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Math Help - integrate e^something

  1. #1
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    integrate e^something

    why does e^2x = 1/2 e^2x

    does e^-x = -e^-x or -1/e^-x ?
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  2. #2
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    Quote Originally Posted by jeph View Post
    why does e^2x = 1/2 e^2x

    does e^-x = -e^-x or -1/e^-x ?
    First,
    e^{2x} DOES NOT EQUAL 1/2e^{2x}

    We need to find,
    \int e^{2x} dx
    Use the substitution t=2x
    \frac{1}{2} \int e^t dt = \frac{1}{2}e^t+C=\frac{1}{2}e^{2x}+C

    Similarly with the other one.
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    i dont get how the 1/2 gets there...
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  4. #4
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    Quote Originally Posted by jeph View Post
    i dont get how the 1/2 gets there...
    Because if t=2x then t'=2 so we need to introduce a factor of two into that expression. Thus, I simply multiply it in and divide it out.
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  5. #5
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    \int e^{kx}\,dx=\frac1ke^{kx}+c

    where k is an arbitrary constant.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    \int e^{kx}\,dx=\frac1ke^{kx}+c

    where k is an arbitrary constant.
    And k\not = 0
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    And k\not = 0
    but of course
    Last edited by Jhevon; May 30th 2007 at 02:54 PM.
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  8. #8
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    Yeah, that's what I said "arbitrary"

    (common sense)
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Yeah, that's what I said "arbitrary"

    (common sense)
    i guess, but mathematicians don't take anything for granted, and therefore, they are in the habit of rigorously defining common sense things.

    for instance, to define a rational number, a mathematician would say:

    a rational number r is one that can be expressed as  r = \frac {a}{b} \mbox { for } a,b \in \mathbb {Z}, b \neq 0

    even though the b \neq 0 is the "common sense" part
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  10. #10
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    Ahahahaha, got it!!

    You're right, we gotta be stricts.
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