Answer
$$M_{5}= 0.387,\ \ \ T_5= 0.384 $$
Work Step by Step
Given $$\int_{1}^{2} \ln x d x ,\ \ \ \ N=5$$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{5}=0.2$
Therefore, the sub intervals consist of
$$[1,1.2],[1.2,1.4],[1.4,1.6],[1.6,1.8],[1.8,2] $$
The midpoints of these sub intervals are
$$\{ 1.1, 1.3,1.5,1.7,1.9\}$$
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(1.1)+ f(1.3)+ f(1.5)+ f(1.7)+ f(1.9)\right]\\
&\approx 0.387
\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(1)+2f(1.2)+2f(1.4)+2f(1.6)+2f(1.8)+f(2)\right] \\
&\approx 0.384
\end{align*}
