What is area y=8cos(6x),y=8-8cos(6x) ?

The answer to area y=8cos(6x),y=8-8cos(6x) is 8/(sqrt(3))+(16pi)/9

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area y=8cos(6x),y=8-8cos(6x)

area

y = 8 cos (6 x), y = 8 - 8 cos (6 x)

\frac{8}{3} + \frac{16 π}{9}

Decimal Notation

10.20385 \dots

Solution steps

Apply the area formula:

\int_{\frac{π}{18}}^{\frac{5 π}{18}} ∣ 8 cos (6 x) - (8 - 8 cos (6 x)) ∣ d x

Solve

\int_{\frac{π}{18}}^{\frac{5 π}{18}} ∣ 8 cos (6 x) - (8 - 8 cos (6 x)) ∣ d x : \frac{8}{3} + \frac{16 π}{9}

The area is:

= \frac{8}{3} + \frac{16 π}{9}

cos (8 x)

y^{^{''}} + \frac{1}{4} y^{^{'}} + 2 y = 2 cos (2 t)

x \to a lim (e^{f (x)})

f (x) = 0.8 x^{2} + 4, g (x) = x, [- 6, 6]

\int \frac{u}{x} d x

What is area y=8cos(6x),y=8-8cos(6x) ?
The answer to area y=8cos(6x),y=8-8cos(6x) is 8/(sqrt(3))+(16pi)/9