You are here:

Calculus/Int. Calculus (Mean value)


Eq of curve is y=b sin^2(pi.x/a). Find mean height of the portion  for which 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[sin(2pi b/a)]/4pi

Mean Value=Integral y/(b-a)=b/2-[ab sin(2pi b/a)]/4pi(b-a).

Ans given in the book is 'b/a'. I am not getting it anyway.

A slight mistake in y=b[1-cos(2pi x/a)/2]/2  where it should be y=b[1-cos(2pi x/a)]/2.

Going beyond that, your workings are fine. You might wish to verify if the problem has made mention of the special nature of a and b, eg they are integers. This would aid in the simplification.

Hope this helps. Peace.


All Answers

Answers by Expert:

Ask Experts


Frederick Koh


I can answer questions concerning calculus, complex numbers, vectors, statistics , algebra and trigonometry for the O level, A level and 1st/2nd year college math/engineering student.


More than 7 years of experience helping out in various homework forums. Latest Presence is over at . You can also visit my main maths website where I have designed "question locker" vaults to store tons of fully worked math problems. A second one is currently being built. Peace.

IEEE(Institute of Electrical and Electronics Engineers )

Former straight As A level student from HCJC (aka HCI); scored distinctions in both C and Further Mathematics B Eng (Hons) From The National University Of Singapore (NUS) B Sc (Hons) From University of London External (Grad Route)

©2016 All rights reserved.