Question

Sketch the region enclosed by the given curves.

Answer 1

Answer:

Area (A) $\simeq$ 0.0682

Step-by-step explanation:

The sketch for the region enclosed by the given curves can be found in the image attached below.

From the image below;

The two curves intersect in the area of tan 9x = 2 sin 9x

Recall that:

$\dfrac{sin \ 9x }{cos \ 9x } = 2 \ sin \ 9x$

$cos \ 9 x= \dfrac{1}{2}$

making x the subject; then:

$x = \dfrac{1}{9} cos ^{-1}(\dfrac{1}{2})$

$x = \dfrac{1}{9}(\dfrac{\pi}{3})$

$x = \dfrac{\pi}{27}$

The subdivision of the domain is in two intervals $[-\dfrac{\pi}{27}, 0], [0, \dfrac{\pi}{27}]$

where;

$x \ \ \varepsilon \ \ [ -\dfrac{\pi}{27},0]; tan \ 9x > 2 sin \ 9x$

$x \ \ \varepsilon \ \ [ 0,\dfrac{\pi}{27}]; tan \ 9x < 2 \ sin \ 9x$

Area (A) = $\int ^0_{-\pi/27}}} ( tan \ 9x - 2 \ sin \ 9x ) \ dx + \int ^{\pi/27}_{0}(2 \ sin \ 9x - tan \ 9x \ ) \ dx$

$= \begin {bmatrix} \dfrac{1}{9} In \ sec 9x + \dfrac{2}{9} \ cos \ 9x \end {bmatrix}^0_{-\pi/27} + \begin {bmatrix} -\dfrac{2}{7} cos 9x - \dfrac{1}{9} \ In \ sec \ 9x \end {bmatrix}^{-\pi/27} _0$

$= \dfrac{1}{9}(1-In 2) + \dfrac{1}{9}(1-In 2)$

$= \dfrac{2}{9}(1 - In 2)$

Area (A) = 0.06818

Area (A) $\simeq$ 0.0682