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FA06Ma.9Hw9Sols

# FA06Ma.9Hw9Sols - Ma 1a – Section 1 Fall 2006 HOMEWORK#9...

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Unformatted text preview: Ma 1a – Section 1 Fall 2006 HOMEWORK #9 Solutions 1. See solutions in Apostol. 2. (a) 6.9.19, p.236. Let f ( x ) = x log( x ) and g ( x ) = x log( x )- x . Then f ( x ) = log( x )+1 and g ( x ) = log( x ). Using integration by parts, we see Z x log 2 ( x ) dx = x log( x )( x log( x )- x )- Z ( x log 2 ( x ) + x ) dx + C, which shows that Z x log 2 ( x ) dx = 1 2 x 2 log 2 ( x )- x 2 log( x )- 1 2 x 2 + C. (b) 6.9.20, 9.236. Let u = 1 + t . Then du dt = 1, and we have Z e 3- 1 dt 1 + t = Z e 3 1 du u = log | u | e 3 = 3 . (c) 6.9.23, p.236. Let f ( x ) = x 2 log( x ) and g ( x ) = x log( x )- x . Using integration by parts we see that Z x 2 log 2 ( x ) dx = x 2 log( x )( x log( x )- x )- Z ( 2 x 2 log 2 ( x )- x 2 log( x )- x 2 ) dx + C, which gives 3 Z x 2 log 2 ( x ) dx = x 3 log 2 ( x )- x 3 log( x )+ 1 3 x 3 + Z x 2 log( x ) dx + C. (1) Using integration by parts again (with f ( x ) = x log( x ) and g ( x ) = 1 2 x 2 ) we find Z x 2 log( x ) dx = 1 2 x 3 log( x )- 1 2 Z ( x 2 log(...
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FA06Ma.9Hw9Sols - Ma 1a – Section 1 Fall 2006 HOMEWORK#9...

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