Answer
$$
\int_{1}^{3} \frac{\ln x}{x} d x
$$
Here $a=1, b=3,$ and $n=4,$ with $(b-a) / n=(3-1) / 4=1/2 $ as the altitude of each trapezoid. Then $x_{0}=1, x_{1}=1.5 , x_{2}=2 , x_{3}=2.5 ,$ and $x_{4}=3 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&\text { Trapezoidal Rule }\\
&n=4, b=3, a=1, f(x)=\frac{\ln x}{x}\\
&\begin{array}{l|l|l}
\hline i & x_{1} & f\left(x_{i}\right) \\
\hline 0 & 1 & 0 \\
1 & 1.5 & 0.27031 \\
2 & 2 & 0.34657 \\
3 & 2.5 & 0.36652 \\
4 & 3 & 0.3662 \\
\hline
\end{array}
\end{aligned}
$$
The trapezoidal rule:
$$
\begin{aligned}
\int_{1}^{3} \frac{\ln x}{x} d x \approx & \frac{3-1}{4}\left[\frac{1}{2}(0)+0.27031+0.34657\right.\\
&\left.+0.36652+\frac{1}{2}(0.3662)\right] \\
=& 0.5833
\end{aligned}
$$
Exact value:
$$
\begin{aligned}
\int_{1}^{3} \frac{\ln x}{x} d x &=\int_{1}^{3} \ln x d (\ln x) \\
&=\left.\frac{1}{2}(\ln x)^{2}\right|_{1} ^{3}\\
&=\frac{1}{2}(\ln 3)^{2}-\frac{1}{2}(\ln 1)^{2} \\
&\approx 0.6035
\end{aligned}
$$
Work Step by Step
$$
\int_{1}^{3} \frac{\ln x}{x} d x
$$
Here $a=1, b=3,$ and $n=4,$ with $(b-a) / n=(3-1) / 4=1/2 $ as the altitude of each trapezoid. Then $x_{0}=1, x_{1}=1.5 , x_{2}=2 , x_{3}=2.5 ,$ and $x_{4}=3 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&\text { Trapezoidal Rule }\\
&n=4, b=3, a=1, f(x)=\frac{\ln x}{x}\\
&\begin{array}{l|l|l}
\hline i & x_{1} & f\left(x_{i}\right) \\
\hline 0 & 1 & 0 \\
1 & 1.5 & 0.27031 \\
2 & 2 & 0.34657 \\
3 & 2.5 & 0.36652 \\
4 & 3 & 0.3662 \\
\hline
\end{array}
\end{aligned}
$$
The trapezoidal rule:
$$
\begin{aligned}
\int_{1}^{3} \frac{\ln x}{x} d x \approx & \frac{3-1}{4}\left[\frac{1}{2}(0)+0.27031+0.34657\right.\\
&\left.+0.36652+\frac{1}{2}(0.3662)\right] \\
=& 0.5833
\end{aligned}
$$
Exact value:
$$
\begin{aligned}
\int_{1}^{3} \frac{\ln x}{x} d x &=\int_{1}^{3} \ln x d (\ln x) \\
&=\left.\frac{1}{2}(\ln x)^{2}\right|_{1} ^{3}\\
&=\frac{1}{2}(\ln 3)^{2}-\frac{1}{2}(\ln 1)^{2} \\
&\approx 0.6035
\end{aligned}
$$
