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Article #15Re: Prove Integration by Parts formula
From: Rob Morewood
Date: Sun, 21 Feb 1999 00:00
Date: Sun, 21 Feb 1999 00:00
26 lines
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1672 bytes
aytcH (pinky@NOSPAM.for-president.com) wants to prove: : Integration by Parts: integral of udv = uv - integral of vdu : I think I find their derivatives are the same, then the formula is proven. <details invoving a double integral omitted.> You have the exact RIGHT idea. Take the derivatives with respect to x. The chain rule is required to differentiate the integrals. (On the left: differentiate with respect to v and multiply by dv/dx). WARNING: Two functions with the SAME derivative do not have to be equal, they can differ by a constant. But the indefinite integrals in your formula are only defined up to a constant (they include an undetermined constant). So this formula IS proven just by showing that the derivatives are equal. If you forget about that hidden constant, you can use the formula to "prove" that 1=0! Consider u=1/x and v=x (so du=dx and dv=-(1/x^2)dx): Integral of (1/x)dx = 1 - Integral x*(-1/x^2)dx = 1 + Integral (1/x)dx Cancel the Integral on both sides to get: 0 = 1. |\/| || Burnaby South Secondary School | |orewood@planeteer.com || Beautiful British Columbia Mathematics & Computer Science || (Canada)
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X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: f7972,5a3a3d43952b6ee9 X-Google-Attributes: gidf7972,public From: see.sig@for.real.address (Rob Morewood) Subject: Re: Prove Integration by Parts formula Date: 1999/02/21 Message-ID: <19990221.7BD2C38.348E@van1p12.planeteer.com> X-Deja-AN: 446672876 References: <01be5c92$59693400$5107400c@default> Organization: UUNET Canada News Transport Newsgroups: can.schoolnet.math.sr