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Math Help - increasing or decreasing

  1. #1
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    increasing or decreasing

    the question is

    Let F(a) be the function which gives the area under the graph of y=x*e^-x between x=0 and x=a for a>0

    i found F(a) to be (-x*e^-x)-(e^-x)
    then evaluated with the given endpoints i got

    (-a*e^-a)-e^a+1
    (not sure if this is correct)

    the second question is

    is F increasing or a decreasing function?
    is it the same that if the second derivative tells the concavity? pointers anyone?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by acosta0809 View Post
    is F increasing or a decreasing function?
    is it the same that if the second derivative tells the concavity? pointers anyone?
    Look at it like any other function.


    \frac{dF}{dx}=x*e^{-x}\Rightarrow\frac{d}{dx}(x*e^{-x})=e^{-x}-xe^{-x}=0

    \frac{1}{e^x}(1-x)=0\Rightarrow\text{ critical values at }x=1
    Last edited by VonNemo19; July 14th 2009 at 06:45 PM.
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  3. #3
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    Hello, acosta0809!

    Let F(a) be the function which gives the area under the graph of y\,=\,xe^{-x}
    between x=0 and x=a for a>0.


    i found F(a) to be: . -xe^{-x} -e^{-x}

    then evaluated with the given endpoints i got: . -ae^{-a} -e^{-a}+1
    (not sure if this is correct)
    It's correct!


    The second question is:

    Is F(a) an increasing or a decreasing function?
    Find the first derivative (slope).

    We have: . F(a) \:=\:ae^{-a} - e^{-a} + 1

    Then: . F'(a) \;=\;ae^{-a} - e^{-} + e^{-a} + 0 \;=\;\frac{a}{e^a}

    Since a > 0, then: . F'(a) \:=\:\frac{a}{e^a} is always positive.

    Therefore, F(a) is an increasing function.



    If we look at the graph, it's obvious . . .
    Code:
    . . . - |
    . . . - |     ..*.
    . . . - |   *::::::*.
    . . . - | *::::::::::*..
    . . . - |*::::::::::::::::*
    . . . - |:::::::::::::::::|        *
    . . . - * - - - - - - - - + - - - - - 
    . . . - |                 a
    As a increases, the area under the curve increases.

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