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

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

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area y=sin(2)(x),y=sin(3)(x),0<= x<= pi

area

y = sin (2) (x), y = sin (3) (x), 0 \leq x \leq π

\frac{π ^{2} ( sin ( 2 ) - sin ( 3 ) )}{2}

Decimal Notation

3.79080 \dots

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} ( sin ( 2 ) - sin ( 3 ) )}{2}

The area is:

= \frac{π ^{2} ( sin ( 2 ) - sin ( 3 ) )}{2}

x - \frac{1}{x}

t e^{- t} sin (2 t)

y^{^{'}} - \frac{4}{t} y = t^{5} e^{t}

x \to 1 - lim (\frac{x - 3}{x - 1})

\int y (2 y + 1) e^{4 y^{3} + 3 y^{2} - 4} d y

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