What is area 6sin(x),6cos(2x),[-pi/2 , pi/6 ] ?

The answer to area 6sin(x),6cos(2x),[-pi/2 , pi/6 ] is 3sqrt(3)+(3sqrt(3))/2

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area 6sin(x),6cos(2x),[-pi/2 , pi/6 ]

area

6 sin (x), 6 cos (2 x), [- \frac{π}{2}, \frac{π}{6}]

33 + \frac{3 3}{2}

Decimal

7.79422 \dots

Solution steps

Apply the area formula:

\int_{- \frac{π}{2}}^{\frac{π}{6}} ∣ 6 sin (x) - 6 cos (2 x) ∣ d x

Solve

\int_{- \frac{π}{2}}^{\frac{π}{6}} ∣ 6 sin (x) - 6 cos (2 x) ∣ d x : 33 + \frac{3 3}{2}

The area is:

= 33 + \frac{3 3}{2}

f (x) = \frac{1}{x + 2}

y^{^{''}} - 4 y^{^{'}} + 4 y = t e^{2 t}

\frac{d}{d x} (x^{\frac{2}{3}} - 3)

\int \frac{1}{x ( 3 x )} d x

\int \frac{4 x}{x ^{2} - 9} d x

What is area 6sin(x),6cos(2x),[-pi/2 , pi/6 ] ?
The answer to area 6sin(x),6cos(2x),[-pi/2 , pi/6 ] is 3sqrt(3)+(3sqrt(3))/2