first was not meant to be defined as written, it's actually, without the square.

let and both continuous, then and now substitute in this integral and that shows that thus is constant and and then obviously then and the conclusion follows.

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- Apr 4th 2010, 10:17 AM #1

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- Apr 4th 2010, 01:47 PM #2

- Apr 4th 2010, 04:52 PM #3

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