Answer
$$
y= x \ln x, \quad \quad 1 \leq x \leq 2
$$
The area of the surface obtained by rotating the curve about the x-axis
$$
\begin{aligned}
S&=\int 2 \pi y d s \\
&=\int_{0}^{1} 2 \pi ( x \ln x )(\sqrt{1+\left(1+\ln x \right)^{2}} dx \\
& \approx 7.248933 \quad\quad\left[\text {by using Simpson’s Rule } \right] \\
&\approx 7.248934 \quad\quad\left[\text {by using a calculator } \right] \\
\end{aligned}
$$
Work Step by Step
$$
y= x \ln x, \quad \quad 1 \leq x \leq 2
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}= 1+\ln x
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\left(1+\ln x \right)^{2}} dx\\
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the x-axis
$$
\begin{aligned}
S&=\int 2 \pi y d s \\
&=\int_{0}^{1} 2 \pi ( x \ln x )(\sqrt{1+\left(1+\ln x \right)^{2}} dx \\
\end{aligned}
$$
Simpson’s Rule to approximate the given integral with the
specified value of $ n=10$ , with $ n =10 , a = 1$, and $b = 2 $ we have
$$
\Delta x=\frac{b-a}{n}=\frac{2-1}{10}=\frac{1}{10}
$$
Let
$$
f(x)=2 \pi ( x \ln x )(\sqrt{1+\left(1+\ln x \right)^{2}}
$$
then
$$
\begin{aligned} S &= \int_{1}^{2} 2 \pi ( x \ln x )(\sqrt{1+\left(1+\ln x \right)^{2}} dx\\
& = \frac{\Delta x}{3}\left[f\left(x_{0}\right)\right. +4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)+\cdots \\
& \quad \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \\
&\approx S_{10}\\
& \approx \frac{1 / 10}{3}[f(1)+4 f(1.1)+2 f(1.2)+4 f(1.3)+\\
& \quad\quad+2 f(1.4)+4 f(1.5)+2 f(1.6)+4 f(1.7)+\\
& \quad\quad +2 f(1.8)+4 f(1.9)+f(2)] \\
&\approx 7.248933
\end{aligned}
$$
But the surface area by using the a calculator is approximately $ 7.248934 $
