What is area 4-4cos(7x),4cos(7x),0>= x>= pi/7 ?

The answer to area 4-4cos(7x),4cos(7x),0>= x>= pi/7 is (16pi}{21}+\frac{8sqrt(3))/7

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area 4-4cos(7x),4cos(7x),0>= x>= pi/7

area

4 - 4 cos (7 x), 4 cos (7 x), 0 \geq x \geq \frac{π}{7}

\frac{16 π}{21} + \frac{8 3}{7}

Decimal Notation

4.37308 \dots

Solution steps

Apply the area formula:

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

Solve

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

The area is:

= \frac{16 π}{21} + \frac{8 3}{7}

\frac{d}{d x} (2 x x + \frac{1}{x})

\int x^{2} + 7 d x

y^{^{''}} = - 3 y - 2 y^{^{'}}, y (0) = 1, y^{^{'}} (0) = 0

\int \frac{4}{1 + x ^{2}} d x

3 y^{^{'}} - 2 y = e^{- \frac{π}{2}}

What is area 4-4cos(7x),4cos(7x),0>= x>= pi/7 ?
The answer to area 4-4cos(7x),4cos(7x),0>= x>= pi/7 is (16pi}{21}+\frac{8sqrt(3))/7