Doug C. answered • 09/21/16

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The following graph shows how to divide the target area into 10 rectangles.

https://www.desmos.com/calculator/0ayh0k33is

If the interval [1,10] is divided into n rectangles all with the same width, then the width of each rectangle will be (10-1)/n or 9/n.

Using right hand end-points along the x-axis to get the length of each rectangle results in something like this where x

_{i }is the x-coordinate of that end-point.x

_{1}= 1 + 9/n (one width from the starting point)x

_{2}= 1 + 2(9/n) (two widths from the starting point)x

_{i }= 1 + i(9/n) (i widths from the starting point)So a formula for the area of the i

^{th }rectangle is (9/n)[(1/10)(1+9i/n)^{2 }+ (1+9i/n)] (length times width)So the total area is the ∑ from i = 1 to n of that expression.

That expression can be simplified to a point where you are using formulas like ∑ from i = 1 to n of i = (n)(n+1)/2. There is a similar formula for ∑ i

^{2}= [n(n+1)(2n+1)/6]^{}The simplification actually gets really messy. The goal is to get a formula for the total area that involves only n, and then find the limit of that formula as n -> ∞.

You end up with something like this:

limit as n-> ∞ [ 729/60 ( 2 + 3/n + 1/n

^{2}) + 972/20(1+1/n) + 99/10]Yikes. Picture those fractions with n in the denominator as n gets large getting closer and closer to 0.

There is not enough room here to show the work involved to getting to the simplified form.