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

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

$\Large{\text{1.}}$ How and why is $(*)$ true, in spite of $ \int{f(t)\text{ }dt} \color{#B53389}{\text{ (ie: the set)}} =\int_{t_0}^t f(s) \text{ } ds \color{#B53389}{\text{ (ie: an element of the set on the left)}} \quad ?$


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

Thanks to your explanations, I cognise 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.
On the contrary, $C$ is ANY real number.

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

$\Large{\text{4.}}$ How to perceive/savvy/seize $(*)$ directly, without rewriting it by dint of FTC or other theorems?

$\Large{\text{5.}}$ What is the intuition behind $(*)$?

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3  
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". –  Pedro Tamaroff May 11 '13 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. –  LePressentiment May 12 '13 at 13:06
1  
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 '13 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. –  LePressentiment May 12 '13 at 13:42
    
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 itself –  rajb245 Sep 9 '13 at 4:18

3 Answers 3

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? –  LePressentiment May 7 '13 at 23:01
    
@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. –  Karl Kronenfeld Sep 9 '13 at 4:07
    
@KarlKronenfeld: Many thanks. –  LePressentiment Sep 10 '13 at 0:46
    
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 defines the indefinite integral as an arbitrary antiderivative. –  user64494 Oct 5 '13 at 10:00

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.

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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? –  William Stagner May 12 '13 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. –  LePressentiment May 14 '13 at 20:36
    
I've updated my answer. Let me know if I didn't explain well enough. –  William Stagner May 15 '13 at 14:33
    
Thank you again. Sadly, I still cannot grok Q3. Could you please explicate some more? –  LePressentiment Sep 3 '13 at 4:37

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