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FoolAroundMath
 one year ago
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?
FoolAroundMath
 one year ago
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I like the cheating method :) Makes life easy. But I can't help you on the real proof..

alekos
 one year ago
Best ResponseYou've already chosen the best response.0in essence they are fractions, infinitesimal fractions. Quite valid to treat them that way. dx/dt = lim t>0 Δx/Δt

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for 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: \[\frac{{dF}}{{dx}} = f\left( x \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. \[dF = \frac{{dF}}{{dx}}\frac{{dx}}{{dt}} = f\left( x \right)dx\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\( F(x) = \int f(x) dx \) 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
 one year ago
Best ResponseYou've already chosen the best response.2but i agree totally with @LifeEngineer, in the physical sciences you won't go wrong just flipping them round like fractions :p

FoolAroundMath
 one year ago
Best ResponseYou've already chosen the best response.0Sweet 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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