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Thread: Area under 1/x

  1. #1
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    Area under 1/x

    Can anyone see a simple way of explaining why the area of the triangle ( regardless of where the tangent is drawn) is always the same? In other words why does it work? I can justify it algebraically.
    Attached Thumbnails Attached Thumbnails Area under 1/x-1-over-x.png  
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  2. #2
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    Re: Area under 1/x

    suppose the point of tangency is $x_0$

    then the tangent line will have slope $-\dfrac {1}{x_0^2}$ and will include the point $\left(x_0, \dfrac{1}{x_0}\right)$

    Given this line see if you can figure out where the x and y intercepts occur.

    The area of the triangle is then trivial to compute.

    Spoiler:
    you should find that $A = \dfrac 1 2$
    Last edited by romsek; Jan 17th 2018 at 09:10 AM.
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  3. #3
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    Re: Area under 1/x

    Yes i can see an algebraic way as you describe above. Is there a geometric way of seeing the result.
    I thought what if we fitted a rectangle under it? Its area would always be one.. can't see why the triangle will always be fixed ?
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  4. #4
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    Re: Area under 1/x

    Quote Originally Posted by rodders View Post
    Yes i can see an algebraic way as you describe above. Is there a geometric way of seeing the result.
    I thought what if we fitted a rectangle under it? Its area would always be one.. can't see why the triangle will always be fixed ?
    It is the reciprocal function of $y=x$. This gives it tremendous symmetry across that line. What you are seeing with this area is one manifestation of that symmetry. I am not sure what you are looking for as a "geometric way of seeing the result". But, I would start with that line of symmetry.
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