MIT18_014F10_pset3sols

MIT18_014F10_pset3sols - 18.014 Problem Set 3 Solutions...

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Unformatted text preview: 18.014 Problem Set 3 Solutions Total: 12 points Problem 1: Find all values of c for which c (a) x (1 x ) dx = 0. c (b) | x (1 x ) | dx = 0. Solution (4 points) (a) Computing, we get c c 1 1 x (1 x ) = ( x x 2 ) = 2 c 2 3 c 3 . Setting the right hand side equal to zero, we get solutions 3 c = and c = 0 . 2 (b) Observe x (1 x ) if x 1 and x (1 x ) if x 1 or x 0. There are three cases. If c < 0, then c c 1 1 | x (1 x ) | = x (1 x ) = 2 c 2 + 3 c 3 which is never zero for c < 0. If c 1, then c c 1 2 1 3 | x (1 x ) | = x (1 x ) = 2 c 3 c which is zero only if c = in the range c 1. Finally, if c > 1, then c 1 c | x (1 x ) | = x (1 x ) + x (1 x ) . 1 The first integral is 6 1 , and the second integral is non-negative by the comparison theorem (Thm. 1.20) since x (1 x ) on the interval (1 ,c ). Hence, this integral is always positive and never zero....
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This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.

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MIT18_014F10_pset3sols - 18.014 Problem Set 3 Solutions...

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