Maths
posted by MS .
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[1cos(2pi x/a)/2]/2
Integral y from a to b=b/2(ba)ab/4pi[sin(2pi b/a)sin2pi)
MV=b/2[ab sin(2pi b/a)]/(ba)
Ans given is b/a. I am not getting further.

y(x) = b sin^2(πx/a)
The mean of the curve over the range b to a is:
y_ave = 1/(ab) ∫(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(ab)/2 + ab sin(2πb/a)/(4π)
1/(ab)∫(x=b to a) y(x) dx
= (b/2) + (ab sin(2πb/a))/(4π(ab)) 
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.

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