(200 terms) equals Let f (x) = 1 + x + x 2 + +x 100 = x101 1

Transcription

1 SECTION 5. PGE 78.. DMS: CLCULUS CHPTE 5. Sectio 5. pge 78 i INTEGTION Sums d Sigm Nottio j j i i j i i j j j i i terms equls i i 99 i i i i i i i i i i i 99 si j sii j i m k 5 m+6 k + i 6 + i +i, j ,, k k +,,, i i l m l + l + +l l! m i e i/ e+/ e / terms equls. { if + if 5. { if + + if 6. Let f if. The f d + d [k k + ] [k k k ] k 9 [k + ] k k Let s The 9, 99 s

2 INSTUCTO S SOLUTIONS MNUL SECTION 5. PGE 78 Subtrctig these two sums, we get s / / Thus s + + / f i + f i im f i + im + jm+ f j f i im f i im f + f m, becuse ech sum hs ol oe term tht is ot ccelled b term i the other sum. It is clled telescopig becuse the sum folds up to sum ivolvig ol prt of the first d lst terms., m j j m m m im m i i + m m + m + mm + m j j + j j To show tht Fig. 5.. i we write copies of the idetit oe for ech k from to : +, k + k k +, ddig the left d right sides of these formuls we get + i +. Hece, i The umber of smll shded squres is Sice ech hs re, the totl re shded is i. But this re cosists of lrge right-gled trigle of re / below the digol, d smll trigles bove the digol ech of re /. Equtig these res, we get i The formul i + / holds for, sice it ss i this cse. Now ssume tht it holds for some umber k ; tht is, k i kk + /. The for k +, we hve k+ i k i+k+ kk + +k+ k + k +. Thus the formul lso holds for k +. B iductio, it holds for ll positive itegers. 77

3 SECTION 5. PGE 78.. DMS: CLCULUS 7. The formul i + + /6 holds for, sice it ss i this cse. Now ssume tht it holds for some umber k ; tht is, k i kk + k + /6. The for k +, we hve k+ i k i + k + kk + k + + k + 6 k + 6 [k + k + 6k + 6] k + k + k + 6 k + k + + k Thus the formul lso holds for k +. B iductio, it holds for ll positive itegers. 8. The formul r i r /r for r holds for, sice it ss i this cse. Now ssume tht it holds for some umber k ; tht is, k r i r k /r. The for k +, we 9. hve k+ r i k r i + r k r k r + r k r k+ r. Thus the formul lso holds for k +. B iductio, it holds for ll positive itegers.. Fig The L-shped regio with short side i is squre of side ii + / with squre of side i i/ cut out. Sice ii + i i i + i + i i i + i i, tht L-shped regio hs re i. The sum of the res of the L-shped regios is the re of the lrge squre of side + /, so. To show tht + i. j , we write copies of the idetit k + k k + 6k + k +, oe for ech k from to : ddig the left d right sides of these formuls we get + j + 6 j + j + Hece, j j so + j +.. The formul i + / holds for, sice it ss i this cse. Now ssume tht it holds for some umber k ; tht is, k i k k + /. The for k +, we hve k+ i k i + k + k k + + k + k + k +. k + [k + k + ] 78

4 INSTUCTO S SOLUTIONS MNUL SECTION 5. PGE 8 Thus the formul lso holds for k +. B iductio, it holds for ll positive itegers.. To fid j , we write copies of the idetit k + 5 k 5 5k + k + k + 5k +, oe for ech k from to : lim lim lim [ ] sq. uits , ddig the left d right sides of these formuls we get j + j + j + 5 j +. Substitutig the kow formuls for ll the sums ecept j, d solvig for this qutit, gives Fig j Of course we got Mple to do the doke work! i i i i We would guess correctl tht Sectio 5. pge 8 i + +. res s Limits of Sums. The re is the limit of the sum of the res of the rectgles show i the figure. It is. This is similr to #; the rectgles ow hve width / d the ith hs height i/+, the vlue of + t i/. The re is lim i + 8 lim i lim sq. uits.. This is similr to #; the rectgles hve width / / d the ith hs height the vlue of t + i/. The re is lim 8 lim + i i lim + + 6sq. uits. 79

5 SECTION 5. PGE 8.. DMS: CLCULUS. This is similr to #; the rectgles hve width / / d the ith hs height the vlue of + t + i/. The re is lim 7 lim + i + i lim sq. uits re lim S + sq. uits The re is the limit of the sum of the res of the rectgles show i the figure. It is lim lim lim [ + [ ] ] sq. uits. Fig The required re is see the figure lim + lim [ [ + + lim sq. uits. ] + ] Fig Divide [, ] ito equl subitervls of legth b poits i i, i. The S [ i + ] i + Use Theorem d c. + + Fig

6 INSTUCTO S SOLUTIONS MNUL SECTION 5. PGE Fig The regio i questio lies betwee d d is smmetric bout the -is. We c therefore double the re betwee d. If we divide this itervl ito equl subitervls of width / d use the distce betwee d for the heights of rectgles, we fid tht the required re is lim i lim lim i sq. uits. Fig. 5.. The height of the regio t positio is. The bse is itervl of legth, so we pproimte usig rectgles of width /. The shded re is lim lim lim 8 i i 8i 8i sq. uits. Fig The height of the regio t positio is. The bse is itervl of legth, so we pproimte usig rectgles of width /. The shded re is lim + i lim lim + i sq. uits. + Fig

7 SECTION 5. PGE 8.. DMS: CLCULUS The height of the regio t positio is +. The bse is itervl of legth, so we pproimte usig rectgles of width /. The shded re is lim lim lim i 6i 6i 6i sq. uits. Now we c use l Hôpitl s rule to evlute / lim / lim Thus the re is [ ] / l lim lim /+ l l. squre uits. l. Divide [, b] ito equl subitervls of legth b b poits i ib, i. The S b e ib/ b b eb/ e b/ i b eb/ eb/ e b/ b eb/ e b e b/. Let r b. re lim S e b lim e b/ i Use Thm. 6..d. lim r + er r + r e r e b lim r + e r eb sq. uits. [ ] b. re lim [ b + Fig. 5.. b + + b lim lim b + b sq. uits. + ] b. The required re is lim lim / [ +/ + +/ + + +/] / lim lim / lim [ + / + / + + / ] / / /. / 5. Let t b b b / d let b Fig. 5.. b b b, t, t,..., t b. 8

8 INSTUCTO S SOLUTIONS MNUL SECTION 5. PGE 8 The i th subitervl [ i, i ] hs legth i t i t. Sice f i, we form the sum ti S t i t t i [ b / t ]. 7. Let r d c b. The re uder the curve is lim S c r [ ] lim r + r l c l squre uits. lim r + c r l c b This is ot surprisig becuse it follows from the defiitio of l. Fig s i represets sum of res of rectgles ech of width / d hvig heights equl to the height to the grph t the poits i/. Hlf of these rectgles hve egtive height, d lim S is the differece of the res of the two trigles i the figure bove. It hs the vlue sice the two trigles hve the sme re Fig b 6. s + i Fig i represets sum of res of rectgles ech of width / d hvig heights equl to the height to the grph + t the poits i/. Thus lim S is the re of the trpezoid i the figure bove, d hs the vlue + 5/ 7/. Fig s i represets sum of res of rectgles ech of width / d hvig heights equl to the height to the grph t the poits i/. Thus lim S is the re of the trigle i the figure bove, d therefore hs the vlue. 9. S j sum of res of rectgles i the figure. Thus the limit of S is the re of qurter circle of uit rdius: lim S. 8

9 SECTION 5. PGE 8.. DMS: CLCULUS 5. f si o [,], 6. P 6 {, 6,,,, 56 }, L f, P 6 [ ] + Sectio 5. pge 9 Fig The Defiite Itegrl U f, P , [ ] 6. f cos o [, ],. [ L f, P cos + cos + cos + cos ]. U f, P [cos + cos + cos ] + cos. cos. f o [, ], 8. P 8 {,,,,, 5,, 7 }, L f, P 8 [ ] 7 U f, P 8 [ ] + 9. f o [, ],. L f, P [ ]. U f, P [ ].. f e o [, ],. L f, P e + e + e + e e e e. U f, P e + e + e + e e ee.8.. f l o [, ], 5. [ L f, P 5 l + l l l l 9 ] U f, P 5 [l l 75 + l 85 + l 95 ] + l.596. / Fig / 7. f o [, ]. P {,,,...,, }. We hve L f, P U f, P, Thus lim L f, P lim U f, P /. If P is prtitio of [, ], the L f, P U f, P + for ever, sol f, P lim U f, P /. Similrl, U f, P /. If there eists umber I such tht L f, P I U f, P for ll P, the I cot be less th / or there would eist P such tht L f, P >I, d, similrl, I cot be greter th / or there would eist P such tht U f, P <I. Thus I / d d /. 8

10 INSTUCTO S SOLUTIONS MNUL SECTION 5. PGE 9 8. f o [, ]. P {,, We hve L f, P,...,, } i + s, U f, P Thus i i d. s. 9. f o [, ]. P {,,,...,, }. We hve usig the result of Eercise 5 or 5 of Sectio 6. L f, P i i s, U f, P i + + s. Thus d.. f e o [, ]. P {,, 6,...,, }. We hve usig the result of Eercise 5 or 5 of Sectio 6. B l Hôpitl s ule, e / lim e/ lim / e / / e / lim / lim Thus. lim. lim. lim. lim lim L f, P lim U f, P e i i d d i si si d l + i l + d i 5. lim t t d Note tht i [ i is the midpoit of, i ]. 6. lim + i lim + i/ e d.. d + 7. Let b d i + i where i. Sice f is cotiuous d odecresig, Thus, L f, P f + f + f + + f b [ ] f + f i, U f, P f + f + + f + f b b [ ] f i + f b. L f, P e / + e / + e 6/ + +e / e / e / e e /, U f, P e / + e 6/ + e 9/ + +e / e / L f, P. U f, P L f, P [ b f i + f b f b f b f. ] f i 85

11 SECTION 5. PGE 9.. DMS: CLCULUS Sice lim [U f, P L f, P ], therefore f must be itegrble o [, b]. 8. P { < < < }, P { < < < j < < j < < }. Let m i d M i be, respectivel, the miimum d mimum vlues of f o the itervl [ i, i ], for i. The L f, P U f, P m i i i, M i i i. If m j d M j re the miimum d mimum vlues of f o [ j, ], d if m j d M j re the correspodig vlues for [, j ], the m j m j, m j m j, M j M j, M j M j... + d 8, + Fig d shded re Therefore we hve m j j j m j j + m j j, M j j j M j j + M j j. Fig. 5.. Hece L f, P L f, P d U f, P U f, P. If P is refiemet of P we c dd the ew poits i P to those i P oe t time, d thus obti 5. b b d L f, P L f, P, U f, P U f, P... Sectio 5. Properties of the Defiite Itegrl pge 96 b c f d + f d + + b c f d + f d c c f d f d f d + f d f d f d f d f d + f d Fig d Fig t dt b 86

12 INSTUCTO S SOLUTIONS MNUL SECTION 5. PGE 96 t 5. d re i figure below Fig d qurter disk si d. The itegrd is odd fuctio d the itervl of itegrtio is smmetric bout. s ds shded re t re of circle re see # below + P s Fig. 5.. s Q O. u5 u + du du 6. Fig d re i figure bove re sector POQ re trigle PO 6 Fig Let +. d shded re. u d 6 d d 6 6 d 7 d Fig t dt t dt.. e e d odd fuctio, smmetric itervl + t 9 t dt 9 t dt t 9 t dt.. v v dv + d

13 SECTION 5. PGE 96.. DMS: CLCULUS si d d + 6 si d d / d l t dt t dt t dt l l l/ l / dt t t dt l l Fig verge vlue / / s ds l l l 6. / s ds / s ds s ds. sg d l l l + l l sg 7. verge + d [ ] + 8. verge b b b + d [ ] b + b + + b b verge + si t dt [ ] dt + si tdt [ + ]. verge d. Let The f Fig. 5.. { + if < if. f d re + re re +. +,. verge vlue / d shded re 5. g d d + Fig. 5.. d

14 INSTUCTO S SOLUTIONS MNUL SECTION 5. PGE I d d + d sg d re re. re 6. re re +. Therefore I / d re re / Fig Fig [] d shded re d re re Fig verge + sg d + d + d +., Fig , Fig

15 SECTION 5. PGE 96.. DMS: CLCULUS b b b b f f d f d fd f k d b b b f f b f d k b d b f b f b f d + k d f d kb f + k b b k f + b This is miimum if k f. f d b f Sectio 5.5 The Fudmetl Theorem of Clculus pge / d d d / 6 / 6 d d d 9/8 8 + d d d / 9 [ 9/ ] [ 9 / ] /6 / / / / cos d si /6 / + / sec θ dθ t θ t si θ dθ cos θ / / + si u du u cos u e d e e e. e e d odd fuctio, smmetric itervl 5. e d e l e l 6. d l l l l 7. d + t 8. / / d si 6 d 9. si si si 6 6. d + t t 8. re d 5 5 sq. uits. 5 Fig. 5.5., 9

16 INSTUCTO S SOLUTIONS MNUL SECTION 5.5 PGE e e. re d l e e l e l e sq. uits. 5. For itersectio of + d, we hve + +. Thus or. The idicted regio hs re Fig re e e. re d 6 sq. uits. re + d + [ ] + + sq. uits. 6 Fig Sice 5 5 +, therefore t 5 d, d > if 5 < <. Thus, the re is 5 d 5 d sq. uits. Fig Sice d itersect where, tht is, t d, thus, re / d 6 6 d sq. uits., / 5 Fig re Fig re shded re d sq. uits. 9

17 SECTION 5.5 PGE.. DMS: CLCULUS, e Fig The two grphs itersect t ±,, thus re d d sq. uits. Fig re cos d si sq. uits. cos re Fig ,. re 7 Fig / d / 7 7/ sq. uits. 9. re / d / d / / sq. uits. / re 7 / / Fig re e d e, e sq. uits... / / Fig cos d cos d / si si + + sg d d + / / / / d cos d 9

18 INSTUCTO S SOLUTIONS MNUL SECTION 5.5 PGE 5. verge vlue d verge vlue e d e e6 e vg. / l d l / l l 8. Sice gt {, if t,, if < t, the verge vlue of gt over [,] is [ dt + / l ] dt [ ] + t. d si t dt si d t d si d d [ t ] si d si t dt t dt t d si t dt d si t dt d t d t si si d d si u u si u u si u u si u u du du + d si u du d u [ si du + ] du + si d t cos cos t d dt + + t e. d cos θ dθ si θ d dθ d [ cos θ d si θ si θ cos θ cos θ si θ si θ csc θ sec θ cos θ t 5. Ft cos d F cosu du d d F cos cos 6. H H e t dt e t dt + e H e t dt + e + e 7. f + f t dt f f f Ce f Ce C e f e. e ] d 8. f f t dt f f f Ce f C f e. 9. The fuctio / is ot defied d therefore ot cotiuous t, so the Fudmetl Theorem of Clculus cot be pplied to it o the itervl [, ]. Sice / > wherever it is defied, we would epect d to be positive if it eists t ll which it does t. si t 5. If F 7 + t dt, the F si + d F7. 5. F cos + t dt. Note tht < for ll t, d hece + t < cos cos + t. 9

19 SECTION 5.5 PGE.. DMS: CLCULUS The itegrd is cotiuous for ll t, so F is defied d differetible for ll. Sice lim ±, therefore lim ± F. Now F cos + + ol t. Therefore F must hve mimum vlue t, d o miimum vlue. [ 5. lim ] re below 5, bove, betwee d 5 d lim si + si + +si lim sum of res of rectgles show i figure si d cos 5. lim + + lim lim... si re below Fig , bove, + betwee d + d t Fig Sectio 5.6 The Method of Substitutio pge 8 e 5 d Let u 5 du d e u du eu + C e5 + C. cos + b d Let u + b du d cos udu si u + C si + b + C. + d Let u + du d u / du 9 u/ + C 9 + / + C. e sie d Let u e du e d si udu cos u + C cose + C. d + 5 Let u + du 8 d u 5 du 8 u + C + + C. si d Let u du d si udu cos u + C cos + C. 9

20 INSTUCTO S SOLUTIONS MNUL SECTION 5.6 PGE e d Let u du d e u du eu + C e + C. + d Let u + du d u du u l + C + l + C. cos + si d Let u si du cos d du + u t u + C t si + C. sec t d du u si u + C Let u t du sec d si t + C. e + e d e / + e / e / e / d Let u e/ e / du e / + e / d du u l u +C le / e / e + C l + e + C. l t t dt Let u l t du dt t udu u + C l t + C. ds 5s du 5 u Let u 5s du 5 ds 5 u/ + C 5 5s + C d Let u + + u du u + C du + d C t t dt du u Let u t du t dt si u + C si t d Let u + 6 du d du + u t t + C. u + C. + C d e e + d + e Let u + e du e d du u l u +C l + e + C. d e + e e d e Let u e + du e d du u + t u + C t e + C. t l cos d Let u l cos du t d udu u + C l cos + C. + d d + d Let u du d i the first itegrl ol du + si u + si + C u + si + C. d d + Let u + + du d du u + u t + C + t + C. d + d 5 du 5 u si u5 + C Let u du d si + C si + C

21 SECTION 5.6 PGE 8.. DMS: CLCULUS.. 5. si cos 5 d si cos 5 cos 7 d u 7 u 5 du Let u cos du si d u8 8 u6 6 + C cos8 cos6 + C. 8 6 si t cos 5 tdt si t si t cos tdt Let u si t du cos tdt u u 6 + u 8 du u5 5 u7 7 + u9 9 + C 5 si5 t 7 si7 t + 9 si9 t + C. si cos d u du Let u cos du si d [ + cos] cos d d [ + cos + cos ] d + si + si 8 + si cos d si + si + C. + C sec 5 t d Let u sec du sec t d u du u5 5 + C sec5 + C. 5 sec 6 t d sec t + t d Let u t du sec d u + u + u 6 du u + 5 u5 + 7 u7 + C t + 5 t5 + 7 t7 + C. t sec d u + C cos + C. t + t sec d u / + u 5/ du Let u t du sec d si si cos d d cos d si + C. 8 cos si 6 d d cos + cos cos d 8 si + + cos d cos si d Let u si 8 du cos d 5 si si + u du si si si + + si + C si si + + si + C u/ + u7/ 7 + C t / + 7 t 7/ + C. si / cos d Let u si du cos d u du u / 7 u7/ + C u / si / 7 si7/ + C. cos si si d Let u si du cos d cos u si udu du + cos u cos u + du u si u si u + + C 8 8 si si si + si si + C. 96

22 INSTUCTO S SOLUTIONS MNUL SECTION 5.6 PGE si l cos l d Let u sil cosl du d u u du u 6 u6 + C si l 6 si6 l + C. si cos d t sec d Let u t du sec d u du u + C t + C / d Let u +, u du d 7 u u / du 7 u/ u / 7 7 e si l / 7 7 d. Let u l du d si udu cos u / si cos d t sec d sec sec t d Let u sec du sec t d u du u u + C sec sec + C. csc 5 cot 5 d csc cot csc csc d Let u csc du csc cot d u 8 u 6 + u du.. / / cos si d / + cos cos + d / / / si 8 si + 6. / si 5 d / / cos si d Let u cos du si d u + u du u u + 5 u5 / u9 9 + u7 7 u5 5 + C 9 csc9 + 7 csc7 5 csc5 + C.. e e dt t l t Let u l t du dt t du u l u l l l. 8. cos si 8 d cot csc d cot + cot csc d u + u du u5 5 u7 7 + C 5 cot5 7 cot7 + C. Let u cot du csc d. /9 si cos d Let u si /6 du cos d / / u du u / l / l / /. 97

23 SECTION 5.6 PGE 8.. DMS: CLCULUS 5. / / / + cos d cos d / / si d / / 6. re cos d si /. cos d Let u cos udu si u du re du d cos u / + 6 d Let u + 6 du d du 6 u l u 6 l l 6 5 l sq. uits. d + 6 Let u du d du u + 6 u 8 t +6 sq. uits. Fig The re bouded b the ellipse / + /b is b d b u du. Let u d du The itegrl is the re of qurter circle of rdius. Hece re b b sq. uits. 9. We strt with the dditio formuls cos + cos cos si si cos cos cos + si si d tke hlf their sum d hlf their differece to obti cos cos cos + + cos si si cos cos +. Similrl, tkig hlf the sum of the formuls we obti si + si cos + cos si si si cos cos si, si cos si + + si. 5. We hve cos cos b d [cos b + cos + b] d cos[ b] d + cos[ + b] d Let u b, du b d i the first itegrl; let v + b, dv + b d i the secod itegrl. cos udu+ cos v dv b + b [ ] si[ b] si[ + b] + + C. b + b si si b d [cos b cos + b] d [ ] si[ b] si[ + b] + C. b + b si cos b d [si + b + si b] d [ si[ + b] d + si[ b] d] [ ] cos[ + b] cos[ b] + + C. + b b 98

24 INSTUCTO S SOLUTIONS MNUL SECTION 5.7 PGE 5. If m d re itegers, d m, the { cos m cos si m si } d cosm ± cosm + sim ± m ±. si m cos d sim + m + d sim + + sim d cosm + cosm + m + m b periodicit. If m the si m cos m d si m d cos m m b periodicit. 5. If m k, wehve f cos m d + + cos m d k cos cos m d k b si cos m d. B the previous eercise, ll the itegrls o the right side re zero ecept the oe i the first sum hvig m. Thus the whole right side reduces to Thus m cos m d m m + cosm m + m. similr rgumet shows tht b m f cos m d. f si m d. d For m wehve f cos m d + f d d k cos + b si d + +, so the formul for m holds for m lso. Sectio 5.7 pge res of Ple egios. re of d sq. uits. 6 Fig. 5.7.,. re of d / sq. uits. Fig. 5.7.,. re of 8 d 6 6 sq. uits. 99

25 SECTION 5.7 PGE.. DMS: CLCULUS 6 +6 Fig Fig For itersectios: 6 8 i.e., or. [ ] re of 6 d 8 d 6 6 sq. uits.,8 6. For itersectios: i.e., or. re of + d + [7 + + ] d 9, + 9 sq. uits. 7 6, Fig Fig For itersectios: } re of d sq. uits. Thus itersectios of the curves occur t d 6. We hve 6 re of + d sq. uits. 6, Fig ,

26 INSTUCTO S SOLUTIONS MNUL SECTION 5.7 PGE 8. Shded re d sq. uits.,,, Fig re of d / 5 sq. uits. Fig For itersectios: 5. Thus 5 +, i.e.,. The grphs itersect t / d. Thus 5 re of d / 5 l / 5 l sq. uits. 8,, / +5, Fig For itersectios: + i.e., or. re of [ ] d [ + ] d + 9 sq. uits.. re of shded regio Fig [ ] d d 5 5 sq. uits. 5 Fig. 5.7.

27 SECTION 5.7 PGE.. DMS: CLCULUS. The curves d itersect t ±. + Thus re of + d t 6 sq. uits. 5 + Fig re si d cos + sq. uits. Fig si. For itersectios: + + i.e., or. [ ] Shded re + d [ l + ] l sq. uits. +,, 7. re of 5/ / Fig si cos d cos + si 5/ / + sq. uits. / si 5/ Fig cos 5. The curves d 5 itersect where 5 +, i.e., where. Thus the itersectios re t ± d ±. We hve re of 5 d 5 + sq. uits. 8. re / / / Fig si d + cos d / sq. uits. + si

28 INSTUCTO S SOLUTIONS MNUL SECTION 5.7 PGE t, si Fig Fig re / si si d cos + si cos / sq. uits.. For itersectios: / t /. Thus ±. re / t d / sec l 8 l l sq. uits. si si Fig / t /. re / / cos si d cos d si si cos / sq. uits. Fig For itersectios: sec. Thus ±/. re / sec d l sec + t / l + sq. uits. Fig For itersectios: t or. / re t d / l sec 8 l sq. uits. sec Fig. 5.7.

29 SECTION 5.7 PGE.. DMS: CLCULUS. For itersectios: cos /. Thus ±. re cos d 8 si 8 sq. uits. cos 7. re of d d Let u du d u / du u/ sq. uits. Fig Fig For itersectios: si /. Thus ±. re si d cos sq. uits. si Fig For itersectios: e +. There re two roots, both of which must be foud umericll. We used TI-85 solve routie to get.86 d.69. Thus re + e d + e.999 sq. uits. 8. Loop re + d Let u + udu d u uu du u 6 u + u du 7 u7 5 u5 + u 56 sq. uits. 5 + Fig The tget lie to e t is e e, or e. Thus + e Fig re of e e d e e e sq. uits.

30 INSTUCTO S SOLUTIONS MNUL EVIEW EXECISES 5 PGE e. i + i/, i,,,...,, i /. e Fig ,e. The tget lie to t, is, or. The itersectios of d this tget lie occur where +. Of course is double root of this cubic equtio, which therefore fctors to +. The other itersectio is t. Thus re of + d , sq. uits. f d lim lim lim lim i i + [ + i + i [ + ] i / + i/ is iem sum for f + o the itervl [, ]. Thus lim + d + /. + i ] + si d si d 5 d / of the re of circle of rdius 5, 8 Fig d re re. eview Eercises 5 pge j j + j + j + j j j + j + j j + j + j j + j j The umber of blls is ii ,5. 8. Fig cos d re re 9. f. h cos Fig si d [ ] d vi #9 5

31 EVIEW EXECISES 5 PGE.. DMS: CLCULUS. f t. f. gs. gθ t si s si d, f t sit + t dt, f + si cos e si u du, e cos θ e si θ l d g s e sis g θ le cos θ e cos θ si θ le si θ e si θ cos θ si θ cos θe cos θ + e si θ 5. f + f t dt f f f Ce / 6. I f + f Ce / C e/ f e/. fsi d Let u d du u f si u du but si u si u f si u du f si d I. Now, solvig for I, we get fsi d I ufsi u du f si d. 9. d itersect where, tht is, t d. Sice o[, ], the required re is d 5 5 sq. uits.. d meet where +, tht is, t d. Sice o[, ], the required re is d sq. uits.. si d cos itersect t /6, but owhere else i the itervl [,/6]. The re betwee the curves i tht itervl is /6 cos si d si + cos /6 + sq. uits... 5 d / meet where 5 /, tht is, where 5 +. There re four itersectios: ±d ±. B smmetr see the figure the totl re bouded b the curves is 5 d 5 + sq. uits d itersect where +, tht is, where +, mel t d. Sice + o[, ], the required re is 5 + d + 9 sq. uits.. 8. The re bouded b,, d is d sq. uits... Fig. -5. cos + d Let u + du 6 d cos udu si u + C si + + C

32 INSTUCTO S SOLUTIONS MNUL EVIEW EXECISES 5 PGE e l d Let u l du d/ udu u 9t + t dt t 9 + t dt Let u 9 + t du t dt 5 5 udu 9 u/ 98 9 si d u du u u l si cos d Let u cos du si d + C cos cos + C e u du Let v eu + eu dv e u du dv + v v t 8 t e t l / d t udu / t u u Let u l du / d / sec u du si s + ds Let u s + s + du ds/ s + si udu cos u + C cos s + + C. F cos t t 5 si 5 dt si t 5 dt cos t dt 8 5 t 5 t si + C t dt. Sice / + t > for ll t, F will be miimum whe is miimum, tht is, whe. The miimum vlue is F dt + t t t. F hs o mimum vlue; F </ for ll, but F / if, which hppes s ±.. f if, d f < otherwise. If < b, the b f d will be mimum if [, b] [, ]; etedig the itervl to the left of or to the right of will itroduce egtive cotributios to the itegrl. The mimum vlue is d.. The verge vlue of vt d/dt over [t, t ]is t d t t dt dt t t t t t t v v. t t t t. If t is the distce the object flls i t secods from its relese time, the t g,, d. tidifferetitig twice d usig the iitil coditios leds to t gt. The verge height durig the time itervl [, T ]is T T gt dt g T T gt T Let f + b + c + d so tht f d + b + c + d. We wt this itegrl to be f + f / for ll choices of, b, c, d d. Thus we require tht + + b + + c + + d f d + b + c + d. 7

33 EVIEW EXECISES 5 PGE.. DMS: CLCULUS It follows tht d must stisf t first glce this sstem m seem overdetermied; there re three equtios i ol two ukows. However, the do dmit solutio s we ow show. Squrig equtio d subtrctig equtio we get /. Subtrctig this ltter equtio from equtio the gives /, so tht / the positive squre root sice we wt <. ddig d subtrctig this equtio d equtio the produces the vlues + / d /. These vlues lso stisf equtio sice Chllegig Problems 5 pge 5. i i/, i, f / o [, ]. Sice f is decresig, f is lrgest t the left edpoit d smllest t the right edpoit of itervl [ i /, i/ ]ofthe prtitio. Thus U f, P L f, P i / i/ i / / / i/ i/ i / / / U f, P /. Now, b l Hôpitl s rule, / lim / lim / [ ] / l / lim / l. Thus lim U f, P lim L f, P l s.. cos j + t cos j t cos jt cos t si jt si t cos jt cos t si jt si t si jt si t. Therefore, we obti telescopig sum: si jt [ ] si t cos j + t cos j t [ ] si t cos + t cos t [ ] si t cos t cos + t. b Let P {,,,,... } be the prtitio of [,/] ito subitervls of equl legth /. Usig t / i the formul obtied i prt, we get / si d lim si j lim cos si/ / lim si/ lim cos cos.. si j + t si j t + cos cos + cos si jt cos t + cos jt si t si jt cos t + cos jt si t cos jt si t. Therefore, we obti telescopig sum: cos jt [ ] si t si j + t si j t [ ] si t si + t si t. 8

34 INSTUCTO S SOLUTIONS MNUL CHLLENGING POBLEMS 5 PGE 5 b Let P {,,,,... } be the prtitio of [,/] ito subitervls of equl legth /. Usig t / i the formul obtied i prt, we get / lim cos d cos j + lim si si si/6 6 6 /6 lim si/6 lim + si si 6 6 si si.. f /, < < < <. If c i i i, the so i < c i < i. We hve Thus f c i i i < i c i < i, d lim i i i i telescopig i i. f c i i. 5. We wt to prove tht for ech positive iteger k, j k k+ k + + k + P k, where P k is polomil of degree t most k. First check the cse k : j P, where P certil hs degree. Now ssume tht the formul bove holds for k,,,...,m. We will show tht it lso holds for k m +. To this ed, sum the the formul j+ m+ j m+ m+ j m+ m + m + + j m + + obtied b the Biomil Theorem for j,,...,. The left side telescopes, d we get + m+ m+ m + + m + m + j m+ j m + +. Epdig the biomil power o the left d usig the iductio hpothesis o the other terms we get m+ + m + m+ + m + + j m+ m + m + m+ m + +, where the represet terms of degree m or lower i the vrible. Solvig for the remiig sum, we get j m+ m + m+ + m + m+ + m + m+ m+ m + + m+ + so tht the formul is lso correct for k m +. Hece it is true for ll positive itegers k b iductio. b Usig the techique of Emple i Sectio 6. d the result bove, k d lim j k+ lim k+ j k k+ lim k P k k+ k+ k Let f + b + c + d. We used Mple to clculte the followig: The tget to f t P p, f p hs equtio g p + bp + cp+ d + p + bp+ c p. 9

35 CHLLENGING POBLEMS 5 PGE 5.. DMS: CLCULUS This lie itersects f t p double root d t q, where q p + b. Similrl, the tget to f t q hs equtio h q + bq + cq + d + q + bq + c q, d itersects f t q double root d r, where r q + b p + b. The re betwee f d the tget lie t P is the bsolute vlue of q p f g d 8 p + 8 bp + 5 b p + b p + b The re betwee f d the tget lie t Q q, f q is the bsolute vlue of r q f h d 8 p + 8 bp + 5 b p + b p + b which is 6 times the re betwee f d the tget t P. 7. We cotiue with the clcultios begu i the previous problem. P d Q re s the were i tht problem, but r, f r is ow the iflectio poit of f, give b f r. Mple gives r b.., which is 6/7 of the re betwee the curve d its tget t P. This leves /7 of tht re to lie betwee the curve, Q, d the tget, so Q divides the re betwee f d its tget t P i the rtio 6/. 8. Let f + b + c + d + e. The tget to f t P p, f p hs equtio g p +bp +cp +dp+e+p +bp +cp+d p, d itersects f t p double root d t the two poits p b ± b c bp 8 p. If these ltter two poits coicide, the the tget is double tget. This hppes if 8 p + bp + c b, which hs two solutios, which we tke to be p d q: p b + b 8c q b b 8c p b. Both roots eist d re distict provided b > 8c. The poit T correspods to t p + q/ b/. The tget to f t t hs equtio h b 56 + b c 6 bd +e+ b 8 bc + d d it itersects f t the poits U d V with -coordites u b b 8c, v b + b 8c. + b Sice p r b + p d r q b + p hve the sme sig, must lie betwee Q d P o the curve f. The lie Q hs rther complicted equtio k, which we wo t reproduce here, but the re betwee this lie d the curve f is the bsolute vlue of q r f k d, which Mple evlutes to be 8 8 p + 8 bp + 5 b p + b p + b, U P T Fig. C-5.8 B Q S V

36 INSTUCTO S SOLUTIONS MNUL CHLLENGING POBLEMS 5 PGE 5 The res betwee the curve f d the lies PQ d UV re, respectivel, the bsolute vlues of q p f g d d v u h f d. Mple clcultes these two itegrls d simplifies the rtio / to be /. b The two iflectio poits d B of f hve - coordites show b Mple to be α b b 8c d β b + b 8c. It the determies the four poits of itersectio of the lie k through these iflectio poits d the curve. The other two poits hve -coordites r b 5b 8c s b + 5b 8c. d The regio bouded b S d the curve f is divided ito three prts b d B. The res of these three regios re the bsolute vlues of α r β α s β k f d f k d k f d. The epressios clculted b Mple for k d for these three res re ver complicted, but Mple simplifies the rtios / d / to d respectivel, s ws to be show.