I found this question and answer: Fundamental Theorem of Calculus: Why Doesn't the Integral Depend on Lower Bound? .
Would anyone be able to explain it words? I don't get the connection between the specific integral property mentioned in the answer and the theorem.
Because $f_1(x)=\int_a^x g$ and $f_2(x)=\int_b^x g$ differ by a constant, namely $f_2-f_1= \int_a^b g$. Hence they have the same derivative.
A derivative is a rate of change. A rate of change depends on what is changing. $$ \int_a^x f(u)\,du $$
When one puts $\dfrac d {dx}$ in front of this, one is viewing it as a function of $x$ with $a$ not changing. One gets $f(x)$.
If one writes $\dfrac d {da}$ in front of it, regarding $x$ as remaining fixed, while $a$ changes, then one gets $-f(a)$.
If $F$ is an antiderivative of $f$, then $$ \int_a^x f(t) \; dt = F(x) - F(a). $$ So if you take $x$ as a variable and take the derivative with respect to $x$, then the constant $F(a)$ disappears: $$ \frac{d}{dx} \int_a^x f(t) \; dt = \frac{d}{dx} F(x) - F(a) = \frac{d}{dx} F(x) = f(x). $$ So the $a$ "doesn't matter".
The key here is that $a$ is constant. If we had, for example, $$F(x)=\int_{x}^{a} f(t) dt$$ then we have $F'(x)=-f(x)$ (the minus sign is because it's the lower bound. Interestingly, we can have more complicated examples like $$G(x)=\int_{x}^{x^{2}}f(t)dt$$ where the integral depends on the upper and lower bounds. This is harder to differentiate, but gives the result $G'(x)=2xf(x^{2})-f(x)$
