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anonymous

  • 5 years ago

Does anyone know the integral of -e^{-2y}

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  1. amistre64
    • 5 years ago
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    i think I can do that....

  2. anonymous
    • 5 years ago
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    (1/2)e^(-2y)

  3. amistre64
    • 5 years ago
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    [S] -e^-2y dy (1/2) [S] -2 e^-2y dy e^-2y ----- is what I get if I did it right :) 2

  4. anonymous
    • 5 years ago
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    We are having a dispute as to whether or not the 2y stays negative

  5. amistre64
    • 5 years ago
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    +C ....dont forget the +C lol

  6. anonymous
    • 5 years ago
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    arrrgh

  7. amistre64
    • 5 years ago
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    yeah thats right....i got it....go me.... its not my birthday but it was at somepoint....uhhunh

  8. anonymous
    • 5 years ago
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    i h8 this someday i'll start writing alculus and someone'll go "+C!!!"

  9. amistre64
    • 5 years ago
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    LOL

  10. anonymous
    • 5 years ago
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    So the integral of e^{-x) = -e^{-x}+c doesnt e^{-2y} end up -(1/2)e^{-2y}+c? Or am I wrong about the -e?

  11. anonymous
    • 5 years ago
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    thats exponential shouldnt change so thats correct, but it is divided by constant -2 so it works out to be positive

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