hw-8-sec-8.8-solns

# hw-8-sec-8.8-solns - 696 CHAPTER 8 TECHNIQUES OF...

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Unformatted text preview: 696 CHAPTER 8 TECHNIQUES OF INTEGRATION \g , \1‘; (c) The total area under the graph off IS lim F(t)— — £11m 10(1 — t‘o'l) = 10. taco The total area under the graph of 9 does not exist, since tlim G (t) —— tlim 100%0 11) : 00. J'\ °° 1 t 1 5. I : d : 1‘ ————- d .N U / <3m+1>2 5‘ 33° 1 <3w+1>2 ”5 W 1 1 1 1 1 —————d 2— ~d — d=d :——— C:————-—— /(3.’L‘+1)2 \$ 3/,le2 u [u— 3x+1, u 3:5] 3u+ 3(3m+1)+C1 1 t 1 1 1 1 1:1 —_——— =1' ——-——— —=0 ——:—. so £1301 3(3x+1)]1 13301 3(31‘,+1)1r 12] 1 12 12 convergent 6 0 1 dw: lim 0 1 doc: lim [llnl2m—5|]0=11m[1n5—llnl2t#5|]=—oo 1 #00 2a: — 5 t-Hoo t 230 — 5 t->#oo 2 t t-ooo ‘ Divergent = lim 2—11) ta—oo t \/2—w 11100 #1 —1 7./ 1 dw 1 dw: 1im [—2x/2—w]:1 [u227w,du=—dw] = lim [—2 \/§ + 2 V2 — t] z oo. Divergent t—>—oo 1 1; O1) 11 d1": lim :0 —————dx= 111111 11 1—1lim —:1—+l 1 ' 0 (m2+2)2 Hoe 01)2(a: +2 1—1002 a32+2 0—2:»00 t2+2 2 '1 — 2 4(0 +% = i.) Convergent t [—Zeﬁy/E] = lim (—Ze’t/2 + 26—2) = 0 + 26”2 : 26—2. 4 t—>oo 9 fooe e—y/2dyz lim ﬁe e‘y/zdyz lirn t-ioo t—aoo 3],: Convergent EDILL 6-21 dt=1im [\$116 '2‘ dt: lirn [—%e#2t];1 = lirn [——%e2 + £64m] = 00. Divergent i °° wdzv 0 mdw °° suds: 11. = d /_001+a:2 [001+x211/0 1+av2an t-r-oo 0 soda: . o . . / 1 + 1:2 — 1.131.1100[%1n(1 + 332)]1 = hm [0 — %ln(1 +t2)] = ~00. D1vergent —OO 12. I = fj°oo(2 —v4) dv = 11 + 12 = [300(2 — 124) do + [5"(2 —v4) dv, but 1 I1 = lim [2v — 1111111 2 lim (— —2t + 1t5) 2 ~00. Since I1 is divergent, I is divergent, and there is no need to t—+——oo 3* 0° Divergent evaluate I 2. 698 CHAPTER 8 TECHNIQUES OF INTEGRATION L, 1 \ oo 2 0 2 oo 2 oo 2 @Lw 9:."56 dm: /‘oO Qimﬁ d\$+/0 9:336 d\$:2/0 9::36 d1: [sincetheintegrandiseven]. Now m2da: uzxi‘l :/ %du uzgv :/%(3dv) :l/ d1} 9+u2 du:3d“ 9+9v2 9 1+1)2 9 + m6 du = 322011 1 1 3 2 g tangl U + C = 5 tan‘1(%> + C : étan-1<——m3> + C, oo 2 t 2 3 i 3 \$02 :1: d1: : 211m a: da: : 211m ltarf1 m— : 211m ltan—1 t— : a - E : Z. 0 9+m6 Moo 0 9+:56 Moo 9 3 0 H009 3 9 2 9 Convergent °° 8““ t e“ 1 e t 1 et 1 24. d3: = 1im dm 2 11m —— arctan -— = ~— lim arctan —— — arctan —— f0 2 2 [x/ﬁ x/ﬁ] x/E H00 < x/ﬁ J3) : 75 5 ﬂ 6 3 9 oo 1 t 1 lnt '3 u=1nI 1 Int 25 / :1:(1na:)3 dm—tlggo :v(1na:)3 (ix—3320A u du [d =dw/9E] _tli>n§o [—2172]1 1 1 1 1 = lim — +— 20+ — : —. Conver ent °° marctana: . t avarctanx :vdm dm 26 /O (1+m2)2 dac : 3111010 0 WC”). Letu: arctanat,dv Z M Then du = 1-1-1132, U—l 2:811:12 _ —1/2 and ’ 2 (1+m2)2 ” 1+x2’ marctanx dac _ *1 arctancc + 1 d3: 3: :tane, (1+\$2)2 — 2 1+m2 2 (1+\$2)2 dmssec20d9 2 1+x2 _ _1arctana: 1 sec 6d0 x _ 2 1+3:2 2 (secZO)2 I _ larctanac 1 2 1 - 2 14—302 +2/cos 0d0 _ _1arctancc + 9 + sine cosO +0 d 2 1-1—9172 4 4 i-larCtanm—Q—larctanm4—l at +0 _ 2 1+ac2 4 41+;c2 It follows that 0° marctanx dm __ 11m -1 arctanx +larctanx+l :1: t 0 (1+ac2)2 1.4» 2 1+12 4 41+x2 0 — 11m _1arctant +larctant+l—t—- —0+l . 1+0— 1 Conver ent ”Moo 21+t2 4 414th2 ‘ 4 2 ’8‘ g 1 1 1 _ 1 . 27.] —3—5dm: 11m 336 5dos: 11m ~—3—4 :—§ 11111 1——4 =00. Dlvergent 0 a: t—>0+ t t—»0+ 4m t 4t—>0+ t SECTION 8.8 IMPROPER INTEGRALS 699 (29/: 1 (1w: lim t(3—x)_1/2dm= 11m [—2(3—m)1/2]t :—211m (meﬁ) =—2(0—1):2. m 1—»3- 2 tA—»3# 2 t—>3” Convergent 14 d1; 14 1 4 4 14 4 :1‘ 2r/d=1' - 23/4L ——1 [163/4~t 23/4] 29' /_2 4334—2 talEH2+ t (\$+ ) a: tﬂ11n2+[3(m+ ) t 3ta13n2+ ( + ) = §(8 — 0) : ¥ Convergent 8 4 da: — lim 8 4(13 — 6)‘3 d3: — 11m [—2(m — (5)4]8 — —2 1’ i ‘ 1 — D' t so. f, M - 116+ , - 116+ r 1351+ 22 (to 6)2 — 00- mg 3 dm 0 dm d3: 0 da: . 90‘3 t 1 1 1 . _= — —bt —=1 —— =1 ~———= . 31, /_2 3:4 /_2 \$4 +/: \$4, u _2 {34 £113: 3 ‘2 H115: 3163 24 oo D1vergent 1 t d | 33. There is an inﬁnite discontinuity atcc = 1. 033(22— 1) 1/5 dm— — f0( 35 ~ 1) 1/5 dac + f33( a: — 1) 1/5 dm. Here t f01(\$—1)_1/5da:: lim f0 (32—1) 1/5dx: lim [gm—1)“? = lim [E(t—1)4/5—%] =#3— and t—vl' t—>1’ 0 t—+1“ 33 f33(x—1)1/5dw=11m ftﬂ(m—1)1/5da§: lim[i(m—1)4/5] :11m[%~16—E(t~1)4/5]:20. 1 1H1+ t—>1+ t t—.1+ Thus, f033(:c — 1)_1/5 day : *3 + 20 : E. Convergent Kr} WW) 2 1/ (4y — 1) has an inﬁnite discontinuity at y : i. 1 1 1 1 1 d : 111m d = lim \$1114 “1 : lim lln3—lln4t#1 :00, /1/4 4y — 1 y 1—»(1/4)+ t 4y — 1 y 11(1/4)+[4 I y ”t t—+(1/4)+ [4 4 ( )] 1 1 1 . 1 . . so / dy diverges, and hence, / dy dlverges. D1vergent 0 1/44yﬁl 4y—1 3 3 1 3 day dzv da: do: 35.1: —= —=1 1: __ __. Aﬂ—ams /(z—1)<x—5> 1+ 2 /(m—1>(x*51+/1<x—1><x—5> Now—#2 A + B :> 1:A(x—5)+B(m~1). (m—1)(m—5) 22—1 :I:—~5 Setac:5toget1=4B,soB=i.Setmzltogetl=—4A,soA=—i.Thus * —l l 1 1 ‘ I :1 4 4 =1” —- - — 1 1 131n— 0 (\$_1+x_5)dw tirln_[ 4ln|x 1|+41n|m 5|]0 = lim K iln|t 1| I iln|t 5|) ( ﬁln| 1|+4111nl 51)] t—>1_ ll 00, since lim ('%1n|t—1|):oo t—bl‘ Since 11 is divergent, I is divergent. 35-1/2cscasdatz lim [7: /2 cscmdw— - lim [1n Icsca: — coth ],,/2— — lim [ln(csct A cot t) ~ 1n(1 v 0)] t—H'r— t—nr— t—HT_ sm t t——»7r‘ = lim Incl—ﬂ) = 00. Divergent or Formula 96 : lim [(u-1)e“];/lt [ “spans ] : lim [4671— G—1>61/t] tv—+0’ L—>0’ 2 . 5 2 . ~ :,_# hm (s~1)e [szl/t] =—~~ hm 5 12 e s—a—oo e saioo 6‘5 : ,_2_ ﬂ : —Z, Convergent e e 1 1/1 1 1 1 _ 38. f e da: : lim 161/35 ‘ — dcc = lim ue" (—du) [u _ 1/7" 0 m3 t—aO‘t’ t a: m2 t——>0+ l/t _ . _ u l/t useparts _ . 1 1/15 F“ 31,131+ [(u De ]1 [orFor-mula96] _ tEIE): [(t — 1)6 = hm (s — 1)es [s = 1/t] 2 oo. Divergent s->oo 3 2 _ 2 2 _ - 2 2 _ ’ Z_ _ integrate by parts 39. I — f0 2 lnzdz — \$331+th 27 lnzdz v 3331+ [32(311195 UL [oruseFormulaIOl] : 3131+ [3(31112— 1) — gt3(3lnt— 1)] : \$1112 _ g — \$351+ [253(311115— 1)] =§1n2 # g _% . 31nt—1 H _ 3/t . = ' 3 _ : 1 _: 1 = _3 = NOWL 3351+ [t (31“ 1)] \$131+ t‘3 \$351+ -3/t4 £1314 t ) 0' Thus, L = 0 and I = gan — %. Convergent @ntegrate by parts withu:1nm,dv=dar/\/E :> du:dm/x,v=2\/E. 11mm 1 11130 . < 1 1 dm ——d =1‘ ——d\$= 11m 2x/Eln w2/ _ =1' /0 x/E :13 Ag: 1 ﬂ t—>O+ [ xL t ﬁ :13: =11m(—2¢£1nt—4+4¢£):_4 t—»0+ , . , Int 5 . _ , smce 3351+ \ﬂ lnt — 3151+ Z372 — \$351+ —t‘3/2/2 — hm (—2 x/i) — 0. 41. (—2x/Z1nt—4[x/E] Convergent 6— lim etze t—v—oo L. 1 t > ...
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