Posted by MS on .
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.

Maths 
Graham,
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)) 
Maths 
Graham,
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.

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