Video Lecture

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We know that our goal is to ultimately get a “good approximation” of the area under the curve using the area of rectangles as a substitute for the difficult to calculate area under a generic curve. To see how this works, we start with an easy to understand shape before moving to more generic and difficult curves. Consider the following area:

There are two obvious aspects to consider when determining how to arrange our rectangles to approximate the desired area. How many rectangles should we use, and where exactly should we put them? Let’s start by tackling the later question first, so for now let’s assume we are going to use three rectangles (we’ll return to the question of how many rectangles we actually want to use a bit later).

Since we are starting with three rectangles, and the total width of our region is six, we can go ahead and make the rectangles all the same width - which makes them two units wide each. It’s worth a mention here that there is no reason we have to make them the same width, but it turns out to be mechanically convenient as we will see later.

Next, there are two obvious ways to arrange the rectangles. We could have the top right corner of each rectangle start on the curve and fill out the rest of the rectangle accordingly, like this:

But we could also make it so the top left corner of the rectangles start on the curve and fill out the rest of the rectangle, like this:

Note here that the left most “rectangle” really has a height of zero, so it doesn’t show up - thus we are only really using two rectangles as our approximation.

There is one more, less obvious way we could go about this though. We could instead have the middle of our rectangle start on the curve and then fill out the rectangle accordingly. Like this:

Believe it or not, these are the three main ways to approximate curves with rectangles - called the right endpoint, left endpoint, and midpoint approximations respectively. We will look at each of these and develop a general formula for each of these approximations.

Before we get to the specific formulas, let’s take a moment to record what is the same in each case. For each of these we want to write out the sum of the areas of the rectangles, which is simply the height of each rectangle times their width. So if we label our three rectangles $R_1$, $R_2$, and $R_3$, then the approximated area will be the area of $R_1$ plus the area of $R_2$ plus the area of $R_3$. The only difference between our three approximations, really, is the actual area of those rectangles. And, since we made their width uniform (the width of each rectangle is 2, regardless of method) the only real difference is the height of the rectangles in each method.

Right Endpoint Approximation:

For our first approximation (the right endpoint approximation), we mentioned that the top right corner of each rectangle would start on the curve. This means that we can use that point to get the height of the rectangle. Let’s take another look at our rectangular approximation, and in particular, the points where the rectangles meet the curve.

These points represent the height of the rectangle to the left of the point, but that height is just the $y$-value of the point, which we can get by determining the $y$-value of the graph! In particular, the height of the first dot is $f(2)$ because the dot is on the graph at the $x$-value $2$. Thus we can find the heights by evaluating the function at the right-hand $x$-value of the rectangle, as we see here:

So, our formula then, would be the width (which is 2) times the height of each rectangle, which we now know as the function evaluation of the right-hand $x$-value. This gives us:

(Approximate) Area	$($Area of $R_1) +($ Area of $R_2) +($ Area of $R_3)$
	$=($ width of $R_1) ($ height of $R_1)$ $+ ($ width of $R_2) ($ height of $R_2)$ $+ ($ width of $R_3) ($ height of $R_3)$
	$= (2) f(2) + (2) f(4) + (2) f(6)$
	$= (2)(2) + (2)(4) + (2)(6)$
	$= 24$

Left Endpoint Approximation:

We can see that this same process applies for the left and center endpoint approximations too, it’s just that the $x$ value that we use to find the height is different for each method. Let’s look at the filled out version of the left endpoint approximation picture:

Here we notice that the leftmost rectangle actually has height zero because the left endpoint for that rectangle is the origin. So we can write the same formula as we did in the last approximation, only now the height is taken by the left-most $x$-value of the rectangle, so the $x$ values where we are evaluating $f$ will be different for each rectangle.

(Approximate) Area	$($Area of $R_1) +($ Area of $R_2) +($ Area of $R_3)$
	$=($ width of $R_1) ($ height of $R_1)$ $+ ($ width of $R_2) ($ height of $R_2)$ $+ ($ width of $R_3) ($ height of $R_3)$
	$= (2) f(0) + (2) f(2) + (2) f(4)$
	$= (2)(0) + (2)(2) + (2)(4)$
	$= 12$

Midpoint Approximation:

Finally, the same pattern holds for the midpoint approximation, which now uses the middle $x$-value instead of either ends. So our finished picture would look like this:

Again we can write the formula out, still by changing only the $x$ value we use to get the height of the rectangle. Thus we have:

(Approximate) Area	$($Area of $R_1) +($ Area of $R_2) +($ Area of $R_3)$
	$=($ width of $R_1) ($ height of $R_1)$ $+ ($ width of $R_2) ($ height of $R_2)$ $+ ($ width of $R_3) ($ height of $R_3)$
	$= (2) f(1) + (2) f(3) + (2) f(5)$
	$= (2)(1) + (2)(3) + (2)(5)$
	$= 12$

There were a lot of pictures and numbers, but the important takeaway is that the only difference between the different endpoint approximations is what $x$-value you use to determine the height of the rectangles. Everything else about the formulas are the same. This is going to be important to remember as we tackle the other question we had at the beginning of this whole thing - what about the number of rectangles?

Trying to allow for an arbitrary number of rectangles (meaning, however many rectangles someone might want to use, without letting us know beforehand) makes writing out a general formula more technical and the resulting formula is more intimidating. Indeed, this is where we need to use our Sigma Notation (if you don’t remember or aren’t clear on sigma notation, you should revisit that material first and then come back to this!) But again, it’s useful to keep in mind that the only real difference between any of the endpoint approximation formulas we are about to get, is what $x$-values we use to find the height, so the formulas should look very similar, but hopefully we have primed the pump with our previous examples with 3 rectangles enough that you can spot the differences (and why they are different).

With our goal in mind then, let’s consider how our formula changes when we move to using $n$ rectangles. This effects two aspects of our situation. First, remember that the region we are estimating has a fixed width, so as we increase the number of rectangles, the width of each rectangle gets smaller. This is where it can be helpful to just assume that each rectangle is the same width, as we can then compute the width of each rectangle by dividing the full interval length by the number of rectangles - in this case “$n$”.

In general then, if our region is spread across the interval $[a,b]$ then the width of each of $n$ rectangles would be the total width of the interval; $b-a$, divided by the number of rectangles used to cover that width, $n$. So each rectangle has width $\frac {b-a}{n}$. By convention, this width is denoted by $\Delta x$. So, for example, if we wanted to write our right endpoint estimate with 3 rectangles, using this new notation, it would be written as follows:

The second impact that shifting to $n$ rectangles has, is how we need to represent the $x$ values that are evaluated to get the height of each rectangle.

With each of our methods, we initially used three rectangles, and as such we could simply look and see which $x$-values would be needed to represent the height of each rectangle. Now, we don’t know exactly how many rectangles are going to be used, or more accurately, we are leaving the number of rectangles as the arbitrary constant “$n$”, which means we inherently cannot calculate concrete numbers for the $x$-values for each of the height calculations. Nonetheless, we can still come up with a nice algorithm for the height. Our resulting formulas will look a bit intimidating, but it is helpful to keep the more concrete forms we started with in mind, to be able to pull apart our formulas into the different parts that represent our concrete geometric elements; heights and widths of specific rectangles.

So now we know the formula for the right endpoint approximation, but what about the left endpoint and midpoint approximations? Well the formulas are remarkably similar, to the point that it is easy to get them confused if you are trying to just memorize the symbols. But if we compare the left and right endpoint approximation pictures we might notice something interesting...

Notice that, the left endpoint approximations are essentially the same rectangles, just moved over one spot to the right. In other words, the first rectangle doesn’t move over any rectangle widths before we get the correct $x$-value to use for the height, and after the first rectangle we do the same process of moving to the right one rectangle width to get the next $x$-value for the height. This gives us the formula:

$$\sum \limits _{k=0}^{n-1} (\Delta x)f(a+ k(\Delta x)) =\sum \limits _{k=0}^{n-1} \left (\frac {b-a}{n}\right )f\left (a+ k\left (\frac {b-a}{n}\right )\right )$$

Notice that this formula looks identical, and it’s very easy to miss the difference. The difference between this and the right endpoint approximation is the start and end values of $k$. Instead of going from $1$ to $n$, we are going from $0$ to $n-1$. This is because the right endpoint approximation started out by needing to move $1$ rectangle width to the right to get it’s first $x$ value for the height, but the left endpoint approximation doesn’t need to move over (i.e. it needs to move over “0” widths) to get it’s first $x$-value for the height.

Similarly, if we recall how the midpoint approximation worked, it used the middle $x$ value, which we can either get by taking the right endpoint approximation and moving backward half a width, or the left endpoint and moving right half a rectangle width. By convention we typically use the form that starts with the left endpoint form and adds half a width to each $x$ value. This gives us the formula:

$$\sum \limits _{k=0}^{n-1} (\Delta x)f\left (a+ \left (k+\frac {1}{2}\right )(\Delta x)\right ) =\sum \limits _{k=0}^{n-1} \left (\frac {b-a}{n}\right )f\left (a + \left (k+\frac {1}{2}\right )\left (\frac {b-a}{n}\right )\right )$$

So, in this segment we have discussed how to estimate the area under a generic curve using rectangles. We started with a concrete number of rectangles and concrete curve, but using that we developed our three estimation techniques, the right, left, and mid-point formulas. In particular, to estimate the area of a region bounded by a function $f(x)$ and the $x$-axis, on the interval $[a,b]$ using $n$ rectangles, we have the following formulas:

Again, this can be a little intimidating, but remember that the “$\Delta x$” is just the width of the rectangles, and multiplying by $k$ is merely the process of moving to the “next rectangle’s $x$-value for height”.