Answer
$$1.59$$
Work Step by Step
Given $$\int_{0}^{2}\left(x^{2}+1\right)^{-1 / 3} d x, \quad N=10$$
Since $\Delta x=\dfrac{b-a}{N}=\dfrac{ 2}{10}=0.2$, 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_{10}&=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5) +2f(x_6)+4f(x_7) +2f(x_8)+4f(x_9) +f(x_{10}) \right] \\
&=\dfrac{1}{15}\left[f(0)+4f(0.2)+2f(0.4)+4f(0.6)+2f(0.8) +4f(1)+2f(1.2)+4f(1.4)+2f(1.6) +4f(1.8)+f(2) \right]\\
&\approx 1.59
\end{align*}
