Saturday

November 29, 2014

November 29, 2014

Posted by **MS** on Wednesday, September 4, 2013 at 1:47am.

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.

- Maths -
**Graham**, Wednesday, September 4, 2013 at 3:36amy(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))

- Maths -
**Graham**, Wednesday, September 4, 2013 at 3:39amAnd, 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 3:52amDoes 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- (...

Maths - a curve ahs parametric equations x=t^2 and y= 1-1/2t for t>0. i)find ...

math - The region R, is bounded by the graphs of x = 5/3 y and the curve C given...

math - The region R, is bounded by the graphs of x = 5/3 y and the curve C given...

Integral - That's the same as the integral of sin^2 x dx. Use integration by ...

MATH-HELP! - The region R, is bounded by the graphs of x = 5/3 y and the curve C...

why won't anybody help me - The region R, is bounded by the graphs of x = 5/3 y ...

calc - find the area between the x-axis and the graph of hte given function over...

K - (a) Find the indeﬁnite integrals of the following functions. (i) f (t...

calculus - Given the curve defined by the equation y=cos^2(x) + sqrt(2)* sin(x) ...