Notion of an integral as a summation

[sachin]

Homework Statement

the notion of an integral is a special form of summation of differentials and an indefinite integral is also an integral with limits 0 to x what is conventionally written without the limits,

Relevant Equations

∫f(x)dx with limits 0 to x = ∫f(x)dx

While analyzing the foundation of calculus,
I am finding that the notion of an integral is a special form of summation of differentials and an indefinite integral is also an integral with limits 0 to x what is conventionally written without the limits,

the notation is given in the image,

Pl confirm if my assumption is correct,
thanks.

 

[FactChecker]

That is not correct. The antiderivative of the function, $f(x)$, with no specified limits indicates a function of $x$ whose derivative is $f$. As such, any constant can be added and it would still be an antiderivative. That constant is sometimes not indicated, but it should be. In other words, if $d/dx\ g(x) = f(x)$, then $\int f(x) dx = g(x)+C$, where $C$ is an (as yet) unknown, arbitrary constant. There are many problems where the correct value of $C$ must be determined.

 

[Frabjous]

It is also poor notation to have the dummy variable identical with a bound.

 

[PeroK]

That's not correct. The indefinite integral is an equivalence class of functions whose derivative is the integrand.

Technically, you cannot use the same variable $x$ in the integrand and as a bound. We have,
$$\int_0^x f(t)dt = F(x) - F(0)$$That is a specific function of $x$. And$$\int f(x)dx = F(x) + C$$That is a set of functions (equivalence class). Where $F(x)$ is any function where $F'(x) = f(x)$. These are both examples of the Fundamental Theorem of Calculus.

 

[pasmith]

And$$\int f(x)dx = F(x) + C$$That is a set of functions (equivalence class). Where $F(x)$ is any function where $F'(x) = f(x)$.

A point which is often missed is that if the domain of $f$ is connected then $C$ must indeed be a constant, but if the domain is not connected (eg. $f(x) = x^{-1}$, $x \in \mathbb{R} \setminus \{0\}$) then $C$ can take different values on different connected components of the domain.