Answer
$$M_{5}=1.117,\ \ \ T_{5}=1.120$$
Work Step by Step
Given $$\int_{2}^{3} \frac{1}{\ln x} d x ,\ \ \ \ N=5$$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{5}=0.2$
Therefore, the sub intervals consist of
$$[2,2.2],[2.2,2.4],[2.4,2.6],[2.6,2.8],[2.8,3] $$
The midpoints of these sub intervals are
$$\{ 2.1, 2.3,2.5,2.7,2.9\}$$
Hence
\begin{align*}
M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\
M_{5}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_4)\right]\Delta x\\
&=\frac{1}{2}\left[ f(2.1)+ f(2.3)+ f(2.5)+ f(2.7)+ f(2.9)\right]\\
&\approx 1.117
\end{align*}
Now find $T_5$
\begin{align*}
T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+.....+2f(x_{n-1})+f(x_n)\right]\Delta x\\
T_{5}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+2f(x_4)+f(x_5)\right]\Delta x\\
&=\dfrac{1}{10}\left[f(2)+2f(2.2)+2f(2.4)+2f(2.6)+2f(2.8)+f(3)\right] \\
&\approx 1.120
\end{align*}
