MATH_122_200702_solution_key

MATH_122_200702_solution_key - Final Exam MATH 122 Winter,...

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Unformatted text preview: Final Exam MATH 122 Winter, 2008 Pg 1 Pg 2 Pg 3 Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Total llpts llpts llpts llpts llpts llpts llpts llpts llpts lOO Name Section Show all your work on the exam paper, legibly and in detail, to receive full credit. N0 Calculators. la)(4 pts)Given the filnction f(x)= I0x(sin(t3)+4)dt,find f(0) and f'(0). 10m : S:(s/n/£J)+l7’)d-é = @ 1b) (4 ptS)Find the antidefivative: (Hint: the denominator does not factor). x —— x L 6L”; 6X +~éx+tf dot: é><+—é= .4690 J [KM o/X ,_ S é/XH) dx 9 i J“, 2 6 L 7L —/ BXLf—éx 7“! 3X +676 ‘7’ in /u/ + c ; anfix‘wxw) 2Q, .- lc)(3 pts) Find the antiderivative: Ixex dx M 5’ 5* ’— ~—v— 2a)(4 pts)Find the antiderivative: I x2 —3x—4 L 49 7:6 X "5X'4 X~€D 60/) E z A + 3 fl/ I) 6& (204057914) (ref) 00.0 = X’L 7‘ '7) 2b)(4 pts)What is the form of the partial fraction guess for the following function (you do not need to evaluate the constants: K 3x4+4x3+l6x2+20x+9 Q (x+2)(x2 +3)2 LS- A + gx+c + bx+€ OWL) [XL+3\ {xiv—Qt °° de 2c)(3 ts)Evaluate the integral: 2 L, 00 [x —3x—4 g S W dx : L/M J xtrsx-"cf S x ’ L ‘300 L L (X A (j l ; (,m/l [n [x40 —~//) [X-H) / r; /4 .L ‘ Lvl/ .4. + > / 2 MM /m X 3 —/ (—— lfl L -1 Lem L411; n g 3a)(4 pts)Find the antiderivative: [sin 46 cos3 60’6 4 g g/n Lia 6057—6) 6059 4’9 : $5M a fl~5m¥> (556 : JB ff 9 M =$M 9 : 8(3/4 6 — Sm 9')[056 /& id“: Cow/fig :: §@4-a£> 0/“; = a5 __ i7“ : (5/46/5/QC/4602 S 5’“ 5’ y 9;,” 3b)(3 pts)Evaluate: Icosz(26)d6 M 3&9 MM = 2/9 d9; /: a? 0 1T L I I/LS 6051MB dm. ‘ i 3 C03 [XV/X: :COSX'SMY +fisi/xj / :0 '— u T—Zl: (osx Smx JTLX = L [LL cums/Mo?) 4% M3 Iv L D 3C)(4 pts)Set up the integral to find the volume obtained by revolving around the x-aXis the region between y=x and y=x2. YOU DO NOT NEED To EVALUATE THIS INTEGRAL. ;X f a” L X: x” é—m a =— X L ‘ x ~>< 2’0 i x(x~/) =0 l L X: 0 Xcl J x —-( /X) 6k \J 2 “n- 80 @4 l) 8” l l L bf _ g (X — K 3 4x 4a)(4 pts)Eva1uate:J: I.dx Z X : m'l—xz I) A) ~' ~ A A SM Bryn/Q— (fig—02((L X- 4b)(3 pts)Eva1uate: I020 31:)2 K ’ \‘6 A g L t g “ o (bx); + ' 2 @ K; I 44/ Q) a;/~X HIM/:— -—6/X "‘10"de i LL=L 3—5!“ K‘Lg du:_/ :../,L_’ “a, “:5 9 00 lfi/Vflégé 4c)(4 pts) Evaluate: Iolxln(x)dx (Hint: you may use the fact that x2 1n(x) : 0) CL w um g X/n/X)fl/X — 2—30 2 L5; ~ X2“ I gx MAG clx: LL?” /n/>() —“ L x ix “7450’1vgdx 2. J l 7— :: LL. /4 b): X 7 \i L. 2. gix MAO/X : \len/ LX 5/ i 51 _ 9L [0 ——l:] ’" Lima 11 2 J) 4 c» —- o - 5a)(6 pts) Transform the polar equation r(6) into a rectangular equation. 6 aracse l ZfS/flé :2 W W >< 9/ Igoévbzgcél 5b)(5 pts)Find the arclength of the polar curve r : e9, where O S 6 S 27: q D L’IT NF 2 g Lew“ 19 = r: S D LTT b 371“ » r2 (‘36 l2: s l: [e H 6 ll —4 K‘ £051: 0/6177; 6a)(5 pts)Sketch the curve r : 3 cos(36). Please use the polar plot provided. 5 " 71? Tfl LS 21'!) B h CflQD/O/b 6b)(6 pts) Sketch the curve r : 2(1— cos 6) . Please use the polar plot provided. 7a)(6 pts)Find the slope of the tangent line to the polar curve r = sin (36) when 6 : 71".9'4 . / (9 m ; F 5/116 +r5w r:§/4/36.> é —— S} g“ r [OJ f‘ 4 r/: 3(05/36) m 2 3 50> [56) 5/46 + ym/se) 609(9— 3 @0536 £059 — SinflSSDS/n/é) 5 + 52;, £2 a (*3) [TE—5&5: 7b)(5 pts)Set up the integral representing the area of the region inside the cardioid r : 2 + 2cos6 and outside the circle r = 3. YOU DO NOT NEED TO EVALUATE THE INTEGRAL. S 8a)(8 pts) A bullet of mass m, fired straight up, is slowed by the force of gravity and a drag force of air resistance kv, where k is some constant and v is the velocity. The differential equation describing the velocity as a function of time is: 111% : —(kv + mg) . In the appropriate system of units, k=l, g = 32, and rn = 1/32. Find the general solution to this differential equation expressing v as a function of l. / ON 7 A / _/—_«.__ - +~3L ~~u+x J; at ‘9 j" V 3 1 JV 3 5.. 32 0/; MW: ~32; +4, {V + l) _. -—‘ L HQ“) 3L£+C 38/5 fig («32. é 90 \/H:— be 4523; iv: be —~i 8b)(3 pts) If f (x) : limiw) : O, and a > O, the show that x~>00 dx Jew = a2 fer/(m — gm) — af(0) (Note: f(x) is some general function. Do not use a SPECIFIC function to show that this ism”) L “4X // €41 jail/>3 dX: £2300 SC) 8 f 50 fl/X L L ad 15%) (5“ / ~ 85 “‘36? 0/”in quF 701/0) 7L 4 5L affix /X -41, / _ ,ibL7LflO /’4> grew/y] I/L) 9 .- 73 h} "a fl/a + ClaIQLf/L) Z ~4X m ——4 + 4 J0 4/36) 8 3 4L 0 7%) 8 xix 4%) 47%) . . 1‘: 3k 3 , The R1emann Sum 11m 2 cos(1 + —)— represents the net s1gned area under a curve k:1 ’7 ’7 f(x) on an interval [61,19]. (9a)(6 pts) For what function and on what interval is this the Riemann Sum? [(241) 84505 [X BJX 5Q l 83Co§ (1+X\d)x b (9b)(6 pts) Find the net signed area that the Riemann Sum represents, using the fundamental theorem of calculus. 6 CO5 (la-XBGPX : 31/) £719 / o b Extra Space judv = uv — jvdu jf'<x>gdx = f(x)g(x) — jf<x>g'<x>dx Isl 2dx_x~¢'ll—x2 1 . C dx I49 —x ——2 +Earcsm(x)+ I II Zarcsin(x)+c X "El—x2 .' I dx a. dy q I a d1” a .'1+' de I—"+—“dr 1’“+—“d{9 [1. W) [\de (6”) IV: (d6) :11; sin(6) + r cos(6) 1 d— — I We fif(f(x)2 —g<x>2>dx 7r cos(6) — r sin(6) 2 d6 . n 1. ",1 11—1 .H Ism xdx:——sm (x)cos(x)+— sm (x)dx n n n 1 n71 ' "—1 n72 Icos xdxz—cos xs1nx+— cos (x)dx n n I dxz :arctan(x)+C I%=%+larctan(x)+C 1+x (1+x) 2(1+x) 2 10 ...
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MATH_122_200702_solution_key - Final Exam MATH 122 Winter,...

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