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Modulus and integration

  1. May 4, 2013 #1
    Q. Show that the area between the positive x-axis, the y axis and the curve [itex] y = ||e^x - 1| - 1| [/itex] is ln4 - 1

    I've drawn the curve:

    http://gyazo.com/cfd52af0f82e0e7d6b063681a73de45a

    I notice for x < 0 (as I drew e^x to start out with, that's how I noticed it):
    y = e^x


    for x>ln(2) y = e^x - 2 (again, as I drew it before hand)

    I can't seem to see what y will be for for 0≤x≤ln(2), as all my other notices are because I drew them beforehand.

    Could anyone explain how I can split up y accordingly for x <0 for 0≤x≤ln(2) x > ln(2) ?
     
  2. jcsd
  3. May 4, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    For 0≤x≤ln(2), which range does e^x cover? In particular, what is its smallest value? Based on that, can you get rid of the inner modulus? In the same way, you can get replace the outer modulus.
     
  4. May 4, 2013 #3

    SammyS

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    Use the definition of absolute value, twice.
    [tex]|e^x-1| =
    \begin{cases}
    e^x-1 & \text{if } e^x \geq 1 \quad \text{i.e. } x\ge 0 \\ \\
    1-e^x & \text{if } e^x < 1 \quad \text{i.e. } x < 0
    \end{cases}
    [/tex]
    [tex]||e^x-1|-1| =
    \begin{cases}
    |e^x-1|-1 & \text{if } |e^x-1| \geq 1 \quad \text{i.e. } x\ge \ln(2) \\ \\
    1-|e^x-1| & \text{if } |e^x-1| < 1 \quad \text{i.e. } x < \ln(2)
    \end{cases}
    [/tex]
    So you want x < ln(2) which gives ##\displaystyle ||e^x-1|-1| =1-|e^x-1| \,,\ ## but also x > 0 which tells you that ##\displaystyle |e^x-1|=e^x-1 \ .##

    Put those together.
     
  5. May 8, 2013 #4
    thank you!
     
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