Notion of an integral as a summation

Click For Summary

Homework Help Overview

The discussion revolves around the foundational concepts of calculus, specifically the relationship between integrals and summation, as well as the nature of indefinite integrals and their notation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to clarify their understanding of integrals as a form of summation and questions the correctness of their assumption regarding indefinite integrals. Other participants challenge this view, discussing the implications of notation and the nature of antiderivatives.

Discussion Status

Multiple interpretations of the concepts are being explored, with some participants providing clarifications on the definitions and properties of indefinite integrals and antiderivatives. There is a focus on the importance of notation and the implications of using the same variable for bounds and integrands.

Contextual Notes

Participants note that the discussion involves assumptions about the nature of constants in indefinite integrals, particularly in relation to the connectedness of the domain of the function.

sachin
Messages
61
Reaction score
7
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.
notion of an integral.png
 
Physics news on Phys.org
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.

UPDATE: See @pasmith 's comment #5 about the possibility of the constant ##C## being different in two sections of a domain that is not connected.
 
Last edited:
It is also poor notation to have the dummy variable identical with a bound.
 
  • Like
Likes   Reactions: PhDeezNutz and FactChecker
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.
 
PeroK said:
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.
 
  • Like
Likes   Reactions: MidgetDwarf and FactChecker

Similar threads

Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 73 ·
3
Replies
73
Views
8K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
4
Views
5K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K