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January 31, 2015

January 31, 2015

Posted by **MS** on Wednesday, September 4, 2013 at 1:47am.

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.

- Maths -
**Graham**, Wednesday, September 4, 2013 at 3:36amy(x) = b sin^2(πx/a)

The mean of the curve over the range b to a is:

y_ave = 1/(a-b) ∫(x=b to a) y(x) dx

sin^2(πx/a) = 1 - cos(2πx/a)

∫y(x) dx

= (b/2) ∫ (1 - cos(2πx/a)) dx

= (b/2) (x - a sin(2πx/a)/(2π)) + constant

= bx/2 - ab sin(2πx/a)/(4π) + constant

∫(x=b to a) y(x) dx

= b(a-b)/2 + ab sin(2πb/a)/(4π)

1/(a-b)∫(x=b to a) y(x) dx

= (b/2) + (ab sin(2πb/a))/(4π(a-b))

- Maths -
**Graham**, Wednesday, September 4, 2013 at 3:39amAnd, that is just about as far as it goes. You can play around with the sine identities, but it doesn't simplify much further.

- Maths -
**MS**, Wednesday, September 4, 2013 at 3:52amDoes 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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