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Need help w/ an integral

  1. Jan 25, 2005 #1
    Ok I know that the [tex]\int e^x = e^x + C[/tex]

    now I dont understand this problem.

    [tex]\int_0^1 e^{-3x} dx = -(1/3)e^{-3(0)} - -(1/3)e^{-3(1)}[/tex]
    [tex]= 1/3(1- e^{-3} )[/tex]

    Where does the 1/3 come from?
  2. jcsd
  3. Jan 25, 2005 #2
    anyone?? have to study for a test any help would be greatly appreciated.

    I just want to know where the constant -1/3 came from.
  4. Jan 25, 2005 #3


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    *I'm using -> to point to the derivative.

    e^x -> e^x

    e^3x -> 3e^3x

    e^-3x -> -3e^-3x

    See how the 3 came up? The -1/3 is used to get rid of the -3 because in the integral we don't have a -3.

    -1/3e^-3x -> e^-3x

    Note: I know it's sloppy, but you should get the idea.
  5. Jan 25, 2005 #4

    wow I feel really stupid to forget that lol, I forgot that when [tex]e^x[/tex] is raised to something like 3x you use the derivative of that times the original [tex]e^x[/tex] function.

    thanks :D
  6. Jan 25, 2005 #5


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    I forget it sometimes to because your so used to copying e^x down as the derivative.

    Good Luck!
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