Asked 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[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.
            
        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.
Answers
                    Answered by
            Graham
            
    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))
    
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))
                    Answered by
            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.
    
                    Answered by
            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.
    
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