll we need to do is divide oth sides y 5 giving. nd then te the reciprocl of the squre root of oth sides. Therefore = 5. We show this using the Comprison Test; tht is we must find function f such tht f ( ) 5 Ech suintervl hs length ; the suintervls re ( ) nd ( 5 ) so the smple points re = nd =. 5 Thus the pproimtion is ln d ( ln + ln ) = ln + ln( ) = ln + ln + ln = 6 ln The integrnd s second derivtive is f ( ) = =. Therefore we re interested in finding n given or. n.5 n ; integer vlue of n so n = 6. therefore ( ) ( ) nd n upper ound for its solute vlue on the intervl [ ].5n or ( ) 5 is 5.. The right side simplifies: n n % n giving 5.8. Nturlly we wnt n 6 This must e done with cylindricl shells. Ech one hs dv = πrhdr = π( ln ) d so = ( ) V π ln d. For simplicity s se we will momentrily ignore the fctor of π. To find n ntiderivtive for this use integrtion y prts with u= ln nd dv = d. This gives du = nd v=. Therefore d d uv vdu = ln + = ln + d = ln + + C. Evluting this t is esy it gives ut t we hve n symptote. Therefore we must concern ourselves with ( ln + ). The second term clerly its to zero nd the first term my e nlyzed with L Hôpitl s Rule. We first rewrite it s ln. Now we hve n indeterminte form of type. Differentiting the numertor nd denomintor
5 Clerly this is lso zero. Therefore evluting the integrl hs given us = which we must multiply y π to give π. gives = ( ). 7 The integrnd s denomintor fctors to ( )( ) eecuted s follows: we epect something of the form the denomintors tht A( ) ( ) + nd integrtion y prtil frctions of A B + = + ( + )( ) + B + =. This is rerrnged to give ( A B) A B d is ( + )( ) so we now from clering + + = which llows A+ B= us to set up the system. This solves to ( AB ) = ( ). Thus the originl integrnd is equivlent to +. Now we cn integrte ech term giving ln( + ) + ln( ) which cn e rewrit A+ B= + ten s ln =. Now we evlute ln + s follows (ignoring the ; we ll put tht c in t the + = end): ln ln ln ln = = ln5 + ln. Since the function f ( ) = ln is continuous for very lrge vlues of we cn pull out the nturl logrithm function from the it: ln5 + ln = ln5 + ln = ln5. Now we need to return the + to its rightful plce; the nswer is ln5. > on the intervl in question mens tht the function is concve up. Therefore secnt line such s tht used in the trpezoid sum will lwys e ove the function itself ming the trpezoid pproimtion too lrge. Therefore III must e one of the correct nswers; options nd c re incorrect. Without nowing nything more out the shpe of the function it is impossile to conclude whether tngent line to the midpoint of suintervl will lie entirely ove or elow the function or neither so we cnnot me conclusion out the ccurcy of the midpoint sum. Therefore d is the correct nswer. 8 The fct tht f ( ) 9 The suintervls re [ ] [ ] nd [ ] ; ech hs length. With the trpezoidl pproimtion we count ech suintervl s endpoint twice ecept for the first nd lst one s outer points nd divide y on the outside to indicte verging ech trpezoid s ses. Thus the integrl s trpezoidl pproimtion is given y e ( ) e e e ( = ) = choice e. ( e e e e ) ( e e e e ) 5 This cn e rewritten s. The division y 5 indictes tht there re 5 suintervls nd the fct tht 5 = 5 the coefficient s numertor is indictes tht the difference of the its is. Rememer tht the summnd tes ( ) the form f + n which tells us tht the lower it of the integrl is. Since we now tht the difference of the its is the upper it is nd since n = 5 the function is f ( ) =. This mens tht the cor rect response is. The rc length formul for y on [ ] is tht y C = + for some C; since ( ) ( ) s = + d. d Here dy dy = 9 so =. This mens d d y = 6 is on the curve c is not possile nswer nd only is correct.
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