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HW20 - > m A = 3(1?“*5 W" M_3 an IL —’ I'i “4...

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Unformatted text preview: > + , m A = 3;: (1%?“ *5: W" M _3_ an IL —’-! I 'i “4 25.) So 9"") é” = 500‘") 30% +511CX-9 a? L : A + 8 5 A : M So CX—ljvq/S’M b—wl‘ : gum ~3LX-I)§fb 4H“ o _ km ’3 , -— ” 3 = 00 .. V b—M' §(b—1)/3 GOV} + 3 L2.) VM :- n‘ 8:QE“X)2M ; ,LLW 1r €019 CC—Ax) °‘-2< b-‘Poo : )‘M TV- 9.ng [‘0 — E' ....n. __1__ ~ 1 be» -2 o ‘ 5—,“; 3: €25 2.” Math 16B Kouba Improper Integrals Determine the following improper integrals. 1.)/ 6-7:” da: 1 M C 3 .l 1/ 1.2%» 2L] 1 KC!) -0200 ._ -— 1 ll A—v 4* ~.. ~ ._ .4 , ,- A _ )9 M... hm. -44 u...“ t.‘ “M; M“; .v )M/A‘ Q=Q~1TY‘:Q.T1'C¢Z)= LIN-m 3 M35:— C=olrrr=2TTCé)=1er—m. J/é—o/ OV‘ 37.71». x3660m.x ”(ix tm. z- m. I A» 1.7, m. .5130 fr. (i. 6k Pi“. unwind) 8 C: 911TIPC‘21T(30)=601TA'M> (fame/“‘32 0" 40W~- ‘38.5m X SLOOMX ‘fl-x zHoM-7( > M T ‘2m.x 5230f)? MANN/4M ELM/Me. a) a p)". m 8r ,(4 - a)" 304w): 3(o-rr~ H3.Im/M .M' 60 a (1*? ERMA x4 - ok— CfCaTrr) - 762mg): '0?er " M w ...
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