quiz4-sol - u du =-2 √ u C =-2 √ 3-x C Therefore lim t...

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Quiz 4 – Math 2153, Calculus II – Sept. 16, 2011 1. Evaluate the improper integral Z 3 2 1 3 - x dx Solution This is an improper integral of type 2. We have Z 3 2 1 3 - x dx = lim t 3 Z t 2 1 3 - x dx Set u = 3 - x , we have du = - dx and Z 1 3 - x dx = Z - 1
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Unformatted text preview: u du =-2 √ u + C =-2 √ 3-x + C Therefore, lim t → 3 Z t 2 1 √ 3-x dx = lim t → 3 (-2 √ 3-x )± ± t 2 = lim t → 3 (-2 √ 3-t + 2 √ 3-2 ) = 0 + 2 = 2...
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This note was uploaded on 09/22/2011 for the course MATH 2153 taught by Professor Staff during the Fall '08 term at Oklahoma State.

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