# quiz 3 - ± ∞ 1 ln x x 4 dx = lim B →∞ ²-ln x 3 x 3...

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Math 1B: Calculus February 10, 2010 Quiz 3 Lecturer: Prof. Mina Aganagic GSI: Gary Sivek Name: Answers 1. (2 pts) Evaluate ± 4 x 2 - 1 x dx . This requires a trigonometric substitution, and since 4 x 2 - 1=(2 x ) 2 - 1wemay choose 2 x =sec θ, 2 dx =sec θ tan θdθ : ± 4 x 2 - 1 x dx = ± sec 2 θ - 1 (sec θ ) / 2 sec θ tan θdθ 2 = ± tan 2 θdθ = ± (sec 2 θ - 1) =t a n θ - θ + C = sec 2 θ - 1 - θ + C = 4 x 2 - 1 - sec - 1 (2 x )+ C. 2. (3 pts) Determine whether ± 1 ln x x 4 dx converges; if so, evaluate it. The easiest way to simplify this is integration by parts with u =ln x , dv = dx/x 4 , du = dx/x , v = - 1 / (3 x
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Unformatted text preview: ± ∞ 1 ln x x 4 dx = lim B →∞ ²-ln x 3 x 3 ³ ³ ³ ³ B 1-± B 1-dx 3 x 4 ´ = lim B →∞ ²-ln B 3 B 3 + ln 1 3-1 9 x 3 ³ ³ ³ ³ B 1 ´ = lim B →∞ µ-ln B 3 B 3-1 9 B 3 + 1 9 ¶ Of the three terms, the second one is clearly zero, and the ±rst one is zero as well by a straightforward application of L’Hopital’s rule (please review this if you don’t remember it). Thus the limit exists, and the integral converges to 1 / 9 ....
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