Definite Integral = Indefinite Integral with Variable as Upper Limit

I don't understand why the following equality in grey is true, from P36 in Elementary Differential Equations, 9th Ed by Boyce, DiPrima et al:

$$\int{f(t)\text{ }dt} = \int_{t_0}^t f(s) \text{ } ds \text{ where t_0 is some convenient lower limit of integration.}$$

$\Large{\text{Question #1.}}$ $\int{f(t)}\text{ }dt$ has no lower or upper limits of integration, so how can you equate it with $\int_{t_0}^t f(s) \text{ } ds$, where $t$ (the upper limit of integration) varies along the $s$-axis and $t_0$ (the lower limit of integration) is a fixed point on the $s$-axis? Is $t_0$ related to the integration constant $C$?

$\Large{\text{Question #1.1.}}$ How is $\int{f(t)\text{ }dt} \color{purple}{\text{ (ie the set)}} =\int_{t_0}^t f(s) \text{ } ds \color{purple}{\text{ (ie an element of the set on the left)}}?$

$\Large{\text{Question #2.1.}}$ Specifically, as you explained, $C = -g\left(t_0\right)$ is required. But how is this true? $g(t)$ is one particular function and $t_0$ is one particular argument. On the other hand, $C$ is any real number.

(Supplementary to William Stagner's answer and Peter Tamaroff's comment)

$\Large{\text{Question #3. (Please see Q#2.1 as well)}}$ Thanks to your explanations, I understand that: $\int{f(t)}\text{ }dt = g(t) + C \qquad \forall \ C \in \mathbb{R}\ \tag{*}$ and $\int_{t_0}^t f(s) \text{ } ds = g(t) - g(t_0) \tag{**}$

$(*) = (**)$ is true $\iff C = -g(t_0).$ But how is $C = -g(t_0)$?

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When the author writes $\int f(t) dt=\int_{t_0}^t f(x) dx$ he is saying that $\int_{t_0}^t f(x) dx$ is an antiderivative for $f$. That is all there is to it. There is no need to be talking about $\int f(t) dt$ being a set. It is true there is some bad notation going on, however, but don't interpret that the author is equating "a set" with "a function". – Peter Tamaroff May 11 at 0:06
@PeterTamaroff: Thank you for your comment. I've updated my original post to avoid the notation insinuating that (a set) = (a function). Unfortunately, I still remain troubled by Question #3 and don't understand why the equality in grey is true. – LaPrevoyance May 12 at 13:06
I believe that the first equality you present is the book's definition of the symbol on the left. So there's nothing to prove or to ask. It's just a convenient way to get one function instead of a set of functions. – egreg May 12 at 13:24
@egreg: Thanks for your comment. However, the aforementioned textbook didn't define the equality in grey. Instead, it just said that these two integrals were the same and I don't understand why. – LaPrevoyance May 12 at 13:42

The first expression $$\int{f(t)\,dt}$$ is the antiderivative of $f$. It is defined up to a constant: $$\int{f(t)\,dt} = g(t) + C.$$ The second expression $$\int_{t_0}^t f(s) \, ds$$ is the definite integral of $f(x)$. By the fundamental theorem of calculus, $$\int_{t_0}^t f(s) \, ds = g(t) - g(t_0).$$

Two expressions are equal when $C = - g(t_0)$. Formally, $\int{f(t)\,dt}$ is a set of functions of the form $f(x) + C$, and $\int_{t_0}^t f(s) \, ds$ is an element of this set.

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 Thank you for your response. Could you please look at 2 suppelementary questions with which I've updated my original post? – LaPrevoyance May 7 at 23:01

As he stated, $$\int f(t)\ dt = g(t) + C$$ is well defined up to a constant. What does this mean? There is a whole class of antiderivatives of $f$ which we can specify exactly, namely $g(t)$ plus some constant. Thus, the set of anti derivatives of $f$ is $$\{ g(t) + C : C\in \mathbb{R}\}.$$ So we're looking at a set of functions. The expression $g(t) +C$ is really just shorthand for the set of functions defined above. Hence, $\int_{t_0}^t f(s)\ ds = g(t) - g(t_0)$ is just an element of this set, where $C=-g(t_0)$.
I think this answers both of your questions. The expression $g(t) + C$ is not the antiderivative of $f$, but represents the set of antiderivatives of $f$, which differ from $g$ by the addition of a constant.
Question 3 If I understand correctly, this is mainly an ambiguity between fixed and free variables. It is not true that "we don't know the value of $t_0$." We do know the value, it is in fact, $t_0$! Although $t_0$ is arbitrary, it is still a fixed variable. So in the expression $$g(t) - g(t_0)$$ the variable $t$ is free to vary over $\mathbb{R}$, but $t_0$ is absolutely fixed from the beginning (although arbitrary).
Also, strictly speaking, it would be more accurate to say that $(**) \in (*)$.
 Thank you very much for your response. Unfortunately, I still don't understand why ($*$) = (**). Could you please look at supplementary question #3 in my original post? – LaPrevoyance May 10 at 22:42 Sorry, I don't think I understand your question. Could you try to rephrase it? – William Stagner May 12 at 23:03 No problem. I've rewritten Question #3 in my original post but I'll rephrase it once more here. I understand that the equality in grey holds $\iff C = -g(t_0)$. But I don't understand how we are given or know $C = -g(t_0).$ The textbook defined $t_0$ as some fixed limit of integration, for which we've no other information. Since we don't know $t_0$, therefore we can't know $-g(t_0)$ either. – LaPrevoyance May 14 at 20:36 I've updated my answer. Let me know if I didn't explain well enough. – William Stagner May 15 at 14:33 Thank you very much for your response once again. Sadly, I still can't figure out Q#3, so I've updated/shortened it in my original post. Moreover, please also see Q #2.1. – LaPrevoyance May 15 at 16:52