What is area 5cos(9x),5-5cos(9x) ?

The answer to area 5cos(9x),5-5cos(9x) is (10)/(3sqrt(3))+(20pi)/(27)

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area 5cos(9x),5-5cos(9x)

area

5 cos (9 x), 5 - 5 cos (9 x)

\frac{10}{3 3} + \frac{20 π}{27}

Decimal Notation

4.25160 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{\frac{π}{27}}^{\frac{5 π}{27}} ∣ 5 cos (9 x) - (5 - 5 cos (9 x)) ∣ d x : \frac{10}{3 3} + \frac{20 π}{27}

The area is:

= \frac{10}{3 3} + \frac{20 π}{27}

f (x) = 8 sin^{6} (x)

x \to 4 + lim (\frac{1}{∣ 4 - x ∣})

\int t - 4 cos (t) d t

y = x^{3}, at x = 5

\frac{2 r - 1}{t} d r + \frac{r - 2 r ^{2}}{t ^{2} - 1} d t = 0

What is area 5cos(9x),5-5cos(9x) ?
The answer to area 5cos(9x),5-5cos(9x) is (10)/(3sqrt(3))+(20pi)/(27)