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Substitution Method on indefinite integral

  1. Feb 20, 2012 #1
    Say we are solving an indefinite integral ∫x√(2x+1) dx.

    According to the textbook, the solution goes like this.

    Let u = 2x+1. Then x = (u-1)/2.

    Since √(2x+1) dx = (1/2)√u du,
    x√(2x+1) dx = [(u-1)/2] * (1/2)√u du.

    ∫x√(2x+1) dx = ∫[(u-1)/2] * (1/2)√u du. <= What justifies this??
    The rest is just trivial calculation.

    The only theorem I could rely on here to solve this problem was,

    -------------------------------------------------------------------
    If u=g(x) is a differentiable function whose range is an interval I, and f is continuous no I, then
    ∫f(g(x))g'(x) dx = ∫f(u) du.
    --------------------------------------------------------------------

    Which is so called the substitution rule.

    How can this theorem justify the process of the solution I wrote above?
     
  2. jcsd
  3. Feb 20, 2012 #2

    chiro

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    Science Advisor

    Hey Chsoviz0716 and welcome to the forums.

    The justification for this is just the substitution rule for integration.

    If we had to derive everything like this all the time we wouldn't get anything done in reality. It is ok to just say 'substitution rule' and then say what substitution you used (like you already did).

    The difference between these kinds of rules for definite and indefinite integrals is that with the definite integral you need to also transform the limits whereas the indefinite one you don't.

    Also make sure to put your constant of integration for indefinite (pain in the arse but yeah has to be done). Remember that you can always find your constant C by using your initial condition at x = 0 for a normal (x,y) function.
     
  4. Feb 20, 2012 #3
    Thank you for the reply.

    But what I'm trying to say is that the kind I mentioned above isn't the type that has the form of ∫f(g(x))*g'(x) dx, to which we can apply substitution rule directly.

    And the solution kind of distorted the theorem and solved it in a weird way.

    If ∫2(2x+4)^5 dx becomes ∫u^5 du, I accept that because the theorem explicitly approves of it.

    However, the one I wrote above is different.
    To be specific, how is ∫x√(2x+1) dx = ∫[(u-1)/2] * (1/2)√u du the consequence of the substitution theorem? What's f, g, g' here?
     
  5. Feb 20, 2012 #4

    pwsnafu

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    You use the substitution rule backwards: right hand side to left hand side. We want to evaluate
    [itex]\int x \sqrt{2x + 1} dx.[/itex]
    We let [itex]f(x) = x \sqrt{2x+1}[/itex]
    We let [itex]x = (y-1)/2[/itex]
    So then [itex]x'(y) = 1/2[/itex]
    So
    [itex]\int f(x) dx = \int f(x(y)) x'(y) dy[/itex]
    [itex]\int x \sqrt{2x+1} dx = \int \frac{y-1}{2}\, \sqrt{y} \, \frac{1}{2} \, dy[/itex]
     
  6. Feb 20, 2012 #5
    Thank you,

    It missed such an easy point.
     
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