The following graph shows how to divide the target area into 10 rectangles.
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 xi is the x-coordinate of that end-point.
x1= 1 + 9/n (one width from the starting point)
x2= 1 + 2(9/n) (two widths from the starting point)
xi = 1 + i(9/n) (i widths from the starting point)
So a formula for the area of the ith 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 ∑ i2 = [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/n2) + 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.