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int {4}^{6} [(3 x^2 + 1) / (x^2)] dx =?

2007-04-14 07:06:37 · 2 answers · asked by bhicks 1 in Science & Mathematics Mathematics

2 answers

= ( 3x - 1/x) between 4 and6


(3x^2+1)/x^2 = 3 +1/x^2 (
=18-1/6-12+1/4 = 6 1/12 =73/12

2007-04-14 07:18:08 · answer #1 · answered by santmann2002 7 · 0 0

= integral [4,6] (3 + x^(-2)) dx
= [4,6] ( 3x -x^(-1) )
= 18 - 1/6 -12 + 1/4
= 6 1/12

2007-04-14 14:12:12 · answer #2 · answered by hustolemyname 6 · 0 0

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