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I would like to know how can someone justify rigorously the following argument, which is often involved in the solutions of many physics problems: given $v(t)$, we know that $v=\frac{dx}{dt}$, so we integrate in the following way: $\int_{x_0}^{x}dx=\int_{t_0}^{t}v(t)dt$. My question is, why are we allowed to "shift" the extremes of integration when moving from $dx$ to $dt$? (obviously, I know that the upper x represents the position at time t, but I'm seeking for a mathematical proof)

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  • $\begingroup$ What are "the extremes of integration"? $\endgroup$
    – Yuriy S
    Commented Mar 19, 2022 at 19:51
  • $\begingroup$ Ultimately it comes down to the fact that the two integrals are over different domains. The first is an integral over $x$, in the $x$ domain. The second is an integral over $t$, in the $t$ domain. Luckily, the two variables $x$ and $t$ are related - precisely by the position function itself $x = x(t)$. The fundamental theorem of calculus (the statement that these two integrals are equal) viewed in this way is essentially the change of variables theorem, as the transformation $t \to x$ is given by the position function, and the jacobian of the transform is precisely $dx/dt = v(t)$. $\endgroup$
    – Rob
    Commented Mar 19, 2022 at 20:11

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The statement that if $v(t) = dx/dt$, then $$ \int_{t_0}^{t_1} v(t) \, dt = \int_{x_0}^{x_1} \, dx, $$ where $x_0:= x(t_0)$ and $x_1 := x(t_1)$, is precisely the Fundamental theorem of calculus, and a mathematical proof can be seen in the linked article. Note that in that article, $x$ is the variable and $f$ is the function, whereas for you, $t$ is the variable and $x$ is the function.

I would also point out that in your question, you use the expression $$ \int_{t_0}^t v(t) \, dt. $$ I would avoid using the integration variable $t$ as the same symbol as one of the limits of integration. I know that it can be understood but it can sometimes lead to confusion when doing changes of variables. So I replaced the upper limits with $t_1, x_1$ as I wrote above.

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