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Find the exact value of the area under one arch of the curve y(t) = V0sin(wt). Assume V0 and w are positive constants. You can use V_0 for V0. Note - V is upper case here
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y=asin(wt) Now we find the period, and we know that half a period is an arch.
The period of y=sin(t) is 2pi, so what would y=sin(wt) be?
@Ldaniel
t/w
2pi/w
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Alright, then we know that we want to integrate the function from 0 to pi/w, right?
@Ldaniel
yes\
So, we are looking for \[\int\limits_{0}^{pi/w}asin(wt)dt\]
where a and w are both constants.
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take out the a, and it becomes \[a \int\limits_{0}^{pi/w}\sin(wt) dt\]
|dw:1335426994623:dw|
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