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-
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.
y(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)
= (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))
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.