Answer
$$
\int_{0}^{1} \sqrt{x^{3}+1} d x, \quad n=5
$$
The Midpoint Rule gives
$$
\begin{aligned}
\int_{0}^{1} \sqrt{x^{3}+1} d x \approx \sum_{i=1}^{4} f\left(\bar{x}_{i}\right) \Delta x \approx 1.10967
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{1} \sqrt{x^{3}+1} d x, \quad n=5
$$
The width of the subintervals is
$$
\Delta x=(1-0) / 5=\frac{1}{5},
$$
so the endpoints are 0, 0.2, 0.4, 0.6, 0.8 and 1, and the midpoints are 0.1, 0.3, 0.5, 0.7 and 0.9.
The Midpoint Rule gives
$$
\begin{aligned}
\int_{0}^{1} \sqrt{x^{3}+1} d x \approx \sum_{i=1}^{5} f\left(\bar{x}_{i}\right) \Delta x =\frac{1}{5}(\sqrt{(0.1)^{3}+1}+\sqrt{(0.3)^{3}+1}+\sqrt{(0.5)^{3}+1}+\sqrt{(0.7)^{3}+1}+\sqrt{(0.9)^{3}+1}) \\
\approx 1.10967
\end{aligned}
$$
