Answers for Ch. 5 Review: The Integral
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1 Answers for Ch. 5 Review: The Integrl. So, I m to sketch grph of. I sketch things with softwre. Four subregions, Δ ½. The region R is onl in the first qudrnt, nd the -intercepts re nd, so the verticl seprtors will be t.5,, nd.5. In order to grph those segments, I hd to clculte the -vlues on the curve, nd I m going to need them lter, so I ll give them to ou here Sketch the left-endpoint rectngles; find the re. I drew in horizontl segment in the leftmost intervl, but rell, tht s not rectngle, becuse its height is zero. In ech cse, the bse of the rectngle is ½. re ½ + ½ ½ + ½.5.75 re under.. Do the sme with right-endpoints. This time, I ll fctor out the ½ to sve tping. re ½( ).75. Creep! But the zeros on the end do ber this out.. Sketch the midpoint rectngles; find the re. For this, I need the heights t the midpoints of ech intervl:.5,.75,.5,.75. Here s tble of vlues The vlues hve si deciml plces, onl three of which I ve put in the tble, but I ll use ll of them in m computtions below. re ½( ).5 5. Sketch the trpezoids nd find the re. Clculte the res of the trpezoids individull nd dd them up. (In this cse, the two on the ends re ctull tringles.) The re of trpezoid is given b A h( b + b ), where the bses re the two prllel verticl sides, nd the heights re the horizontl distnces between them. re.5( +.875) +.5( ) +.5( +.5) +.5(.5 + ).75. Ch. 5 review nswers p. of 9 AP Clculus AB
2 . Find the ect re of R with definite integrl nd the FTC. I ll need n ntiderivtive for this. ( ) Then evlute this t nd t, nd subtrct. (8 ) 7. Estimte the surfce re of the pond pictured using the trpezoidl rule. pictured below. ) Fill in the tble of vlues of g for from to. Tht s going to men evluting different integrls. Before ou give up on me, though, remember tht these re just res tht I m dding up. To ese things long, I m going to reproduce the drwing, this time with the intervls shown grphicll f t f t The mesurements cross the pond re - vlues, but in order to get the whole re, I need vlues t ech end of the pond, too. Clerl, those should be zero. In other words, here s tble of vlues. The -vlues chnge b ft ech time Since the heights re the sme ever time ( feet), nd the re formul hs ½ ever time, I ll fctor those out. (( + 5) + (5 + 5) + (5 + 8) + (8 + 8) + (8 + 7) + (7 + 75) + (75 + 8) + (8 + )) 9,9 ft. 8. To enter the definite integrl on TI-nspire, ou use the commnd from either Menu, Clculus, Numericl Integrl, or through the templte menu on the kepd (just to the right of the 9 ke). Then fill in the blnks in the templte. The nswer is.98 to three deciml plces. 9. I like this question. Thus ou re subjected to it. g( ) f ( t), where f is the function So now ou cn see the vlues I got. But here s n eplntion of ech vlue. g() f ( t). No wih, no re. f t ; the vlue is negtive g() ( ) becuse the height is negtive. g() f ( t) f ( t) + f ( t) + ; it s the previous vlue plus the net rectngle. g() f ( t) f ( t) + f ( t) + g() f ( t) f ( t) + f ( t) + ½.5. This time the etr piece is tringle. 5 5 g(5) f ( t) f ( t) + f ( t).5 + ½. Agin, the new piece is tringle, but this time it s bove the -is, so it s positive increment. 5 g() f ( t) f ( t) + f ( t) 5 Ch. 5 review nswers p. of 9 AP Clculus AB
3 + ½ ( + ).5. I used the formul for the re of trpezoid, ½ h (b + b) for the piece from 5 to. 7 7 g(7) f ( t) f ( t) + f ( t).5 + ½ ( +.5) g(8) f ( t) f ( t) + f ( t) ½ (.5 + ) g(9) f ( t) f ( t) + f ( t) ½ ( +.5).5 9 g() f ( t) f ( t) + f ( t) ½.5.5 And there ou go. b) Plot the points. I hve tble; this won t be hrd c) Where does g hve its minimum? Well, it certinl ppers to be t. If ou look bck t the grph of f, tht s where the height chnges from negtive to positive; the re dded stops decresing the totl, nd begins rising it gin. The clculus nswer is tht g( ) f ( t) indictes tht g () f (), nd tht since g chnges signs from negtive to positive t, tht must be the loction of minimum. d) The four points on g tht re colliner re the first four: (, ), (, ), (, ), (, ). I cn see this both from the grph nd from the fct tht the chnges in nd re identicl ech time. e) The gretest rte of increse of g is between nd 7, since tht s where the intervl of gretest positive re is in the digrm. Evlute ech integrl using u-substitution if needed. I m going to go stright through these; if I find n ntiderivtive worth commenting on, I will. But in ech cse, ou cn check m ntiderivtive b tking derivtive of our own.. 5 5] 5() 5( ) + You cn lso think of this geometricll; it s rectngle with bse nd height Ch. 5 review nswers p. of 9 AP Clculus AB 5 π / sin ] π /. cos π sin sin. ( + 7) + 7. ( + 7 ) (( ) ( ) + 7( )) ( + 7) ( 7) ( ) / d / π /. sec π θ dθ tnθ] / tn π tn Perhps ou re thinking tht ou hve no ide wht tn(π/) is, nd re hoping ou get clcultor for this stuff. Well, mbe but mbe not. You cn do it s sine divided b cosine, but ou d hve to know those vlues. Mbe it s time to brush up, eh? (I m feeling Cndin.) 7. + ( + ) + 7
4 ( ) cos This one looks little crz. You don t know function whose derivtive is cos. Or do ou? Check this out. sec cos Tht s little better. You re supposed to be wre tht the derivtive of tngent is the squre of secnt. Let u. Then du. sec sec sec tn tn( ) u du u + C + C 9. 9r dr r Since there is n inside prt here, I ll let tht be u nd see how it goes. Let u r. Then du r dr. So insted of coefficient of 9, I need coefficient of. Since 9, I ll fctor out, like this: 9r r dr r r dr. And now for the substitution: du u u du r + C u / + C r r dr sec. tn Well, given tht I know tht the derivtive of tn is sec, I m thinking it s nice u- substitution problem... Let u tn ; then du sec. du u u / du u/ + C u + C tn + C Ch. 5 review nswers p. of 9 AP Clculus AB r r dr Prett. There s n inside; u-substitution. Let u r ; then du r dr. I ll need. r r dr r r dr Since this is definite integrl nd we re chnging the limits of integrtion, I need to find new limits to mtch u. When r, u ; when r, u. So the integrl becomes u / u / du. u / du [ / / ] Note: It s not problem tht the upper limit is less thn the lower limit. The Fundmentl Theorem of Clculus works in n cse. This cn t possibl be nother substitution problem, cn it??? Let u + 5; then du. I cn bring out from the numertor to leve the I d like there. + 5 du ln u + C ln ( + 5) + C. u Before ou tr to bust me for leving off the bsolute vlue in the lst step, think bout it.. Looks like power (½), but there s not just vrible there. It s u-substitution once gin. Let u ; du. I ll put in the I need for du nd tke it out gin with coefficient of ¼.
5 . u/ du u / + C ( ) / + C ( ) This is nother simple u-substitution. Let u ; du. All I need is the negtive sign. ( ) ( ) u + C + C + C u 5. ( + 5). 7. u du At first, it might be little tempting to multipl the binomil out. But don t. Yes, it would work; no, ou probbl wouldn t get the nswer right. Tr u-substitution, with u s the prt inside the prentheses. Let u + 5; then du. Not too bd. I ll need to put in nd tke out. ( + 5) ( + 5) u du u7 7 + C + 5 ( ) 7 + C More u-substitution. B the number of problems like tht, ou might be led to believe tht it s n importnt topic. Let u e + (the inside); du e. / / du u du u + C u e + + C There re couple of ws to go bout this one. Me, I memorized the ntiderivtive of tngent. But I don t rell think tht ou need to. Insted, ou cn rewrite the integrnd s frction nd go from there. 8. sin First, rewrite tngent. tn. It cos looks like u-sub gin, but which one is u? Sine nd cosine both hve derivtives in the integrnd. The ke is tht du (or, or, or whtever) cn never end up in the denomintor of the integrnd. Tht mens u cos, nd du sin. sin sin cos cos du ln u + C u ln cos + C If ou ve looked up the ntiderivtive formul for tngent someplce, ou ll probbl notice tht it looks different thn this, nd involves secnt function. But tht s the sme thing. Wtch. ln cos + C ln cos + C ln + C ln sec + C. cos The re both right. Oooh. Ugl. It looks s though the u-substitution might involve u, but tht s not the onl option. There re two resonble possibilities for u: either or rcsin. Let s look t the du choices. If u, then du. If u rcsin, then du. Yep, I think we hve winner. The first option does nothing to help the rcsine, but the second one works beutifull.?? u du Yes, there re question mrks for limits of integrtion. I need to figure out wht vlues of u correspond to the originl limits in. If, then u rcsin. (How do I know tht? Since sin, rcsin, too.) If, then u π rcsin. Ch. 5 review nswers p. 5 of 9 AP Clculus AB
6 So m integrl becomes π π π u du u π π θ 9. ( + θ / ) dθ θ + θ / ( ) dθ I pulled the out front first just to get it out of the w. This hd better be u-substitution, too, becuse I don t know wht else would work. Let u + θ /. Then du θ/ dθ θdθ. So I ll need to del with the frction there. θ dθ + θ / ( ) And, since it s definite integrl, I ll be chnging the limits of integrtion, s well. When θ, u + / ; when θ, u + /. Substituting ll of tht, I get u u du. + This initill looks like mbe u-sub would work, but if the denomintor is u, then du ( ), nd tht s nowhere to be seen. In fct, this denomintor would benefit from completing the squre. See how + is lmost, but not quite, perfect squre trinomil? You cn see it s + 9 +, which is ( ) +. + ( ) + Then if we let u, du, nd the integrl becomes u + du, which I suspect... ou m recognize s rctn u + C rctn + C. ( ) This is going to require division; ou cn tell becuse the degree of the numertor is t lest s lrge s the denomintor Formtting tht in MthTpe ws remrkbl nnoing, nd I m not ll tht hpp with how it cme out. So the originl integrl is now + + ln 5 + C This is nother long division, for the sme reson s the lst one. I hve decided not to tpe out how to do the ctul division. Becuse there re no odd degree terms, ou cn ctull get b without putting in the terms. Here s how it turns out: rctn C 5 + Prett. Not u-substitution, s du is not pprent nwhere. Completing the squre hs potentil, nd it would look better to me if the sign on were positive. I ll fctor negtive out of the trinomil. I cn t tke it out of the rdicl, though; tht would chnge the sign of the rdicnd nd give imginr numbers not something we re deling with in this course. Ch. 5 review nswers p. of 9 AP Clculus AB
7 ( 5) lred looks little better. See how similr the trinomil is to perfect squre? + 9 ( ) ( ) 9 + ( ) 9 This is vrition on the derivtive of rcsine. But if tht s going to work, we need where the 9 is. So I ll fctor tht out. Wtch. 9 ( ) 9. Whew. I shll let u, so du. Substitution gives du rcsin u + C u rcsin + C. No, I don t nticipte nthing quite tht involved on the test. But if ou cn do this, won t ou be well-prepred? For the net si problems, ou get grph of function, f, nd some informtion: the shded region A hs n re of.5, nd f ( ).5.. f () The re of this region is given s.5, but in the grph, it s below the -is. So the integrl s vlue is f () represents the re bove the is in region B. It s tempting to subtrct the two given vlues nd get n re of. But tht s just wrong. Here s wh: f () f () + f (). Obvious, right? Now I ll substitute vlues f () And now solve for the integrl. f () See? The net re is.5, but since prt is negtive, the rest must be greter thn.5.. f () The bsolute vlue hs the grphicl effect of reflecting the portion of the grph tht s below the -is in tht line. It mkes ech of the individul res positive. So f () f () f ().5 7 The coefficient just comes outside of the integrl. No problem. 8. ( + f ()) + f () Remember tht propert? The second prt hs known vlue, but wht bout the first one? It s rectngle units wide nd units tll. So + f () Ch. 5 review nswers p. 7 of 9 AP Clculus AB
8 9. ( ) ( ) f f 5. Switching the limits of integrtion chnges the sign of the nswer, nd we found the integrl bck in problem. Find d for ech function. These re emples of the ppliction of the prt of the fundmentl theorem of clculus tht ss if f is d continuous, then f ( t) f ( )... + cos t d cos + The integrl looked just like the theorem, so the onl thing I hd to do ws substitute for t nd lose the trppings of the integrl. (Heh. Trppings.) 7 + cos t Hmmm. The difference here is tht the upper limit of integrtion is not just. This is job for dut-duh-dh! the Chin Rule. The inside prt of the function of is 7, so tht s wht I ll need to multipl b the derivtive of. It s esier to show thn to s. d cos ( 7 ) + I replced the t b the vrible upper limit of integrtion, nd multiplied on the end b the derivtive of 7. I cn t do much with tht. I suppose it would be resonble to put the in front of the rdicl, but I don t wnt to. It s fine s it is.. + t This one differs from the theorem in tht the limits of integrtion re bckwrd. There s rule bout definite integrls tht mkes this strightforwrd thing to fi. It goes like this:. b f ( ) f ( ) + t + t form of the theorem. d + + b, so now it fits the t + Oooh. Worse. The problem is not so much the, becuse tht s just the chin rule gin. No, this time there re two vrible limits of integrtion. Wht does tht men? I m going to hve to split this up into two different integrls in order to differentite. The rule for this is s follows: b c c f ( ) + f ( ) f ( ) b. I m going to use it in reverse. I cn pick n constnt I like to be the dividing vlue. I like zero. You cn choose something tht ppels to ou. + t + t + t + Notice tht m constnt doesn t hve to be between the limits of integrtion. In this cse, it s rell unlikel tht is between the limits; if it were, would be zero, nd tht s not vrible n more, is it? Bck to the problem. I now need to differentite tht sum of integrls. I ll use some nottion to mke this esier for me to write. d + t + t + For the first prt, I ll echnge the limits nd chnge the sign, s in problem. d + t + t + Then differentite, using the chin rule for the second one. Ch. 5 review nswers p. 8 of 9 AP Clculus AB
9 d ( ) Tht won t esil simplif, so I m done. For the lst seven problems, I m given tht f() hs positive derivtive for ll vlues of nd tht f() nd tht g( ) f ( t). The questions re true/flse.. g is differentible function of. True. The FTC prt bout derivtives ss: If f is continuous on [, b] then g( ) f ( t) is differentible function, nd g () f (). Your book doesn t cll the function g, but this problem does. In n cse, g is differentible. Cse closed. zero. Therefore f must chnge signs from negtive to positive t. Flse. 8. g hs locl minimum t. True. As rgued in the previous question, g f chnges from negtive to positive t, so there s minimum. 9. The grph of g hs n inflection point t. Flse. There will be point of inflection if g chnges signs. Since the first derivtive of g is the function f, the second derivtive of g is the function f. And f is positive for ll. Remember? 5. The grph of dg crosses the -is t. True. Duh. dg is f(). And f (), nd f is lws positive. So f increses right through tht point t (, ), crossing the -is. 5. g is continuous function of. True, becuse differentibilit implies continuit, nd we ve just estblished tht g is differentible.. The grph of g hs horizontl tngent line t. The grph of g? Well, the slope of tngent line is determined b the vlue of the derivtive, so I should find the derivtive of g. And tht s up there in the bo in 5: g () f (). So if g hs horizontl tngent, t, tht would men tht g (). So, wht is g ()? It s f (), nd tht is given to be. TRUE. 7. g hs locl mimum t. Well, I know now tht g (). There will be locl mimum if g chnges signs from positive to negtive. (Tht s clled the first derivtive test; ou lerned it bck in the fll.) Since g () is f (), I d like to know if the vlue of f chnges from positive to negtive when. There doesn t seem to be n informtion bout tht. Ecept this: f () hs positive derivtive for ll vlues of. So f is incresing. And t, it s Ch. 5 review nswers p. 9 of 9 AP Clculus AB
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