Why doesn't the fundamental theorem of calculus depend on the lower bound? 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.
 A: 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: 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 $-f(a)$.
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".
A: 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 $G'(x)=2xf(x^{2})-f(x)$
