A.6 Upper & Lower (Darboux) Sums

On this screen we'll explore Upper and Lower Sums, also known as Darboux Sums. These are fundamentally important since they give us a new way to bound the true value within these two sum values we can calculate.

You might have noticed that each of the Contexts we've used in this section has been either strictly decreasing or strictly increasing:

In Context 1, the water drains out of the container at a rate that decreases linearly with time.
In Context 2, the pressure inside the water tower increases linearly with depth.
In Context 3, as Gabriel falls his velocity only increases during the 10 seconds of interest.

We could thus easily develop our overestimates and underestimates by simply choosing whether to use the Left Sum or Right Sum in each circumstance. That's because the function's minimum and maximum value always occurred at either the start or the end of each subinterval.

Graph of a bell curve with maximum at the point (0,1) On this screen we're going to see how to handle situations in which the max or min does not necessary occur at either endpoint of every subinterval. To do so we're going to use as our example the perhaps-familiar bell curve function $𝐁 𝐞 𝐥 𝐥 𝐜 𝐮 𝐫 𝐯 𝐞 𝐟 𝐮 𝐧 𝐜 𝐭 𝐢 𝐨 𝐧 : 𝑦 (𝑥) = 𝑒 - 𝑥 2 / 6$ shown to the right.

Quick note on abstraction

In each of our previous three Contexts, we presented a physical scenario with real, tangible quantities, to make it easy (we hope) to envsision how much water, or force, or distance we "accumulated" as we summed over our subintervals. By contrast, the bell-curve function we're using on this screen has no such associated physical scenario. We're thus now taking another abstracting-step toward more general use of our new tools.

That is, we're generalizing. Imagine that the quantity y tells you how much the output-variable "something" changes based on a change in input-variable x. Our goal remains to estimate how much of that "something" accumulates over each subinterval interval change $Δ 𝑥 .$ The ideas are all the same; we now just don't have physical quantities to attach to the output and input variables.

If this shift toward abstraction is uncomfortable for you, that's ok. Embrace the challenge! Delving further into layers of abstraction is often very difficult, and yet crucial work to develop deeper understanding.

We are going to to approximate the total accumulation for our bell-curve function over the interval $[- 2, 4] .$ To do so, we will use our same basic approach that you now know so well, using $𝑛 = 4$ subintervals:

1. Approximate the accumulation during each subinterval:

A p p r o x a c c u m u l a t i o n d u r i n g s u b i n t e r v a l 𝑖 : (r e p r e s e n t a t i v e f u n c t i o n v a l u e f o r s u b i n t e r v a l) \cdot (s u b i n t e r v a l d u r a t i o n)

2. Sum accumulations for all of the subintervals.

Since $𝑛 = 4$ subintervals, the subinterval width is $Δ 𝑥 = 𝑏 - 𝑎 𝑛 = 4 - (- 2) 4 = 64 = 32 = 1.5$ The graph below shows the function again, now with the four subintervals marked: $[- 2, - 0.5],$ $[- 0.5, 1],$ $[1, 2.5],$ $[2.5, 4] .$ Same graph, now with dashed lines showing the subintervals [-2, -0.5], [-0.5, 1], [1, 2.5], [2.5, 4]

Developing an overestimate

Let's first develop the overestimate for the total accumulated amount.

Choosing an approach

As Question 1 illustrates, we need to develop a new way to choose $𝑦 𝑖$ so that we are guaranteed that we develop an overestimate for each subinterval. What ideas come to mind for you? (Take a moment....)

The key realization here is that, to develop our guaranteed overestimate, we need to find a way to determine the function's maximum value on each subinterval. Let's introduce a bit of new notation to capture this idea:

We need to choose $𝑥 𝑖$ in each subinterval such that $𝑦 (𝑥 𝑖) = 𝑀 𝑖$ ,
where $𝑀 𝑖$ is the maximum value of y in the i-th subinterval.

For instance, looking at the graph above you can see that for subinterval $𝑖 = 2,$ which spans $[- 0.5, 1],$ the maximum occurs at $𝑥 2 = 0$ where the function's value is $𝑦 (0) = 𝑀 2 = 1 .$ You can imagine similarly determining the maximum value for each of the other three subintervals.

The Upper Sum, $𝑆 u p p e r,$ uses the function's maximum value on each subinterval, wherever it occurs on that subinterval.

With this choice, we are guaranteed an overestimate, because by definition $𝑀 𝑖$ is greater than (or equal to) any other output value in the subinterval. We call the resulting sum of all of the subintervals' overestimates the Upper Sum, and we use $𝑆 u p p e r$ to represent it:

𝑆 u p p e r = 𝑀 1 \cdot Δ 𝑥 + 𝑀 2 \cdot Δ 𝑥 + 𝑀 3 \cdot Δ 𝑥 + 𝑀 4 \cdot Δ 𝑥

where

Δ 𝑥 = 32 .

Let's again emphasize that in our earlier Contexts 1-3, either the Left Sum or Right Sum was automatically the Upper Sum because the maximum occurs at either the leftmost or rightmost on each subinterval since those were all strictly increasing or strictly decreasing functions. It's only as we broaden to be able to work with other functions, including our bell curve here, that we need to extend how we craft the Upper Sum more generally.

With that said, what comes to mind for you as an approach or approaches for how to determine the maximum output value on each and every subinterval, no matter how many of those there are? (Again, please take a moment....)

We'll look at two possibilities; there are certainly others, which you can explore in depth (among other topics!) in a "Numerical Methods" math course if you'd like. Since this is not a numerical methods course, we're not going to dive into any of these in-depth; instead, we want to make sure you understand the basic ideas behind the approach to finding an Upper Sum more than actually learning how to implement them. With that said, we could:

Use our Calculus tools to find the global max (and min) on each subinterval. For our given function $𝑦 (𝑥),$ we can certainly take our usual approach and find its derivative $𝑦' (𝑥),$ find the local max (or min), and then determine where each subinterval's max (or min) is. We're actually only using the bell curve function to illustrate some key points here and don't care enough about it to proceed with this approach, but know you can imagine how you would proceed . . . although you may be daunted by the idea of having to find the global max (or min) for hundreds or thousands (or even more) subintervals. And if you aren't daunted by the idea for this particular function (since it actually only has one local max at $𝑥 = 0,$ as you can verify) you can imagine more complicated functions where this idea quickly becomes unfeasible.
Determine "by hand" the location of the subinterval's max (or min). You could, for instance, use the interactive calculator below to simply read off the function's maximum value for each subinterval and use that. This is a perfectly reasonable method to use with a small number of subintervals (like $𝑛 = 4$ ), but again becomes unfeasible for large n.

We'll introduce another method to find the function's maximum value on each subinterval below. But for now, let's use the approach we just discussed because, well really because it's so easy we'd might as well.

Build Upper Sum

You can drag the red dot that's on the x-axis to determine the function's value $𝑦 (𝑥)$ at any point on the interval $[- 2, 4] .$ Use this capability to determine the function's maximum output value for each of the four subintervals.

Then drag the appropriate pieces to their correct locations in the sum-builder below.

Now that we've built the sum using the function's maximum value, let's complete the calculation:

𝑆 u p p e r = 𝑀 1 \cdot Δ 𝑥 + 𝑀 2 \cdot Δ 𝑥 + 𝑀 3 \cdot Δ 𝑥 + 𝑀 4 \cdot Δ 𝑥 = 𝑦 (- 0.5) \cdot Δ 𝑥 + 𝑦 (0) \cdot Δ 𝑥 + 𝑦 (1) \cdot Δ 𝑥 + 𝑦 (2.5) \cdot Δ 𝑥 = (0.9592) (1.5) + (1) (1.5) + (0.8465) (1.5) + (0.3529) (1.5) = 4.7379 ✓

Hence we have determined so far that

A c t u a l a c c u m u l a t i o n \leq 𝑆 u p p e r ⏞ 4.7379

Note that little has changed from our calculations in the preceding Contexts: the only new bit is that our representative value for each subinterval is the function's maximum value on that subinterval. This means that, as before, we can visually represent each term of the sum on a graph of y versus x as a rectangle of height $𝑀 𝑖 = 𝑦 (𝑥 𝑖)$ and width $Δ 𝑥,$ as shown below.

Same graph, now showing the upper sum rectangles for the four subintervals.

We'll look at another way to find all of the $𝑀 𝑖$ values below. First, let's move on and determine our guaranteed underestimate.

Developing an underestimate

The Lower Sum, $𝑆 l o w e r,$ uses the function's minimum value on each subinterval, wherever it occurs on that subinterval.

To calculate the guaranteed underestimate, we can similarly choose $𝑥 𝑖$ for each subinterval such that $𝑦 (𝑥 𝑖) = 𝑚 𝑖$ , where $𝑚 𝑖$ is the minimum value of y in the i-th subinterval. With that choice, we are crafting the Lower Sum, represented by $𝑆 l o w e r$ : $𝑆 l o w e r = 𝑚 1 \cdot Δ 𝑥 + 𝑚 2 \cdot Δ 𝑥 + 𝑚 3 \cdot Δ 𝑥 + 𝑚 4 \cdot Δ 𝑥$

Build Lower Sum

Figure repeated from above showing the subintervals [-2, -0.5], [-0.5, 1], [1, 2.5], [2.5, 4]

You can use the figure below to determine visually where the minimum occurs on each subinterval. Or, if you'd rather, you can use the interactive calculator in the 'Build Upper Sum' activity above.

Either way, as before drag the appropriate pieces below to build the correct Lower Sum for n = 4.

We'll use the sum you just built to find the Lower Sum's value for $𝑛 = 4 :$ $𝑆 l o w e r = 𝑚 1 \cdot Δ 𝑥 + 𝑚 2 \cdot Δ 𝑥 + 𝑚 3 \cdot Δ 𝑥 + 𝑚 4 \cdot Δ 𝑥 = 𝑦 (- 2) \cdot Δ 𝑥 + 𝑦 (1) \cdot Δ 𝑥 + 𝑦 (2.5) \cdot Δ 𝑥 + 𝑦 (4) \cdot Δ 𝑥 = (0.5134) (1.5) + (0.8465) (1.5) + (0.3529) (1.5) + (0.0695) (1.5) = 2.6735 ✓$ The figure below shows each term of the sum represented as a rectangle of height $𝑚 𝑖 = 𝑦 (𝑥 𝑖)$ and width $Δ 𝑥 .$ Same graph, now with dashed lines showing the subintervals [-2, -0.5], [-0.5, 1], [1, 2.5], [2.5, 4] Having found the Lower Sum value, we have now bounded the true value such that $𝑆 l o w e r ⏞ 2.6735 \leq A c t u a l a c c u m u l a t i o n \leq 𝑆 u p p e r ⏞ 4.7379$

Focus on Error Bound

As usual, let's now consider the error bound and see how we can improve it.

Error bound value

The way in which we calculate the error bound hasn't changed at all from previous screens.

As we've done in each of our Contexts, now that we have our initial calculations in place we can work to reduce the error bound and squeeze in on the actual accumulation that occurs over $[- 2, 4]$ .

In our earlier Contexts, since we were using Left and Right Sums we could develop nice formulas to specify $𝑥 𝑖$ within each subinterval. For example, for the Left Sum we start at $𝑎 = 𝑥 1$ (the first point of the overall interval) and then simply increment by $Δ 𝑥$ to find $𝑥 𝑖$ for the next subinterval. Then computing the sum with upwards of 1000 (or even more!) terms was doable as long as our calculator had some sort of summation capability, which Desmos has.

Now, however, the maximum $𝑀 𝑖$ and the minimum $𝑚 𝑖$ for each subinterval can occur anywhere in that subinterval, depending on the specific function. While above we proceeded "manually" for $𝑛 = 4$ subintervals, as we noted, as we increase n this approach is no longer viable.

What other option(s) come to mind for how to proceed for any function you are given, especially with the computing power that lies in your browser? (If you code, how might you find the function's maximum and minimum values on every subinterval?) Again take a moment to think this through for yourself . . .

Once again, if you are interested in how to proceed in clever ways that maximize computing resources and minimize computation time, consider taking a numerical methods course. There you will learn various effective and clever ways to deal with questions such as this.

For our purposes, behind-the-scenes we are going to take a "brute force" approach, and in Step 1 of our approximation method we use your browser (with JavaScript as the coding language) to simply compute the function's values at a large sampling of points over each subinterval, and thereby determine the maximum and minimum values for that subinterval. We then take those values and use them when we build our sum in Step 2.

This will all be invisible to you when you use the interactive calculator below: Use the sliders to set the values of the interval's start-pointa and end-point b, and the number of subintervals n. Then hit the "Calculate Upper & Lower Sums" button, and your browser will start computing.
Warning: Depending on your device, these calculations may take several seconds or more. Please be patient.

You might start with n = 4 to recreate the sum values we found above. Note that you must set the values for a and b to match our interval of $[- 2, 4]$ as well.

Necessary number of subintervals and corresponding subinterval width

We could of course choose a smaller tolerance level and have you find the new number of required subintervals, but you get the idea.

Stepping Back: The Big Picture

What we've done here may all seem rather routine to you by now, so let's stop and acknowledge an enormous accomplishment, one that actually marks a major milestone in the development of Calculus: starting with (almost) any rate-function, we can compute the resulting "accumulation" that happens over a given interval to within whatever tolerance we desire. We do this by squeezing in on the true value by the Lower and Upper Sums, and choosing an appropriate number n of subintervals, as we have just done.

This clever formulation of approximation sums is attributed to Jean Gaston Darboux (pronounced "Dar-boo"), and so these kinds of sums are aptly named Darboux Sums:

DEFINITION

Darboux Sums, or Upper and Lower Sums

U p p e r S u m : 𝑆 u p p e r = 𝑛 \sum 𝑖 = 1 𝑀 𝑖 \cdot Δ 𝑥 𝑖 L o w e r S u m : 𝑆 l o w e r = 𝑛 \sum 𝑖 = 1 𝑚 𝑖 \cdot Δ 𝑥 𝑖

where $𝑀 𝑖$ is the function's maximum value on subinterval i, and $𝑚 𝑖$ is the function's minimum value on subinterval i.

In an unusual move, we're not going to provide any further Practice Problems for you to develop your skills at computing Darboux Sums. (The reality: since this isn't a coding class, what could we do other than provide you with more calculators and sliders to implement code we've written? There's just not a lot to practice here.) That said, the understanding of Darboux sums you've developed will be foundational as we transition to integrals — and if you continue to more advanced mathematics, you will learn more about the important role that Darboux sums have.

The Upshot

We are still following the same basic approach we have been throughout this Section. That is, to approximate the accumulation of something that varies over the interval of interest, we divide the interval into n subintervals, and then first (1) approximate the accumulation during each subinterval: $A p p r o x i m a t e a c c u m u l a t i o n d u r i n g s u b i n t e r v a l 𝑖 : (r e p r e s e n t a t i v e f u n c t i o n v a l u e s u b i n t e r v a l) \cdot (s u b i n t e r v a l d u r a t i o n)$ Having done that, we (2) sum the accumulations of all of the subintervals.
The only new — but crucial! — ideas on this screen are:
For the "representative function value for the subinterval," we can choose the Maximum value $𝑀 𝑖,$ in which case we develop the Upper Sum, $𝑆 u p p e r,$ or we can choose the minimum value $𝑚 𝑖,$ in which case we develop the Lower Sum, $𝑆 l o w e r .$
The exact accumulated value is guaranteed to lie between the Lower and Upper Sum values. Furthermore, we can determine our approximation to the true value to within whatever tolerance we choose by making the number of subintervals, n, sufficiently large.

On the next screen we will summarize our work in this Section, at which point we'll be ready, finally, to take the big step of defining the Definite Integral.

For now, what thoughts or questions do you have about Upper and Lower (Darboux) Sums? Or about any other Calculus-related items? Please head on over to the Forum and share with the rest of our learning community!