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1 min read•june 18, 2024
Emery
Emery
Welcome back to AP Calc with Fiveable! In the last guide, we explored accumulations of change and thought about how to compute them. Here, we’ll take a deep dive into Riemann sums, a method for approximating the area under the curve to find the accumulation of change.
As we delve into integral calculus, our goal has shifted from computing the instantaneous rate of change to computing the area under the curve. Riemann sums are a useful tool for approximating this. Take a look at this graph:
You would have a hard time computing this geometrically. But, we can approximate it using familiar shapes, like this:
We can see that this isn’t exact, but that our approximation improves if we use more and more rectangles:
The use of these rectangles to approximate the area under the curve is called a Riemann sum. There are four main Riemann sums:
Now, let’s get into each of these in detail!
There are two basic types of Riemann sums, called “left endpoint” and “right endpoint.” Here is an example of the same curve with a left Riemann sum, versus one with a right Riemann sum:
You can see that the left and right refer to which points we use to determine the height of our rectangles. Left Riemann sums touch the curve with their top left corners, and right Riemann sums touch the curve with their top right corners.
Another thing that we can vary when deciding to take a Riemann sum is the width of our base. Our base length can either be uniform or non-uniform, as illustrated here:
The base lengths formed by dividing up the total interval are called “subdivisions.”
We know that a Riemann sum isn’t a perfect calculation of our area under the curve, but are we overestimating or underestimating when we use one? Try these problems and then fill in the blanks to see if you can figure out the pattern!
Fill in the Blanks! 🧠
⚡ Answer Key:
If you haven’t already guessed the pattern, here are the rules:
Now that we know which types of Riemann sums make which types of errors in their estimations, how might we make our approximation more accurate?
If you guessed by averaging the two estimations, you’d be correct! We can do this in one of two ways.
If you’ve already calculated the left and right Riemann sum for a graph, you can simply take their average. But, there’s an even faster way to do this, using another simple geometric shape—the trapezoid! Instead of finding just the left or right endpoints, we mark every point over the interval we’re interested in. Then, we connect pairs of points together:
This creates trapezoids! Instead of using the area formula for a rectangle, we now take the average of the two side lengths and multiple them by the base, using this formula:
Midpoint Riemann sums have the same goal as trapezoidal Riemann sums—to minimize the amount of error. They accomplish this by using the middle point of the rectangle as its height, as illustrated here:
Just like left and right Riemann sums, the trapezoidal and midpoint Riemann sums may be an over- or underestimation depending on the type of graph.
Be familiar with the following:
Sometimes we’re not given a graph to base our estimates on! Let’s work through a sample problem to learn how to approach these types of problems.
Given the function , find and over the interval [0,4].
The very first thing you should do in this type of problem is draw a graph!
Then, pick one of the Riemann sums to calculate. We’re going to solve for and first.
First, figure out how wide each rectangle will be by dividing the total interval by the number of subdivisions we want:
Then, draw our rectangles on the graph for each type of endpoint:
We can also make a table of values from the graph:
x | y (height) |
0 | 0 |
1 | 0.5 |
2 | 2 |
3 | 4.5 |
4 | 8 |
Based on our graphs, to find we need to use the left endpoints, which are . Since our base width is 1, we can just sum their values. Thus,
Let’s do the same thing for , using :
Now on to !
We know that there will be four trapezoids since we are using the same number of subdivisions. We can draw it like this:
We can also use the same table of values from the first part of the problem to find our values for a and b.
x | y (height) |
0 | 0 |
1 | 0.5 |
2 | 2 |
3 | 4.5 |
4 | 8 |
Let’s plug these values into our equation to find the area of our first trapezoid:
We can do this for all four trapezoids and add them up to get our total area:
Finally, let’s find , starting by drawing our rectangles onto the graph:
First, we can plug in our x-values from the graph to find the height of each rectangle:
x | y (height) |
0.5 | 0.125 |
1.5 | 1.125 |
2.5 | 3.125 |
3.5 | 6.125 |
Since our the width of each base is just 1, we can simply sum the four values to get our area under the curve estimated by :
Bonus Question: Are and overestimates or underestimates?
Solution: is an overestimate and is an underestimate, which means the true area under the curve lies somewhere between 10.5 and 11!
Now take some time to practice these on your own!
Problem 1 Solution****
First, let’s graph this function over the given interval:
Then, let’s figure out what the width of our rectangles will be and graph with the right endpoints: .
Now, let’s find their heights numerically:
x-value | y (height) |
2 | 5 |
3 | 7 |
4 | 9 |
5 | 11 |
Now, let’s solve by summing the areas of each rectangle:
Problem 2 Solution****
First, let’s graph this function over the given interval:
Then, let’s figure out the width of our rectangles and graph with the left endpoints:
Now, let’s find their heights numerically:
x-value | y (height) |
0 | 0 |
1/3 | 10/9 |
2/3 | 22/9 |
1 | 4 |
4/3 | 52/9 |
5/3 | 70/9 |
Now, let’s solve by summing the areas of each rectangle:
We can rewrite this by factoring out our base length like so:
Problem 3 Solution****
First, let’s graph the function over the given interval:
Then, let’s figure out the width of our rectangles and graph with the trapezoids:
Let’s start by solving for by creating a table of values:
x-value | y (height) |
2 | 8 |
2.5 | 14.75 |
3 | 23 |
3.5 | 32.75 |
4 | 44 |
4.5 | 56.75 |
5 | 71 |
5.5 | 86.75 |
6 | 104 |
6.5 | 122.75 |
7 | 143 |
7.5 | 164.75 |
8 | 188 |
Now, let’s plug it into our equation:
And simplify!
Let’s do the same with midpoints:
x-value | y (height) |
2.25 | 11.1875 |
2.75 | 18.6875 |
3.25 | 27.6875 |
3.75 | 38.1875 |
4.25 | 50.1875 |
4.75 | 63.6875 |
5.25 | 78.6875 |
5.75 | 95.1875 |
6.25 | 113.1875 |
6.75 | 132.6875 |
7.25 | 153.6875 |
7.75 | 176.1875 |
Now, let’s plug in and solve:
Here are the key takeaways you should now understand about Riemann sums:
Best of luck! 🍀
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