Is the integral of $\frac{dx}{dt} \cdot f(x) dt$ equal to the derivative of $f(x)$ with respect to $x$?

Question

I'm trying to figure out a proof and I cant understand if this is true.

Is

$$\displaystyle \int {\frac{dx}{dt} \cdot f(x)dt} = \int f(x) dx$$

If so why? Thanks!

Answer 1

There are a lot of ways to approach this depending on how you define $\text dt, \text dx(t)$, and $\int f(x) \text dx$. Using the standard definitions in introductory calculus, $\text dt$ is a differential, and if $x(t)$ is a function of $t$, then $$\text dx(t) = \frac{\text dx}{\text dt}(t)\text dt = x'(t)\text dt$$ For a given $x$ with differential $\text dx$, we say $$x = \int \text dx$$ Thus, the statement $\int f(x)\text dx$ means to find the $F$ whose differential is $$\text dF = F'(x)\text dx = f(x)\text dx$$ then $$\int f(x)\text dx = F(x)$$

Thus,

$$ \int \frac{\text dx}{\text dt}f(x)\text dt = \int f(x) x'(t)\text dt = \int f(x) \text dx $$

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Edit to note that this definition of $\int$ is the "antiderivative" or "indefinite integral". This is different from the definite integral $\int_\Omega$ for some domain $\Omega$. Showing the relationship between $\int_\Omega$ and $\int$ is essentially the fundamental theorem of calculus, and is not what is shown here.