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Very Simple Question, please help: Integral of a derivative squared |
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| Jun5-08, 01:19 PM | #1 |
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Very Simple Question, please help: Integral of a derivative squared
Hello,
I am trying to figure out how to integrate this, I know it must be simple but I am not sure how to do it. [tex]\int(\frac{dx}{dt})^{2}dt[/tex] Thanks a lot! |
| Jun5-08, 01:40 PM | #2 |
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I think partial integration can work.
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| Jun5-08, 01:43 PM | #3 |
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You just need x defined in terms of t
x = x(t) then you can differentiate with respect to t, then you square dx/dt then you integrate that across t from t1 to t2 right? |
| Jun5-08, 01:53 PM | #4 |
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Very Simple Question, please help: Integral of a derivative squared
Wait a second is x(t) explicitly known?
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| Jun5-08, 02:26 PM | #5 |
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hi, thanks for your answers. x(t) is not known that is why I am not sure how to do it. Otherwise what Nick mentioned would be easily applicable.
Any other ideas? What do you mean by partial integration dirk_mec1? Thanks! |
| Jun5-08, 02:59 PM | #6 |
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He means "integration by parts". Do you have any reason to think that there is any simple answer to this question? I can see no reason to assume that
[tex]\right(\frac{dx}{dt}\)^2[/tex] even has an elementary anti-derivative. |
| Jun5-08, 02:59 PM | #7 |
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I dont think you can just integrate [tex] \int (f'(x))^2 \mbox{d}x[/tex], right? The integration by parts(thanks hallsofIvy ) however gives:
[tex] x (x'(t))^2 - \int x \cdot 2x' \cdot x''\ \mbox{d}t [/tex] |
| Jun5-08, 04:27 PM | #8 |
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I thought that there was an easy equivalence like:
[tex]\int(dx/dt)dt = x[/tex] I guess not! thanks for your help in any case. |
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