Very Simple Question, please help: Integral of a derivative squared


by gilgtc
Tags: derivative, integral, simple, squared
gilgtc
gilgtc is offline
#1
Jun5-08, 01:19 PM
P: 6
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!
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dirk_mec1
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#2
Jun5-08, 01:40 PM
P: 664
I think partial integration can work.
Nick R
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#3
Jun5-08, 01:43 PM
P: 70
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?

dirk_mec1
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#4
Jun5-08, 01:53 PM
P: 664

Very Simple Question, please help: Integral of a derivative squared


Wait a second is x(t) explicitly known?
gilgtc
gilgtc is offline
#5
Jun5-08, 02:26 PM
P: 6
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!
HallsofIvy
HallsofIvy is offline
#6
Jun5-08, 02:59 PM
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Thanks
PF Gold
P: 38,879
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.
dirk_mec1
dirk_mec1 is offline
#7
Jun5-08, 02:59 PM
P: 664
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]
gilgtc
gilgtc is offline
#8
Jun5-08, 04:27 PM
P: 6
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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