dummy variables (1 Viewer)

[PhysicsMaths]

A typical example of this would be to prove that

int(f(a-x)dx) =int(f(x)dx)

when using the substitution u = a-x and progressing to int(f(a-u)du), why is u interchangable with x?

the main dispute I have is that it appears to conflict with the relationship u = a-x, which suggests that u does not equal to x.

 

[anomalousdecay]

no

u = a - x

x = a - u

du/dx = -1

du = -dx

So int(f(x)dx) = - int(f(a-u)du)

Also, int(f(a-x)dx) = - int(f(u)du)

Substitute them in appropriately and it should make sense.

 

[Drsoccerball]

 

[leehuan]

You meant the definite integral. Because it does this after the u-sub

 

[porcupinetree]

Think about it this way: if we have a function f(x), and we wish to find the area between the curve and x-axis between x=0 and x=a, then surely the area wouldn't be different if we switched the label 'x' with the label 'u'? The label 'x' is simply an arbitrary symbol; it could be anything, e.g. elephant as Drsoccerball suggests.

 

[Carrotsticks]

You are understandably confused.

I've had many discussions, some getting quite heated actually, with many teachers over this exact same concept and I still hold to this.

The problem comes from the way functions are taught. They are explained as 'functions of a variable', which is a hugely misleading way of explaining it.

A function is just that.... a function. It is NOT a 'function of a variable'.



The very last expression is precisely what we want. Throw away the idea of the variable, and focus on the FUNCTION.

Going by this, provided that the proper restrictions are met etc, then integration by substitution now becomes



Once integration is understood in this way, then your question becomes a non-issue.

 

[Trebla]

Keep in mind this notion of a dummy variable makes sense because the integral is definite and always evaluates to a constant. You can't argue the same thing if you are doing a substitution on an indefinite integral.

 

[InteGrand]

Just look at any specific example and you should be convinced too; obviously the following two are equal (when you evaluate them, the variables "go away"):

and

 

[Carrotsticks]

Just look at any specific example and you should be convinced too; obviously the following two are equal (when you evaluate them, the variables "go away"):

and

That's not really where students have issues.

As separate things, as you have presented above, both integrals are clearly the same.

However, give a relationship between the two variables, and now you've got a whole lot of students saying "Can we just do that...? But there's a relationship between them".

This is why freeing the function from the notion of the variable is advantageous here.