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B Integration by substitution

  1. Feb 27, 2017 #1
    I have seen the wikipedia's proof which can be found here: https://proofwiki.org/wiki/Integration_by_Substitution
    However sometimes, we have problems where you have a ##d(x)## times ## f(g(x))## times g prime of x where we use substitution and it works but the proof didn't prove this condition..

    I was wondering if you can prove why it works through infinite sums like for example
    ##( f(g(x)) g^{'}(x) dx + f(g(x+dx)) g^{'}(x+dx) dx + ..... ) ##
    If I can change that to this
    ##( f(y) dy + f(y+dy) dy + .... ## where you set y = g(x)
    It would statisfy me.. But is it possible to prove it using this?
     
  2. jcsd
  3. Mar 1, 2017 #2

    Ssnow

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    Substitution works because the chain rule for derivative works:

    ## [f(g(x))]'=f'(g(x))g'(x) ##

    if you integrate both sides, you will obtain

    ## \int [f(g(x))]' dx=\int f'(g(x))g'(x) dx ##

    that is the same of ##\int f'(g(x))g'(x) dx = f(g(x)) + c ## because the integral is the antiderivation so ## \int [f(g(x))]' dx=f(g(x))+c##. This formally can be done substituting in the integral ##\int f'(g(x))g'(x) dx## the function ## g(x)=t## so ##g'(x)dx=dt## (we do a substitution on differentials) differentiating and ##\int f'(g(x))g'(x) dx=\int f'(t)dt=f(t)+c##, returning in ##g## we have: ##\int f'(g(x))g'(x) dx=f(g(x))+c##.

    Ssnow
     
  4. Mar 1, 2017 #3
    Oh, Can I think of it as this?
    ## [f(g(x))]'=f'(g(x))g'(x) ##
    This is just df/dx So If I make a variable called u= g(x) and then du/dx = g(x) now flipping this over gives you dx/du = 1/g(x) and we can then multiply it to df/dx to get df/du by chain rule and then we integrate with respect to u.

    What bothers me is: Why is dx substitute-able rather than a notation to let us know what we are integrating in respect to?
    I know that it refers to something infinitesimal and the integration just means the infinite sums of f(x) multiplied by dx which would make sense in these cases
    du/dx = 1/u
    u du = 1 dx
    ##\int u du = \int 1 dx ##

    Now I can apply the same logic to the substitution rule. I can assign a value u and say dx = du/g'(x) which then substitute it and integrate to get the infinite sum of ## \int f(u) du ## But I still prefer the first argument.

    Excuse me if I am talking nonsense just getting started at integration
     
  5. Mar 1, 2017 #4

    Ssnow

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    Gold Member

    Yes.

    I don't know if can help but the fact is that the presence of derivative inside the integral permit you to substitute ''the differential'', that is the espression ##g'(x)dx## (that you can see as an ''incremental quantity'' ) with another much simpler as ##du##. Consequence of this: your integral will be simpler then the first (this is similar to the integration by part rule)

    No absolutely, your talk is perfectly reasonable :smile:.

    Ssnow
     
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