Physical interpretation of this integral equation involving distance and time?

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SUMMARY

The discussion centers on the interpretation of the integral equation ##\int_0^x (1-t^2)/(1+t^2) \, dt##, particularly when considering ##x## as position and ##t## as time. Participants emphasize that while ##\frac{dx}{dt}## represents instantaneous change in position over time, the integral signifies a net change. It is concluded that not all calculus problems require a physical interpretation, and sometimes the mathematical formulation stands alone without direct application to physical concepts. The importance of distinguishing between dummy variables and defined variables in integrals is also highlighted.

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Irishdoug
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Homework Statement
I have been given a generic question, however I don't know how to interpret it if the variables had an assigned meaning.
Relevant Equations
f'(x) = ##\frac{d}{dx}## (##\int## ##\frac{1-t^2}{1+t^2}## *dt). The integral has the limits 0-x.
I am able to solve the problem however if x was position and t was time how is this problem interpreted?

I know, for example that ##\frac{dx}{dt}## tells us how the position of something changes as time changes (or instantaneous change) and an integral gives a net change so to speak.
 
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Irishdoug said:
I am able to solve the problem however if x was position and t was time how is this problem interpreted?

I know, for example that ##\frac{dx}{dt}## tells us how the position of something changes as time changes (or instantaneous change) and an integral gives a net change so to speak.

You can easily write ##\int_0^x (1-t^2)/(1+t^2) \, dt##; just right-click on the image to see the LaTeX commands.

Anyway, why should ##f(x)## and/or ##f'(x)## have any particular "interpretation"? Maybe this is just a problem in calculus without any interpretation or without any relation to distance and time.
 
Im asking more so about the RHS of the equation. The reason I want to know how to interpret it as I do physics and equations require a physical interpretation.
 
Irishdoug said:
Im asking more so about the RHS of the equation. The reason I want to know how to interpret it as I do physics and equations require a physical interpretation.

If the integral arises in some "application" in physics, engineering, chemistry, biology, economics, ... then you may be able to come up with a reasonable interpretation. If it is just a calculus practice-example there may not be any interpretation, and you are wasting your time looking for one. It would be much, much better to just realize that calculus is a branch of mathematics that is widely applicable to numerous fields and situations, and that not everything need be related to physics. Equations definitely do NOT need a physical interpretation, although you may find it difficult to accept this. Basically, I am saying that you are hindering your own learning process by constantly trying to give a physical interpretation to everything. (I say this as somebody who earned a PhD in physics from MIT.)
 
Ok thanks for the heads up!
 
Irishdoug said:
I am able to solve the problem however if x was position and t was time how is this problem interpreted?

I know, for example that ##\frac{dx}{dt}## tells us how the position of something changes as time changes (or instantaneous change) and an integral gives a net change so to speak.

##t## looks more like a dummy integration variable here than time. For example, you can (almost) write:

##f(x) = \int_0^x g(x) dx##

But, technically, you are using ##x## as both the dummy integration variable and the variable for which your function is defined.

In general it's better to use different symbols for the two variables. For example:

##f(x) = \int_0^x g(x') dx' = \int_0^x g(u) du##

Or, in your case, ##t## is used as the dummy variable.
 
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