Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 11

You need to use the midpoint rule to approximate the interval. First, you need to find Delta x , such that:
Delta x = (b-a)/n
The problem provides b=2, a=0 and n = 5, such that:
Delta x = (2-0)/5 =2/5
Hence, the following 5 intervals of length 2/5 are: [0,2/5], [2/5,4/5], [4/5,6/5], [6/5,8/5],[8/5,2].
Now, you may evaluate the integral such that:
int_0^2 x/(x+1) dx = Delta x(f((0+2/5)/2) + f((2/5+4/5)/2) + f((4/5+6/5)/2) + f((6/5+8/5)/2) + f((8/5+2)/2) )
int_0^2 x/(x+1) dx = 2/5(f(2/10) + f(6/10) + f(1) + f(14/10) + f(18/10))
int_0^2 x/(x+1) dx = 2/5(2/(10(2/10+1)) + 6/(10(6/10+1)) + 1/2 + 14/(10(14/10+1) + 18/(10(18/10+1) )
int_0^2 x/(x+1) dx = 2/5(2/12 + 6/16 + 1/2 + 14/24 + 18/28)
int_0^2 x/(x+1) dx = 2/5(1/6 + 3/8 + 1/2 + 7/12 + 9/14)
int_0^2 x/(x+1) dx = 2/5(1/6 + 3/8 + 1/2 + 7/12 + 9/14)
int_0^2 x/(x+1) dx = 2/5(0.1666 + 0.3750 + 0.5 + 0.5833 + 0.6428)
int_0^2 x/(x+1) dx = 0.9070
Hence, approximating the definite integral, using the midpoint rule, yields int_0^2 x/(x+1) dx = 0.9070.