Problem set 4 -Solutions

Problem set 4 -Solutions - 6.3 Solutions{I.1 1 24 The...

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Unformatted text preview: 6.3 Solutions {I .1: 1 24. The expaneien is 2 :——1 5—4 C. 15—4.: 4fll EH 3:1 # A + B + e + D a 1 A B {xi—11rx1—4)_x—1 x+1 x—E x+2 5—12: «VF—'1 +fi+x ‘4— 1_ I: l l #(A+B)~/§+(A—B)x xL-flfx+l}(x3—4}= 2—(—3}_ 5 _ 5— 3 3 l l ______ —I—1-— I— I — — 2‘5 I 4 1 AFB=fl CZIIII‘ll—z—z— 1 1 1 1 1—}! {x — 1H: + 2] 3(4) 3 dI=— + )IETI 2 ffi—IE 25/3 (wig—I «JE—FI fl: lim fi: fi:—%_ 1 1—}— 111.1 —- I— [—J =—(—1I1|~/§—xl+h1]~/§+xl)+fl 2‘05 Therefere l 5 =filfl :g+l +C. I2 l 1 —_1: —[I2_n(x2_4)dx=~—Ehl[x—ll+Elfl[I+ll+ x A B l l '“- mix—1+”; ih'xtz'fih'xtg'tfi- _(A+3B}x+[3A—BJ 26. Wehave _ 3x1+3x—3 :11: ed: A+BB=1 l 3 f— I zf— Letu=x—i =':-'- =§-A=—.B=—. 1— 2 - 1 l {SA—B=fl In 1D [I *IH) [ix—t11+% du=dx f xdx 2i ( l + 3)::1'3: _. udu +1 {in 3x1+3x—3 1n 3x~—l x+3 _ {1.2+}? E (“brill l 3 =—ln3 —l —1u 3 C. 3 3E! Ix |+ JI+ |+ Letu=§tanm 22. Here the expmtsienis 3 dz: = g SEC2 1: {iv in the second integral. 2 ;.; +1: A + Bx—t—C fl 2 d x3+3 x+2 .x1—2x+4 1 1 1 ?SEH’” _mix?—2x+4}+B{x3+2x)+C(x+21 “5 “2+; +Ef 9 .1 —— —seev 13+E 16 ... ... =— — ease I: :—2A+23+C=fl=}A=EBmE.Cm-Fga 2[IE_;.;+1) 3fi 4A+2C=l _1 2 =— [ta—f—sinveequ—l—C sewe have fix: _I +1} +3‘fi 2 —1 + 2 m_i1.r—1+2 1—[x—EJa/i +C [I +1dx=— +— f— hf— _1"IE—I+11' 3J3 J2? +3_~./§(—2~/x1—x+1 n1 x3+3 x+2 12 I(.1I:—l)-—+—3flT 2 l2I_1 x—E C ._ — =—ta.n_ +—+ :Etul:x l Jfi 4/? 3(x3—x—f-l} u: x 5 l T 3 =—1111;..:+;:|+E ”Jr in 12 H1+3 4H3 1'? + in =—1n 2 —1n 1—2:: 4 12 11+ |+24 (I +) 1 —l «.5 +—ta.n_lx +5: Mi J3 ' Fig. 3.25 6.5 Solutions '5'" l 2. f —a'.1: Leta=2x—1 lit—1333 3 I: a) du=2dx DD (3' E —f —H =1 lim H_2fla'a 2 5 HEIE fut—me 5 E :0: (diverges) 5 ‘1 d1 _ C ah: ti. = 11n1 fl I122 --I1 C—m— g. a3 "-II n— n— 1 = — lirn Jam 2 fit—me The integral diverges tn infinity. DD 11]. f aa—I iii: a R =: lim Ste—x if; R—EDCI a. U2): dV=e_‘dx dU=dx V:—e_r R E ljrn (—Jre_Jr + f a”[ tilt) R—rm G I] l' R 1 —|- l l =: 1111 —-— -— — =: . R—mu e3 a3 The integral eenvergea. H 5" d1 [15. f Let n! =-. 111.1: I? .xflnx} a: d1: = —I .1" [HR if lnfl : lin1 _" = lirn 111M R—pnn ] H R—snn l r: lirn ln(1nR)—ln1== GD. R—rm Thisr integral diverges te- infinity. I]! 31]. Since 5 — far all J: 13 fl, therefnre x5 + l 3 an it I:f I ax I] 15-1-1 I II! E. DD I1 d F_/-fl15+lx+_/; 15+1I {Ifl I1 d +fflfl til I — — fl. IS—l—l i I3 : I] +I1. Since I1 is a proper integral (finite) and I; is a eenvergent impreper integral, {see Theereni 2), therefere I eenvergea. ...
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