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Source: P36 of Elementary Differential Equations, 9th Ed by Boyce, DiPrima et al.

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

$\Large{\text{1.}}$ I now know: $ \int{f(t)\text{ }dt} \qquad$ is $\color{green}{\text{ a set of functions $\qquad$ (**)}} $
and $ \int_{t_0}^t f(s) \text{ } ds \qquad$ is $\color{#B53389}{\text{ an element of set (**) above.}} \quad $
So how and why is $(*)$ true? How can a $\color{green}{\text{ a set of functions}}$ = $\color{#B53389}{\text{ an element of the same set}} $ ?


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

Thanks to your explanations, I now know that: $\int{f(t)}\text{ }dt = g(t) + C \qquad \forall \ C \in \mathbb{R}\ \tag{$\natural$}$ $\int_{t_0}^t f(s) \text{ } ds = g(t) - g(t_0) \tag{$\blacklozenge$}$

Since $g(t)$ is one function and $t_0$ is one arbitrarily chosen argument/number,
thus $-g\left(t_0\right)$ is ONE FIXED number.
In contrast, $C$ is ANY real number.

$\Large{\text{3.}}$ So $(\natural) \mathop{=}^{?} \, (\blacklozenge) \iff C \mathop{=}^{?} -g(t_0).$ But how and why is : $ C \mathop{=}^{?} -g\left(t_0\right) \; $?

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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".2013-05-11
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    @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.2013-05-12
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    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.2013-05-12
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    @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.2013-05-12
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    It's really an interpretation of what the indefinite integral symbol means. You usually think of it representing a family of curves with the same derivative, indexed by a parameter $C$. The right hand side definition get's you a family of curves as well if you let $t_0$ be arbitrary too, and these are indexed by $t_0$. Generally, the right hand side will be a subset of the left hand side. However, if you DEFINE the indefinite integral sign to have the property above, and let $t_0$ vary, then at most you lose some of the antiderivatives but the new definition has useful properties in itself2013-09-09
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    @rajb245: Thanks. $t_0$ is "some convenient lower of limit of integration", so it's one arbitrary but fixed argument. How can it vary? I still don't comprehend.2013-09-10
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    I apologize for my sloppy use of the word "vary". A more precise way of saying this is: "If you DEFINE the indefinite integral sign to have the property above, then for arbitrary but fixed $t_0$, the new definition generates a subset of the functions under the old definition. This new definition has more useful properties than the more general $+C$ form."2013-09-10
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    @rajb245: Thanks again. No apologies necessary! Does your comment revert to and resurrect Question 1.1 above, which I still don't savvy? Also, my textbook never defines the grey box to be an equality, so can it be examined without surmising it as such?2013-09-13
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    Suppose you knew nothing of indefinite integration, and the symbol had no meaning to you. But you understand definite integrals in terms of convergent sequences of Riemann sums. Now someone defines the indefinite integral to you in terms of the definite one as above. Then question 1.1 presupposes that the symbol on the left represents an entire set, when we never established this at all, under the way of thinking that I'm proposing. As for your text not introducing that as a definition, then the answer is that they were sloppy to introduce the anti derivatives first.2013-09-14
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    Another way of thinking of it is to interpret the equality given as: "An anti derivative of $f(t) is given by the following convergent limit of a sequence: ..." This is absolutely a true statement.2013-09-14
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    @rajb245: Many thanks. I'll mull this over. Would you like to conjoin your comments into an Answer for which I will upvote?2013-10-03
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    There are other good answers here, so I'll refrain from a full answer. I would also point you to this: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_second_part2013-10-03
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    @LePressentiment: There is unfortunately no way to edit just the bounty text. Since it was a new bounty I have removed it and refunded your reputation. Feel free to add the bounty again, but do be careful in the phrasing of your bounty text: I doubt a moderator will provide this service the next time around.2013-10-05
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    @ArthurFischer: Thank you very much.2013-10-06
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    Don't you have "definite" and "indefinite" switched around in the title of your question?2013-10-11
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    @bof: Thanks! Yes. Amended.2013-10-26

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