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hw-1-sec 8.8-solns - 71.33 “~31?“me CHAPTER 8...

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Unformatted text preview: 71.33%..- “~31?“me CHAPTER 8 TECHNIQUES OF INTEGRATION 696 E (c) The total area under the graph of f is tlim F(t) : tlim 10(1 — F“) = 10. ago 400 The total area under the graph of 9 does not ex1st Since 11m G(t) — tlim 10(1:0 1 1) = 00 °° 1 t 1 I = d — 1 d N C9 / (3m+1)2 06 1220/1 (3M1 w °W 1 1 1 1 d :— ——d : 1,d :3d :——— C: , /(3$+1)2 9” 3 u? u [u 3“ u “”1 311+ 3(313+1)+C 1 ‘ 1 1 1 ”I 3320i 3(3m+1)] tiniloi 3(3t+1) 12] 0+ 12 12 convergent 0 1 dw: 11m [51n|2cc—5l]t ¢1_i_,120[%1n5—%1n|2t—5H:_00 00 a: dwzlim m dm=1im-1- -1 zll'm :-+1 0 ($2+2)2 t—eoo 0 (m2+2)2 t—aOOZ $2+2 Zt—aoo 152+? 2 1 (0 + l) : i. Convergent t 9. I400 e’y/Z dy = tlim f: 6”” dy = tlim [—25‘9/2L : t1im(—Ze-t/2 + 26—2) = 0 + 26’2 = 2e‘2. aoo 40° 4°C Convergent .f::o 3“” dt = 11m fag—1 6'“ dt = lim [—ée”2t];1 = lim [—162 + ée‘h] : oo Divergent 11—>——oo $——>—oo I-+~oo °° nude: 0 mdat °° ccdm 1 . :— d 1 [1001“? [N1+x2+/0 1+x2an O xdcc : lim [1 1n(1 + m2)]0 : lim [0 - l111(1 + 3)] = ~00 Divergent —oo 1+1? t—->—oo 2 t t—>—‘oo 2 ' 12. I : £242 — 11“) d1) = I1 + [2 : IEOOQ — v4) d'u + fow (2 — U4) d1), but I1 = t lirn [2v — 211115]: = t 11111 (—215 + $5) : —oo. Since 11 is divergent, I is divergent, and there is no need to evaluate 12. Divergent 13. If; me'x2 dac = ono (L'CIIQ d3: + I? 336””2 die. 0 42 , - 42 0 _ ‘ 1 —2 _ _ 1 f_oo:ce dm—tgljloo(*%)[e L ~t11r31m(—§ (1-6 t ) _—%.1—~5,and oo -052 . 42 t - 1 — 2 , 1 _ 1 10 we mange-lime 0:33.147 (e‘ -1)——§-<—1>—§. Therefore, [3°00 cue—“”2 day = —% + % : 0. Convergent ‘ SECTION 8.8 IMPROPER INTEGRALS I: 697 ie . — uzfi ' V; taco V; dcc: 11m 1 e (2du) [du2dx/(2fi)] 1 _ 2t1irgo(‘e*”+ 6‘1): 2(0+ 64) 2 26‘1. Convergent 15 f2 s1n6d9 : tlim f S1116 d6 = hm [— cos 6] 7r ~ tIim (— cost + 1) This limit does not ex15t so the integral IS divergent. Divergent oo 0 oo . 1 . O . 1 . 16. I : f_oo cos 7rtdt = [1 + [2 = fgoo cos7rtdt + f0 cos 7rt dt, but I1 = hm — s1n7rt : hrn —— Sln7rt and sa—oe 71’ S‘*—00 7 t 1) 1 [In (fi) — In (5)] =In1+ln2 = In 2. Convergent —5s __ - _l —53 _ i —53 by integration by 58 d8 _ tlirg: [ 586 25 6 1 [parts with u = s] 1 W = tlil’glo (—%t€_5t — fies—5t + i) = 0 — 0 + % [byI’HospitaI’s Rule] 25 . Convergent 6 6 _ V 20 NET/3 dr 2 Iim T‘BT/a dr : Iirn [37‘er/3 — {Jar/3] 6 by magnum by I ~—oo t—>—oo t t—a—oo L partswithu : 7- _ Iim (1832 — 962 — 3tet/3 + get/3) = 9e2 — 0 + 0 [byI’HospitaI’s Rule] tfl—oo “ ‘ g = 962. Convergent \ 00 111 a: (In (I?) 2 t by substitution with (111 02 . 1; ,1 @ / ‘ dm 2 Iim ] : Iim 2 oo. Dlvergent 7' 1 32 1—.00 2 l u = In 2:, du = dx/$ taco 2 22, 1 2 [:0 x3e‘x4 dz = [1 + 12 = ffoo $36—$4de + f000 3 4 t t4 4 _ . 3 _x4 _ . _u l u:z , Iz—tlinoio Owe day—th’rgo 0 e (4du) [du=4z3dz] ‘1‘ _11~ ut4_11- —t4 1_10 I)“; “ fig: ‘8 0—2411; -6 + -z< + -4- Since f (at) = m3e‘z4 is an odd function, 11 2 ~41, and hence, I = O. Convergent I {1 :9?! l' ail 698 CHAPTER 8 TECHNIQUES OF INTEGRATION 00 $2 0 $2 00 1‘2 oo .232 @100 9+$6 d3: : /_OO 9+16 d.iL‘+/O 9+1’6 dm 2 2A 9+$6 dIE [smcethemtegrandrs even} Now We m3 1 édu ”:31, _ %(de)-_ 1 dv 9+x6 du=312dz _ 9+u2 du=3dv “ 9+9v2_9 1+7}? 3 = lTall—I'U—i-C': ltan—l(§) +0: 1tan‘1<-$—) +0, 9 9 9 3 °° 272 . t x2 , 1 _1 1133 t . 1 11 t3 2 7r 7r 802/0 9+m6dx—2tlino10 0 9+$6da§=2tlingo[§tan <?)L=2tlir§o§tan (3)25525 Convergent °° em t ea” [ 1 er}: 1 ( e‘ 1 24. / dac : lim d2: : lim —— arctan— = — Iim arctan —— — arctan— o aw+3 “W10(af+%v§f t~w v5 v§o vfihwv v5 v 7r 7r\/— _):7g. Convergent 00 1 t 1 Int ‘1 1 Int 25. / dw: lim 03x: lim msdu [1“ “7“ j = lim [——] 23(ln $>3 ) taco e x(lnm)3 t—roo 1 du: dz/w t_,oo 2112 1 1 1 1 1 ~ lim 1 — — 0+ — — —. Conver ent Hui 2(mt)2 2] 2 2 g °° marctanx . t xarctanm mdm d1: 26./0 mdflf 2 tliglo 0 de. Letu 2 arctanx, (1’1) 2 m. Then du Z 1 +m2, 1 2mdm —1/2 : — ‘ 2 , d 2 (1+ac2)2 1—i—x2 an marctanxd _ larctanx 1 dzv zztane, (1+.T2)2 $_—§ 1——:L’2 +5 (1+x2)2 dmzseczede 1+ 2 _ 1 arctanm + 1 sec2 19(149 x x h 2 1—— m2 2 (36(326)2 I _ larctanzz: 1 2 1 -— 2W+2/COS ng 1 arctanx 0 sin<9 €080 —‘§T:??+z+“‘i——+C 1 arctanzc 1 1 a: = ———— — t — C 2 1+ac2 +4arc anx+41+w2 + It follows that °° xarctanm 1 arctanm 1 1 a: t /0 (1+m2)2 d9” ti‘i‘o[ 2 1+:c2 + 46‘rc any” 41+:c2]0 . larctant 1 1 t 1 7r 7r 2 hm <~§ 1+t2 +Zarctant+11+t2) =0+Z-§+O:§. Convergent 1 3 1 1 3 1 27. / ——dac = lim Bat—5 dm = lim [——] = —— lim (1 — —) = 00. Divergent o 4 t—>0+ t4 SECTION 8.8 IMPRCPER INTEGRALS 699 ./ dx: lim t(3~x)‘l/2dm: lim [—2(3*m)1/2]t ——2 lim (F— f)=— 2(20—1) 2. — 5c t—>3’ 2 1—3; 2 t—v3’ _‘ Convergent "' ,i 14 dill 14 4 14 4 ,‘ = 1. 2 ~1/4d : 1. _ 2 3/4 _ __ 1' [1 3/4 _ 3/4] 29. /.2 m 133+ t (x + ) w Pliny 3(a:+ ) 15—3H11n2+ 6 (t+ 2) = §(8 — 0) = 33—2. Convergent 8 .———4 dcc , lim 84(33 — 6)_3 dz — Iim [—2(:c A 6)’2]8 — -2 lim i — 1 — D‘ t , 30. 6 (m _ 6)3 t_‘6+ t — ta“ t — MM 22 (t _ 6)2 ‘ oo. 1vergen 3dm—/Od—JC+/39l£but/O—d—m—lim —E:t —1im —171 “ D' t p 31. '2 $4 " _2 $4 0 $47 _2 1:4 — t_>0_ 3 _2 — “'07 3153 24 — 00. 1vergen : 1 d3: - t d”: - - —1 t . . —1 7r 32 0 m = 33%: 0 W = in; [51“ do = $2“— t: 5- Convergent 33. There is an infinite discontinuity at m = 1. 33(27— 1) 1/5 das— ~ 10 (ac — 1) 1/5 dzz: + I133( :13 — 1) 1/5 dar. Here O 1 —1 5 5 . fo (1* / (”11:11: fa 06"” ” drill?i§(x‘1)4/5io:tli‘l‘1i%(t‘1)4/5‘§i:“gand 33 #15 _- 33 715 _~ 5 4533,. A L (1-1) War—11:: t (H) Ww—ngflw-D/L 221111-16 EU—W/fii—m Thus, f033(3c — 1)’1/5 d1: = —% + 20 = 74—5. Convergent ~ .f(y) : 1/ (4y — 1) has an infinite discontinuity at y = % 1 1 1 . 1 . 1 . ———-dy 11111 d 2 11m lln4 —1 = 1 l13—114t~1 = , 1/44y~1=t~<1/4)+ t 4y—1 y w:—1<1/4)+[‘1 I y IL Miriiw [4 n 4 I“ H 1 d diverges and hence /1 1 dy diverges Divergent 1/4 43/ — 1 y i , 0 4y “’1 I dx 3 d1: 1 (ix 3 d3: 2—— = —— = Ii + 12 — l . cc —6a:+5 o(m—1)(27—5) 0 (:c—1)(a:—5) 1(:1c»1)(23—5) Now 1 — A : B => l:A(m—5)+B(:c—1). (m—1)(m—5) _:c—1 m—5 Setw=5toget1=4B,soB:i.Setx:1t0get1=—4A,soA=—%.Thus t i _l l [1: 11m 4 + 4 dmz lim [——ln|:1r:—1l+zl1 1n|x—5| p41- 0 x—1 zit—5 t—»1' 0 H lim [(a-‘lI 1n|t — 1| + §1n|t — 5|) — (film—1| +§1ni—5|)] tal— = 00’, since iylim (—i 1n|t — 1|) = 00 t—41’ 0° 11 is divergent, I is divergent. hesczvdat— — 11m LIE/2 cscmdm— — lim [In |csccc — cot :r:[ ]w/2 = lim [ln(csct — cot t) — In(l — 0)] t—nr— t—>7r t—r7r— : lim ln(1—_——C-(fl) = 00. Divergent t—nr— sint "‘4'": . 700 3 CHAPTER8 TECHNIQUES OF INTEGRATION ~ _ . u 71 usepans _ ~ —1 ’ hm [(u— De ]1/t [orFormula96] — £115: [—26 _ < o l/w tv 1/t 37. / 6 dm : 11m lei/”3 ~ —1— dm = lim ue“ (—du) [u 2 UL ]: W 1 $3 t_.0* _1:L’ 1:2 t—r0* _1 t—>O# 2 , . 2 , ~ 1 2 :——— hm (3—1)? [szl/t] :——— 11m 5 g——— lim 1 e 5—4700 6 sH-oo 6-5 e s—v—~oo —e*3 2 2 = #— fl 0 : ——. Convergent e e 1 1/z 1 321/ 6 dm: lim 3 0 {E3 t—>0+ t $ 1 1 61/1'7 dac : lim ue“ (—du) [ I: t—>0+ l/t _ . __ u l/t usepans __ . l —_ l/t ‘ 1311,1314. [(u De 11 [orFormula 96] _ 31,151+ [(15 1)e = lim (5 ~ 1)es . [s = 1/1] = 00. Divergent s—§OO _ 0 3 2 39.1: 222lnzdz: lim ft2z21nzd2: lim l%(31nz#1)] [ uzl/m, du: Adm/12 1 integrate by parts Formula 101 :\ t-»0+ t-»O+ t oruse ‘ 3 8 - 3 : 351+ [3(31112 — 1) — gt (3lnt — 1)] = 31112— 5 ~%111131+[t(31nt# 1)]:31112—3— . . 31nt—1 H . 3/t , — 3 —— : : —— : — 3 : NowL — £1115: [t (3lnt 1)] 11—13): t—3 3—1.1ng —3/t4 121351+< t ) 0. Thus, L = 0 and I = g 1112 — %. Convergent Integrate by parts with u = H 1nm,dv = dcc/fi => du : den/110,1) = Zfi. 1lnm . 1lnar . 1 1 dm , Afidm 333+ t fidm—tgrél+([2filnw]t—2fl E>—tgrg+<—2fllnt—4[fi] ll t—>D+ since lim fl lnt = lim i——>O+ t—>0+ 41. 42. lim (-2\/E1nt—4+4\/i) =—4 lnt fl . 1/t _ . _ tel/2 ‘ $133+ —t-3/2/2 - 113351ku J?) _ 0‘ 1 t Areazfimemdmz lim [em] t—»—oo t—aoo Area = f: e—x/2 dm : —2 lim [6 Convergent 26— lim et:e t——>—oo t—~>oo 1 t tolH .b‘ > t W2] = —2 lim (ft/2 + 2e = 2e —2 ...
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