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?