The Riemann Sum is a way of approximating the area under a curve on a certain interval [a, b] developed by Bernhard Riemann. The way a Riemann sum works is that it approximates the area by summing up the area of rectangles and then finding the area as the number of rectangles increases to infinity with an infinitely thin width. The following gif may give you a better visual example:

Riemann sums are useful because if you're dealing with a complex function, the area underneath it can be approximated with just a few rectangles. As the number of rectangles increases, you can see that the accuracy of the approximation increases. A perfect Riemann sum has infinitely many rectangles with an infinitely thin width, so of course, an infinite Riemann sum cannot be summed up by hand (unless you want to sit there for a VERY long time, and by that I mean literally forever). The formal definition of a Riemann sum is:

Woah! That looks really really scary, doesn't it? Well, it's actually quite simple, so let's break it down. The large E looking symbol, if you are unfamiliar, is the Greek letter sigma and is used for summing together a bunch of elements or numbers. 

The use of sigma here makes sense since we're summing together a bunch of rectangles. Now, let's tackle the inside of the sum, the f(xk)*Δx. Remember, we're summing together the 

, so this must be the area of some rectangle. Recall that A = bh (or alternatively A = lw) for rectangles. So one of these must be the height of our rectangles, the other must be our width.

You can see in the image that Δx = (b-a)/n. What does this mean? Our area is being taken on the interval [a, b], so b-a is just the length of our interval. For example, if we were finding the area on the interval [3, 8], b - a = 5. Similarly, n is the number of rectangles we have (in a perfect Riemann sum, n approaches infinity). Therefore, (b-a)/n is the equivalent of saying we're taking our interval and chopping it up into n sized pieces on that interval. Here's a visual:

In this example, we are finding the sum from 0 to 8, so b - a = 8. We have n = 4 rectangles, so Δx = (b - a)/n = 8/4 = 2. This makes sense, we took a range of 8 and cut it into 4 

 rectangles (The rectangles in a Riemann sum will have the same width if you are just given a number of rectangles and an interval).

If Δx is the width of our rectangles, f(xk) must be the height! This makes pretty intuitive sense since, for some x value, f(x) tells you essentially the height of the graph at that x. What is k, though? k tells us essentially what rectangle we're on. So for example, at k = 1, we are looking at the first rectangle. Well, how do we find f(xk)? First, we must find xk itself. You'll see in the formula above that xk = a + kΔx. Let's think about this for a second. a is our initial value, our starting point on our interval. k is what rectangle we're on and Δx is the width of each rectangle. xk, therefore, is a way of calculating the x value of each rectangle. In this instance, we're looking at the right x value of each. If we start at k = 0, we have a 

So imagine you are given this equation: f(x) = x^2. Your interval is [0,5] and n = 5. For the purposes of simplicity, I am going to demonstrate a Left Riemann Sum.

1. Understand the information the question gives you

Equation: f(x) = x^2
Interval: [a,b] and in the case of this problem [1,5] - these are the parameters for the section under the curve that you are estimating
N: n=4 - this number determines how many rectangles you are splitting the section under the curve into. (Note: More rectangles = more precision, but also means more work)

2. Find Δx - the width of each of your rectangles

The top left (for a Left Riemann Sum) corner of each rectangle meets the curve (If you were to take a Right Riemann Sum you would use the top right corner of each rectangle and if you were to use a Midpoint Riemann Sum the height would be where the middle of each rectangle hit the curve)

Our heights for this specific problem would be 1, 4, 9, and 16 

4. Find the area of each rectangle and add them together

^In simplest terms, this equation will help you solve any Riemann Sum. Note that all the steps are the same for Right Riemann Sums except for #3. Just remember to use the top left corner of your rectangles for each Left Riemann Sum and the top right corner for each Right Riemann Sum.

For some quick background, when you use the areas of rectangles to estimate the area under a curve, that estimate is a Riemann Sum. When using this method you have two options: a Left Riemann Sum or a Right Riemann Sum. But what's the difference and when should you use each one?

Right Riemann Sum: The right corner of each rectangle touches the curve.

Left Riemann Sum: The left corner of each rectangle touches the curve.

In all honesty, there are no set restrictions that determine when to use each one. Neither method is strictly better than the other.  It is up to you to look at how the rectangles fit within the curve to determine whether a Right or Left Riemann Sum will provide you with a more accurate estimate, but for the most part, the AP exam will tell you which approximation to use.

👉 2007 AP Calculus AB Right Reimann Sum FRQ

2011 AP Calculus AB Trapezoidal Reimann Sum FRQ

2012 AP Calculus BC Midpoint Reimann Sum FRQ

You may have noticed earlier that we discussed whether or not a Riemann sum is an overapproximation or an underapproximation. This depends on two things: if the function is increasing or decreasing, and if we are using a righthand Riemann sum or a lefthand Riemann sum. If a function is increasing, lefthand is an underapproximation and righthand is an overapproximation and vice versa.

Riemann sums are an incredibly powerful tool when approximating areas, and they can even be used, when infinitesimal, to calculate exact areas. Approximating areas is useful, especially when you're dealing with a function where finding an anti-derivative is difficult or would take too long. Using Riemann sums in calculus opens up a whole new world of integration

With Reimann Sums, it is important to master all Applications of Integration.

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AP Calculus: Riemann Sums

, also known as integration by substitution, is one of the key components of integrals. Chances are, you've come across an integral like the one below and been completely lost on where to start.

Fear not! In this article, we'll go over what U-Sub is and how and when you should use it! For the sake of learning the concept first, all of the examples we're going to be using will be indefinite integrals.

Before using u-sub, you want to be able to write your integral as

If we look at the function above, there are a couple things that we should take note of

f(x) - this would be represented as cos(x) in the example

f(g(x)) - in the example, this would be cos(x^2)

g(x) - the 'x' of 'f(x)', in the example at the top of this article, this would be x^2

g'(x) - the derivative of g(x), in the example, this would be 2x

U-substitution is all about making taking the integral of a function easier. To do this, we need to substitute a part of the function with '

' so we can be left with something easier to work with.

 of g(x) changes as well. G'(x) becomes the derivative of '

This example is perfect because we can clearly see what the derivative of g(x) is but it doesn't always work out so easily. To ensure that you're correctly finding g'(x), simply take the derivative of g(x).

In the case of the example, this would play out something like this

Now that we have 'u' and 'du', we can substitute them into our original integral (remember, the original integral was the integral of cos(x^2) * 2x dx).

Next, all we have to do is take the integral in terms of 'u', which basically just means taking the integral but instead of 'x' as you would usually put, 'u' would take its place.

Our last step is to substitute back in g(x) for 'u'.

While this example worked out pretty well, as mentioned before, the majority of the integrals you're going to be taking won't. This means that it might take a little bit of extra work getting the integral into this type of format

Say, we have the another integral like the one below

It looks very similar to the one that we just worked with! The only problem, is that pesky '12x'. So, how do we go about solving this?

We're going to go through the same steps that we did last time.

 g'(x). 12x is simply 2x multiplied by 6 right? So, if we use the rule of constants we can pull that 6 up front and be left with the 2x we need.

Now, we can take the integral as usual, but make sure not to forget about the 6 we took out!

And that's it! Now you know how to use u-sub like a pro! If you need some extra help and practice with this concept check out these awesome links too!

U-Substitution

This video reviews the two parts of the Fundamental Theorem of Calculus (FTC) and how they are related. It reviews strategies for solving AP problems using the FTC, and how to apply them to different types of AP problems.  We then review practice problems using both parts of FTC and what to look out for when taking the AP Exam. 

Using FTC and Integral Defined Functions

Presentation Slides, FTC and Integral Defined Functions

A look at the different techniques required to integrate in AB Calculus.

Integration Techniques Part II

 is a method of integration that transforms an 

 into an integral of the product of one function’s derivative and the others antiderivative. Although this sounds confusing at first, we can build this concept from the product rule

, one of the functions will be f(x) and the other will be g’(x). If other methods like u-substitution don’t work, 

Sometimes, you might see the equation written like this:


This is the same concept in a different form.

. However, we still need to know what dV and dU actually mean. If U is a function of x, then U = f(x). Taking the derivative of both sides, we find that dU = f’(x)dx. Same thing for V: If V = g(x),  then dV = g’(x)dx. This new way of writing the integration by parts formula is just a 

sneakier version of the original formula.

How and Why Would I Use This Formula? 🤔

The best way to explain this formula is through an example. Take the following integral:

At first glance, you may try u-substitution, but this method will take you nowhere. Instead, notice that the integrand is a product of two functions. This is an opportunity to use integration by parts. Let’s recall the formula:

The first step should be to figure out what f(x) and g’(x) should be. This step is probably the most challenging of this method, because you have to choose which is which in order to make the problem work. 

This is where a handy acronym comes into play: 

The function closer to the “E” side should be g’(x) and the function closer to the “L” side should be f(x). 

We know we need g(x) and f’(x) to use the integration by parts formula, so let’s find those now:

 You may have to use integration by parts more than once if your resulting integral is also a product of two functions that can't be solved using u-substitution. 

Integrating Using Integration by Parts

Integrating Rational Functions Using Partial Fractions 💡

The integrand is a rational function with two linear factors in the denominator. With our current techniques, we can’t solve this integral. However, by using partial fractions, we can split the rational integrand into two separate fractions that we know how to handle.

For the AP Calculus BC exam, you will only be dealing with examples where the denominator can be factored into non-repeating linear factors. A non-repeating linear factor is a term of the form (Ax+B) where A and B are numbers, and each term only shows up once in the denominator.

Steps to Using Partial Fraction Decomposition 🏠

 To find the values of A, B, C…, you may need to use algebra techniques such as a system of equations, a matrix, or matching of coefficients. 

 Sometimes the numerator may not be just a number. It may be a linear or quadratic term. If the degree of the numerator is less than the denominator, you can use partial fractions immediately. If the numerator is the same or a higher degree, you will need to use polynomial long division or synthetic division until your numerator is one degree less than the denominator.

Using Linear Partial Fractions

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