Answer: (a) The required surface area is \(\pi\left(\mathrm{e}^{2}-\frac{1}{16}\mathrm{e}^{-2}+\frac{1}{16}\right)\).

(b) The Maclaurin series up to and including the term in \(x^2\) is \(\frac{5}{4}+\frac{3}{4}x+\frac{5}{8}x^2\).

(a) For rotation about the \(x\)-axis, the surface area is

\(S=2\pi\int_0^1 y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\).

Here \(y=\mathrm{e}^x+\frac14\mathrm{e}^{-x}\), so

\(\frac{dy}{dx}=\mathrm{e}^x-\frac14\mathrm{e}^{-x}\).

Then

\(1+\left(\frac{dy}{dx}\right)^2=1+\left(\mathrm{e}^x-\frac14\mathrm{e}^{-x}\right)^2\)

\(=1+\mathrm{e}^{2x}-\frac12+\frac1{16}\mathrm{e}^{-2x}\)

\(=\mathrm{e}^{2x}+\frac12+\frac1{16}\mathrm{e}^{-2x}\).

Also,

\(\left(\mathrm{e}^x+\frac14\mathrm{e}^{-x}\right)^2=\mathrm{e}^{2x}+\frac12+\frac1{16}\mathrm{e}^{-2x}\).

Since \(\mathrm{e}^x+\frac14\mathrm{e}^{-x}\gt 0\), it follows that

\(\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\mathrm{e}^x+\frac14\mathrm{e}^{-x}\).

So

\(S=2\pi\int_0^1 \left(\mathrm{e}^x+\frac14\mathrm{e}^{-x}\right)^2\,dx\)

\(=2\pi\int_0^1 \left(\mathrm{e}^{2x}+\frac12+\frac1{16}\mathrm{e}^{-2x}\right)\,dx\).

Integrating,

\(S=2\pi\left[\frac12\mathrm{e}^{2x}+\frac12x-\frac1{32}\mathrm{e}^{-2x}\right]_0^1\).

Now substitute the limits:

\(S=2\pi\left(\frac12\mathrm{e}^2+\frac12-\frac1{32}\mathrm{e}^{-2}-\left(\frac12-\frac1{32}\right)\right)\)

\(=2\pi\left(\frac12\mathrm{e}^2-\frac1{32}\mathrm{e}^{-2}+\frac1{32}\right)\)

\(=\pi\left(\mathrm{e}^{2}-\frac1{16}\mathrm{e}^{-2}+\frac1{16}\right)\).

(b) Using the standard expansions,

\(\mathrm{e}^x=1+x+\frac{x^2}{2}+\cdots\)

and

\(\mathrm{e}^{-x}=1-x+\frac{x^2}{2}+\cdots\).

Hence

\(\mathrm{e}^x+\frac14\mathrm{e}^{-x}=\left(1+x+\frac{x^2}{2}+\cdots\right)+\frac14\left(1-x+\frac{x^2}{2}+\cdots\right)\).

Collecting terms gives

constant term: \(1+\frac14=\frac54\),

coefficient of \(x\): \(1-\frac14=\frac34\),

coefficient of \(x^2\): \(\frac12+\frac18=\frac58\).

Therefore the Maclaurin series up to and including \(x^2\) is

\(\frac54+\frac34x+\frac58x^2\).