Answer
$$T_6=22.216,\ \ \ \ M_6= 15.89 $$
Work Step by Step
Given $$\int_{-2}^{1} e^{x^2} d x ,\ \ \ \ N=6$$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{3}{6}=0.5$
Therefore, the sub intervals consist of
$$[-2,-1.5],[-1.5,-1],[-1,-0.5],[-0.5,0],[0,0.5],[0.5,1], $$
The midpoints of these sub intervals are
$$\{ -1.75, -1.25,-0.75,-0.25,0.25,0.75\}$$
Hence
\begin{align*}
M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\
M_{6}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_6)\right]\Delta x\\
&=\frac{1}{5}\left[ f(-1.75)+ f(-1.25)+ f(-0.75)+ f(-0.25)+ f(0.25)+f(0.75)\right]\\
&\approx
\end{align*}
Now find $T_6$
\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_{6}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+2f(x_4)+2f(x_5)+f(x_6)\right]\Delta x\\
&=\dfrac{1}{4}\left[f(-2)+2f(-1.5)+2f(-1)+2f(-0.5)+2f(0)+2f(0.5)+f(1)\right] \\
&\approx 22.216
\end{align*}
