7_Problem CHAPTER 9

7_Problem CHAPTER 9 - PROBLEM 9.6 CONTINUED For the second...

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PROBLEM 9.6 CONTINUED For the second integral use substitution 1 1, ( 1 ) y ud u d y y b u bb = −⇒ = = − 01 0 yu ybu = = = = Now 1 2 22 2 00 1( 1 ) y yd y b u u d u b  −=   ∫∫ 35 3 5 7 1 2 2 2 2 0 0 33 1 1 242 2 357 buu u d u b u u u =− + + 3 7 08 43 0 1 6 1 0 5 1 0 5 b −+ =+ + = = Then 3 16 51
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This note was uploaded on 07/11/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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