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Why is this proof wrong |
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| May13-09, 10:19 PM | #1 |
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Why is this proof wrong
Another one of those 0 = 1 proofs.
If you dislike them, please navigate away to cats doing things with captions. Otherwise stay tuned. Tell me what's wrong (obviously I know what's wrong but I just want to put it out there for people to mull over). Note, you need to know basic calculus. [tex] \int \frac{1}{x} dx = \int \frac{1}{x} dx [/tex] [tex] u = \frac{1}{x} [/tex] [tex] dv = dx [/tex] [tex] du = \frac{-1}{x^2} dx [/tex] [tex] v = x [/tex] [tex] \int \frac{1}{x} dx = u * v - \int v du [/tex] [tex] \int \frac{1}{x} dx = \frac{1}{x} * x - \int x * \frac{-1}{x^2} dx [/tex] [tex] \int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx [/tex] [tex] \int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx [/tex] [tex] 0 = 1 [/tex] |
| May13-09, 10:28 PM | #2 |
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Right off the bat, [tex]u = \frac{{dv}}{x} = dx[/tex] makes no sense. The second line makes no sense either.... am i completely missing something?
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| May13-09, 10:32 PM | #3 |
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Sorry those are supposed to be separate. I will edit that in.
This is what it should look like [tex] u = \frac{1}{x} [/tex] [tex] dv = dx [/tex] I have also added an extra step just to show that I am going to be using integration by parts to do the "proof" |
| May13-09, 11:25 PM | #4 |
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Why is this proof wrong
[tex]\int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx[/tex] isn't valid. Integration by part goes like:
[tex]\int\limits_a^b {udv} = [uv]_a^b - \int\limits_a^b {vdu} [/tex] The term you think is 1 is actually 0. |
| May14-09, 02:51 AM | #5 |
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We are dealing with antiderivatives. So, given F(x) and G(x) that are antiderivatives of 1/x, it is true that F(x) = G(x) + C, for some constant C. For example, the second to last line you have can be written as
ln(x) + C = 1 + ln(x) + D for some constants C, and D. We do not then conclude that 0 = 1. The integral of 1/x dx is a FAMILY of functions that all differ by a constant. |
| May14-09, 04:04 AM | #6 |
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I only got to the end of the second sentence
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| May14-09, 04:44 AM | #7 |
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Mentor
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Russell has the right answer. To be strictly correct, the rule for integration by parts is better written as
[tex]\int u dv = u*v - \int v du + C[/tex] We drop the arbitrary constant because it is implied by the very use of indefinite integrals, or antiderivatives. The inverse of the derivative is not unique. |
| May14-09, 04:46 AM | #8 |
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Russell and D_H are correct, the constant term is missing. Gj guys :)
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| May16-09, 03:16 PM | #9 |
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| May16-09, 03:38 PM | #10 |
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Right, but since I defined the problem in the start as indefinite integrals, I was looking for the answer that was related to indefinite integrals. Anyways. It's just a simple problem
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