Answer
$$0.746855$$
Work Step by Step
Given$$ \int_{0}^{1} e^{-x^{2}} d x, \quad N=4$$
Since $\Delta x=\dfrac{b-a}{N}=\dfrac{1}{4}$ , then by using Simpson’s rule
\begin{align*}
S_{n}&=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3).....+4f(x_{n-1})+f(x_n)\right]\\
S_{4}&=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+ f(x_{4}) \right] \\
&=\dfrac{1}{12}\left[f(0)+4f(1/4)+2f(2/4)+4f(3/4)+ f(1) \right]\\
&\approx 0.746855
\end{align*}
