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Definite Double Integral of a single variable

  1. Aug 18, 2014 #1
    1. The problem statement, all variables and given/known data
    This isn't actually homework. I was messing around in my notebook trying something when I ended up writing something to the effect of this:

    [itex] dT = \frac{V^{2}}{R(1+α dT)}dQ[/itex]
    [itex]R(1+α dT) dT = V^{2}dQ[/itex]
    Where α and V are constants.

    Now, I'm fairly sure what I had done made no sense physical and I screwed up, which is fine, but as I wrote this, I realized something: I'm not sure how to solve this, or if this is even a valid equation


    3. The attempt at a solution

    So, I tried integrating with a naive and hopeful glimmer in my eye:
    [itex]RΔT+\frac{a}{2}ΔT^{2}= V^{2}Q[/itex]

    Where [itex]ΔT=T_{2}-T_{1}[/itex]
    You see, I'm interested in a definite integral, I'm interested in the change between [itex]T_{1}[/itex] and [itex]T_{2}[/itex].

    Now, I'm pretty sure this is wrong enough to make someone cry, so I'll just leave it at that as ask where I went wrong and why?
     
  2. jcsd
  3. Aug 18, 2014 #2
    Having no reference for what the constants ##\alpha## and ##V## and the variables (?) ##Q##, ##R##, and ##T## are intended to represent, there is no reason why the equation $$\Delta T = \frac{V^{2}}{R(1+α \Delta T)} \Delta Q$$ or its "differential brother" $$dT = \frac{V^{2}}{R(1+α dT)}dQ$$ can't model some physical system.

    If your end goal it to integrate something (and it looks like it is) then the hand-waver would look at $$RdT+\alpha R dT^2=V^2 dQ$$ and say something along the lines of,

    The key reason why the "negligible term" business works is because, in the case where ##f## is an integrable function, we have

    $$
    \lim_{n\rightarrow\infty}\sum_{i=1}^n\left(f(x_i)(\Delta x)^2\right)
    =\lim_{n\rightarrow\infty}\left(\Delta x\sum_{i=1}^nf(x_i)\Delta x\right)
    =\lim_{n\rightarrow\infty}\Delta x\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x
    =0\cdot\int_a^bf(x)\ dx=0$$ for ##\Delta x=\frac{b-a}{n}## a regular partition.
     
  4. Aug 19, 2014 #3
    Thanks. But, I'm not really interested in handing waving the odd term away. I'm interested in how I'd actually solve the equation exactly.
     
  5. Aug 19, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What gopher_p did is NOT "hand waving" he showed why [itex]\int\int dx dx= 0 [/itex] for any variable x.
     
  6. Aug 19, 2014 #5
    Sorry, I miss-read it. That said, there are two things I do not understand:
    In the following:
    $$
    \lim_{n\rightarrow\infty}\sum_{i=1}^n\left(f(x_i)(\Delta x)^2\right)
    =\lim_{n\rightarrow\infty}\left(\Delta x\sum_{i=1}^nf(x_i)\Delta x\right)
    =\lim_{n\rightarrow\infty}\Delta x\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x
    =0\cdot\int_a^bf(x)\ dx=0$$ for ##\Delta x=\frac{b-a}{n}## a regular partition

    Why does ΔX come out of the sum? Why can I write
    [itex]\lim_{n\rightarrow\infty}\Delta x\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x[/itex]

    I can't write [itex]\lim_{n\rightarrow\infty}\Delta x^{2}\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)[/itex] can I?

    And secondly, wouldn't [itex]\int\int dx dx= \frac{1}{2}x^{2}? [/itex]
     
  7. Aug 19, 2014 #6
    If you have a regular partition, the ##\Delta x## is constant with respect to the sum; ##\frac{b-a}{n}## is independent of ##i##.

    You can write it the second way, but that doesn't help you evaluate the limit. It's likely that ##\sum\limits_{i=1}^nf(x_i)## is not a convergent sequence.
     
  8. Aug 19, 2014 #7

    Zondrina

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    Homework Helper

    If ##\Delta x = \frac{b-a}{n}##, then as ##n → ∞##, ##\Delta x → 0##.

    Notice ##\Delta x## is independent of the summation index ##i##, and can therefore be pulled outside of the sum.

    The other part is a Riemann sum, which converges to an integral:

    $$=(\lim_{n\rightarrow\infty}\Delta x) (\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x)
    =0\cdot\int_a^bf(x)\ dx=0$$

    More to the point though, without knowing what ##T## is exactly, the problem hasn't been stated clearly.
     
  9. Aug 19, 2014 #8
    T is an independent variable of the system in question. Actually, the equation was dependent on the change in T, ΔT, not T itself, which is why the integral was definite.

    And thanks for clearing everything up everyone!
     
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