Prove Integration by Parts formula

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2 total messages Started by aytcH Sat, 20 Feb 1999 00:00

#14

Author: aytcH
Date: Sat, 20 Feb 1999 00:00

10 lines
784 bytes

Integration by Parts: integral of udv = uv - integral of vdu
I think I find their derivatives are the same, then the formula is proven.

if I let 'S' stand for the symbol of integration, S udv = uv - S vdu
I started by solving Sudv by doing a double integral : SSudv and got
(u*u/2)dv + uv
and so on, but I don't know where to go from there or if I'm approaching it
correctly.

Please help.

Re: Prove Integration by Parts formula

#15

Author: Rob Morewood
Date: Sun, 21 Feb 1999 00:00

26 lines
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
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Mathematics & Computer Science  ||             (Canada)

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