Answer
$$ 0.7468$$
Work Step by Step
Given $$\int_{0}^{1} e^{-x^{2}} d x $$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{4}=0.25$ , then by using Simpson’s rule, we get
\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(0.25)+2f(0.5)+4f(0.75)+ f(1) \right]\\
&\approx 0.7468
\end{align*}
