Posted by MS on Wednesday, September 4, 2013 at 1:47am.
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, Wednesday, September 4, 2013 at 3:36am
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, Wednesday, September 4, 2013 at 3:39am
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, Wednesday, September 4, 2013 at 3:52am
Does it indicate that the answer 'b/a' given in the book may be wrong? I tried many times but could not get it.
Answer This Question
Related Questions
 calculus  Find complete length of curve r=a sin^3(theta/3). I have gone thus (...
 calc  i did this problem and it isn't working out, so i think i'm either making...
 Calculus help??  I'm not sure how to solve this and help would be great! d/dx [...
 Integration by Parts  integral from 0 to 2pi of isin(t)e^(it)dt. I know my ...
 maths  Please can you help me with this question? Choose the option which is a ...
 Math  1. On the interval [0, 2pi] what are the solutions to the equation ...
 Calculus  1) The period of a trig. function y=sin kx is 2pi/k. Then period of y...
 Trig  Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < ...
 math  Evaluate. 1. sin^1(1/2) 2. cos^1[(root 3)/2] 3. arctan[(root3)/3] 4...
 Trigonometry  Solve the equation for solutions in the interval 0<=theta<...
More Related Questions