
William W. answered 02/20/20
Experienced Tutor and Retired Engineer
For the right endpoints the picture looks like this:
The rectangles divide the length "3" into 4 equal pieces so each rectangle is 3/4 units wide (the x values at each rectangle break occur at x = 0, x = 0.75, x = 1.5, x = 2.25, and x = 3)
The first rectangle has a height of f(0.75) or 1/(0.75 + 1) = 1/1.75 = 4/7
The second rectangle has a height of f(1.75) or 1/(1.5 + 1) = 1/2.5 = 2/5
The third rectangle has a height of f(2.25) or 1/(2.25 + 1) = 1/3.25 = 4/13
And the fourth rectangle has a height of f(3) or 1/(3 + 1) = 1/4
So the area Rn = 0.75(4/7 + 2/5 + 4/13 + 1/4) ≈ 1.14684
Notice that this is an underestimate since we are missing area that the rectangles are not covering.
The same holds true for Ln expect now the first rectangle has a height of f(0), the second has a height of f(0.75) etc. The result is: 0.75(1 + 4/7 + 2/5 + 4/13) ≈ 1,70934
This would be an overestimate since the rectangles will be above the function graph line. The true are is somewhere in the middle.