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Thread: Area

  1. #1
    Oct 2008


    Find The Area Of The Region:

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

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

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

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

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

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

    But the integral is

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