Answer
$$T_{5}=0.744,\ \ M_5=0.74$$
Work Step by Step
Given $$\int_{0}^{1} e^{-x^2} d x ,\ \ \ \ N=5$$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{5}=0.2$
Therefore, the sub intervals consist of
$$[0,0.2],[0.2,0.4],[0.4,0.6],[0.6,0.8],[0.8,1] $$
The midpoints of these sub intervals are
$$\{ 0.1, 0.3,0.5,0.7,0.9\}$$
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_5)\right]\Delta x\\
&=\frac{1}{5}\left[ f(0.1)+ f(0.3)+ f(0.5)+ f(0.7)+ f(0.9)\right]\\
&\approx 0.74
\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(0)+2f(0.2)+2f(0.4)+2f(0.6)+2f(0.8)+f(1)\right] \\
&\approx 0.744
\end{align*}
