Quiz 3.2 - ± ∞ 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

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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 .Th i ssugg e s t s the substitution x - 2=2sec θ , dx =2sec θ tan θdθ : ± dx x 2 - 4 x = ± 2sec θ tan θdθ 4sec 2 θ - 4 = ± 2sec θ tan θdθ 2tan θ = ± sec θdθ =ln | sec θ +tan θ | + C =l n | sec θ + sec 2 θ - 1 | + C =l n ² ² ² ² ² ² x - 2 2 + ³ ´ x - 2 2 µ 2 - 1 ² ² ² ² ² ² + C =l n ² ² ² ² x - 2 2 + x 2 - 4 x 2 ² ² ² ² + C. 2. (3 pts) Determine whether ± 0 e x e 2 x
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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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This note was uploaded on 10/20/2010 for the course MATH 53903 taught by Professor Christ during the Spring '09 term at University of California, Berkeley.

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