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Antiderivative question

  1. Oct 22, 2006 #1
    antiderivative of e^-x

    I get -e^-x

    antiderivative of cos(mx) where m is a constant

    I get [tex] \frac {1}{m} \sin x [/tex]

    Please let me know if my attempts are correct. Thanks alot!
     
  2. jcsd
  3. Oct 22, 2006 #2

    quasar987

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    If you can calculate an ordinary derivative, you can always check whether the anti derivative you found is correct. For instance,

    [tex]\frac{d}{dx}(-e^{-x}) = (-1)(-e^{-x})=e^{-x}[/tex]

    So your antiderivative is correct. Actually, the most general antiderivative to e^-x is -e^-x + C, where C is a constant. Do you see why?
     
  4. Oct 22, 2006 #3
    because the constant C changes the y value by a certain amount moving the functuion vertically on the graph?

    This is what I know of the top of my head and am not quite sure how it relates to your question though.:confused:
     
  5. Oct 22, 2006 #4

    quasar987

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    An antiderivative of a function f(x) is a function F(x) such that F'(x) = f(x). Is that your definition of antiderivative also? If it is, then suppose you found just such an antiderivative F(x). Then F(x) + C is also an anti derivative of f(x) because

    [tex]\frac{d}{dx}(F(x)+C) = F'(x) + \frac{d}{dx}C = F'(x) + 0 =f(x)[/tex]
     
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