Maths

Eq of curve is y=b sin^2(pi.x/a). Find mean value for part of curve where 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/4pi[sin(2pi b/a)-sin2pi)

MV=b/2-[ab sin(2pi b/a)]/(b-a)
Ans given is b/a. I am not getting further.

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asked by MS
  1. y(x) = b sin^2(Ī€x/a)

    The mean of the curve over the range b to a is:
    y_ave = 1/(a-b) âˆĢ(x=b to a) y(x) dx

    sin^2(Ī€x/a) = 1 - cos(2Ī€x/a)

    âˆĢy(x) dx
    = (b/2) âˆĢ (1 - cos(2Ī€x/a)) dx
    = (b/2) (x - a sin(2Ī€x/a)/(2Ī€)) + constant
    = bx/2 - ab sin(2Ī€x/a)/(4Ī€) + constant

    âˆĢ(x=b to a) y(x) dx
    = b(a-b)/2 + ab sin(2Ī€b/a)/(4Ī€)

    1/(a-b)âˆĢ(x=b to a) y(x) dx
    = (b/2) + (ab sin(2Ī€b/a))/(4Ī€(a-b))

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    posted by Graham
  2. And, that is just about as far as it goes. You can play around with the sine identities, but it doesn't simplify much further.

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    posted by Graham
  3. Does it indicate that the answer 'b/a' given in the book may be wrong? I tried many times but could not get it.

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    posted by MS

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