Sect. 8.8 49-54 - 49. Forz > X 0, x3 + 1 < :r...

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Unformatted text preview: 49. Forz > X 0, x3 + 1 < :r J;3 1 = x2' ,1 f= ? :r- 1 GX IS convergent I. by E" quation 2 with I r. = 2 > x- 'foo 1, so ,1 -3-- x X +1 , d;r; IS convergent by the Comparison convergent. 2 Theorem, X {I -,-3- - [« x +1 dx is a constant, so j'OO _, 0 -3-;];- :c + da: = lox j'l -.-3- +1 dx + l, 1'00 -3-XX +1 dx is also 50. For x ;:: 1, + e-'" x >- 2 [since e-r > 0] > 1 ;" 1'00 l, ~ ,1; d» is divergent by Equation 2 with p = 1 ::; 1, so X .f/,00 2 +a:e1 x dx is divergent by the Comparison Theorem . 51, Forr > 1, f(x) = ~> J. 'I; +1 - x x fA V X4 + 1 > ,2 :T. x = -:' so X 1 1= 2 f(x) dx diverges by comparison dx with ~'OO ,2 1 -: dx, which diverges X by Equation 2 with p = 1 ::; 1. Thus, J~oof(l;) dx = JI2 f(l;) + J~oof(x) ds: also diverges. 52 , FOl' z >_ 0, arctan ~ x < 7r -2 < 2, so arctanx 2 + e- < __2_._ < ~ 2+& ~ ct. = '2c'-x, Now 1= 1 o 00, 2e-'dx= lim t~oo. /'t..2e-"d~;= 0 lim [-2e 1.-00 - ']0= lirn I-ex) (2 -/+2 e ) = 2. so I is convergent, and by comparison, '00 ,0 / arctan x . -.,.--dx IS convergent. 2 + eX -> 53, For 0 < 'l x ::; 1, see x 1; Vx > / Now X3/2 1= j o X-3/2 dx = Jim t~O+ 1. t X-3/2 dx = lirn t~O~ [-I/?] - 2x - 1 = lirn ( -2 . t t~O+ + 2 ) = 00, so I is divergent, r; V /. and by comparison, 1 sec2 x. 1 , o x vx IS divergent. . 54, For 0 < x ::; 1, sin ' x , ::; vx Vx X-1/2 1 " Now 1= ./0 r" ~ dx = Vx lim I~O+ j'IT I dx = lim [23)/2]" I~O+ I = lim (27r - 2.fi) 1-0+ = 271' - () = 27r, so I is convergent, and by comparison, j 'r. sin ' x . --;==- dx IS convergent. o vx ...
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This note was uploaded on 12/08/2010 for the course MATH 3B taught by Professor Castroconde during the Spring '09 term at Irvine Valley College.

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