# Calc08_3day2 - 8.3 day 2 Tests for Convergence Greg Kelly...

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8.3 day 2 Tests for Convergence Greg Kelly, Hanford High School, Richland, Washington

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Review: 1 P dx x 0 P 1 P x dx - 1 lim b P b x dx - →∞ 1 1 1 lim 1 b P b x P - + →∞ - + 1 1 1 lim 1 1 P P b b P P - + - + →∞ - - + - + If then gets bigger and bigger as , therefore the integral diverges . 1 P 1 P b - + b → ∞ If then b has a negative exponent and , therefore the integral converges . 1 P 1 0 P b - + (P is a constant.)
1 x e dx - 1 lim b x b e dx - →∞ 1 lim b x b e - →∞ - ( 29 1 lim b b e e - - →∞ - - - 1 1 lim b b e e →∞ - + 0 1 e = Converges

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Does converge? 2 1 x e dx - Compare: to for positive values of x . 2 1 x e 1 x e For 2 2 x 1 1 1, e x x x x e e e <
2 1 x e 1 x e For 2 2 x 1 1 1, e x x x x e e e < Since is always below , we say that it is “bounded above” by . 2 1 x e 1 x e 1 x e Since converges to a finite number, must also converge! 1 x e 2 1 x e

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Direct Comparison Test: Let f and g be continuous on with for all , then: [ 29 , a ( 29 ( 29 0 f x g x x a 2 ( 29 a f x dx
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