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Math Help - Area

  1. #1
    Member
    Joined
    Oct 2008
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    109

    Area

    Find The Area Of The Region:

     y = \frac{\pi}{4x} and y = tan x between x = \frac{\pi}{6} and the first intersection point of the curves to the right of x = 0

    At the moment I have the graph drawn and I need to find the first intersection point to the right of x = 0
    I also have the integral wrote up along with the antiderivative so I hope its right.

    A = \int_{\frac{\pi}{6}} \left( (tan x) - (\frac{\pi}{4x}) \right) dx

    A = -ln ( cos(x) ) - \frac{\pi}{4} ln x \mid_{\frac{\pi}{6}}

    All I need is the right value and im set
    thanks to anyone who helps : )
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    Hi

    The value you are looking for is \frac{\pi}{4} since \frac{\pi}{4\:\frac{\pi}{4}}=1=\tan\left(\frac{\pi  }{4}\right)

    But the integral is

    A = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \left(\frac{\pi}{4x}- \tan x \right) dx
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