What is area f(x)=8sin(x),g(x)=8cos(2x) ?

The answer to area f(x)=8sin(x),g(x)=8cos(2x) is 18sqrt(3)

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area f(x)=8sin(x),g(x)=8cos(2x)

area

f (x) = 8 sin (x), g (x) = 8 cos (2 x)

183

Decimal Notation

31.17691 \dots

Solution steps

Apply the area formula:

\int_{\frac{π}{6}}^{\frac{3 π}{2}} ∣ 8 sin (x) - 8 cos (2 x) ∣ d x

Solve

\int_{\frac{π}{6}}^{\frac{3 π}{2}} ∣ 8 sin (x) - 8 cos (2 x) ∣ d x : 183

The area is:

= 183

f (x) = \frac{1}{6} (2 x + 1)^{3}

\int \frac{( x ^{5} - x ^{3} + 6 x )}{x ^{4}} d x

x^{3} - 11 x^{2} + 30 x, - x^{3} + 11 x^{2} - 30 x

f (x) = (x - 2)^{3} (x - 1)

y^{^{''}} + 36 y = - 24 sin (6 t)

What is area f(x)=8sin(x),g(x)=8cos(2x) ?
The answer to area f(x)=8sin(x),g(x)=8cos(2x) is 18sqrt(3)