Answer
$$
\int_{0}^{2} \frac{x}{x+1} d x, \quad n=5
$$
The Midpoint Rule gives
$$
\begin{aligned}
\int_{0}^{2} \frac{x}{x+1} d x \approx \sum_{i=1}^{4} f\left(\bar{x}_{i}\right) \Delta x \approx 0.9071
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{2} \frac{x}{x+1} d x, \quad n=5
$$
The width of the subintervals is
$$
\Delta x=(2-0) / 5=\frac{2}{5},
$$
so the endpoints are $0 ,\frac{2}{5} ,\frac{4}{5} ,\frac{6}{5}, \frac{8}{5},$ and $2 $ and the midpoints are$ \frac{1}{5 } ,\frac{3}{5 },\frac{5}{5}, \frac{7}{5}$ and $\frac{9}{5}$
The Midpoint Rule gives
$$
\begin{aligned}
\int_{0}^{2} \frac{x}{x+1} d x \approx \sum_{i=1}^{5} f\left(\bar{x}_{i}\right) \Delta x \\
=\frac{2}{5}\left(\frac{\frac{1}{3}}{\frac{1}{5}+1}+\frac{\frac{3}{5}}{\frac{3}{5}+1}+\frac{\frac{5}{5}}{\frac{5}{4}+1}+\frac{\frac{7}{5}}{\frac{7}{5}+1}+\frac{\frac{9}{8}}{\frac{8}{5}}\right) \\
=\frac{2}{5}\left(\frac{127}{56}\right)=\frac{127}{140} \approx 0.9071
\end{aligned}
$$
