Don't understand why an indefinite integral is valid only on a interval

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SUMMARY

The discussion centers on the concept of indefinite integrals as presented in Stewart's Calculus, specifically regarding their validity on specified intervals. The theorem states that the general antiderivative of a function, such as f(x) = 1/x², is valid only on intervals like (0, ∞) or (-∞, 0). This is due to the fact that the constant of integration, denoted as C, can differ across these intervals, necessitating separate constants (C1 for x < 0 and C2 for x > 0) to accurately represent the antiderivative. Understanding this convention is crucial for correctly interpreting indefinite integrals in calculus.

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  • Basic knowledge of piecewise functions
  • Concept of continuity and differentiability in calculus
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Students of calculus, particularly those studying indefinite integrals, educators teaching calculus concepts, and anyone seeking to deepen their understanding of antiderivatives and their properties across different intervals.

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Don't understand why "an indefinite integral is valid only on a interval"

Hi I'm using Stewart's Calculus, in the section of indefinite integral, they say:

"Recall from Theorem 4.10.1 that the most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative. We adopt the convention that
when a formula for a general indefinite integral is given, it is valid only on an interval.
Thus, we write

\int 1/x^2 dx = - (1/x) + C

with the understanding that it is valid on the interval (0, \infty) or on the interval (-\infty, 0). This is true despite the fact that the general antiderivative of the function , f(x)=1/x^2, x\neq0 , is:

F(x):
- (1/x) + C1 if x<0
- (1/x) + C2 if x>0"

Well, I don't understand this convention, don't know if it is something too obvious and I'm complicating myself, like, "there could be points where the indefinite integral isn't defined" or maybe there's a subtle point behind this, maybe they want to say that a indefinite integral that holds for any interval, no matter how small, it's a valid indef. int. of the function.

Please help me to understand this, it seems that it doesn't affect too much the rest of the topics, but I don't like to skip things that I don't understand.

Thank you & excuse me if my english isn't very clear.

Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution

 
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I don't really get it, I think all he is trying to say, is that he will always denote the arbitrary constant that can be added by C, even though the integral may have to be split up in multiple regions (like in the example x < 0 and x > 0) in each of which C can have a different value (and therefore should technically be written in separate cases like
F(x):
- (1/x) + C1 if x<0
- (1/x) + C2 if x>0
So you, as reader, should be careful with these kinds of functions and keep in mind that on different intervals, the constant may be different.
 
mmm...

Yeah, maybe it's just what Compuchip says, because if we use the same constant in both intervals it could be thought that a particular primitive must have the same constant in both intervals. Right?
 
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?

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