A few topics tend to slip through the holes in Calculus. The course lavishes attention on derivatives and integrals, and students become comfortable with them-but they forget their foundations. Limits and Riemann Sums are how we evaluate derivatives and integrals by hand, without any fancy rules, and both show up prominently on the AP exam.

The formula for Riemann Sums looks tremendously difficult to memorize at first glimpse-but if we understand what it really means, learning it is perfectly reasonable.

#### Understanding the Formula

Integrals are used to find the area between a curve and the x-axis, as shown, and Riemann sums are our method to approximate this area. We do this by breaking the shape up into multiple rectangles of equal width, then finding the area of each of the rectangles.

Of course, this isn't a perfect estimate of the area beneath the curve-but it's pretty close. As we break the area beneath the curve into more and more rectangles, our estimate becomes more and more precise, until, eventually, it exactly equals the area beneath the curve.

If you're working with a relatively small number of rectangles, it's possible to calculate the Riemann Sum by hand, even without a formula-but that formula will become *very* important once we try to get the area *perfectly*. Let's start building up to the formula!

First things first, we know that we want to find the area prescribed by these rectangles. Of course, that's pretty easy; the area of a rectangle is width times length!

A = Î”*x* f(*x*1) + Î”*x *f(*x*2) + Î”*x* f(*x*3)

Here, the we'll call the width of each our rectangles Î”*x*: the change in the *x*-displacement between each of our rectangles. The height of our rectangles will be dictated by our function; the height of the first is the value of the function at *x*1, the second is the value of our function at *x*2, and so on and so forth. All of this is just basic geometry so far!

From this point, we can factor Î”*x* out of each of our terms:

A = Î”*x* [f(*x*1) +* *f(*x*2) + f(*x*3)]

From here, we can express the sum of f(*x*1),* *f(*x*2), and f(*x*3) as summation notation-and we've made it right back to our original formula!

We've got three problems left before our formula is usable, however:

We need to make sure that Î”

*x*is the width of each of our rectangles.We need to make sure that our summation starts at the appropriate

*x*-coordinate.We need to make sure that our summation aligns with the width of each rectangle.

The first problem is nice and easy to solve! The distance between A and B can be expressed as A - B, and we want to split it into *n* equal sections. Therefore, we can express Î”*x* as:

Figuring this out also helps us solve our remaining problems. Normally, summation notation increments by 1. The summation of 3*x* + 2 from 1 to 3, for instance, would be [3(1) + 2] + [3(2) + 2] + [3(3) + 2], for instance.

This presents us with a problem, however: what if our rectangles don't have a width of 1? Luckily, it's easy to solve-if we multiply *x* by the width of each rectangle (Î”*x*), then our sum automatically aligns itself with the width of each rectangle!

From there, all we need to do is to add A to make sure that we start our summation from the correct *x*-coordinate, and we're good to go; we've rediscovered our original equation!

Let's take a look at how each of these might be represented in a sample graph:

Everything checks out; our formula multiplies the sum of the heights of our three rectangles by their widths. We're good to go!

#### Right, Left, and Midpoint Riemann Sums

The formulas we've been working with so far have all been for what's known as right Riemann Sums-that is to say, we use the right side of our rectangles to calculate their height.

On the AP exam, you might also encounter left or center Riemann Sums:

*Left Riemann Sum*

*Midpoint Riemann Sum*

Luckily, we don't need to change much in our formula to account for these! The only thing that we need to tweak is *x**i*. For left sums, we simply need to replace *i* with (*i - 1*) in our *x**i* formula. Midpoint sums are halfway between the two, so we replace *i* with (*i* - 0.5).

Aside from that, everything is just as normal!

#### Infinite Riemann Sums

Many Calculus AB courses don't cover infinite Riemann Sums, but they're the foundation that leads to integrals. The line of logic that leads us to use them runs something like so:

We can approximate the area beneath a curve using Riemann Sums.

The more rectangles we use, the more accurately we approximate the area.

Therefore, if we use infinitely many rectangles with a width that approaches 0, then we can

*perfectly*find the area beneath the curve.

This line of logic leads to the following adjustment to our formula:

*Infinite Riemann Sums formula*

Almost everything is just like normal; we've just added a limit that makes us approach infinitely many rectangles. Most students are unlikely to be asked to evaluate such a limit, so we won't cover the topic in-depth here and now. However, if you find yourself in need, you'll need a few formulas to be able to evaluate the limit of sums as *n* approaches infinity.

These are some of the longest problems you can face in high school-make sure that you don't make any small algebra errors along the way!

Riemann Sums are an oft-forgotten topic, as integrals provide a much easier method to evaluate the same math. However, the AP exam doesn't shy away from asking you about them; make sure that you don't get caught unawares!

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