Simple question II

Discussion in 'Math' started by boks, Dec 13, 2008.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    Is my book right to write

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

    ?

    In case, why?
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Remembering that exp(x) and exp (-x) are mirror image functions and that both are positive definite you should be able to see that the dotted area is your first integral and the shaded one the second.
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    boks,

    For the same reason that -a|x| = -a(-x) = ax when x <= 0, which is the range of the integral. In other words |x| = -x when x is negative or zero.

    Ratch
     
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