## 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?

1. anonymous

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

2. alekos

in essence they are fractions, infinitesimal fractions. Quite valid to treat them that way. dx/dt = lim t->0 Δx/Δt

3. 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: $\frac{{dF}}{{dx}} = f\left( x \right)$

4. Michele_Laino

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

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

6. IrishBoy123

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

7. 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.