Answer: (a) The mean value of a function on an interval \([a,b]\) is \(\frac{1}{b-a}\int_a^b f(x)\,dx\).
So for \(y=\sec x\) on \(\frac{\pi}{6} \le x \le \frac{\pi}{3}\),
\(\text{mean value} = \frac{1}{\frac{\pi}{3}-\frac{\pi}{6}}\int_{\pi/6}^{\pi/3} \sec x\,dx = \frac{6}{\pi}\ln\!\left(\frac{2+\sqrt{3}}{\sqrt{3}}\right).\)
(b) The arc length of \(y=-\ln(\cos x)\) from \(x=0\) to \(x=\frac{\pi}{3}\) is
\(\displaystyle \int_0^{\pi/3} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \ln(2+\sqrt{3}).\)
(a) The mean value of \(y\) with respect to \(x\) over \(\frac{\pi}{6} \le x \le \frac{\pi}{3}\) is
\(\text{MV}=\frac{1}{\frac{\pi}{3}-\frac{\pi}{6}}\int_{\pi/6}^{\pi/3}\sec x\,dx.\)
Using \(\int \sec x\,dx=\ln(\sec x+\tan x)+C\),
\(\text{MV}=\frac{1}{\pi/6}\Big[\ln(\sec x+\tan x)\Big]_{\pi/6}^{\pi/3}.\)
Now
\(\sec\frac{\pi}{3}+\tan\frac{\pi}{3}=2+\sqrt{3},\)
and
\(\sec\frac{\pi}{6}+\tan\frac{\pi}{6}=\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}=\sqrt{3}.\)
So
\(\text{MV}=\frac{6}{\pi}\left(\ln(2+\sqrt{3})-\ln(\sqrt{3})\right)=\frac{6}{\pi}\ln\!\left(\frac{2+\sqrt{3}}{\sqrt{3}}\right).\)
(b) For \(y=-\ln(\cos x)\),
\(\frac{dy}{dx}= -\frac{-\sin x}{\cos x}=\tan x.\)
The arc length from \(x=0\) to \(x=\frac{\pi}{3}\) is
\(s=\int_0^{\pi/3}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=\int_0^{\pi/3}\sqrt{1+\tan^2 x}\,dx.\)
Since \(1+\tan^2 x=\sec^2 x\) and \(x\in[0,\pi/3]\), \(\sqrt{\sec^2 x}=\sec x\). Hence
\(s=\int_0^{\pi/3}\sec x\,dx=\Big[\ln(\sec x+\tan x)\Big]_0^{\pi/3}.\)
At \(x=\frac{\pi}{3}\), \(\sec x+\tan x=2+\sqrt{3}\); at \(x=0\), \(\sec 0+\tan 0=1\). Therefore
\(s=\ln(2+\sqrt{3})-\ln 1=\ln(2+\sqrt{3}).\)
