Notion of an integral as a summation

Click For Summary
The discussion clarifies that the integral is not merely a summation of differentials but represents an equivalence class of functions whose derivative is the integrand. An indefinite integral, lacking specified limits, indicates a function whose derivative matches the integrand, allowing for the addition of an arbitrary constant, C. The notation used must differentiate between the integrand and the limits, as using the same variable for both is incorrect. Additionally, if the function's domain is not connected, the constant C can vary across different sections. This understanding is crucial for accurately applying the Fundamental Theorem of Calculus.
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 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 MidgetDwarf and FactChecker
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
20
Views
2K
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
7K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
4
Views
5K
  • · Replies 4 ·
Replies
4
Views
6K