What is area 3cos(x),3cos(2x),0<= x<= pi ?

The answer to area 3cos(x),3cos(2x),0<= x<= pi is (9sqrt(3))/2

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area 3cos(x),3cos(2x),0<= x<= pi

area

3 cos (x), 3 cos (2 x), 0 \leq x \leq π

\frac{9 3}{2}

Decimal Notation

7.79422 \dots

Solution steps

Apply the area formula:

\int_{0}^{π} ∣ 3 cos (x) - 3 cos (2 x) ∣ d x

Solve

\int_{0}^{π} ∣ 3 cos (x) - 3 cos (2 x) ∣ d x : \frac{9 3}{2}

The area is:

= \frac{9 3}{2}

(1 + y^{2} sin (2 x)) d x - 2 y cos^{2} (x) d y = 0

\frac{d y}{d x} = x y^{3}

\int \frac{1}{x ^{2} + 12} d x

3 x, 9 x^{2}

n = 1 \sum \infty \frac{2 n}{1 1 ^{n}}

What is area 3cos(x),3cos(2x),0<= x<= pi ?
The answer to area 3cos(x),3cos(2x),0<= x<= pi is (9sqrt(3))/2