What is area 7cos(9x),7-7cos(9x) ?

The answer to area 7cos(9x),7-7cos(9x) is (14)/(3sqrt(3))+(28pi)/(27)

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area 7cos(9x),7-7cos(9x)

area

7 cos (9 x), 7 - 7 cos (9 x)

\frac{14}{3 3} + \frac{28 π}{27}

Decimal Notation

5.95224 \dots

Solution steps

Apply the area formula:

\int_{\frac{π}{27}}^{\frac{5 π}{27}} ∣ 7 cos (9 x) - (7 - 7 cos (9 x)) ∣ d x

Solve

\int_{\frac{π}{27}}^{\frac{5 π}{27}} ∣ 7 cos (9 x) - (7 - 7 cos (9 x)) ∣ d x : \frac{14}{3 3} + \frac{28 π}{27}

The area is:

= \frac{14}{3 3} + \frac{28 π}{27}

\int \frac{x ^{2}}{x ^{2}} d x

\int (2 x e^{x^{2}}) d x

y = x^{3}, y = x^{2} - 2 x, [0, 1]

\int_{0}^{1} \frac{y}{e ^{8 y}} d y

\int \frac{ln ( 7 x )}{x} d x

What is area 7cos(9x),7-7cos(9x) ?
The answer to area 7cos(9x),7-7cos(9x) is (14)/(3sqrt(3))+(28pi)/(27)