Midterm Exam 1 - X) : fie/100 -MATH 2153 : Midterm I (50 pt...

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Unformatted text preview: X) : fie/100 -MATH 2153 : Midterm I (50 pt x 2 = 100 pt). Namezr V" Show your work for full credit. / 1‘2 In mdsc. (hint: use integration by parts) 9 ‘7 3 ‘ -1 J ix‘lnmix: 335 ’w alpxge My 1. (8 pt) Evaluate 1,1: /nx dv= x; l x dm 3 H I ng jww: uv—fwim 2. (8 pt) Evaluate / tan3 mdz‘. f—LLan 3xdr =j1‘éom2x - {anx Ax : f(;¢cz>< 4) {Mk d): M"; £0~HK fanx ' tanfldK Aw = 5552259): '3 fSeczx‘éamcdx ' ffam ><cb< 3. (8 pt) Evaluate 27 ————d. /\/3—2a:—w2_ m ‘ X /J3*2x-x" B—lx -x‘= 3- (mm = 3- Lmzx +1) +4 = 4-way: f‘t’ ‘I'me" U: "H' d“ 5 5"“ xru" ud\ fm M=ZS?/\(9 a UK: Zwsa (fie 2scne TL 2195949 4(Zsin9)‘: 1} 4- ‘kzn‘d 3 Wu'ss’fl‘ :’ ' ‘1‘“2‘9 : 24059 V4 "‘ (253A&)I’ b :a “s z 25“".9 "'g 5A 58 f2$1n9§19 ‘ fd9 1, 4. (8 pt) Evaluate fx—ll’rixé’K 5. (8 pt) Determine Whether the given integral is convergent or divergent. Evaluate if it is convergent. 00 _ 2 f :ce 1 das. 0 i- 1. r ->< J"; I} Xe d): q we ow: Zfl‘” “M” a”\ I; I.“ H75; e k .1: NM ( away-t M LEM —e_xz> : I, e_o& + £0 ’ B I 14.3” 0 «Ver (4% alwmdx 6. (5 pt) Find the arc length function for the curve y = x2 — %lnx taking Pg(1, 1) as the starting point. (10093? K" ‘93:“ (K) a . ¥b®=2x-é(%)32x~g; 1/l 4/ Hm =7)! +(zx— gm" [/243HZx-QLK‘AK . (.m «z _ 9/2 “‘1)‘ «slew { Wm"; 0‘“ “t 4 I” M fd: fly I M=2+éx ., jazz Jduidx i rm 4 £ 2 2+in Jul-6:17;;wa ="f, w—m m “L “(4.1mm -I £4.41. *1M -1 gzwi)‘ a? - : m =— —————— ‘KH 3) 4/0”) L’Lhni) =’ ,1 (M43 .2 I.» rm fuzz“! \;=u~l “F”, 0 2W 01“ dV‘J‘“ { I .M'M *asw: _; MW Jfézv 1f"- g'é'fi’” o 24v ’23; 2W 3 0.2V}: 4 2 0 24V 1 {' . Jc J ‘f’fi" 0% i 9 vi '1 ’ J \F’\ I ,‘ + 2 6"?” J NM (é? 293+ " 2 «we 6 0 7. (5 pt) Consider the curve below. 3; = we“, 1 S a: 5 8. Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the y—axis as O 8. (Bonus 2pt , {1% partial credit) Find the centroid of the region bounded by the curvesy=cosm,y:0,x:0, and m‘ZW/Q. I it A: joviéoSXfl‘X 2’ “3"”X 03. :5/h{§l’)v$§n[¢>: / Some trigonometric identity. . 1 . . smA cos B = 5 (511101 — B) + s1n(A + B)), sinAsinB = é<cos(A — B) — cos(A + B)>, 1 2 cosAcosB = (COS(A — B) + cos(A + $2+a2_a / dw - ltan-1(E)+C. ...
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Midterm Exam 1 - X) : fie/100 -MATH 2153 : Midterm I (50 pt...

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