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jok3333 [9.3K]
4 years ago
5

Sketch the region enclosed by the given curves.

Mathematics
1 answer:
Karo-lina-s [1.5K]4 years ago
8 0

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

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