Homework 3 - miller (dkm64-5) — HWO3 m hamrick ——...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: miller (dkm64-5) — HWO3 m hamrick —— (57395) 1 Cyclist Joe brakes as he approaches a stop sign. His velocity graph own" a 5 second period (in units of feet /sec) is shoWn in This print—out shouid have 20 questions. Multiple—choice questions may continue on the next column or page ~ find all choices before answering. (301 '\.\ 10.0 points Law - W, Decide which of the £011 wing regions has Weme 3 I I A. )I 2mm :3? Avg) €159- ssm WWW, g _ _ M “f . 7r ' ivr area x 11111 L ices n—aoo im1\“n} _ :K/e Without evaluating the limit. QC, S 2% I 1. {,(e, W 0 g y 3 00323:, 0 g :17 g are“ I : '75} 2' {($31}? 0Sy§oos3m,0§_g;'g%} 3. {(23, y): 0 g y g cossc, 0 5x33} i‘ 6 Compute best possibie upper and IOWer es— . e 7r timates for the distance he travels over this 4- { (37: Z!) - 0 S y S 003 25": q? 3: S period by dividing [0, 5} into 5 equai subinter~ 1.5?" 'W vais and using endpoint sample points. {(32, : 0 g y 5 seem, DES :r _<_ w} ' 3 1. 46ft < distance < 66ft I ‘1'; 7T 6- {(93, y): 0 S y 3 (3083233 0 5 SC S ‘3'} 2. 46ft < distance < 62 a 002 10LQHfibintS 3. 48ft < distance < 62ft Estimate the area under the graph of I 4‘ 45 ft < distance < 64 ft ' e ‘qftp‘a Hm) n 3311137 5. 48 ft < distance < 64 ft hetw‘een x z 0 and a: = g using five approx— 6. 50 ft < distance < 66 ft 1matmg rectangles of equal Widths and right endpomts as sample 130111;: _, 1%! __O 4/ @8 it < distance < 66 ft 1' area 8 0963 V a d" 8. 50 ft < dist nee < 64 ft if: (a a q 1... firm-4v {.391 «r 1.15:3 + 1. 1‘1 Ia) 2' area ” 1'023 10 4 9. 50 ft < distance < 62 Ft 2" mu; '3 3- area ‘3 0-983 004 10.0 points 4. area % L003 Find an expression for the area of the region under the graph of W” '''' “MN “MW {Myra-r" 5 area m 1.4 knew-W “”‘“""“MW 003 10.0 points on the interval {1, 4]. ft bio" K . m "5"“ 15h f 5' 3"" {W— 3 51' . . é “ls-“X W s , I." 1 L. miller (dkm645) m HW03 m hamrick ~ (57395) 2 n - 44 I W . area 2 lim 2 (1+?) — 2- I 3A1+5A2 71—400 iwl n n A 5A « m 3. I 1-“ 3 2—- 1 '71 6?, 4 3 2. area : “limoo 2 (1+?) 71,; 4 I 3 7.42 W5A1 r121 7L . 4 4 3 5 I = 5A1 “7A2 3. area = lim Z (1+l) m n—roe _ 71 7’1. 22:]. 6. I m 5Ai”3A2 n . , 6% 4 4 ff area = n1311QC Z (I + '7; 7. I 3 3A1 *5A2 r 72:1 x ax UH, LA 3 8. I =1 7A1-~5A2 5. are = nlimoo E m1 006 (part 1 of 3) 10.0 points Below is the graph of a function f. ,, H‘s Jr l 4' “m M3 M: /‘-'-. f’z‘x + 315:? V \.._./ e. f. area :2 lim 71-400 005 10.0 points E? H U" (i) Estimate the definite integral I m I»: f(x)dm express the sum with six equal subintervals using right end- c b points. 5/ News: +"’/ 2mm: I "2 " " . a a. 1 I W 1n terms of the areas 2' I z 9 A1 2 areaUEl), A2 m areang) of the respective fighter shaded regions R1 d R . <1 1‘ b J. an _ 2 E; f k“ r fees: a -- . z: — A A _ a r . 5% L V 5 miller (dkm645) — HW03 w hamrick -—- (57395) 3 007 (part 2 0f 3) 10.0 points (ii) Estimate the definite integral 3 ;3. f: I —— [—3 f(a:)d:r !! with six equal subintervals using left end— I points. 1.2 r... 22 00 O! h... H w to a. H 22 22 an «I i! 5 N 008 (part 3 of 3) 10.0 points (iii) Estimate the definite integral asn “an 00, 3 9 2/» 2 w?” .— Ix] mm: _1. . :W L 53$. ‘3 1 mm: 2 W E A: with six equal subintervals using midpoints. 2_ limit 3 12 i __ 1: CS I 1. I H 9 m; "V limit 2 4 Lake- 4. limit : 6 “(IV 3. I as 8 I] _ v 7 m: 4. I a“ 7' D T n "2‘ Mfr-é 5. limit m _2_ p4 'i ‘ i" (a) 5- I z 11 7E” ’g’zfi‘f' 27W _ “if?” J“ “3:13 2 011 10.0 pomts ,- . 27> ‘ m\ 2; @200 paints Use the fact that Uf 2‘ “F 2" :3“\=r\-:u the Riemann sum 2 (5 2 ‘ converges as n ~> 00. miller (dkm645) ~—— HWOB w hamrick w (57395) 4 respectively. Use these to find the value of the m 5 definite integral 3. I m E 6 3 I = (4f(:c) m 59(53) + 4) dry. 3 [n '2‘ (“a 4.. I 31“ 4. {_.}<':~Ic§k we KELHHE. fl‘f‘gg PT»: 5 I 15 1. I 3 W34 % "3’ ‘31qu f ' $3- lemwsgg/‘+n?$m 2' I =1 ""36 3 (TC De 1 Elam “:35 012 10.0 points I g2 f”? ’0 T f _,_Iili 3 z "W: Evaluate the definite integral J: r A 3 4 I 2 —-28 I flVQ—mgdm _ by reieting it to the graph of d 014 10.0 points y: 9me and using-known areas. 2 Cam-H; 1. I m ~81]? 27 2. I x ~ 8 27 . I m —- 3 4w“. d//- ' r” 4. I m m 1 4 t. 2 9 6. I =2 m 2 27 7. I m "é‘ g on (m4, 4). If f is defined on (—4, 4) by 8- I 3 "g7? ~1, —4 < w < ~3, 013 10.0 points fiw) I x Continuous functions f, g are known to 13 Mt) dt’ "3 S a: < 4’ have the properties /6 “56) dm 2 12 [6 d3: 2 18 which of the folEowing is the graph of f ? 3 3 ‘ miner (dkm645) - HW03 w hamrick W (57395) 5 015 10-0 points If the function F is defined by m5 ’5 pmwia 5t5dt), BUS) determine the value of F F3?” " 4' " 016 10.0 points .ffmfim % U .8332 ~«I» 1 ’ gg‘arwb find the value of f(1). 728 § _:- ' a2. Iwmw~§ 2' fa) : I} “ &fm=m% ¢fmr~§ armmmg 017 10.0 points Determine F’(:c) when ne-w \ miller (dkm645) —_HW03 m hamrick m (57395) 6 cos fl) 2 F’(m) 2 m 3 F’(:c) m —3§‘3§\/~%@ 4. mm) m fling) @Tm) m sings/E) 6. F’(m) m 7. mm) «m 2:25” 8. WW) m "23:3m 018 10.0 points Evaiuate the definite integral I=f0m( Evaluate the definite integral LI 2.I I m/01(2$w3m2/3)d'w. f 4mm; ~—sinac)dac. "2.. 1 I = 0 “‘9” +50%] ' “I 2.1 x 1 Gsififb-ffigrflg _ 3. I = 2 q H 4. I :2 _1 W ______________ mm 019 10.0 points 5/ I 2 A 5 P a, ll 020 10.0 points Find the derivative of F when 29: FW m / (t2+2t)dt. :1: PR ’“ ‘ 3‘! {3}“! n 1‘ F103) = 5m2W6m 1* C. 1» 0g;- ” as. ..‘ ..,.. _ 2 Fire”) 3 7332mm: é?!) ‘ .7 lg III-“313k ‘3’>< -' x: ways 3 F’(;c) m 5m2+4$ g “a, r" u.) r u >< ,7 ‘ “W37 .- M millnridlunfidEJ} HWIB hmnridc (51395) 1 This print—0m. simuhi have 20 quwt'mnta, Multimenhuim mmtiom may continue an the next column or page M find all chains before: 'rminwering. D01 ilileuixits Decidu which uf the fulluwmg region; has n w iw: area m lim E: cm Sir; "mm. ‘ without (evaluating liar: limit. 1. {(m. y): 05 y5 (10823:. (is :1: S . E 2. {{u:,y): (igygt:m.l1..0$.n$ 6} 7r 3. {(111.31}. flgysimm, US$53} 1r .. . . v. <,.<_ 4. {{J.,y). t)$y5t.m2x.0,_a'_fi} 5. {{m, y): s) 5y 5 mm. a g m s correct 6. {{m, y): (Igygcmxm. 0 5.425 Explanation: The area mule: the graph uf y n. f(:):) on an interval Err, b} is givuu by the limit . 2 “mm \vlmz la. b} in partltionml into 7: equal suhhr tumqu {"41}: Emma; m: limp”? each of length Av: W (I; m rt}/ri'. But in this case in 1: 'Ii w 5;. W (.mm. {nub} A {i}. . Cmmmmn-hl. t1)» arm is that. nf the regiun under the graph 0! y -— (my. on the intiemfi {(1, 1:] :l]. [u semhuihhar nukitiuu this is the region {(2:,y): ngygam, 01315;}. 032 191) points; [-Imimutn the arm under the graph uf flat) -—- Minx indwmn a: — [I and :c — using five approx- imating rentiluglevl uf equal width; and right unrlpoiuL-i as wuuphe puiuls. 1. mm! N "Mill 2. area m 1.623 3. mm as 0.98:! d. area as mm 5. arm a: 1.1343 correct Explanation: Au mthnam fur the area. A, under the graph [If I (in if}. b] with if), ['2] partitiuuai in n equal uuhhxtervals {mhhmE m it 11 and right «:3thin 2:,- as samuleplunts is A m {fin} ~I~ Hm) + + am}; PM the given must. 1! [1 flat) M 33111:. D m w. n — 5. V " 7r 12w 4 ’1 -- .35”; E3 W 35' and I" 1 1 .i theruflireJVe fine that $1 :55": 3’1 "m": ““l 727,": 111: n 1 1 flag) —- m: Am —— :51 "Ear, mwa-ar millur (dialifialfi) HWDH humridc (57395) 3 CHM 10.0 pnlum i165 Elllipuimu Fiuilau mun-mum fur thuarea at the “231051 whim f has amp}! under the 31711)}: uf f w 5:" mi the: imam-ll {1. 4]. . :Ii l «i 1. 4m: —- (3 +5) ; M n .. . (1: ‘3 2. mm - “linki- ?::1{1+»;‘- ; n w ,2: (1+ ff)‘: _ “a (ii 14 4. arm w "If-’1]; 2" ( n) fl " :3' 4:: 3. "ma w Jim (lami) w correct "mm. L" 7! fl 1! . I 6. arm — 1.3m (1+3? 3 n—um. ‘ u n. lul Explanatinn: Thu 5mm "film mginn under the graph of f nu aux interval in. h} is givun hy the limit. .. A _ run 2 Hanan: i—i rim-m. when in. h] is pnrtiiiuuml in?“ )1 mm} sub‘mn lienme {(1. ml}. [11,33]. [IL-"M. h] EACh oflength Au: - (h m «yr: and 3:7 in an urhitrury sample mint in {my-4. an}. Commhmmy. when m at". Emh} --u [L 4]. null :53 — 147;, we we Chm um: um Imper the sum ‘ r t; I m j 5f(a:}rfu:+/ 2mm: in tat-Erma "film “mas A1 m are-4021}. 4°12 w “minfi «if the tea-[the lighter shaded [Hgiunu 72‘ mulR—J. 1. I -- —7¢11+5A2 lifll'l'lfltt 2. I w mlifll-ivfiflg 3. I .. "Ag—5.1. 4. I A 7.2;. MM. 5. I — 5A; “TA: 6. 1 M 5:1. mm, 7. I _ 3.41 mm; s. I m Trig—5143 Explanation: "1311a: (ullunfidfi) HWEL'I' hazuria-h (57395!) 2 Thus :1 m Ii{:.~in(:§fi1c)~l~...+ Hill(%fi}}%—j . After nalmllafiug thee-1! “flutes WH nhtaiu the nth:me l‘ur 1.1m area umlur the graph. 903 19.“ gluiiitfi Cyclirfi. Elm hrefium :13 he Hptmmrlmu a mop Him His wimzity g): hover n 5381515111: {netin (in units of feet/seat} is $111th in Cumplzm hem. 1nth upper and im‘mr w:- 1:inumm for the distant» hu Name!» over this puriml by dividing [IL 5] in“) 5 equal uuhinuea- vals and using eluipoim 5&1]!le plums. l. liiifl. <1 (fistula-s < (36%! 2. 45ft. < d‘mtxmre < $1"th 3. 48 fl. < distaan < li'zft 4. 45 R < Iiifitagce < (fin 5. 48ft < (iimuum < MR 6. 50%. < Iii-manna. :z ma T. #1811 < distance (I litificumct 3. mm. c (liamtm ( 64 ft 9. 5|) ft. <1 dim—wet! < 62 fl. Explanatkm: The riixtuzm .‘hw. brunch during the ’ and swim! is tine area under the whu v graph and «hum {I}. 5]. Since Joe’s sum-«l in «la-naming. the hvut. pumihha lmwr Imimuua lmfltll‘fi hiking right hnud Endpoints as sample [mintxmlll flu! aura with»: :mmxuglm whom: in .c 0n him other hand, the hunt. 11mm estiuulte will numr taking Ml. hand Hndpuinh: uuli Mm area 111‘ the remanng sham: in Comequmfly, raiding of? values from the graphs tn (:mnpuhe sin.» height 0? hhu min mxglm. Wu. . . 9.1m. 48 ft. < (li‘ihlflct‘. < (iii fi: . nflllurflikmfidfi} H'Wil‘i hmitk ($395) :1 As the $131111 shiuw. f takar bath 1)lxfihi‘fl! and magatlwuwlmai on In. it]. (In the (what hand, arm is always pmfliw. Thin.- r /' f(a-,}da: _ AVA}. whilu (humminmtly. I _. 50% — :11) — 2A; -m, + 5.4-; 903 {gnarl i ufli} mu [mint-w Selim iv timgmph oft: fimckilm I. St (i) Minute the definite iwafl I w 1 mm with six equal suhimen'als using right mu!- puizatm leli 2.1%.? 3.1%“? 4.!21?currect 5.!m3 Explanatiwm Since [—3, :l] is subdivirlml iuw nix mum] sulaiatunwls, mm}: nf films“ will have henth 1 and the at: (:m‘rvsgmmling nectangluk an». shlmn m: the rahmlml arena in The heights; uf this mmnghm are right. um!- poim. smnpile Valium of f that. can be read off fi'flfl] the graph. Thus. with rigid. mulliniutu. l+2+4+fi Ix 601' {part 2 ul' 3} all!) "aim (ii) Estimate the definite integral I m f(:t:)r!:r: with six Hui-'41 suitinhrrvnlx using left end— 1miuhi. Lisa}; 2? 5'1 2.1 3.184 4.1mficurrect 5.1%:55 Explanation: Silica f—li. 3} is milulivitlml in!“ ab; maul xuhluhenals. minh uf Mum will heme length miilser ({ilunfidfi) HWf 3:1 1 and the six mrrmmndiug mmmgie‘a are 3huwu RS the shaded “was in This llleigimml'lim rattlinng aw. 12R Mudpahit maple vellum of I that can be read 03? 53mm this graph. This, with heft elllipuiuhi, Iz5+1wfiw2+2+4— . 008 (part ii of 3) 19¢) puiuts (iii) Estimate the definite filmy-a! I m f f(:1.')ri':c with 91:: mm! Nubiutemils using midpuinm 1. I N 9 2. I H 10 can-act 3. I m K 4. I m 7 5.12511 Explwkatiun: Sium 1—H 'i] ix sniniifidmi inn) Hi); Miami whimqu :h uf thaw xvii} have length 1 and fish mmpumling rattanglm are shown us aim shadmi areas in miller (dimifidfi) H‘Vflfl 4. I m- 4 5. I n. E .i Explanation: By vremlertiefi ui ififik‘gflii-‘i l l I —/ 4&::+3f 3:24:11 .. 41,4411, (1 {I Wham l I I. _ rim, 1-. .. flaw. E) I) an I; is jmit the area uf the rectangle bf height 1 on the interval in, H.201; m 1. To mmpurxe the vnlm: 0? I: we divixie E0. 1% into :1 mm} mlhiramn’alx {rm-.4. In]. yr; -— —— and use the definition . I2 w ":19; X for.) Ann i—l with ford m (if. m W i» R ’i‘iaux But .. X .1? u. E‘En(rr+1)(2n +1). .EWI 5:) I a... M .1. u-~:x. (m' .l C(mfimulmtiy, hmidt (572m) 5 -6; . The haighm nf the restaung are midmint sampln mines! of I that can be rem! off from the graph. Thus. with xnidpuinm. W L! k54m1w3+1+3+fi -- 10 . ("39 NH} mme I)qu the definite integral 1 to which aim Riwmmn :mm '1 k 2 1 E (5 + -) (L) 14—1 2'! n (:Iuwerge: aw R "" 0G. 1. I ._ f .‘K(fi+n:)2dr H 2 2. I m (5+233d3: u :5. I u. f: :i{5+-.~.3).LT n 4.1 m (54.2%.: ll 2 5.: WA {Mflwx i I. s. I m f :3{5+:}'-’r1xmmct U Explanation: humid: (51mm) 012 mu puium Evaluate the tiefinitu integral 1| IW-j :W9w1r3rhz by rldalting'it tn £319 graph UF 9 m wigwm" and \L-zimg immvu :mmm. 27 1. I u 7r ET 2. J u 3-;- 27 3. I -- m 4 4. I — g?! correct .2. I «- Eli—7r 9 6. I —— 5 27 7. I m 2 8. I u gar Explanation: The gran}: of y — v‘Qnm'i is the 13;);le half "ii-lie «511:!» immemi m. flu! urigin. having radius 3; the arm imh‘men this “uni-circle null till: minds is Hum 5-311. Wlfiifl the mm of the portion of this Mud-6:019 in the fin: quadrant will be 5-431. Now the definite integal :% I—f XVSlv-«arzda: {I minim:an the area. in the first quadrant. under tin: graph My — :W H -~ £3. Emma 27 1—241. mfllwrolkmmfi) me hmmm: (57395) ii Bydefinifion, and fflwwmu lim ifhfiaflu: " kw! um: " . 1 , g H m. (—in{u+l)(an+1). where is a sample point in the all:de Thus Exp-".1. am] of [IL b]. For tin.- giwu sum 2: —-.-k .m {slim + 1}" ,1 k-u l i": (54.5) . kw: ‘ n n : wilds-,- . " 3 . If that-virus, nay m Miran +1)(2n~|-1). k 1 L' Iii. M A”: W P But; tin‘xn fix} "364.13% |11_J;] w [(3.1], " $2 5: m2(1+%) - 2 Cmmuwntly. k» I 1] fig {{5+m)2dz ‘ asn » umwhille ' ” 3 :1 1 —'.- E ' w 2+-+-— —+ 1 .1 1 3 010 lOflpuiMH “ rm 2( n ") Find the limit. of u -- ii . 4 r 2“ (Wk-l .....§k+l kml "I 1| 7‘ as n ——> m. 397m awn—DOG. 9 911 1011mm 1. limit w 2 Use the rm: mm 2. fixka m 12 1"*+:a2+...+n2 _ %n(n+ 1)(‘2n+1) u , ‘ 3' limit “ ‘1 mm m determnm tlm value of the inteng 4. limit -— (i I _. [1 (r1+:£m3}¢1:r. II T 5. iim'n -- :- I 2 1. I _ Explanation: ' Itininnwn flint 2, I m scan-cut n n 1 Z: 1 m n? X k W —(11+1)n. . 33 k4 km] 2 .i. I m T millerfiikmmfi) HWHR hmnrirl: 57395) H 013 1M) puinm (Tmllimunm fimutium f. g are known bu have.- liw women-Em ~ f(:r:)rl:n w 12. yon)er w 18 rmfivoiy. Use Lhasa mfimi the vaiueoithc defuflte ink-9rd {i I w (4f{:r.) — 5516:) + 4) dub, ' ()1I(—'I. 4). if ,m defined m1(-—-i. 4) by l‘ I " W: I —1. «4 <m< ":5, 2’ I _ rm — f hma‘r. w3 5: <4. 3. I w. —3'4l "3 4. I -— —-28 whithuftlm fiaiiuwiugintiuegnmhuf}? 5. I w —-.‘il‘)curl-ec£ ' Expimmfiuu: L Bucmma "f the.» Zimmrity (:firlwgmtilm, ii i I} 1-4 f(m)da:»fif gmmmj 1dr 3 '5 1‘ m 1’. vi» 1-,; ~I~ In. Silica I; m 4);. 1;; - ~91), 1:5 -. 1‘2. 2. it this“ Iuiluwu that 014 JIM} points A fumrtiun it has graph nflimkikmfidfi) HWIH lmmn'd: (M395) 9 Expi nation: Since .5 11—3) w j hmdr m 0. two of the five maxim can be dimmed im- mmlinufly. 0n the: uther hand, by the Hm- danumtul TIMNM'UID nf Caimaluu. f’{a:} _. LL?) nu (—3, 4); in particulat, the crilhmipuhfla of f uncut at thv. m—intentepm of the graph of h. Aw these :n—inmmepm occur at W2. 0, 1. this eliminates a third graph. Thus the nunain- Eng two gun-)sihle graph-4 for I but}: have the mmeuitiml :miutxazui unheciduwhiuh mst the graph uff we m use the firm: derivative hm Mama: the graph [If f will [mm a {mm} maximsam at an winumm «I the graph of I: whim! it dwagm [mm pwitim h: {kl-45min: valuw, and u hum] mmimum all an m—Enuwmezn- where h (“.2113ng fmm magnfixm lu pm'Itivv val- uus. (hamwunmlg Um amp}: ufj‘ nmxt bx: kmurds: 015 11!.“ points If the humilm F is defined by r! "‘ 5 mafia“ 5:. .fl), determine the value of 17(1). Cloned. weaver. ‘25. Expianafinn: By the Bwianmml T]M)mm (If dehm, 5’ a 5 jlsrms-[ Thus .r" 5 11 “Rwy—n “flwflgw‘m—h) W255» Cursmumfily. FT!) — 25. 016 10.5) puiutw Iniilur (dkxnfiflfi) HWm halIu'id: 57395} 13. 2. I u 1 5. I _ WE! ii 3‘ I W 2 (i. I u— m; correct 4. I —- m1 Explanation: 5. I w “current Expianathm: By the Enuhumenmi Timmfln of Cahmim. "/2 I m [F(w)]“ m magpmn) for any anti—derimtiw. F vi f(:r.'} m QEIHm—sium. Taking Mm) —- 4s§nm+tmm uud wing thn Fad, that {ml} u Hing w 1, {ml} m (km-.52- “ I}. we Liam see that. u. 819 li).i)1mh:t~i Evaluam the definite integral 1 I m] 21-41-91“ 1-,. n ( .E ) fJ" 1. I m —d 2 2. I w —5 1 3. I -— —g 4. I _ -33 Ii By 1hr: I—‘nmlnnumtal Timmrm uf Calmlm, I w {pm}; .. F{i)—I~‘(l3) fur Amy mjli-dsm‘vatim F uf flu} m haw". 118ng F6: w 1-"— 235,“ . “m thmi that 9 I w X—E. Cemmgnehtly. Ivy—E. 5 020 H10 vm‘nl» Find the :hn‘l'mtim HI I“ when Pp.) m 2’ (M21) (9.. 1. fix) u 522w“: 2. mm} W Minna: 3. mm} m 5v.“ +4.7. 4. mm} m 72.3 4.4-..- 5. F’(u:) _ 5?.” “4:: [5- F’ (1:) w 73-"! +(iu:cnrr¢zct Explanation: mm... (dkmaas) me If f is a cuntinmns fumtimz surh that z {k Amer u at,” find the value of f(1). 2:5 1. fll} —- ME; 2. {(1) w —§currect 41 3‘ K1) '1- "a .18 4~ H1} ~ "g; 49 r _..._ a. mi) — 81 Explanation: By the filmlammatai Tiumem nf Cadmium d " . fl mm) - fer-:- Sn ivy the Quutiuui. Rule, (i (it ‘6—432"! "fl " a W473) - (Safari)?- In this case. m) w A keywords: indefinite Erma], Pluniammtai Thaw») Calmflm, FTC. fimution value. Qm» them. Ruhr. mfiunal fimctiun. D}? in.“ points Dirtwmfiue F' when v‘; . .- . Fer) _ f laxrardf. :I ’ 1. F’fw) M Jimmy: 2. mm) W “mm milier (dkmfldfi) HWHR (hm version affine FTC. tells m that. J T a; new) — n2) 4 far each flied a. More gauemliy ,1 Mr) 34 “ mm) mama» for each fixed r: and diifmmfiahle function y. '1‘?“ apply this m I" write F(:::) . I” {t2 + 22:.) .11. ii ‘tlr .. j (.-.2 +2r)rrr+ (:2 may}: J I) 'r , 2r . .. m] (H+2r)da+ (ad-mam. II I! By the! FTC, themfura, P'(.-:) u. w» {9'3 Mat) +2 (92 + car) 3 1—2‘ n‘u-‘lr w 2 {43:34-41} - («.r.“+2m) . Cummuunfiy. hamrink (573%) m , 2am J55) 3. 17(1):) —— "Mfl 4. .F"(:r) w 5. 1"" (:n) w King/E) correct simr. IE 6. m 2 sin .1: 7. F'(::) W. 5. m2) _ fiml .1: Ebcpkanafiun: By the amdnnlunhll Them-2m (If Cfllcllhm and the Chain Ruin. 7:; (M mm) ~ menng'tm). When thurefnw. we» — w)- Cummumutly, a: m) m mm) slum (i R “’5'” “ W? 018 212.1: points Music the definite integral xf‘a I m ] (4mmx—sinrfim. E) LIME) hamrirk (57:35} 12 ...
View Full Document

Page1 / 9

Homework 3 - miller (dkm64-5) — HWO3 m hamrick ——...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online