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Homework Help: Simple question II

  1. Dec 13, 2008 #1
    Is my book right to write

    [tex]\int ^{0}_{- \infty}e^{-a|x|}dx = \int ^{0}_{- \infty}e^{ax}dx[/tex]

    ?

    In case, why?
     
  2. jcsd
  3. Dec 13, 2008 #2
    If x is negative, what is |x|?
     
  4. Dec 13, 2008 #3
    positive, but -a is still negative.
     
  5. Dec 13, 2008 #4

    HallsofIvy

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    That was not the question. If x is negative, then |x|= -x. For x negative, -a|x|= (-a)(-x)= ax.
     
  6. Dec 13, 2008 #5

    gabbagabbahey

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    errmmm...maybe it's best to think of an example...suppose you have x=-2, what is |x|?
     
  7. Dec 13, 2008 #6
    |-2| = 2
     
  8. Dec 13, 2008 #7

    gabbagabbahey

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    Right, and if x=-2, what is -x?
     
  9. Dec 13, 2008 #8
    Should be 2. I can simply put the minus outside the brackets?
     
  10. Dec 13, 2008 #9

    gabbagabbahey

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    Well, if |x|=2 and -x=2, then surely you can say that |x|=-x?:wink:


    So...what does that make -a|x|?
     
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