FoolAroundMath
  • FoolAroundMath
Can someone help me prove that \[\displaystyle \int f(x)\left (\frac{dx}{dt}\right )dt = \int f(x) dx \] What I would do is treat the differentials as fractions and just cancel the \(dt\) on numerator and denominator. However, I feel that it is cheating somehow. Is there a "formal" proof for this?
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I like the cheating method :) Makes life easy. But I can't help you on the real proof..
alekos
  • alekos
in essence they are fractions, infinitesimal fractions. Quite valid to treat them that way. dx/dt = lim t->0 Δx/Δt
Michele_Laino
  • Michele_Laino
for example if we can write the function F as follows: \[F\left( {x\left( t \right)} \right)\] then applying the chain rule we get: \[\frac{{dF}}{{dt}} = \frac{{dF}}{{dx}}\frac{{dx}}{{dt}}\] where F is such that: \[F\left( x \right) = \int {f\left( x \right)\;dx} \] Next multiplying both sides of the last equation, by dt, we get: \[dF = \frac{{dF}}{{dx}}\frac{{dx}}{{dt}} = f\left( x \right)\frac{{dx}}{{dt}}\] since, by definition, we can write: \[\frac{{dF}}{{dx}} = f\left( x \right)\]

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Michele_Laino
  • Michele_Laino
oops.. \[dF = \frac{{dF}}{{dx}}\frac{{dx}}{{dt}} = f\left( x \right)dx\]
IrishBoy123
  • IrishBoy123
\( F(x) = \int f(x) dx \) ie \(\frac{dF}{dx} = f(x)\) then chain it wrt t, ie \(x = x(t), \frac{dF}{dt} = \frac{dF}{dx}.\frac{dx}{dt}\) so \(F(x) = \int \frac{dF}{dx}.\frac{dx}{dt} dt = \int f(x) \ \frac{dx}{dt} dt\)
IrishBoy123
  • IrishBoy123
but i agree totally with @LifeEngineer, in the physical sciences you won't go wrong just flipping them round like fractions :p
FoolAroundMath
  • FoolAroundMath
Sweet thanks, this is exactly what I was looking for. I don't deny that I'll be using them as fractions when applying them :p, it's just that I needed a formal proof. It has been bugging my mind and I couldn't prove it myself.

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