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After performing indefinite integration on both sides, an equation becomes:

$$\ln(|v|) + C = \frac{-t}{RC} + C $$

This then simplifies to:

$$\ln(|v|) = \frac{-t}{RC} + C $$

Where did the $+ C$ go on the left hand side and why?

Some Guy
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threebeesdizzy
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    The first equation is technically incorrect because as written it implies that you have the same constant of integration from both integrals, which is not necessarily true. – David K Apr 02 '21 at 03:52
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    We have three different numbers here that are all called the same name $C$. – littleO Apr 02 '21 at 04:11
  • Right, sorry about the confusion. For verification, the C in the demoninator represents capacitance (for convenience you can think of this as any other arbitrary letter/number) and the + C on both sides is of course representative of the CoI. – threebeesdizzy Apr 02 '21 at 04:34
  • I wish indefinite integrals and the +C were banned from calculus courses. They do more damage than good in learning the key concepts of calculus. Only definite integrals should ever be used. You’ll never see an indefinite integral or the +C again after you finish Calculus 2. – Deane Apr 02 '21 at 04:56
  • And I do recommend changing a variable name if you find yourself using the same name for two different things. – Deane Apr 02 '21 at 04:57
  • And especially the constant of integration should not be called $C$ because it is not the same as the $C$ in the denominator! – Troposphere Apr 02 '21 at 04:58
  • This is very duly noted and I will definitely be more conscious (and less lazy) in choosing variables when posting. Thanks so much! – threebeesdizzy Apr 02 '21 at 05:37
  • @Deane: I agree with you that the "+C" is taught wrongly, but it would be worse if you only deal with definite integrals. – user21820 Apr 02 '21 at 12:32
  • @user21820, why? When using the Fundamental Theorem of Calculus, you need only one antiderivative, not all of them. So you teach students how to find antiderivatives (so I would call that techniques of antidifferentiation). Once you've found one, you can calculate the definite integral. The definite integral is what always arises in practice, never the indefinite integral. The definition of an indefinite integral, because it is meant to represent all possible antiderivatives of a given function and therefore requires the "+C" term, leads to logical inconsistencies like the one here. – Deane Apr 02 '21 at 13:51
  • The region of integration should always be part of the definition of an integral. Note that in Calculus 3, this is taken for granted. It should be the same in Calculus 2. Most students think the indefinite integral is the main topic in Calculus 2 and the definite integral is an application of it. The emphasis should be different. The definite integral should be viewed as the main point of the course, and antidfferentiation is a key tool for computing the definite integral. And we should remove the word "definite". It's an integral, period. – Deane Apr 02 '21 at 13:55
  • @Deane: The fact that you say "leads to logical inconsistencies" just shows that either you were taught it incorrectly or you do not know how to teach it correctly without logical errors. – user21820 Apr 02 '21 at 15:16
  • @user21820, thanks for the feedback. How about the fact that indefinite integrals and the +C convention are never used after Calculus 2? – Deane Apr 02 '21 at 15:55
  • @Deane: I never said anything about teaching "+C". Teaching real analysis well necessarily must come after teaching students basic FOL (first-order logic with a full deductive system), because ultimately anything more complicated than rote calculation needs a proper grasp of FOL, including indefinite integrals. If you think otherwise, please go and try to solve the differential equation given in this post I wrote rigorously. After that, compare with the sketch I gave in the comments. – user21820 Apr 02 '21 at 16:24
  • As for your repeated claim of "never using indefinite integrals after Calculus 2", it's both irrelevant and not true. The fact that you can compute definite integrals (which are defined analytically) using algebraic computation of indefinite integrals (which I prefer to call antiderivatives) already is fundamental, and for this reason alone we need no further justification for the usefulness of antiderivatives. Same in complex analysis, where every holomorphic function $f$ on an simply connected open $D⊆ℂ$ has an antiderivative on $D$ as well. – user21820 Apr 02 '21 at 16:48
  • @user21820, at this point, I have no idea what we're arguing about. I mentioned explicitly the need for antiderivatives when computing a definite integral. What I dislike is the term "indefinite integral" (which you also seem not to like ) and the symbol $\int$ without endpoints to represent the set of all possible antiderivatives, which is why when a student writes the formula, they are obligated to write the "+C". You need to know how to compute antiderivatives in subsequent courses, but the term "indefinite integral" is never needed. – Deane Apr 02 '21 at 17:29
  • @Deane: I repeat that saying things like "write the +C" is logically wrong, so of course you get logically nonsensical results. If you don't understand FOL, you'll naturally not understand my point. And if you don't understand the full rigorous solution to the differential equation in the linked post, then you of course don't understand the need for such rigour involving such constants that arise. – user21820 Apr 02 '21 at 17:42
  • @user21820, I still don't understand our disagreement. I have been saying repeatedly that "write the +C" is wrong. It appears only because it is imposed on students by the textbook or the instructor. – Deane Apr 02 '21 at 18:28
  • I solve ODEs and PDEs for a living, so I don't have much difficulty understanding your point in the other post. – Deane Apr 02 '21 at 18:29
  • @Deane: Ok I think it's a miscommunication. So we agree that those textbooks and instructors that say "write +C" are wrong. What my original comment meant is simply that we can easily and correctly express and reason about indefinite integrals based on antiderivatives. For example, $∫(f(x)+g(x))\ dx = ∫f(x)\ dx + ∫g(x)\ dx + C$ for some constant $C$, for any variable $x$ such that $∫f(x)\ dx$ and $∫g(x)\ dx$ are both defined. And also, $∫z·\frac{dy}{dx}\ dx = ∫z\ dy + C$ for some constant $C$, for any variables $x,y,z$ such that $∫z\ dy$ and $\frac{dy}{dx}$ are both defined. – user21820 Apr 02 '21 at 18:42
  • These facts are what you know as the algebraic properties of the indefinite integral, where the indefinite integral here is defined as some antiderivative (if it exists). The point is, if you don't even define any notion of "indefinite integral", it becomes very unwieldy to even express the algebraic structure of indefinite integration, and so we lose the connection between the algebraic and analytic aspects of integration. Without such syntax and definitions, you simply can't prove algebraic results of the sort I just wrote above. Is my point clearer now? – user21820 Apr 02 '21 at 18:46
  • @user21820, I still think it is much better, including when you're solving an ODE, to always use only definite integrals, where one of the end points might be a variable rather than a constant. And to always use the fundamental of calculus: $\int_a^b f(x),dx = F(b)-F(a)$ for any antiderivative $F$ or $f$. This will be more verbose but reduce the chances of misunderstanding. The other important point that's too overlooked but emphasized in your other post is the need to make sure that, when you do division, make sure the denominator cannot be zero. – Deane Apr 02 '21 at 19:24
  • Let us continue this discussion in chat. – user21820 Apr 02 '21 at 20:05

2 Answers2

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Two constants $C_1$ and $C_2$ can be represented as a single constant $C=C_1+C_2$ or $C=C_1-C_2$.

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Ishraaq Parvez
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  • You're not explaining why there ought to be two constants, nor why you can represent them as one. I know the answer, but you should be explaining to the asker. – user21820 Apr 02 '21 at 12:33
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Let's say you have two functions $f(x)$ and $g(x)$, with $F(x)$ and $G(x)$ being their antiderivatives.

So, if you have $f(x) = g(x)$, and you integrate both sides, you get $F(x) + C_1 = G(x) + C_2$. Both $C$ are undetermined constants, because they are the constants of integration, they can be any constant. Moving the constants over to one side, we get $F(x) = G(x) +C_2 - C_1$.

An undetermined constant minus an undetermined constant is just another undetermined constant, so we can replace it with one undetermined constant $C$. So, we get $F(x) = G(x) + C$.

user182601
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Some Guy
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