`int_0^2x/(x + 1)dx, n = 5` Use the Midpoint Rule with the given value of `n` to approximate the integral. Round the answer to four decimal places.

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Chapter 5, 5.2 - Problem 11 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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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.`

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