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Math Help - Sketch the region and find the area.

  1. #1
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    Sketch the region and find the area.

    Sketch the region and find the area enclosed by the curves.

    x = 1/y which is y=1/x (right?)
    x=0
    y=1
    y=e ( this is the one I'm having trouble graphing/figuring out.)

    Any help please? Thanks
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  2. #2
    MHF Contributor Unknown008's Avatar
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    e is the natural constant, and if you type it in your calculator, you'll have the value of e as 2.718281828

    So, you graph y = 2.7
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  3. #3
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    it would be something like this :
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  4. #4
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  5. #5
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    Quote Originally Posted by action812 View Post
    Sketch the region and find the area enclosed by the curves.

    x = 1/y which is y=1/x (right?)
    x=0
    y=1
    y=e ( this is the one I'm having trouble graphing/figuring out.)

    Any help please? Thanks
    y = e is just a horizontal line ... you do know the value of e, correct?


     <br />
A = \int_1^e \frac{1}{y} \, dy<br />
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  6. #6
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    Duh! Thanks Unknown, I can't believe I didn't put 2 and 2 together. Gonna work on it now...

    And goroner, you didn't post anything?

    EDIT:: Nevermind, I see it now.
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  7. #7
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    Solution

    Use \int_1^e\int_0^\frac{1}{y}\! \,dx\,dy=1.
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  8. #8
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    i post just below sry i misclicked the button and i replayed new one so dumb but you got it check out
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  9. #9
    MHF Contributor Unknown008's Avatar
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    In fact, you are being asked to find:

    \int^e_1 \frac{1}{y} dy

    EDIT: This post would have been after goroner's first post in this thread, but you all posted while I was typing the LaTeX
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  10. #10
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    I got the area to be approximately 1. Can anyone confirm this? Thanks for all the help guys.
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  11. #11
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    Quote Originally Posted by action812 View Post
    I got the area to be approximately 1. Can anyone confirm this? Thanks for all the help guys.
    not approximately ... exactly one.
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  12. #12
    MHF Contributor Unknown008's Avatar
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    Yes, integrating, you get ln e - ln 1 = 1 - 0 = 1 units^2

    EDIT: Yes, if you use a graph instead of the integration, you would get an approximation.
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