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In the book Thomas' Calculus, the Indefinite integral is defined as follows:

we defined the indefinite integral of the function ƒ with respect to x as the set of all antiderivatives of ƒ, symbolized by $\int{ƒ(x)dx}$. Since any two antiderivatives of ƒ differ by a constant, the indefinite integral $\int$ notation means that for any antiderivative F of ƒ , $$\int{f(x)dx=F(x)+C}$$

This directly implies that:

$$\frac{dy}{dx}=f(x) \implies y(x)=\int{f(x)dx} = F(x)+C \tag{1}$$

However, in the book Differential Equations With Applications and Historical Notes, Simmons notes:

We solve it by writing: $$y(x)=\int{f(x)dx} + c \tag{2}$$

The 'it' referring to the differential equation in (1).

Question: Isn't the $c$ redundant? Since the Indefinite integral already contains the collection of antiderivatives, one would think that would be the case, however, from the first fundamental theorem of calculus we get: $$\int_{a}^{x}{f(t)dt}=\int{f(x)dx} \tag{3}$$

and by substituting (3) in (1), $$y(x)=\int_{a}^{x}{f(t)dt}=F(x)-F(a) \tag{4}$$

And now by comparing (1) and (4) we come to the conclusion that: $F(x)+C=F(x)-F(a)$, which clearly has to be False, since a is fixed but C is an arbitrary constant. The only way to make it balanced is to add another arbitrary constant to make: $F(x)+C=F(x)-F(a) + C_1$. But in order to make this happen we have to add the $C_1 to \int{f(x)dx}$ in (1) and hence confirming the validity of (2). So the $c$ in (2) seems necessary again.

How to resolve this contradiction?

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    $\begingroup$ Why are you comparing equations of two different books, both tell same thing but in different way $\endgroup$ Aug 24, 2021 at 17:42
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    $\begingroup$ @LalitTolani While they are talking about the solutions of the same differential equation, the two approaches yield different results, and that is a problem. $\endgroup$ Aug 24, 2021 at 17:47

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Careful, all the second fundamental theorem says is that $$\int_{a}^{x}f(t)\,dt$$ is an antiderivative for $f(x)$. That is not the same as saying that $$\int_{a}^{x}{f(t)\,dt}=\int{f(x)\,dx}$$ which is not true. What you can write is $$\int f(x)\,dx = \int_{a}^{x}f(t)\,dt + C.$$ In any case, yes it is generally considered redundant to write $$\int f(x)\,dx + c,$$ but you would do well to double check how Simmons defines $$\int f(x)\,dx.$$ I would be surprised if their definition varies from the standard, so it may be the case that writing the $+C$ is just their way of making sure that the reader remembers the need for an arbitrary constant. There is certainly nothing incorrect in writing $$\int f(x)\,dx + c$$ because as you have pointed out, if $F(x)$ is an antiderivative of $f(x)$ then $$\int f(x)\,dx = F(x) + C,$$ and since the sum of two arbitrary constants is again an arbitrary constant it is still valid to write $$\int f(x)\,dx + c.$$

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  • $\begingroup$ Since $\int_{a}^{x}{f(t)dt}$ is a function of $x$, consider $y(x)=\int_{a}^{x}{f(t)dt}$ and use (1). This will prove that $\int_{a}^{x}{f(t)dt}=\int{f(x)dx}$ $\endgroup$ Aug 24, 2021 at 19:10
  • $\begingroup$ Excuse me I wanted to write and was in fact referring to the first theorem and I have corrected the post $\endgroup$ Aug 24, 2021 at 19:15

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