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

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 is

The period of

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

On substituting

in the  given function

we get

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

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

link

step-by-step explanation:

y=mx+b

answer is:

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