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Change of variable theorem.

  1. Oct 5, 2005 #1

    quasar987

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    There's this thing I've always wondered about but never really found the answer. However, I'm sure it's obvious to many of you. It goes as follow: How come the change of variable theorem (for simple integrals) is so complicated but the fact is that it never fails to say "let's set u =... Then, du = ...", which is not at all what the change of variable theorem is saying imho.
     
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  3. Oct 5, 2005 #2

    TD

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    Perhaps I'm not understanding you, do you mean the substitution rule?

    [tex]\int_{\varphi \left( \alpha \right)}^{\varphi \left( \beta \right)} {f\left( x \right)dx} = \int_\alpha ^\beta {f\left( {\varphi \left( t \right)} \right)\varphi '\left( t \right)dt}[/tex]
     
  4. Oct 5, 2005 #3

    quasar987

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    Yeah, that's totally what I'm talking about. Why is it formulated in terms of function phi of a variable t? For exemple, compare the two following ways of changing variables for this integral:

    [tex]\int_a^b e^{k(x-2)}dx[/tex]

    The "fast-substitution" way everyone use:

    Set t = x - 2. then by definition of differential of a function of one variable, dt = (dt/dx)dx = (1)dx = dx. The bounds transform like so: x = a --> t = a-2, x = b --> t = b - 2. Hence, plugging all these substitutions in the original integral yields

    [tex]\int_a^b e^{k(x-2)}dx = \int_{a-2}^{b-2} e^{kt}dt[/tex]


    The "rigourous-substitution" way goes:

    Set [itex]\varphi (t) = t + 2[/itex]. Then [itex]\varphi '(t) = 1[/tex]. [itex]\varphi (t_1) = a \Leftrightarrow t_1 + 2 = a \Leftrightarrow t_1 = a - 2[/itex]. [itex]\varphi (t_2) = b \Leftrightarrow t_2 + 2 = b \Leftrightarrow t_2 = b - 2[/itex]. So now according to the change of variable theorem,

    [tex]\int_a^b e^{k(x-2)}dx = \int_{a-2}^{b-2} e^{k(\varphi(t)-2)}(1)dt = \int_{a-2}^{b-2} e^{k(t)}dt [/tex]

    It's so much complicated. SO why is it formulated in this complicated way rather than in the simplest way, which seem to always work anyway?
     
  5. Oct 5, 2005 #4
    I believe it is just a generalization. The second way in which you did it was practically the same as the first, you just showed all of your work.

    Alex
     
  6. Oct 5, 2005 #5

    quasar987

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    What do you mean precisely in this case by "a generalisation", and what do you mean by "you just showed all your work"? It seems to me I did all the steps for both ways only the "rigourous way" has naturally more steps then the "usual way".
     
  7. Oct 5, 2005 #6

    Hurkyl

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    The statement you're using of the change of variable theorem is the one that looks most like a direct application of the chain rule.

    I.E. it's more natural to apply the transformation df(x) → f'(x) dx than it is to transform f'(x) dx → df(x)

    But, I seem to remember at least one calculus text that stated the reverse formula as a corollary.


    As to how people actually do the integral by hand, the forward direction would be written as the substitution x = t + 2, and worked from there.


    Sometimes, the forward direction is most convenient. For example, when integrating √(1-x²), the most natural thing to do is say x = sin u. Sometimes the reverse direction is more convenient, as you've observed.

    Sometimes, it's most convenient to do it neither way -- I've had situations where using an implicit relation, like t² = x² + 2, gives the simplest integrand.
     
  8. Oct 5, 2005 #7

    quasar987

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    Wow that clears some things up!



    P.S. Is there a calculus book you've not read Hurkyl? :tongue:
     
  9. Oct 5, 2005 #8

    Hurkyl

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    I've read all but ε of them for some ε > 0...
     
  10. Oct 5, 2005 #9

    quasar987

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    Yep, that's definitely too much of them... :devil:
     
  11. Oct 8, 2005 #10

    HallsofIvy

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    Fortunately calculus books have the property of all containing exactly the same thing!
     
  12. Dec 9, 2010 #11
    most of the calculus book contain the same example i have read a few.
     
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