Question

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?

Answer 1

I like the cheating method :) Makes life easy. But I can't help you on the real proof..

Answer 2

in essence they are fractions, infinitesimal fractions. Quite valid to treat them that way.

dx/dt = lim t->0 Δx/Δt

Answer 3

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:
\[\frac{{dF}}{{dx}} = f\left( x \right)\]

Answer 4

oops..
\[dF = \frac{{dF}}{{dx}}\frac{{dx}}{{dt}} = f\left( x \right)dx\]

Answer 5

\( F(x) = \int f(x) dx \) 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 \(F(x) = \int \frac{dF}{dx}.\frac{dx}{dt} dt = \int f(x) \ \frac{dx}{dt} dt\)

Answer 6

but i agree totally with @LifeEngineer, in the physical sciences you won't go wrong just flipping them round like fractions :p

Answer 7

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.