finalsol - Math 41 Autumn 2009 Final Exam — December 7...

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Unformatted text preview: Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 1 of 18 1. (9 points) Find each of the following limits, with justification. If there is an infinite limit, then explain whether it is 00 or —00. x+g (a) miiing+ sinm+1 g—r (g’gmw’fis’fls‘z L’%%/§w&) l :2 [I'M ‘ . ‘ X‘a ’W+ J Ink/Gives ’W‘é‘yué/ Lecgase "gig nuwrwfar‘ Z “Sham W/li/Q ‘fl‘e ‘Zehw’lmf‘fm‘ l5 @Wmacflj Z€r0‘ '- f r + 31L ' m) Y") fig. COSX l5; POSi’ij‘C) M‘m( 30 yrs I J msx ‘ flog I‘M x+ Tr/z : Y9 Jyzl’l shim” [E ' im (65— ) (b) $LO+ x 1 {OD ’Erm) whats. is inwermém’ée) win ex. Lef Lelof>x 0; ‘HLeiA M WW a (€3an 6» m at) : [I'M [m3 22" éwrwm’lf’p Xaai‘ {810 ‘x W") (L, L’Mpt/g rule) ‘ .. a Z 1‘ lm (e l) ina’Hirmn-e 6 ’lg’w) . M w x . = /,m 52606: (i, L’ Qp'rtn/érule.) HZ +X€ ~ II'M “62(871! '9‘0‘ A a,“ D ,. xeot ><+/ 0+1 O J Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 2 of 18 2. (14 points) In each part below, use the method of your choice to find the derivative. (a) Find h’(z) if h(z) = zcos(z3) + csc2 z. ( / ~ #1 ink) * @96chng « j‘ogécosfav) + QC‘SC—E 2" :(%mS(E3))Fy1~ (I~Cos(23)+ %’(Lsm/%3)) “35) f 02““ '(“QC’EC‘fiLa ' (b) Find % if my = y””. Film? ‘93 0‘? boil/n SWQS Qua/Jigg’mrrfi'a‘fe 5 I” (Xy) 5]” X) x. =7 yjnx 1‘ x31 7 a? 9 dye-X) " flx‘finr) A X e ? - 7 " New [1% GM; 44 = t1} 12/ng 50+ +5 1,0,1 4/26 «fishermextmu we Mvs‘ll [ff (Ase/2% W! filo/J), #4 Jam m/e { FTC): i x+ 2 __ 0V 4 0/ a dx OCH—[OH ‘ g; SWdH= WWJJ‘ 0 S w 52 = KW ms PW): RW ~W. Q Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 3 of 18 3. (8 points) Suppose all we are given about the function g(:r) is that 9(2) 2 7 and g’(ac) = V 1 + x3 for all :17. (a) Use a linear approximation to estimate g(2.01) and g( 1.98); show all the steps of your reasoning. Lifie’amza’lrovx QS'I-IMr-flée Says that 369 1Q+ argofxfia) 4gp x Wk (2 . J Emma 373% W: W»; :3» 410:) 362.0064 7+3‘(0.0i)= 223 W? 2 30:93)? 7+3 (“0.602) (b) Are your estimates in part (a) too large or too small? Explain. Since SIYkJ=éfi+XZJ%{%xZ) J We 4an 3150*er Mr Q rte.) 1444 yde 6103 '75 Confirm UEWQI. 47w) ‘ijm/ol all 7% 2 tamwa MW 1%) a x22 W/ t; at the fidluajfiwfifk p93 1 ml (2,3th%; qe7w/ (xx/£7 #9 WWW/W . u 2 6O («NI/Le MIL #ygfiwd) sz Math 41, Autumn 2009 Final Exam ~— December 7, 2009 Page 4 of 18 4. (10 points) A child’s kite is maintaining a constant height of 100 feet above the ground. There is a strong wind blowing the kite away from the child, in such a way that the kite is moving horizontally at a speed of 7 ft / sec. (a) At what rate is the child releasing string when 300 feet of string have already been released? Le+ 'f. Leth army/11L Shivg Nimséiij 'n“ is, ‘Hua OpiS’I'an’e MIA/86h ’HLQ K chi/fie hypafenuse "m 11% Jfajmmf [.511 x be ’f/Li horizow’fa, LJ/S'f‘anté Wm #45:, md #49, Poin‘f‘ filmed/y Lelaw 44¢ on Jmuwf. gem fluz Prob/Pm) 31. ,— 7 fazed) I 14‘ A K By R/fhcejmg) Drfiferenharhfi wi’fla +9 fir/2e 1‘) We 62.2% 2 9X? ) 50 iii/0+ 0'3 X 69% A» Fafic a+‘ 2 etc When E5300)’H10V\ x: wieoofwfiioof = MOW: mi?) 30 dg’ipflg. _7—.§_ Vii—2,437 J?” 300 '7‘ ’3‘“[? 5111‘ (b) At what rate is the angle between the string and the horizontal decreasing at this same moment? (Assume the child has negligible height herself.) gm fire 4%va Wthe 9 ’Hw 03/6. in W05) we lave dam, 5: Log— . 77mg} Hpk’r aflfiremtmjj we seczé) r ~J'QQ , Jx X2‘ :75) .X. Se M 3295 4632,073ny No’ia C03 5 1;; so ’MI-‘l‘f in ’Qc‘f ) NOW gag-’17) (3er will?" 59300) M AWE Mismn also [’20. oh’fameo/ sing: lie-q 011401 01%“thij Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 5 of 18 5. (18 points) Let f = 2:4 — 31; + 1. For this problem, we will investigate certain positive solutions to the equation f = 0. (Note: you don’t have to solve the earlier parts of this problem to solve the subsequent ones; just cite any part’s stated result if you need it.) (a) Show that 304 — 3:10 + 1 = 0 for at least one at-value in the interval [1, 2]. Explain your reasoning completely; however, you don’t have to find the exact value. Node 5—‘ l#3+/;:{<O‘J 01W}? flQ)218/3-;Z+l= H >0. EMMUy) ’P is COKhMUo‘L/S) Srmc€ “LI/S QfOIyWDI/Vlla/j 50 #fl MILE/[Zeava QFf/feg')‘ we can COM/Wit” ’{Ed’ ’Hzem is f) [,4 [I3] swat, :0. (b) Let h(a:) be a differentiable function, and suppose that Ma) 2 h(b) = 0 for two different values a and b. Show that there exists a value 0 between a and b with h’(c) = 0. SINCE h l8 [email protected]%a£/r) ‘HQ MW W06 7240mm riff/21:75 0r! #4) 113%!“1/4/ [hill]; [37* WSUEL W’US‘F Va/Vfi C [It/1 SW34 I, L ’1» A] H We): L5} : %TE 5 O , as Jess/801).. (Weém‘fa) (c) Now show that f = 3:4 — 3m + 1 = 0 for exactly one m—value in [1, 2]. (Hint: find f’ and use the result of part (b) to explain why there can’t be more than one value in this interval.) By $(A);m):0 ’per ’i’wo Va/ues a XL!) 175‘” )5 C WM a (Ami I? SUCL! f {(0)119‘ (Nah! f ,’g Smoe q loo/)wyrmia , 3 3 , ‘ BUJY £/&)~ [ix ~31 all")! +146 75 0 WA” X: Al): MW" 7%”! TI: ‘l ) 17mg) W914]? have 1% Valmas a K1: in [QR] ML since ’HpM‘HLC Value—writ: WoU/Mn/Sc have ’fo [i6 Elk/Ben I m C fx'm‘g for Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 6 of 18 (d) Based on part (c), the equation x4 — 3x + 1 = 0 has exactly one solution x in [1,2]. Use Newton’s method with initial guess :01 = 1 to produce two successive approximations to the solution, namely $2 and 333. Show all of your steps, and simplify your answers as much as you can ' , , ,. PM By [UBW‘l’OMS Mimi) Xmfl" X4 .1 Kr Xanfxm-f ‘ #9743 I fo~3 Ll’g ' anal x”: Xafl xii-27$?! : 32—— [656% leg»3 9-3~5 - x H ‘/7 _. OaA a f A Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 7 of 18 6. (10 points) A silo consists of a cylindrical tower of height h and radius 7", capped by a hemisphere of radius 7". (There is no material used for the bottom.) Thus, a formula for its surface area is A5110 2 27rrh + 27rr2. Suppose there is enough material available to build a silo with surface area equal to 500 square feet. What is the maximum possible volume of such a silo? Justify your answer. (Note: if you use the volume formulas on the reference sheet, remember that a hemisphere is half a sphere.) We (We Wat/#5500; We "f’i’uim P>O mi ‘20. ‘He ’Qrmla Or #43 V0(UW1L is V2 iwmf . C)”. ‘ ’ig-Vrnkfirlq,‘ . _ _ 2 (19134“? W“ eyL’mL'bI/r] we A: M 2‘ Big ... owl QTrr 77? Com)!” 0‘4 h>O MA I)?“ 2+ Z a)" / _J I. P ’ J ‘ ‘ n F ST”. ) l-E‘ O<r~\< .250 . Thug) a ’grmda ’Qr ’H’AQ Valor/m: M +erm5' of (,1 giggle. Varij/e‘ P [S ' C ,2. 3 2 5250 3 c 523 —-J- 3‘ 0r 3771*) ’Hn's ’5 ’HM ruddy We, wish ’fo maximiza? OW» ,fim 40mm” O<r< 535-0 / \ 77—7-7“ ' We begin Cmtml FMS; V/fr) :QSO’TTF?" i3 ngver wwwmai) AWL Ills Zara wLem QSO~7rr-“=OJ [-6. when rig“; wczm worbj % m a (“madam WW h )3 fie/{hem 0M7) [Ne—$5 ’HV'S “Him! I'IWAISK’I‘ ls a‘f‘qn en/Pczfa-H] To Chemcfiing Crrbml MOWEW‘) WE Make, VII/1‘)£ #07177" < O ’19” a" .—‘ PagJ’NL YK) W4 #WS fivffi P i” ’H& dew-WM O<Kfig ‘ 17403) W Cl‘Hlel HUM-Lei” rzggé a and géssa/{fiz‘e Maxi/yam / 3600M Dewang Teri ’Q’r Absolute 5mm. Dtflei [720 Mm mfg-,3 : Jig 63,54 is] 5 simply 0t hemisphere .’ WEE) : 42?, Maximum Vo/um 'I Cuék’: ‘ Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 8 of 18 7. (6 points) Mark each statement below as true or false by circling TRUE or FALSE. N0 justification is necessary. The statement “ling—10 + 6:12) : 2” can be rigorously justified by TRUE 13—) saying that for any positive 6, we have {(—10+ 6m) —2| < 6 whenever lac — 2| < 6/2. jig—X4 “wayska is aljebmkaily eyyimkmf to 1*?! <‘ ‘79 L04” you con/fl Say “ Ix‘gkg/Q‘ WIVWW’V “‘314 ff: 0 [mews-Q 3/6 is smaller ’quM £72. - The statement “lim(—10 + 62:) = 2” can be rigorously justified by \ TRUE FALSE :13—>2 saying that for any positive 6, we have [(—10 + 6m) — 2] < 6 whenever Ix ~ 2| < 6/6. [$341 ’niese ’h/vo, heralds-'95; 4N, aijeLrQ)Ca/// swim/wt) and inc/€93! ’HLZ Swath?!“ sidemefl i5 evinc‘Hy +144 Njnroos Mart/m" or? a hmi’l’. The statement “lim (—10 + 6x) = 2” can be rigorously justified by @ FALSE :z:—>2 saying that for any positive 6, we have |(—10 + 6x) — 2| < 6 Whenever la: — 2| < 6/60. E1251 Since 8/60 < 5/6 WMer Ixaakiflzoj We lva lx"2'<€/€ “3 We") “We ’x—‘1k E/C I'S (B70Nfl’fmt’fo [(#10+€x)»-2I<E_ [A]an IF.’0+QX)mQI r." ("lafgx i 2’ /G(7<"&){ 7-” G'X”3'.} If the differentiable functions f and g($) are positive and increas— TRUE FALSE ing, then the product f must also be positive and increasing. filth”) 2 W36“) [email protected] i We a" at W WNW/37x) ave given +0 bet Post/1w) $0 7‘5 g‘éH‘Mrjwj 3 this means 1069360 is- III3(7VP0!S;I:’j; 3M4 also POSWLQ [Ninaon «(T/7‘) my! 65“) 5W. If the differentiable functions f and g(a:) are positive and concave TRUE FALSE up, then the product fg(x) must also be concave up. Lat tfoeX mo? <36d=><z+lj Lott: 0N, pos’rr’ive am! 09mm of, H’MV‘BI‘) i744; frog/act (x403 is Conan/g U20 WWW/we) since #4 game teamed”? 55 5547(8)?) WNch is né‘flcfl’ive when X: ~52) 19w emf/e. If the differentiable function f is odd, then f’ must be even. FALSE 1? x) 2 "4360) flash A’FZ‘X) fl ")0 [2y 11%I‘6n7lm‘fil3 {mail note ’flg WW— Clvam ma). M is) tax) #70) We at w Mm Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 9 of 18 8. (5 points) Verify the following indefinite integral expression by differentiating, showing your steps. (Here a and b are constants.) cam (12+b2 /e‘m sin bx d3: = (a sin bx —- b cos M) + C’ [24 6D" ‘ We, Levi 6:2 [91’3th — Eros j— ag? L1 :2 effl- panxA bros Kn ( . “ ,——— . OM . M24491 @8, ‘ (“Sn/rim“ £60be 1‘“ Bax ” $(ZS‘II4ZX‘A603LX)) g “Eh: ' ( axe“, (AS/144$" [masz + QW‘(abrcosb2< + CD‘- :’ 8 : ., 2 . . a? 51 ( azsmbx a a osz thx +£5,M_[,7<) ‘4“ (H c e -? Clark»; ‘ (“i-bl)31»1b>< Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 10 of 18 9. (9 points) The function f is differentiable. In the table below are some values of f and its derivative 1” m||.1|2 345|6 fx) 3 2 145 4 0 flow I 3 a Approximate 6 3: dm by a Riemann sum using three rectangles, with midpoints as sample . 0 pomts. flf-B/ ASO)L2€ J fiug AX; fl: 5.; W401 leg) MIJP0im£50m is Va 2 it???“ + W?"- Aw + W“) ex 2 (r Nerve» Ax < (WWWDAX 2 WWCE b Approximate 6 x dx by a Riemann sum usin six rectan les, with right endpoints as sample 0 g g points. Viz-QMOJL‘Q; +403 Axcégcgrl me J XL“: an'szc 4% (:G)() Iva Thus; #4 PijlvtAenJPofi/IF sum is Re ‘ Wx,)éx+ {MAX +~'+F{XG)A>< : +‘p(8?)+~ wimp/6])AX =2 (3+;Z+0+/+a/+5)gl : ,5 F (c) Find the exact value of f14 f’ dm. Show your steps. g in [ta/Valli [WWW .—.~ NM) = I-sfl. Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 11 of 18 10. (14 points) (a) Let f = 332. Let R be the region in the :cy-plane bounded by the curve y = f and the lines y = 0, m = 1, and a: = 4. Find the area of R by evaluating the limit of a Riemann sum that uses the Right Endpoint Rule. (That is, do not use the Fundamental Theorem of Calculus.) Ax For n Soloimlerl/a/s $970M Mam: Ax) we Law: 7 x lgxo X1 Y2. ~Ain”! AX: fl: 2 a”)! V7 VI _= e A 3" " ' X4 HzAx— [+11 Q: c¢O)l};?}_..Jt/z‘ 71 fl Rmmm SUM ’Q-n +144 PULL ls Rh: «HEJAX It] 1:] M M‘- “3 <- 3 i8 3 7 n ' 1:; Ugh/j ’Hvz SUMMIz/l’lém ’lormula; 02/! #W WWW/me {30133 M +42% 74% E16144me 50W! {5— R '2 ’3‘0'4 + aflI’H-I + a; v V}{m+l)(2n+./) V’ V‘. M1 ,2 n“ 6‘ : 3+ a VH‘ .rt LRZ'(NI)QM/) ’ 4 L 87 A m e “a a éflmlrafialel L! I’i' / f/MQlI/v) [g XZ’OVX :: [M4 is] Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 12 of 18 (b) Show that n . , 7T 7T Z7T nlzazgtan (7+ it) ‘0 1:1 (Hint: first show how to express the limit as a definite integral, then justify the value of the integral.) It /‘ - ' ’ - - lo envision ’(l’é SUM g’i’qm "g’r—Zfl‘ is A R'émw/y 50m 4Q? a QMC’FIKBM ’P) n is! 2n ‘ we Leah/x flyinj Xi: ‘gT “For [cl]? ’1 " SM‘CQ 7954447578 we must Lama 01': "1'74 am! Axsi mar! 7%6’5 '— QM ) h: arwa (C X») can, an _ ’Emzfl.’ ’1 2M L’. 9‘ VLI " .__ 47ml makes m» 5""4 We A ' is]. X £2,468) ) Whit/Li Q g“ fin ,. fl” 2 ’Peri 3 +0 /* . 714M» f5) 9 774 “II; a 0:31 whim/4;) s 42,” (1X Btfl' ’HHLS‘ M’lrzjml €71,745 28m) AEOWSQ 472V“ l5 0'4 fwcfiOM) awp/ ‘F We lime Seem 711ml My igmfiéjd Cowfimcw (rm 7% [MPH/M Eb) Lj l 50.051111: c 742‘s Wflt’iés 1%, MP --. Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 13 of 18 11. (10 points) A hybrid vehicle is traveling north from Los Angeles on Interstate 5; suppose we measure its position by the number of miles north of LA it is located. Let f (2) be the rate, in gallons per mile, at which the vehicle is consuming fuel with respect to position when the vehicle is located 2 miles north of Los Angeles. (a) What does the quantity f3500 f (z) dz represent? Express your answer in terms relevant to this situation, and make it understandable to someone who does not know any calculus; be sure to use any units that are appropriate. (0 i5 1%, 64¢“) Miami Bl) 4276/ Con/’5qu fin Sal/0‘15) MVP/6h so) Polnfi wkrnih Car is 30 miles marflx LA anal, ’ll'f P0011 WM“ 1”]!— qu l5 Wily/es nay-+11, 010 LA, (b) Let s(t) be the position, in miles north of Los Angeles, of the vehicle at time t (in hours), and suppose 3(0) = 0. Define the function 3(t) H(t) = 0 f(z) d2. What does H (25) represent? Again use simple terminology relevant to this situation, and any appropriate units. 1L6 films {We amemmL 0‘? 48d cal/wwa j“ WA WV, 441.9, P0180 e‘p ‘fllé Cm" 0140? falhé (All/WM ’Hu. car is locatvl’ t” harms fill-eh 3541' . {More myycisalf) 35 ‘HR nunlzercrpaallohssp'gel cox:wa hour; [fife {a J’GUFW‘) (c) Find an expression for H’ (t) in terms of the functions f and s (and possibly their derivatives). What are the units of H’ (t)? Cami" Rate] Z 565) (i { r 070:0 that = 3 64) We; £08 that, Id mm. ten 35%: 3g» =n%)°3’{€) (W979 q Uni’l’s Hag 0- S’LHMSS ~ It; a? "i “or . r: H Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 14 of 18 12. (14 points) Suppose that f has a positive derivative for all values of :1: and that f (1) = 0. Finally, let g<x> = [Om N) at Each of the statements about 9 below is either always true (“T”), or always false (“F”), or sometimes true and sometimes false, depending on the situation (“MAYBE”). For each part, decide which. You must give a brief justification in order to receive credit. 9 is a differentiable function of x. @ F MAYBE Simth has a Pc?3l+lv‘1 Jewva'l‘lifi? n‘l Why ’plf (I’Q’dml‘mué Mrywlmw) what means- +“ is W5 Elwyn/MN" . it: 4mm) 27 F79, 360 is 'Hlfi ambulmvafive 09 P) NIH'CI’J I'mpltesg is g is a continuous function of .73. @ F MAYBE By alcove) a )5 ) 50 5 l3 QV’liMfilieallé/ Cam’bnuous" The graph of g(x) has a horizontal tangent at m = 1. @ F MAYBE By FTC) 3’60470) SO 370:9[9‘0 g has a local maximum at cc : 1. T (F) MAYBE has Q cr‘i’lblfal flUmlOPr a‘l‘ x:l (3’30)- 8%) fire/M) 50 8/7,); 97,) '> (3 137 léejaZ/w M109- Thus) [’7 1"? Seamd [43»:va +6314) 3 has 6" LC”) Ml” 0+ 7‘: (J am; £391! a [(703] max" 9 has a local minimum at m = 1. p F MAYBE See Ollwctl‘] above . The graph of g has a point of inflection at x = 1. T @ MAYBE above) 3//{,) >0 ’l‘kufi) 3 is c/aawjlvj (mam/51%, at >6: I. The graph of %,% crosses the at—axis at a: = 1. @ F MAYBE Al Xfil) é» 17'? i3 61 ’anff‘lem I’d/HQ“ 6200/3 {em (2149/ 6.013 fOSIIfive‘fiflpli/HIW; Le. ‘P/I)£O anal, ‘P is inC’Mas/lyj; a‘l‘XcL 77,03) fire (masses +46“ X41903 (In Hm‘l' flu; graft! lye/GW’HAQY-axij Jagf [9H 6107(21') (anal above Jvil’fo‘lée Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 15 of 18 13. (23 points) Show all reasoning when solving each of the problems below. (a) Find a formula for f if f is continuous, f (0) = 1, and 3 ifx<1 I _ f(m)_{7—2m ifm>l mud" 1060=3><+C| WhIMeWr x<() {(0)21 imflfiej Cl: H’ also Mus‘l Sa’hfg/ 7)("><2’r CZ whewever- )(> (J 'Qv C2 . Bul also) Mt vewiv‘e [W +0 ex'zgx‘J Web ‘1» ’ft/H’l wiuilvj [W 4' (W / x91 x-ar mp £60 l Wehave ("Hook I’lfléxfl)‘ ll, W12 X—9( H i' .. ng «C(x) : lg? (7x—x“+—CZ): 6+CL / :0 we W02 Q2”; *2 ‘ (nee hm c X-N Lemwe we’ve inc/M 7<=Il in ’lht domain (7" $1) Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 16 of 18 3 6 d3: 4‘ (C) /e a: Ina: 61::ij C33 .3 ,. ,L . - d“- X Dix Jx 0/“ USm‘ke . F f) ” ==> XI“: = {7 gum;ch M “ M“ e 1’“ , Schema : X263 ‘9 a‘3 3 : gmflyztlu l (d) /t2e_t dt @ij L7 was r (5 M2152 [m 9,2916} dv=efil¥ v: ~e”{ i) 856915 2 ~ i? ‘1 gaggeflpg = #9201 pm _ WWW or A, 0% 2 => Sfie’oflfs +2fl;[4gt ,, ,. g/f - 4 ‘9 "Qfe + Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 17 of 18 (e) fem/E) dac . , g (/—~> : Z 651/- S()]DS’H’ILU+8 f M ‘5? x a we; 09% ea Mam (r) SSM(\R)0/xc SQMSMHOILL ’5‘ QSmS’mua/q g lgasmuc/n ‘ 3\(~-[4C03q “5,3,5!”th > ’ "chosa «,- Q Sagan/(4 ,\ -—. “20160501 +525)” £1 ’1“ C C? SSM(&)JX :(QQQCOS ‘1" 93m R—f“ Math 41, Autumn 2009 Final Exam — December 7, 2009 Page 18 of 18 14. (10 points) 1 1 (a) Show that O 3/ x4 sinxdm g 0 Firs‘f‘ hate ‘Mmi O S’ Smxé I A» a” x 3335:1941 O$x<£ . .) IQJI'SHMQ O<f<i§§J #30150 4M3 44,5], C333,;7X<(1pw OSXSI‘ MUM/36‘ij lay {’20) We oM’au‘m OSX‘I'SMx \(X‘I 4; 09(4,‘ 87 Properties In+ejm!) WP 0M Com/Ufa g g, I 0 qumxa/x g g‘qu’x O o ) i r ( ( I C C ‘l g x3 A ( Gwd nit/we USOJK 0 am! (5‘49 A 5’0‘I’D—L We have ( ) O O s on‘EWat 55L as mama / COS as (b) Suppose f’(x) = m , and let f (—) = a and f = b for constants a and I). Find the value of the integral Your answer will involve a and b. (/56 (Maj 101417 Park) with £45969 é dv Cdx V 2 X gn/z Err/2. 377/2 M We 10w) 17§Wx)ok 2 ’“ g X?’{X)ol7)< . /z. 7V2, 17/1 ) 1mm 107,956,0‘” I )P 331/2. I 317/: r X (7‘) WI W5 (03X 01K 317/7. 2 A 3m “(g ’S’Mg-T» = §E~£—~.iz~r.o<+;2 ...
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finalsol - Math 41 Autumn 2009 Final Exam — December 7...

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