Calculus 2 - Assignment 2 Solutions

# Calculus 2 - Assignment 2 Solutions - MAT1332 Spring/Summer...

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MAT1332 Spring/Summer 2009 Assignment 2 Solutions 1. Long division gives 6 x 3 - 4 x + 8 3 x 2 - 9 x = 2 x + 6 + 50 x + 8 3 x 2 - 9 x . The denominator has two distinct real roots and in particular 3 x 2 - 9 x = 3 x ( x - 3). We would like to ﬁnd A and B such that 50 x + 8 3 x 2 - 9 x = A 3 x + B ( x - 3) . Taking x : = 0 and x : = 3 gives A =- 8 3 and B = 158 9 . Thus, Z 6 x 3 - 4 x + 8 3 x 2 - 9 x dx = x 2 + 6 x - 8 9 Z 1 x dx + 158 9 Z 1 x - 3 dx + C = x 2 + 6 x - 8 9 ln | x |+ 158 9 ln | x - 3 |+ C , and we conclude that Z 2 1 6 x 3 - 4 x + 8 3 x 2 - 9 x dx = 9 - 1 9 ( 8ln | 2 |+ 158ln | 2 | ) = 9 - 166 9 ln | 2 | which is ≈- 3.7847. 2. The integral converges. This can be seen using the comparison test. First, observe that, for any x , x 6 < x 6 + 1, whence 1 1 + x 6 < 1 x 6 . As such, since R 1 1 x 6 dx converges to 1 5 it follows that R 1 1 1 + x 6 dx converges. 3. First, note that lim x 0 x 2 = 0 = lim x 0 1 - cos( x ). Thus, we may apply l’Hôpital’s rule to conclude that lim x 0 1

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## This note was uploaded on 01/16/2011 for the course MAT 1332 taught by Professor Munteanu during the Fall '07 term at University of Ottawa.

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Calculus 2 - Assignment 2 Solutions - MAT1332 Spring/Summer...

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