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When the formula of kinetic energy is derived using calculus, the differentials in the integral are rearranged like below. Could you point to how to understand this rearrangement of differentials ? o_O I could not understand what is going on behind the scene even after reading posts about separating x and y here and here...

$$ \begin{array}{l}{\Delta K=m \int \frac{d \mathbf{v}}{d t} \cdot d \mathbf{r}} \\ {\Delta K=m \int \frac{d \mathbf{r}}{d t} \cdot d \mathbf{v}} \\ {\Delta K=m \int \mathbf{v} \cdot d \mathbf{v}}\end{array} $$

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    $\begingroup$ For your further interest, kinetic energy can be related as: $$\mathbf{v}=\frac{\partial E}{\partial \mathbf{p}}$$ or alternatively, $$\Delta E=\int \mathbf{v} \cdot d\mathbf{p}$$ See also another answer of mine here. $\endgroup$ – Ng Chung Tak Nov 26 '19 at 2:59
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Funny, we did this same exact lesson today in my AP physics class. I didn't think I'd be applying it so soon.

Anyways, the key is understanding the chain rule, namely (in general): $$\frac{dF}{dx}=\frac{dF}{du}\frac{du}{dx}$$

That's what's going on here. We have:

$$W=m\int_{x_0}^{x_f}{\frac{dv}{dt}{dx}}$$

Which can be rewritten as: $$W=m\int_{x_0}^{x_f}{\frac{dv}{dx}\frac{dx}{dt}dx}$$

Since $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$$

Recall now that $v=\frac{dx}{dt}$, this becomes: $$W=m\int_{x_0}^{x_f}{v\frac{dv}{dx}{dx}}$$

$dx$ cancels leaving you with $$W=m\int_{v_0}^{v_f}{vdv}$$

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Perhaps it is easier to convince if you don't use differentials: $$\begin{align}ma~dr\approx \dfrac{m\Delta v\Delta r}{\Delta t} = m\dfrac{\Delta r}{\Delta t}\Delta v \end{align}$$ $\Delta v, \Delta r, \Delta t$ are real numbers so regular arithmetic can be performed on them. In the limit you replace $\Delta$ by $d$.

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