133quiz6bsolution - Quiz 6 solution 2 1 Evaluate dx (4 x 2 0 2 π dx (4 x 0 2 2 3 = 2 −1 ⎛ x ⎞ ⎪ x = 2 tan θ θ = tan ⎜ 2 ⎟ ⎝

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Quiz 6 solution 1. Evaluate 2 3 2 2 0 (4 ) dx x + 1 2 2 2 tan ; tan 2 2sec 42 s e c x x dx d x θθ θ ⎛⎞ == ⎜⎟ ⎝⎠ = += 2 2 44 4 3 3 2 2 0 00 0 1 1 1 1 1 cos sin 0 (2sec ) 4 4 4 24 ) dx d d x ππ π = = = + ∫∫ 2 2. Determine whether the improper integral converges or diverges. 22 2 0 11 1 18 1 1 lim lim lim ln(1 4 ) lim ln(1 4 ) 14 814 8 8 t tt t t xdx xdx xdx xt xx x →∞ →∞ →∞ →∞ = = =+ = ++ + + Alternatively, use the limit comparison test:
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This note was uploaded on 02/05/2010 for the course MATH 133 taught by Professor Wei during the Fall '07 term at Michigan State University.

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