Answer
$$S_{4} \approx 5.2522$$
Work Step by Step
Given$$\int_{0}^{4} \sqrt{x} d x, \quad N=4 $$
Since $\Delta x=\dfrac{b-a}{N}=1$ , 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}{3}\left[f(0)+4f(1)+2f(2)+4f(3)+ f(4) \right]\\
&\approx 5.2522
\end{align*}