# Let R be the region in the first quadrant that is enclosed by the graph of y=tan x, the x-axis, and the line x=pi/3, how do you find the area?

Nov 5, 2016

ln2

#### Explanation:

The area is given by integration with appropriate bounds

$\therefore A = {\int}_{0}^{\frac{\pi}{3}} \tan x \mathrm{dx}$
$\therefore A = {\int}_{0}^{\frac{\pi}{3}} \sin \frac{x}{\cos} x \mathrm{dx}$ .... [1]

Which we can integrate using a substitution

 { ("Let "u=cosx,=>,(du)/dx=-sinx), ("When "x=0,=>,u=cos0=1), ("And "x=pi/3,=>,u=cos(pi/3)=1/2) :}

$\therefore \int \ldots \mathrm{du} = \int \ldots - \sin x \mathrm{dx}$

Substituting into [1] we get

$A = {\int}_{1}^{\frac{1}{2}} \left(- \frac{1}{u}\right) \mathrm{du}$
$\therefore A = {\int}_{\frac{1}{2}}^{1} \frac{1}{u} \mathrm{du}$ (using ${\int}_{a}^{b} \ldots = - {\int}_{b}^{a} \ldots$)
$\therefore A = {\left[\ln u\right]}_{\frac{1}{2}}^{1}$
$\therefore A = \ln \left(1\right) - \ln \left(\frac{1}{2}\right)$

 :. A = 0-ln(1/2)_  (as $\ln \left(1\right) = 0$)
$\therefore A = \ln \left(2\right)$