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Homework Help: Indefinite Integration by exchange of variables

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    This is only an example but I do not understand what they are doing...

    [tex] \int (4x+1)^{3} + (4x+1)^{2}+(4x+1) dx [/tex]


    2. Relevant equations

    [tex] \int f(u) du = F(g(x)) + C [/tex]

    3. The attempt at a solution

    Let

    [tex] u = 4x+1 [/tex]

    then

    [tex] du = 4 dx [/tex]

    and

    [tex] dx = \frac {1}{4}du [/tex]

    how did they get that dx was 1/4? There are no steps to explain this, it just lists them as having that value.

    So after substituting the problem should look like this:

    [tex] \int (u^{3} + u^{2} + u) * \frac {1}{4} du [/tex]

    which is this:

    [tex] \frac{1}{4}(\frac{u^{4}}{4} + \frac {u^{3}}{3} + \frac {u^{2}}{2}) + C [/tex]

    [tex] \frac{1}{4}[\frac {(4x+1)^{4}}{4}+\frac{(4x+1)^{3}}{3}+\frac{(4x+1)^{2}}{2}] +C [/tex]

    So I suppose the only portion that I really don't understand is how they got the 1/4 dx value out of seemingly nothing....
     
  2. jcsd
  3. Apr 26, 2010 #2
    [tex] du = 4 dx [/tex] solve this for [tex] dx [/tex]
     
  4. Apr 26, 2010 #3
    I guess I just didn't really think about it all that hard... I got another question though, but Ill post it in a new thread
     
  5. Apr 26, 2010 #4
    Your solution is correct.
     
  6. Apr 26, 2010 #5
    There appears to be many notations and definitions of "dy" where y is a dummy variable. Some authors chose to define dy, while others leave it at "no meaning". I came across one interpretation that might help you...

    du/dx = 4 = 4 dx/dx

    So, (1/4) du/dx = dx/dx

    By suppressing the denominator, we get,

    (1/4) du = dx

    Keeping in mind that the denominator is still there. That is, "dy" is shorthand for "dy/dx".

    I like this interpretation because it does not appear to manipulate du/dx like a fraction.

    Hope this helps!
     
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