Respuesta :

ivycoveredwalls
The period of the function is [tex]\frac{2\pi}{w}[/tex] so the first arch extends over the interval [tex]\left[ 0,\frac{\pi}{w} \right][/tex]. The area is

[tex]\int_0^{\pi/w} \sin{(wt)} dt[/tex]

The integral can be found by substitution.
[tex]u=wt \\ du=w \; dt \\ \frac{du}{u}=dt[/tex]

The limits of integration change, too.

[tex]t=0 \Rightarrow u=0 \\ t=\frac{\pi}{w} \Rightarrow u=\pi[/tex]

[tex]\frac{k}{w} \int_0^\pi \sin{u} \; du = \frac{k}{w} \left[-\cos{u} \right]_0^\pi = \frac{k}{w} \left[ 1-(-1) \right] = \frac{2k}{w}[/tex]
Ver imagen ivycoveredwalls

The area under one arch of the curve y(t) = ksin(wt) for t ≥ 0 where k and w are positive constants is 2k /w

The  period of [tex]\rm sin \; \theta[/tex] is [tex]\rm 2\pi[/tex]

The period of [tex]\rm k \; sin (wt) \; is \; 2\pi /w[/tex]

According to the given function

y(t) = ksin(wt) for t ≥ 0

Where k is the amplitude of  the  given function

One arch of the  given function extends from [tex](0 , \pi /w)[/tex]

On substituting

[tex]\rm wt = p ......(1)[/tex]

in the  given function

we get

[tex]\rm y = k \; sin p[/tex]

On differentiating equation (1) with respect to t  we get

[tex]\rm w = \dfrac{dp}{dt}\\dt = dp /w[/tex]

[tex]\rm when \; t= 0\; the \; value\; of \; p\; is\; 0.\\when \; t = \pi/w \; the \; value\; of \; p\; is\; \pi.\\[/tex]

[tex]\rm \dfrac{k}{w} \int\limits^\pi_0 {sinp} \, dp = \dfrac{k}{w} (-cos \;p\; )_{0}^{\pi}\\\\-(k/w) (-1-1) = 2k /w[/tex]

So the area under one arch of the curve y(t) = ksin(wt) for t ≥ 0 where k and w are positive constants is 2k /w

For more information please refer to the link given below

https://brainly.com/question/15122151

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