Answer: (i) \(\frac{12}{5\ln 5}\)

(ii) \(\frac{12}{5}\)

(iii) \(\pi\left(\frac{156}{25}+\ln 5\right)\)

(i) The mean value of \(y\) on \(0 \leqslant x \leqslant \ln 5\) is

\(\displaystyle \frac{1}{\ln 5-0}\int_{0}^{\ln 5} \frac{1}{2}\left(e^{x}+e^{-x}\right)\,dx\).

Now

\(\displaystyle \int \frac{1}{2}\left(e^{x}+e^{-x}\right)\,dx=\frac{1}{2}\left(e^{x}-e^{-x}\right)\),

so

\(\displaystyle \text{mean value}=\frac{1}{\ln 5}\left[\frac{1}{2}\left(e^{x}-e^{-x}\right)\right]_{0}^{\ln 5}\).

Since \(e^{\ln 5}=5\) and \(e^{-\ln 5}=\frac{1}{5}\), this gives

\(\displaystyle \text{mean value}=\frac{1}{\ln 5}\cdot \frac{1}{2}\left(5-\frac{1}{5}\right)=\frac{12}{5\ln 5}.\)

(ii) For arc length,

\(\displaystyle \frac{dy}{dx}=\frac{1}{2}\left(e^{x}-e^{-x}\right).\)

Then

\(\displaystyle 1+\left(\frac{dy}{dx}\right)^2 = 1+\frac{1}{4}\left(e^{x}-e^{-x}\right)^2 = \frac{1}{4}\left(e^{x}+e^{-x}\right)^2,\)

because \(\left(e^{x}-e^{-x}\right)^2=e^{2x}-2+e^{-2x}\) and \(\left(e^{x}+e^{-x}\right)^2=e^{2x}+2+e^{-2x}\).

Hence

\(\displaystyle \sqrt{1+\left(\frac{dy}{dx}\right)^2}=\frac{1}{2}\left(e^{x}+e^{-x}\right).\)

So the arc length is

\(\displaystyle s=\int_{0}^{\ln 5}\frac{1}{2}\left(e^{x}+e^{-x}\right)\,dx = \frac{1}{2}\left[e^{x}-e^{-x}\right]_{0}^{\ln 5}.\)

Therefore

\(\displaystyle s=\frac{1}{2}\left(5-\frac{1}{5}\right)=\frac{12}{5}.\)

(iii) The surface area when the curve is rotated about the \(x\)-axis is

\(\displaystyle S=2\pi\int_{0}^{\ln 5} y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.\)

Using \(y=\frac{1}{2}(e^x+e^{-x})\) and \(\sqrt{1+(y')^2}=\frac{1}{2}(e^x+e^{-x})\),

\(\displaystyle S=2\pi\int_{0}^{\ln 5}\frac{1}{4}\left(e^x+e^{-x}\right)^2\,dx=\frac{\pi}{2}\int_{0}^{\ln 5}\left(e^{2x}+2+e^{-2x}\right)\,dx.\)

Integrating,

\(\displaystyle S=\frac{\pi}{2}\left[\frac{e^{2x}}{2}+2x-\frac{e^{-2x}}{2}\right]_{0}^{\ln 5}.\)

Now \(e^{2\ln 5}=25\) and \(e^{-2\ln 5}=\frac{1}{25}\), so

\(\displaystyle S=\frac{\pi}{2}\left(\frac{25}{2}+2\ln 5-\frac{1}{50}\right).\)

Hence

\(\displaystyle S=\pi\left(\frac{156}{25}+\ln 5\right).\)