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Definite Integral: Exponential

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\int^{\infty}_{0} e^{-y} dy = 1[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Why is this equality true? I understand that the integral and derivative of e^y is always e^y, but I can't make out why this definite integral is equal to 1. I graphed it using a graphing application and can see why it converges to a finite number, but I can't work it out step by step.

    Thanks, I appreciate your help.

  2. jcsd
  3. Sep 18, 2010 #2


    Staff: Mentor

    How about this integral? Can you evaluate it?

    [tex]\int_0^b e^{-y} dy[/tex]
    If you can get that, then just take the limit as b approaches infinity.

    BTW, this should have been posted in the Calculus & Beyond section.
  4. Sep 18, 2010 #3


    User Avatar
    Science Advisor

    Start by letting u= -x.
  5. Sep 18, 2010 #4
    ok, suppose u=-y, then

    [tex]\int^{\infty}_{0} e^{u} dy = e^{u} ] ^{\infty}_{0} = e^{-y} ] ^{\infty}_{0} [/tex]

    By the fundemental theorem of calculus

    [tex] e^{-y} ]^{\infty}_{0} = 0 - e^0 = -1 [/tex]

    Ok, so where in my solution did I go wrong?
    Last edited: Sep 18, 2010
  6. Sep 18, 2010 #5
    Um, I wouldn't make that substitution. Mark44's suggestion is the easiest way.
  7. Sep 18, 2010 #6
    I got it. Thank you!
  8. Sep 19, 2010 #7


    Staff: Mentor

    Here's where you went wrong: If u = - y, then du = -dy. You just replaced dy with du.
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