Difference between integrand's variable and variable in integration's limits

Question

If a function $G(x)$ is defined by $$G(x)=\int_a^xf(t)dt$$ Then by the Fundamental Theorem I can see that $G(x)$ is an antiderivative of $f(x)$ and thus $$G'(x)=f(x)=\frac{dG(x)}{dx}$$ My confusion starts when I look at the integrand $f(t)$ and its variable $t$. The integrand $f(t)$ must have an antiderivative $F(t)$ as well according to: $$f(t)=\frac{dF(t)}{dt}$$ The $G$ and $F$ are both antiderivatives of $f$ and therefore must differ by a constant $c$ at most, which cancels out if you take $dF$ and $dG$, leading to $dF=dG$. Furthermore, if the variables $t$ and $x$ take on the same value $a$ then this would give $$dG(a)=dF(a)$$ $$f(x=a)=f(t=a)$$ $$\frac{dG(x=a)}{dx}=\frac{dF(t=a)}{dt}$$ The last equation surprises me because I considered the numerators to be equal ($t$ takes on the same value as $x$) but I'm not sure if the intervals $dt$ and $dx$ in the denominators are equal for this equation to be valid. These infinitesimals have different meanings according to the integral so they might not be equal.

If $dt \neq dx$, can someone pinpoint where exactly I went wrong in this reasoning?

Answer 1

The definition of derivative is $$ F'(a):= \lim_{x \to a} \frac{F(x)-F(a)}{x-a}=\frac{dF}{dx} $$ Changing $t$ with $x$ in the limit does not change the value of the limit. In other words $$ \frac{dF}{dt}(t=a):=\lim_{t \to a} \frac{F(t)-F(a)}{t-a}=\lim_{x \to a} \frac{F(x)-F(a)}{x-a}=\frac{dF}{dx}(x=a) $$ You are basically changing the name of the variable, but actually the name that you give to a variable of a function tell you nothing about the function itself

Answer 2

I would say that the problem is "leading to $\mathrm dF=\mathrm dG$". You can't really use $\mathrm d$ if you don't have varying quantities to apply them to. Something like "$\mathrm dF$" rather than $\mathrm d\left(F(t)\right)$ doesn't really make sense. If the input didn't matter, then you'd get ridiculous claims like "$F(5+s)$ and $G(7+s)$ must differ by a constant because their derivatives (with respect to $s$) are the same function with different inputs."

You can say something like "Let $y$ be $F(s)$ and $z$ be $G(s)$. Then $y$ and $z$ differ by a constant, and $\mathrm dy=\mathrm dz=f(s)\,\mathrm ds$." But then the problem disappears because $s$ is basically a dummy variable, and there's nothing wrong with equations like $$\left.\dfrac{\mathrm{d}G\left(x\right)}{\mathrm{d}x}\right|_{x=a}=\left.\dfrac{\mathrm{d}G\left(s\right)}{\mathrm{d}s}\right|_{s=a}=\left.\dfrac{\mathrm{d}F\left(s\right)}{\mathrm{d}s}\right|_{s=a}=\left.\dfrac{\mathrm{d}F\left(t\right)}{\mathrm{d}t}\right|_{t=a}$$