Area

**qzno** · February 19th 2009, 11:54 AM

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 : )

**running-gag** · February 19th 2009, 12:07 PM

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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