Business Calc Homework w answers_Part_68

Business Calc Homework w answers_Part_68 - 336 Section 8.3...

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8. E 1 2 1 } x d 2 x /3 } 5 E 0 2 1 } x d 2 x /3 } 1 E 1 0 } x d 2 x /3 } E 0 2 1 } x d 2 x /3 } 5 lim b 0 2 E b 2 1 } x d 2 x /3 } 5 lim b 0 2 3 3 x 1/3 4 5 lim b 0 2 (3 b 1/3 1 3) 5 3 E 1 b } x d 2 x /3 } 5 lim b 0 1 E 1 b } x d 2 x /3 } 5 lim b 0 1 3 3 x 1/3 4 5 lim b 0 1 (3 2 3 b 1/3 ) 5 3 E 1 2 1 } x d 2 x /3 } 5 3 1 3 5 6 9. E 4 0 } ˇ 4 w dr 2 w r w } 5 lim b 4 2 E b 0 } ˇ 4 w dr 2 w r w } 5 lim b 4 2 3 2 2 ˇ 4 w 2 w r w 4 5 lim b 4 2 ( 2 2 ˇ 4 w 2 w b w 1 4) 5 4 10. E 1 0 } r 0 d .9 r 99 } 5 lim b 0 1 E 1 b } r 0 d .9 r 99 } 5 lim b 0 1 3 1000 r 0.001 4 5 lim b 0 1 (1000 2 1000 b 0.001 ) 5 1000 11. E 1 0 } ˇ 1 w d 2 w x x w 2 w } 5 lim b 1 2 E b 0 } ˇ 1 w d 2 w x x w 2 w } 5 lim b 1 2 3 sin 2 1 x 4 5 lim b 1 2 (sin 2 1 b 2 0) 5 } p 2 } 12. E 2 2‘ } x 2 2 1 dx 4 } 5 lim b 2‘ E 2 b } ( x ( / 1 2 / ) 2 2 ) 1 dx 1 } 5 lim b 2‘ 3 tan 2 1 } 2 x } 4 5 lim b 2‘ 1 } p 4 } 2 tan 2 1 } b 2 } 2 5 } p 4 } 1 } p 2 } 5 } 3 4 p } 13. E 2 2 2‘ } x 2 2 2 dx 1 } 5 lim b 2‘ E 2 2 b 5 lim b 2‘ E 2 2 b 1 } x 2 1 1 } 2 } x 1 1 1 } 2 dx 5 lim b 2‘ 3 ln ) x 2 1 ) 2 ln ) x 1 1 ) 4 5 lim b 2‘ 3 ln ) } x x 2 1 1 1 } ) 4 5 lim b 2‘ 1 ln 3 2 ln ) } b b 2 1 1 1 } ) 2 5 ln 3 2 ln 1 5 ln 3 14. E 2 } t 2 3 2 dt t } 5 lim b E 2 5 lim b E 2 1 } t 2 3 1 } 2 } 3 t } 2 dt 5 lim b 3 3 ln ) t 2 1 ) 2 3 ln ) t ) 4 5 lim b 3 3 ln ) } t 2 t 1 } ) 4 5 lim b 1 3 ln ) } b 2 b 1 } ) 2 3 ln } 1 2 } 2 5 3 ln 1 1 3 ln 2 5 3 ln 2 15. E 1 0 } ˇ u u w 2 1 w 1 w 1 2 w u w } 5 lim b 0 1 E 1 b } 1 2 } } ( ˇ 2 u u w 1 2 w 1 w 2) 2 w d u w u } 5 lim b 0 1 3 ˇ u w 2 w 1 w 2 w u w 4 5 lim b 0 1 ( ˇ 3 w 2 ˇ b w 2 w 1 w 2 w b w ) 5 ˇ 3 w 16. E 2 0 } ˇ s 4 w 1 2 w 1 s w 2 w } ds 5 lim b 2 2 E b 0 1 } ˇ 4 w s 2 w s w 2 w } 1 } ˇ 4 w 1 2 w s w 2 w } 2 ds lim b 2 2 E b 0 1 } ˇ 4 w s 2 w s w 2 w } 1 } 2 ˇ 1 w 2 w 1 ( w s / w 2 w ) 2 w } 2 ds 5 lim b 2 2 3 2 ˇ 4 w 2 w s w 2 w 1 sin 2 1 } 2 s } 4 5 lim b 2 2 1 2 ˇ 4 w 2 w b w 2 w 1 sin 2 1 } b 2 } 1 2 2 5 sin 2 1 1 1 2 5 } p 2 } 1 2 b 0 1 b b 2 b 2 3[ t 2 ( t 2 1)] dt }} t ( t 2 1) 2 2 b 2 2 b [( x 1 1) 2 ( x 2 1)] dx }}} ( x 1 1)( x 2 1) 2 b b 0 1 b b 0 1 b b 2 1 336 Section 8.3
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17. First integrate E } (1 1 d x x ) ˇ x w } by letting u 5 ˇ x w , so du 5 } 2 ˇ 1 x w } dx . E } (1 1 d x x ) ˇ x w } 5 E } 1 2 1 du u 2 } 5 2 tan 2 1 u 1 C 5 2 tan 2 1 ˇ x w 1 C Now evaluate the improper integral. Note that the integrand is infinite at x 5 0. E 0 } (1 1 d x x ) ˇ x w } 5 E 1 0 } (1 1 d x x ) ˇ x w } 1 E 1 } (1 1 d x x ) ˇ x w } 5 lim b 0 1 E 1 b } (1 1 d x x ) ˇ x w } 1 lim c E c 1 } (1 1 d x x ) ˇ x w } 5 lim b 0 1 3 2 tan 2 1 ˇ x w 4 1 lim c 3 2 tan 2 1 ˇ x w 4 5 lim b 0 1 (2 tan 2 1 1 2 2 tan 2 1 ˇ b w ) 1 lim c (2 tan 2 1 ˇ c w 2 2 tan 2 1 1) 5 1 } p 2 } 2 0 2 1 1 p 2 } p 2 } 2 5 p 18.
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