Which function suits this description?

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Discussion Overview

The discussion revolves around identifying a function whose definite integral from a to b is equal to its definite integral from b to b + (b-a)/2. Participants explore various functions and their properties, including continuity and specific behaviors of exponential functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the exponential function e^x could be a candidate, suggesting a relationship between the areas of the integrals over specified intervals.
  • Another participant introduces a piecewise function that could satisfy the condition without continuity, defining it as f(x) = c for a ≤ x < b and f(x) = 2c for b ≤ x ≤ b + (b-a)/2.
  • A later reply emphasizes the preference for a continuous function, arguing that it should exhibit consistent behavior over any interval.
  • One participant challenges the claim regarding the exponential function, providing a detailed calculation of the integrals and questioning the assumptions about the relationship between a and b.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of continuity for the function. There is also disagreement regarding the properties of the exponential function in relation to the integrals, with some participants questioning earlier claims without reaching a consensus.

Contextual Notes

There are unresolved assumptions regarding the relationship between a and b, as well as the implications of continuity on the proposed functions.

lapo3399
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Which function's definite integral from a to b will be equal to its definite integral from b to b + (b-a)/2?

[tex]\int_{a}^{b} f(x) dx = \int_{b}^{b+\frac{b-a}{2}} f(x) dx[/tex]
[tex]\int_{a}^{b} f(x) dx = \int_{b}^{\frac{3b-a}{2}} f(x) dx[/tex]

I know that the exponential function e^x will have a definite integral for the second area that is twice as large as the first, and that the next one ((3b-a)/2 to (9b-3a)/2) will be twice as large as the second... so would this function be the exponential function divided by some function like 2^(n-1), where n is the number of the term in the sequence of definite integrals? How would I represent this function without the use of n?

Thanks for any help!
 
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Does it have to be a continuous function? If not, the following piecewise function will work:

Define [tex]f(x) = c[/tex] for [tex]a \leq x < b[/tex] and

[tex]f(x) = 2c[/tex] for [tex]b \leq x \leq b + \frac {b-a}{2}[/tex]

Where c is any constant. If you want a continuous function, I'll have to do more thinking (as will you...)

(of course, if c = 0, then you've got your answer, albeit a trivial one)
 
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Well, a continuous function would be preferable, if possible, as this function should have the same behaviour over any other interval (e.g. from one point to any other point, the area under the first half of that domain will be equal to the area between the halfway point and the 3/4 point, etc).

Thanks for any further suggestions.
 
lapo3399 said:
I know that the exponential function e^x will have a definite integral for the second area that is twice as large as the first ...

This is not correct in general.

If [itex]f(x) = \exp x[/itex] then
[tex]\int_{a}^{b} f(x) dx<br /> = e^b-e^a<br /> = e^a\left(e^{(b-a)}-1\right)[/tex]
[tex]\int_{b}^{\frac{3b-a}{2}} f(x) dx<br /> = e^{(b+\frac{b-a}2)}-e^b<br /> =e^b\left(e^{\left(\frac{b-a}2\right)}-1\right)[/tex]

The ratio of the two integrals is thus

[tex]\frac{e^b}{e^a}\,\frac{e^{\left(\frac{b-a}2\right)}-1}{e^{(b-a)}-1} = \frac{e^{(b-a)}}{e^{\left(\frac{b-a}2\right)}+1}[/tex]

Is there some relation between a and b that you haven't told us about?
 
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