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Integral of 1/2^x

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    1 1/2x

    2. Relevant equations

    2-x = (eln 2)-x

    Could possibly be relevant, I don't know.

    3. The attempt at a solution

    y = 2-x

    ln y = -x ln 2

    1/y dy/dx = -ln 2 -x/2

    ...
     
  2. jcsd
  3. Oct 25, 2015 #2

    HallsofIvy

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    Better is [itex]2^{-x}= e^{ln(2^{-x})}= e^{x ln(2)}[/itex]

     
  4. Oct 25, 2015 #3

    Samy_A

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    Quite relevant, try to build on that.

    ETA: never mind, didn't see the previous post.
     
    Last edited: Oct 25, 2015
  5. Oct 25, 2015 #4
    ∫(1/2x) dx = ∫(1/ex ln 2) dx

    u = ln 2 x → du = ln 2 dx

    1/ln 2 ∫ du/eu = ln eu / ln 2 = ln ex ln 2 / ln 2 = x ln 2/ln2 = x

    Okay, what did I do wrong?o_O

    Edit: I was thinking of u'/ln u, my bad!

    1/ln 2 ∫ e-u du = -e-u/ln 2 = -e-x ln 2/ln 2 = -2-x/ln2

    Is this correct?
     
    Last edited: Oct 25, 2015
  6. Oct 25, 2015 #5
    Wouldn't that be e-x ln 2?
     
  7. Oct 25, 2015 #6

    Samy_A

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    You lost me in this last equality, I guess here something is wrong.

    Also, don't forget to adapt the limits of your definite integral when you apply the substitution (not that I think that you really need that substitution).
     
  8. Oct 25, 2015 #7

    Samy_A

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    Yes, this is correct for the indefinite integral.
     
  9. Oct 25, 2015 #8
    Yes I made an edit on my post, I was thinking of u'/ln u. I think I've got it now.

    Screenshot_2015-10-25-14-23-21.png
     
  10. Oct 26, 2015 #9

    HallsofIvy

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    Yes. Thanks.
     
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