Definite Integral = Indefinite Integral with Variable as Upper Limit

Question

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 $?

---

(Supplementaries to Yury's answer)

$\Large{\text{Question #1.1.}}$ 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 $C = -g(t_0)$?

Answer 1

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.

Answer 2

Since Yury answered your first two questions, I'll attempt your supplementary ones.

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).

Does this clear things up?

Also, strictly speaking, it would be more accurate to say that $(**) \in (*)$.