Can you check to make sure I am completely understanding the Fundamental Theorem of Calculus? I am not sure that I am completely understanding FTC in terms of how the derivative and definite integral are connected. I drew/wrote this depiction of how I currently think they are connected in terms of velocity and position. I was wondering if anyone could point out any potential gaps in my understanding.

 A: I'd say your geometric picture is a good start for understanding the fundamental theorems of calculus. (I say theorems because really there are two of them.) Say the velocity of a car that has been travelling for one second can be modelled by the equation
$$
v(t)=\sqrt{t+1} \quad{1\leq t} \, ,
$$
The velocity-time graph looks like this:

Since the horizontal axis represents time, and the vertical axis velocity, the area under the curve represents the $\text{velocity $\times$ time $=$ displacement}$ of the car. This means that the displacement of the car (from its position at $t=1$) after $T$ seconds is given by
$$
\int_{1}^{T}v(t) \, dt = \int_{1}^{T} \sqrt{t+1} \, dt \, .
$$
But we also know that the derivative with respect to time of displacement is the velocity, meaning that the antiderivative of velocity is displacement. That last sentence was a little sloppy: every function has infinitely many antiderivatives which differ by a constant. But we can use the information given in this example to narrow things down to one antiderivative:
$$
s(t) = \int v(t)\, dt \, = \int \sqrt{t+1} \, dt = \frac{2}{3}\left(t+1\right)^{3/2}+C
$$
When $t=1$, the displacement of the car is zero. Hence,
$$
s(1) = 0 = \frac{2}{3}\left(t+1\right)^{3/2}+C \implies C = -\frac{2}{3}2^{3/2} \, ,
$$
meaning that
$$
s(t) = \frac{2}{3}\left(t+1\right)^{3/2} - \frac{2}{3}2^{3/2} \, .
$$
Hence, after $T$ seconds, the displacement is given by
$$
s(T) = \frac{2}{3}\left(T+1\right)^{3/2} - \frac{2}{3}2^{3/2}
$$
But we also know that
$$
s(T) = \int_{1}^{T} \sqrt{t+1} \, dt \,
$$
which gives us a clue as to how integrals and antiderivatives are related. In more general terms,
$$
s(T) = \int_{1}^{T} v(t) \, dt = \int v(t) \, dt \bigg|_{t=T}
$$
(Here, I'm using the notation
$$
\int v(t) \, dt \bigg|_{t=T}
$$
in a slightly unorthodox way, to refer to one particular antiderivative rather than many.) We are almost there: if we differentiate both sides of
$$
\int_{1}^{T} v(t) \, dt = \int v(t) \, dt \bigg|_{t=T} \, ,
$$
we get
$$
\frac{d}{dt}\int_{1}^{T} v(t) \, dt = v(T) \, ,
$$
which is the first fundamental theorem of calculus. We have at last discovered a concrete connection between integrals and antiderivatives. Again, there is a nice geometric interpretation. Recall that by definition,
\begin{align}
\frac{d}{dt}\int_{1}^{T} v(t) &= \lim_{\Delta t \to 0}\frac{\int_{1}^{T+\Delta t} v(t) \, dt-\int_{1}^{T} v(t) \, dt}{\Delta t} \\[6pt]
&= \lim_{\Delta t \to 0}\frac{\int_{T}^{T+\Delta t}v(t) \, dt}{\Delta t} \, .
\end{align}
As $\Delta t$ shrinks towards zero, the area (marked in green below)
$$
\int_{T}^{T+\Delta t}v(t) \, dt
$$
is better and better approximated by the rectangle with width $\Delta t$ and height $v(T)$.

Since
$$
\lim_{\Delta t \to 0}\frac{\int_{T}^{T+\Delta t}v(t) \, dt}{\Delta t}
$$
is, roughly speaking, the ratio of the area of the rectangle to its width, this limit must be equal to the height of the rectangle $v(T)$, which is what we wanted. If I find the time, I'll try to update this post with more information about the second fundamental theorem of calculus.
