# Using integrals in physics in a rigorous way

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)

• What are "the extremes of integration"? Commented Mar 19, 2022 at 19:51
• 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 $dx/dt = v(t)$.
– Rob
Commented Mar 19, 2022 at 20:11

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.