Question on integral notation (dt, dx, etc.)

hulgy
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Say the limit of the integral is from 0 to t and the integral is ended with a dt. Is this okay?

Generally, all the integrals I see with a variable limit end with a d-letter that is not the same as the variable in the limit. ie: limit is from 0 to t, ends with du
 
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Welcome to PF!

Hi hulgy! Welcome to PF! :smile:
hulgy said:
Say the limit of the integral is from 0 to t and the integral is ended with a dt. Is this okay?

Noooo …

If you have an ∫ f(t) dt, that has to be between limits of t …

t can't depend on itself. :wink:

(once you've integrated over a variable like t, that variable should disappear, not come back to haunt you!)
 


tiny-tim said:
Hi hulgy! Welcome to PF! :smile:


Noooo …

If you have an ∫ f(t) dt, that has to be between limits of t …

t can't depend on itself. :wink:

(once you've integrated over a variable like t, that variable should disappear, not come back to haunt you!)

Hmmm... what if t stood for a value like 1<t<3 which you would have to place into the 0 to t limit (this would technically eliminate the variable...right?). Would the integral as a whole still be wrong then?
 
hulgy said:
Hmmm... what if t stood for a value like 1<t<3 which you would have to place into the 0 to t limit (this would technically eliminate the variable...right?). Would the integral as a whole still be wrong then?

Not following you :confused:

can you write the integral out in full? :smile:

(you can write integrals like this: [noparse]∫ab[/noparse] … ∫ab :wink:)
 
tiny-tim said:
Not following you :confused:

can you write the integral out in full? :smile:

(you can write integrals like this: [noparse]∫ab[/noparse] … ∫ab :wink:)


0t (50)dt for 1<t<3
 
hulgy said:
0t (50)dt for 1<t<3

hmm …

If you mean ∫0s (50)dt for 1<s<3, then that's fine.

But no, you can't use t for two different things.
 
That is a bit of abuse of notation -- not that you won't see such abuse.

The dt inside the integral is a dummy variable. It might as well be tau, or x, or anything. It has nothing to do with the upper limit of the integration.

This is at best confusing:

f(t) = \int_0^t \cos t\, dt

What is df/dt?

If the above integral is expressed as

f(t) = \int_0^t \cos \tau\,d\tau

then it is pretty clear that f(t)=\sin t and that df/dt = \cos t.


Occasionally you will run across things like

f(t) = \int_0^t \cos (t-\tau)\,d\tau

The t inside the above integral is not a dummy variable. As far as the integration is concerned that t inside the integral is a constant.
 
I guess I'm going to get points off for notation on my AP calc test then. Well thanks for all the help, really cleared things up.
 
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