Find the exact area under one arch of the curve
y(t) = ksin(wt)
where k and w are positive constants. Note: Use lower case k and w.

Question

Find the exact area under one arch of the curve 
y(t) = ksin(wt)
 where k and w are positive constants. Note: Use lower case k and w.

anonymous · Answer

im stuck

zepdrix · Answer

Hmm let's start by finding the correct boundaries for one arch.

zepdrix · Answer

$$\Large\bf\sf y(t)\quad=\quad k \sin(\omega t)$$The period is given by:$$\Large\bf\sf P=\frac{2\pi}{\omega}$$

zepdrix · Answer

Which consists of BOTH arches, right?
So we want half of that.

zepdrix · Answer

Sine goes through the origin, ok good we can start at 0.

So our boundaries will be from $\Large\bf\sf0$ to $\Large\bf\sf \dfrac{\pi}{\omega}$

zepdrix · Answer

Any confusion there? :o

zepdrix · Answer

$$\Large\bf\sf \int\limits_0^{\pi/\omega}k \sin(\omega t)\;dt$$

zepdrix · Answer

Bring the k outside, it's just a constant.$$\Large\bf\sf k\int\limits\limits_0^{\pi/\omega} \sin(\omega t)\;dt$$From there you can apply a u-substitution if the omega factor on t is going to confuse you.

zepdrix · Answer

Oh if it's supposed to be a w, then my bad.

anonymous · Answer

thank you sorry I was working it out on my own and I managed to get it without your help my bad i am so used to people ignoring my questions sorry