Answer
$\int_{0}^{2}\frac{x}{x+1}~dx \approx 0.9071$
Work Step by Step
$\Delta x = \frac{b-a}{n} = \frac{2-0}{5} = 0.4$
We can find the midpoints of the five subintervals:
$x_1 = 0.2$
$x_2 = 0.6$
$x_3 = 1.0$
$x_4 = 1.4$
$x_5 = 1.8$
$\int_{0}^{2}\frac{x}{x+1}~dx \approx \sum_{i=1}^{5} f(x_i)~\Delta x$
$\int_{0}^{2}\frac{x}{x+1}~dx \approx 0.4\cdot (\frac{0.2}{0.2+1}+\frac{0.6}{0.6+1}+\frac{1.0}{1.0+1}+\frac{1.4}{1.4+1}+\frac{1.8}{1.8+1})$
$\int_{0}^{2}\frac{x}{x+1}~dx \approx 0.4\cdot (2.26786)$
$\int_{0}^{2}\frac{x}{x+1}~dx \approx 0.9071$
