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(1) A question in the book is as follows and answer given is 'b/a'.:

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

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).

I am not getting ‘b/a’ anyway. Where am I going wrong? Please advise.

(2)Another question is:

"Find arc length of curve y=log sec x for x=0 to x=pi/3."

The answer in the book is 1.732. I proceeded thus-

dy/dx=1/sec x*sec x tanx= tan x; (dy/dx)^2=tan^2 x

Arc length=Int Sqrt(1+tan^2 x)dx=Int sec x dx from 0 to pi/3

=log(sec x+ tan x)from 0 to pi/3

=log(sec 60 +tan 60)- log(sec 0 + tan 0)

=log(2+1.732)- log 1=log 3.732 - 0 =1.317.

I am not getting 1.732. Am I wrong somewhere or is it a misprint in the book? Please advise

(1) The function y=b sin^2(pi.x/a) is periodic of period a. So you can see one cycle of the function from x=0 to x=a. (If you do not want to see the graph, imagine y = sin (pi.x/a), it runs one cycle from x=-a to x=a, so if you square this function, it's enough to go half the cycle.)

So for that reason, you you need to find the mean, you need to do Integral y from 0 to a and not Integral y from a to b. And your mean value should be Mean Value=Integral y/(a-0) from x=0 to x=b.

I see you "wrong answer contains sin(2pi b/a)], so you were probably on the right track if you integral sin^2 by using the power-reducing formula : sin^(A) = (1-cos^2(2A))/2.

If you follow this, you should get the right answer of b/a. The part with sin(2pix/a) should be zero when substituted with 0 and a. Good luck on this!

(2) Yes, the answer should be ln(2+sqrt(3)). So the answer in the book is incorrect.

Calculus

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