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quiz 3.2 - ± ∞ e x e 2 x 3 dx = lim B →∞ ± B e x e...

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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 dx x 2 - 4 x . First, complete the square: x 2 - 4 x = ( x - 2) 2 - 4 = ( x - 2) 2 - 2 2 . This suggests the substitution x - 2 = 2 sec θ , dx = 2 sec θ tan θ d θ : dx x 2 - 4 x = 2 sec θ tan θ d θ 4 sec 2 θ - 4 = 2 sec θ tan θ d θ 2 tan θ = sec θ d θ = ln | sec θ + tan θ | + C = ln | sec θ + sec 2 θ - 1 | + C = ln x - 2 2 + x - 2 2 2 - 1 + C = ln x - 2 2 + x 2 - 4 x 2 + C. 2. (3 pts) Determine whether 0 e x e 2 x + 3 dx converges; if so, evaluate it. This is improper only because of the upper bound. Substitute u = e x and evaluate:
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Unformatted text preview: ± ∞ e x e 2 x + 3 dx = lim B →∞ ± B e x e 2 x + 3 dx = lim B →∞ ± e B 1 du u 2 + 3 = lim B →∞ ¶ 1 √ 3 arctan u √ 3 · e B 1 = lim B →∞ ´ 1 √ 3 arctan e B √ 3-1 √ 3 arctan 1 √ 3 µ Now arctan(1 / √ 3) = π/ 6, and e B / √ 3 goes to in±nity as B does, so we take lim x →∞ arctan x = π/ 2. Then the improper integral does converge, and it con-verges to (1 / √ 3)( π/ 2-π/ 6) = π 3 √ 3 ....
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