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.
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)
∫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))
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.
To find the mean value of the curve where x lies between b and a, we need to calculate the integral of the curve over that interval and divide it by the width of the interval (a - b).
Given the equation of the curve as y = b * sin^2(pi * x / a), you correctly simplified it to y = b[1 - cos(2pi * x / a)] / 2.
Now, let's proceed further to calculate the integral over the interval [b, a]:
∫[b, a] y dx = ∫[b, a] [b(1 - cos(2pi * x / a)) / 2] dx
To integrate this, we distribute b/2 and integrate each term separately:
∫[b, a] (b - b * cos(2pi * x / a)) / 2 dx
= (b/2)∫[b, a] dx - (b/2) ∫[b, a] cos(2pi * x / a) dx
The first term is simply (b/2) * (a - b), which simplifies to (ab - b^2)/2.
For the second term, we need to find the antiderivative of cos(2pi * x / a). The antiderivative of cos(x) is sin(x), so we can divide by the derivative of the argument and multiply by the derivative of the argument to compensate:
(b/2) ∫[b, a] cos(2pi * x / a) dx
= (b/2) * (a/pi) ∫[b, a] cos(2pi * x / a) * (pi/a) dx
= (b * a * sin(2pi * x / a)) / (4pi) [b, a]
Evaluating the second term at a and b gives:
(b * a * sin(2pi * a / a) - b * a * sin(2pi * b / a)) / (4pi)
= (b * a * sin(2pi) - b * a * sin(2pi * b / a)) / (4pi)
= (b * a * 0 - b * a * sin(2pi * b / a)) / (4pi)
= - (b * a * sin(2pi * b / a)) / (4pi)
Now, to find the mean value (MV), we divide the integral by the width of the interval (a - b):
MV = (1 / (a - b)) * (∫[b, a] y dx)
= (1 / (a - b)) * [(ab - b^2)/2 - (b * a * sin(2pi * b / a)) / (4pi)]
= (1 / (a - b)) * (ab - b^2) / 2 - (b * a * sin(2pi * b / a)) / (4pi * (a - b))
Now let's simplify it further:
MV = (ab - b^2) / (2 * (a - b)) - (b * a * sin(2pi * b / a)) / (4pi * (a - b))
= ab / (2 * (a - b)) - b^2 / (2 * (a - b)) - (b * a * sin(2pi * b / a)) / (4pi * (a - b))
Combining the terms with the same denominator:
MV = (ab - b^2 - (b * a * sin(2pi * b / a))) / (2 * (a - b) * 4pi)
Now, to simplify even further, let's multiply the numerator:
MV = (ab - b^2 - b^2 * sin(2pi * b / a) / a) / (2 * (a - b) * 4pi)
= (ab - b^2 - b^2 * sin(2pi * b / a) / a) / (8pi * (a - b))
From here, we can't simplify it further because there is no additional information about the variables a and b. Therefore, we cannot conclude that the mean value is b/a based on the given equation.