Answer
Using the left endpoint of each subinterval:
$R_{100} = 0.8946$
Using the right endpoint of each subinterval:
$R_{100} = 0.9081$
$0.8946 \lt \int_{0}^{2}\frac{x}{x+1}~dx\lt 0.9081$
Work Step by Step
We can compute the left Riemann sum for the function using the left endpoint of each subinterval.
$\Delta x = \frac{b-a}{n} = \frac{2-0}{100} = \frac{1}{50}$
$x_i = \frac{i-1}{50}$
$R_{100} = \sum_{i=1}^{100}f(x_i)\Delta x$
$R_{100} = \sum_{i=1}^{100}\frac{x_i}{x_i+1}\Delta x$
$R_{100} = \frac{0}{0+1}(\frac{1}{50})+\frac{1/50}{1/50+1}(\frac{1}{50})+...+\frac{99/50}{99/50+1}(\frac{1}{50})$
$R_{100} = 0.8946$
We can compute the right Riemann sum for the function using the right endpoint of each subinterval.
$\Delta x = \frac{b-a}{n} = \frac{2-0}{100} = \frac{1}{50}$
$x_i = \frac{i}{50}$
$R_{100} = \sum_{i=1}^{100}f(x_i)\Delta x$
$R_{100} = \sum_{i=1}^{100}\frac{x_i}{x_i+1}\Delta x$
$R_{100} = \frac{1/50}{1/50+1}(\frac{1}{50})+\frac{2/50}{2/50+1}(\frac{1}{50})+...+\frac{100/50}{100/50+1}(\frac{1}{50})$
$R_{100} = 0.9081$
Since the function is increasing, an estimate using the left endpoint of each subinterval is less than the exact value of the integral.
Since the function is increasing, an estimate using the right endpoint of each subinterval is greater than the exact value of the integral.
Therefore:
$0.8946 \lt \int_{0}^{2}\frac{x}{x+1}~dx\lt 0.9081$
