What is area y=sin^2(x),y=sin^3(x),[0,pi] ?

The answer to area y=sin^2(x),y=sin^3(x),[0,pi] is pi/2-4/3

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area y=sin^2(x),y=sin^3(x),[0,pi]

area

y = sin^{2} (x), y = sin^{3} (x), [0, π]

\frac{π}{2} - \frac{4}{3}

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{π} sin^{2} (x) - sin^{3} (x) d x : \frac{π}{2} - \frac{4}{3}

The area is:

= \frac{π}{2} - \frac{4}{3}

n = 0 \sum \infty (x + 1)^{n}

\frac{d}{d x} (2 x y^{4} - 3)

\frac{\partial}{\partial x} (2 x^{3} y + sin (x^{3}))

y^{^{'}} + 4 y = sin (3 t)

What is area y=sin^2(x),y=sin^3(x),[0,pi] ?
The answer to area y=sin^2(x),y=sin^3(x),[0,pi] is pi/2-4/3