Answer
$$M_5= -0.691,\ \ T_5=-0.695$$
Work Step by Step
Given $$\int_{-2}^{-1}\frac{1}{x}d x,\ \ \ N=5 $$
Since $\Delta x=\dfrac{b-a}{N}=\dfrac{3}{6}=0.2 $
Therefore, the sub intervals consist of
$$[-2,-1.8],[-1.8,-1.6],[-1.6,-1.4],[-1.4,-1.2],[-1.2,-1]$$
The midpoints of these sub intervals are
$$\{-1.9, -1.7, -1.5, -1.3,-1.1\}$$
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_5)\right]\Delta x\\
&=0.2\left[ f(-1.9)+ f(-1.7)+ f(-1.5)+ f(-1.3)+ f(-1.1)\right]\\
&\approx -0.691
\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(-1.8)+2f(-1.6)+2f(-1.4)+2f(-1.2)+f(-1)\right] \\
&\approx -0.695
\end{align*}
