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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

3 Answers 3

7

On the LHS of the grey box, $\int{f(t)\,dt}$ is the antiderivative of $f$.
It is defined up to a constant: $\int{f(t)\,dt} = g(t) + C.$

On the RHS of the grey box, $\int_{t_0}^t f(s) \, ds$ 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?2013-05-07
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    @LePressentiment As one of the reviewers of your suggested edit, I would like to point out how cool it is that you updated the references here for Yury.2013-09-09
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    @KarlKronenfeld: Many thanks.2013-09-10
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    If the indefinite integral is defined as the set of all antiderivatives, then we meet the difficulties with the definition of the sum $\int f(x)\,dx+\int g(x)\,dx$. In order to avoid that, [V. Zorich](http://books.google.com.ua/books?id=qA5FTMT7HE4C&printsec=frontcover&dq=Zorich&hl=en&ei=NMJNTZ2wFIyTswbx6rSQDQ&sa=X&oi=book_result&ct=result&redir_esc=y#v=onepage&q&f=false) defines the indefinite integral as an arbitrary antiderivative.2013-10-05
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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 $(**) \in (*)$.

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    Sorry, I don't think I understand your question. Could you try to rephrase it?2013-05-12
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    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.2013-05-14
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    I've updated my answer. Let me know if I didn't explain well enough.2013-05-15
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    Thank you again. Sadly, I still cannot grok Q3. Could you please explicate some more?2013-09-03
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I edited some of rajb245's comments to unify them. ---

The question's really about interpreting what the indefinite integral symbol means. The LHS of $(*)$ is the old definition of the indefinite integral and the RHS of $(*)$ is the new definition but expressed as a definite integral. You think of the LHS of $(*)$ representing a family of curves with the same derivative, indexed by a parameter $C$ . If you let $t_0$ be arbitrary too, the RHS of $(*)$ gives a family of curves as well. These curves are indexed by $t_0$. Generally, the RHS of $(*) \subseteq$ LHS of $(*)$.

However, if you DEFINE the indefinite integral sign to have the property $(*)$, then for arbitrary but fixed $t_0$, RHS of $(*)$ generates a subset of the functions under the LHS of $(*)$. At most, you lose some of the antiderivatives. This RHS of $(*)$ has more useful properties than the more general $+C$ form.

Another way to fathom this ----

Suppose you knew nothing of indefinite integration, and the symbol had no meaning to you. And suppose you understand definite integrals in terms of convergent sequences of Riemann sums. Now someone defines the indefinite integral to you in terms of the RHS of $(*)$.

Then question 1.1 presupposes that the symbol on the left represents an entire set. But under the way of thinking that I'm proposing, we never established this at all.

As for your text not introducing $(*)$ as a definition, then the answer is that they were sloppy to introduce the anti-derivatives first.

Another way of thinking of $(*)$ is to interpret the equality given as: "An antiderivative of $f(t)$ is given by the following convergent limit of a sequence: ..." This is absolutely a true statement.