When we evaluate an indefinite integral of one variable, what area does this yield? For example, take the function $ v(t) = t(8 - t) $ from Grant Sanderson's video on integration. Then its antiderivative is $ x(t) = -\frac{1}{3}t^3 + 4t^2 + C $.
If I evaluate this antiderivative at $ t = 2 $, so $ x(2) = \frac{40}{3} + C $, does it make sense to talk of this quantity as an area? Is there an implicit lower bound $ c $, and if so, what is that bound?
If we are given that $ \int_a^b f'(x) \mathrm{d}x = f(b) - f(a) $, how can we be assured that $ c \le a \le b $ for some lower bound $ c $ that is the lower bound of the areas $ f(b) $ and $ f(a) $?
I apologize if this question is nonsensical. I've made it through Calculus III in a somewhat rote fashion, and I still struggle with developing an intuition for integration.
 A: Here is what integration tells us. Integration is essentially antidifferentiation, but while differentiation tells us the gradient of a curve integration tells us the area under a curve. However, it is meaningless to try to evaluate the integral at a specific point and associate it with area-the area under the curve from where until where (as you seem to be confused about)?
However, we can use the value of the integral at $2$ different $x$-coordinates to determine the area under the graph between those $2$ different $x$-coordinates. For example, if you wanted to find the area under the curve $v(x)=x(8−x)=8x-x^2$ between $x=0$ and $x=2$ you find the defnite integral
$$\int_{0}^28x-x^2dx$$
and the reason why we don't have any constants of integration with definite integrals is because they cancel each other out. To explain: in the above (your) example, we'd soon get
$$(4(2)^2-\frac{2^3}{3}+C)-(0-0+C)=(4(2)^2-\frac{2^3}{3})-(0-0)$$
so it's convenient to always leave out the constant of integration when dealing with defnite integrals.
I hope that helped. If you would like some help understanding why integration tells us the area under the curve, or for any more help, add a comment and I will edit my answer.
EDIT:
Here is a complete explanation on integration to help you:
Before we begin, I just need to make sure you know the definition of a derivative. A derivative tells us the gradient of a function at any particular point and is equal to, for a function $f(x)$:
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
and if we want have a derivative and we want to find the original function then we integrate.
Now, we are going to attempt to find a supposed area accumulator function, ie a function that will tell us the area underneath a graph between the origin and a particular point. Let this supposed function be called $F(x)$.
For any graph of a function $f(x)$, the area under the graph betweeen $2$ points, let's call them $x$ and $x+h$ (where $x+h>x$), will approximately be equal to a trapezium (or trapezoid if you're American:) ) of height $(h)$ and base length and parallel length  $f(x+h)$ and $f(x)$ respectively, ie of area $\frac{h(f(x+h)+f(x))}{2}$. It may help you to draw a graphg to help you fully understand my meaning. The smaller the gap between the $2$ $x$-coordinates the more closely the area under the curve between will be approximately equal to the trapezium. Let's make the the gap between these $2$ $x$-coordinates indefnitely small, so the area, which can be written as $F(x+h)-F(x)$ is equal to
$$\lim_{h\to0}F(x+h)-F(x)=\lim_{h\to0}\frac{h(f(x+h)+f(x))}{2}$$
Now, if we divide both sides by $h$ we obtain the following:
$$\lim_{h\to0}\frac{F(x+h)-F(x)}{h}=\lim_{h\to0}\frac{f(x+h)+f(x)}{2}$$
but we can see that the expression on the left hand side is the definition of the derivative, $F'(x)$, for $F(x)$, our area accumulator function. So we can write
$$F'(x)=\lim_{h\to0}\frac{f(x+h)+f(x)}{2}=\frac{f(x)+f(x)}{2}=\frac{2f(x)}{2}=f(x)$$
So in an example of mathematical beauty we can see that
$$F'(x)=f(x)\implies F(x)=\int f(x)dx$$
So this is why integration tells us the area underneath a curve. This result is known as the Fundamental Theorem of Calculus. If you have any more questions don't hesitate to ask :)
A: A nice way of visualising why the integral in terms of one variable gives area is to view it as this:
$$\int_0^tf(x)dx=\int_0^t\int_0^{f(x)}dydx$$
and since this is a Riemann integral we can think of this as the sum of lots of small rectanges, of area $dydx$ and the area if bounded by the four sides: $[0,t]\times[0,f(x)]$
In your case you have a function $v(t)$ and you want to know the displacement at a given time. The reason why we do not have this unknown constant term is we have defined the range over which this occurs (it is a definite integral), so the change in displacement (distance travelled since $t=0$) can be represented as:
$$x(2)=\int_0^2t(8-t)dt$$
That is to say, we are assuming here that the displacement at 0 is 0 i.e. $x(0)=0$
