Answer
$$ 0.74805$$
Work Step by Step
Given $$\int_{0}^{1} e^{-x^{2}} d x $$
We have:
$\Delta x=\dfrac{b-a}{n}=\dfrac{1}{5}=0.2$
Therefore, 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\\
&=(0.2)\left[ f(0.1)+ f(0.3)+ f(0.5)+ f(0.7)+ f(0.9)\right]\\
&\approx 0.74805
\end{align*}
