Two limited integration questions

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Homework Help Overview

The discussion revolves around two integration conditions involving a function f(x). Participants are exploring the implications of these conditions in the context of electric fields and potentials.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are examining the conditions under which the integrals hold true, with some suggesting that the properties of the function f(x) may relate to its symmetry (even or odd functions). Others are questioning the sufficiency of the conditions provided and seeking clearer definitions or additional criteria.

Discussion Status

The conversation is ongoing, with participants actively engaging in reasoning about the conditions for the integrals. Some guidance has been offered regarding the relationship between the function and its integral, but there is no explicit consensus on the necessary conditions.

Contextual Notes

Participants express confusion regarding the application of these conditions to problems related to electric fields and potentials, indicating a need for clarity in the definitions and assumptions involved.

hhegab
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Peace!

I want to know the conditions that must be satisfied by a function
[tex]f(x)[/tex] for any of the following two cases to be true (each case independent from the other);

1- [tex]\int^a_{-a} f(x) dx = 2 \int^a_0 f(x) dx[/tex]

2- [tex]\int^a_0 f(x) dx = \int^0_a f(x) dx[/tex]

They gave me confusion when I was solving problems related to electric field and electric potential.
 
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I don't know if this condition is sufficient but a condition would be, if we write the integrals in terms of their primitives,

[tex]\int^a_{-a} f(x) dx = 2 \int^a_0 f(x) dx \Leftrightarrow \mathcal{F}(a) - \mathcal{F}(-a) = 2\mathcal{F}(a) \Leftrightarrow \mathcal{F}(-a) = -\mathcal{F}(a)[/tex]

The condition is that it is true iff the primitive of f is a function F such that F(-a) = -F(a)
 
First case...

[tex]\int^a_{-a} f(x) dx = \int^0_{-a} f(x) dx + \int^a_0 f(x) dx[/tex]

Then use the fact that :

[tex]\int^0_{-a} f(x) dx = - \int^{-a}_0 f(x) dx[/tex]

and replace x by -x...the limit -a will then change to a because of this substitution. and dx will become -dx. Now f(x) becomes f(-x) and there are two possibilities. Either f(-x) = -f(x) or f(-x) = f(x)...you know what you will need to achieve so which one of the two is it...


Question 2 :

Just put the integral in right hand side to the left hand side and use the above property to get rid of the minus-sign...what do you get ?


regards
marlon
 
Can you put like , first case is true if f(x) is even and if such and such...
I shall study your answer. But from my first reading I need more.
I need a condition to apply to f(x) so I can use each of the above properties.

hhegab
 

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