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Calculus/Int. Calculus (Mean value)

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Question
Eq of curve is y=b sin^2(pi.x/a). Find mean height of the portion  for which x lies between b and a.

I have gone thus far-
y=b[1-cos(2pi x/a)/2]/2
Integral y from a to b=b/2(b-a)-ab[sin(2pi b/a)]/4pi

Mean Value=Integral y/(b-a)=b/2-[ab sin(2pi b/a)]/4pi(b-a).

Ans given in the book is 'b/a'. I am not getting it anyway.

Answer
A slight mistake in y=b[1-cos(2pi x/a)/2]/2  where it should be y=b[1-cos(2pi x/a)]/2.

Going beyond that, your workings are fine. You might wish to verify if the problem has made mention of the special nature of a and b, eg they are integers. This would aid in the simplification.

Hope this helps. Peace.

Calculus

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