Approximation methods can be tricky. In this video, I go through the method behind Riemann sum problems using an actual exam solution (a right-hand sum).
Finding the area under the curve is something that we talked about all the time in Calculus 1 but when we’re finding that area, why do we actually do it?
One of the best ways to understand this is by looking at something called the Riemann Sum, something that looks like this:
Problem is, it’s really hard sometimes to understand how that approximation relates to the integral that we’re actually setting up.
Let’s jump in.
As coal deposits are depleted, it becomes necessary to strip-mine larger areas for each ton of coal. The graph of g(t) below show the number of acres of land per million tons of coal that will be defaced during step-mining as a function of the number of million tons removed.
(a) Estimate Int[g(t)dt] from 0 to 4 by constructing the right hand sum with n=4 (4 rectangles).
(b) What is the unit of the definite integral Int[g(t)dt] from 0 to 4
(c) What is the practical meaning of the definite integral Int[g(t)dt] from 0 to 4 in the context of this problem?
0:58 – Dissecting the practical meaning of the definite integral of g(t): We can ask ourselves what’s this relationship? What does the number of acres of land have to do with the amount of coal removed and how is this related to the function G of T and the integral?
2:04 – Using units of the definite integral to determine what type of function g(t) is: Moving up to part B of the question we see that we’re using this “acres” unit again, and that’s the unit of the integral itself. So it’s not the function but it’s actually the total of the sum of the area underneath the curve that is represented in acres. Acres is a quantity so we know that the integral is representing a quantity which means that the function that we’re taking the integral of g(t) has to represent some sort of rate. The problem statement also says that g(t) shows the number of acres of land per million tons of coal that will be defaced during strip-mining. This “per” wording gives us another hint that it is a rate, so it’s going to be a rate represented by the ratio of the number of acres to the number of tons.
3:13 – Dissecting the definite integral of g(t): The bounds of this integral represent the number of tons of coal being mined, represented by t. The integral itself represents the total number of acres that are required to extract this 4 million tons of coal. Like we said, our g(t) function is going to represent the number of acres required per the number of tons of coal mined.
4:00 – What the graph of g(t) means and its relationship to the coal mining problem context: Another thing that we can notice about g(t) from the graph is that it’s increasing as more coal is mined. As we progress along the t-axis, g(t) is starting to increase exponentially. What does this mean? As we start mining, our first chunk of area is going to produce 1 million tons of coal. But as we progress to the 2nd million tons of coal, we’re going to require an additional amount of area on top of that initial amount of area so our 2nd square here is larger. Then as we get to our 3rd million ton of coal, we have an even larger amount of acres. And so on.
5:08 – Understanding the right-hand Riemann Sum approximation of the g(t) integral: With a right hand sum, we are constructing an evenly distributed set of rectangles where the height is determined by taking the intersection between the right side and the function. This ends up being four rectangles (n=4) and we add up the areas to get the total area under the curve approximation. In our case, this area approximation is going to be larger than the actual integral itself.
7:36 – How the Riemann Sum estimate relates to the definite integral: why is this an estimate? How is this relate to what we’re doing when we find the actual integral? As your n increases with this right-hand sum, the area approximation becomes more and more accurate. Eventually, if you keep going, you end up with n approaching infinity, and as that happens your area starts to fill in perfectly and you end up with a “definite” solution. This definite solution is what’s represented by the integral.
Animation for Riemann sum “n” increase: “Riemann sum (rightbox)” by 09glasgow09
All right, so that’s what a Reverse Learning breakdown of a Riemann sum area approximation problem looks like. I hope you pulled some useful insights from this that you can apply to any area-under-the-curve approximation problem you come across.