# Taking the limit of the sum of approximating rectangles

I have been trying to tackle this question, and although I think I've found the answer, it seems either too obvious or I'm understanding the question incorrectly.

It goes as follows:

"Find the area of the region that lies under the graph f(x) = x between [0,2] by taking the limit of the sum of approximating rectangles whose heights are the values of the function at the right end point of each interval.

Included is a graph with 6 rectangles.

Now, if you just look at the darn thing, it's easy to see that the area under the graph would be 2 (because it's a right angle triangle). However, using the method asked (if I understood this correctly), I took the sum of the areas of each rectangle (so, each rectangle would be width=1/3, and height=f(x)) to which I got 7/3. Then it asked to take the limit of that number, which is a constant, so it would just be 7/3.

So am I missing something extremely obvious here?

• It says "limit of approximating rectangles", not just the ones shown, so you need to find a general formula for the case where there are $n$ equal-width rectangles and make $n \to \infty$ so that they become finer and the discrepancy in area tends to $0$. – user21820 Mar 21 '15 at 3:39
• To be fair, the quoted sentence talks about the limit of a sum but fails to say it is the limit as what goes to what. We how understand it, understand it only because we already know the purpose of this familiar exercise. – David K Mar 21 '15 at 4:19

The wording of this problem seems a little misleading. But what it seems like is that they want you derive the area under the curve by taking the limit of a summation that you can fit to any number of rectangles you want.

So for instance, let's say that we use n rectangles to approximate the area under the curve. *The width of all rectangles would be 2/n

Now we need to determine a closed form way to determine the right endpoints. Since we're using the right endpoint of the rectangle, we're starting at 0, and the width of the rectangle is 2/n, the first right end point we'll use is (2/n)*1, the next will be (2/n)*2 (2/n+ 2/n), the next will be (2/n)*3,... and so on until we have the nth rectangles right endpoint at (2/n)*n.

So to get the height of the rectangles we'll plug our set of right endpoints {(2/n)*1, (2/n)*2...,(2/n)*n)} into the function f(x)=x. Which gives us a set of heights {(2/n)*1, (2/n)*2...,(2/n)*n)} for the n rectangles.

Multiplying each height by the width (2/n) and summing we get the approximated area of the rectangles. Doing this we get (2/n)((2/n)*1 +(2/n)*2 +...+(2/n)*n))= (4/n^2)(1+2+...+n) =(4/n^2)(n(n+1)/2) =2+(2/n)

So for whatever number of n rectangles we use, this expression 2+(2/n) will gives us the approximated area under the curve using these n rectangles.

Now if we want to find the exact area under the curve we use infinitely many rectangles so that the rectangles fit every nook and crany under the function. In other words we use the limit as n goes to infinity.

So we get lim as n goes to infinity of 2+(2/n) is 2.