What is area 4cos(7x),4-4cos(7x) ?

The answer to area 4cos(7x),4-4cos(7x) is (8sqrt(3))/7+(16pi)/(21)

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area 4cos(7x),4-4cos(7x)

area

4 cos (7 x), 4 - 4 cos (7 x)

\frac{8 3}{7} + \frac{16 π}{21}

Decimal Notation

4.37308 \dots

Solution steps

Apply the area formula:

\int_{\frac{π}{21}}^{\frac{5 π}{21}} ∣ 4 cos (7 x) - (4 - 4 cos (7 x)) ∣ d x

Solve

\int_{\frac{π}{21}}^{\frac{5 π}{21}} ∣ 4 cos (7 x) - (4 - 4 cos (7 x)) ∣ d x : \frac{8 3}{7} + \frac{16 π}{21}

The area is:

= \frac{8 3}{7} + \frac{16 π}{21}

(t^{2} - 2 t) (t^{2} + 2 t)

x \to \infty lim (6 x + 5)

\frac{\partial}{\partial x} (9 y^{4} cos (x^{2}))

f (x) = \frac{1}{x ^{2} - 9}

\frac{1}{2}

What is area 4cos(7x),4-4cos(7x) ?
The answer to area 4cos(7x),4-4cos(7x) is (8sqrt(3))/7+(16pi)/(21)