What is area y=7cos(2x),y=7-7cos(2x) ?

The answer to area y=7cos(2x),y=7-7cos(2x) is 7sqrt(3)+(14pi)/3

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area y=7cos(2x),y=7-7cos(2x)

area

y = 7 cos (2 x), y = 7 - 7 cos (2 x)

73 + \frac{14 π}{3}

Decimal

26.78512 \dots

Solution steps

Apply the area formula:

\int_{\frac{π}{6}}^{\frac{5 π}{6}} ∣ 7 cos (2 x) - (7 - 7 cos (2 x)) ∣ d x

Solve

\int_{\frac{π}{6}}^{\frac{5 π}{6}} ∣ 7 cos (2 x) - (7 - 7 cos (2 x)) ∣ d x : 73 + \frac{14 π}{3}

The area is:

= 73 + \frac{14 π}{3}

\frac{d}{d x} (5 x^{2})

\frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

\frac{\partial}{\partial x} (e^{- 3 x^{2} - 2 y^{2}})

\frac{d}{d x} (- 84 x + \frac{37}{2} x^{2} - \frac{x ^{4}}{4})

\int cos (x) sin (3 x) - 4 sin (x) d x

What is area y=7cos(2x),y=7-7cos(2x) ?
The answer to area y=7cos(2x),y=7-7cos(2x) is 7sqrt(3)+(14pi)/3