Introduction to Riemann Sums
I’m convinced the reason they teach you Riemann Sums and other methods to find areas is to have you “appreciate” what our former mathematicians had to go through before things got easier. I’m the first to admit that I’m not a fan of working with them, just because they are so tedious.
These methods can be used to approximate the area between a curve and the $ \boldsymbol {x}$-axis, which can be acquired much easier by just taking the integral of the function between two different $ x$-values (we’ll do this in the Definite Integration section). But, alas, we have to learn these more difficult methods first.
Important note: “Area under a curve” assumes the curve is above the $ x$-axis. For curves that are both under and above the $ x$-axis, the areas must be computed separately and added together. A Definite Integral computes the area above the $ x$-axis minus any area below the $ x$-axis.
Let’s say we wanted to get the area of the region between $ x=1$ and $ x=2$ for certain “curved” functions above the $ x$-axis. To get an approximation, we could just add up little rectangles that we can form between $ x=1$ and $ x=2$ that are under the curve and above the $ x$-axis. And do you see how the more rectangles we use (the “$ n$”), the more accurate the total area will be?
Do you also see how, depending on whether the upper left or upper right (or midpoint) of the rectangles touch the curve, we’ll get slightly different areas? For example, for some functions, the “right-hand” sums (where the right-hand sides of the rectangles touch the function) over-approximates the area (upper sum, meaning the rectangles go above the actual curve), and for others, it under-approximates the area (lower sum, meaning the rectangles go below the actual curve). Note that “$ n$” represents the number of rectangles.
How does all this relate to Calculus? Soon we’ll learn to use Integration to get the area between a curve and the $ x$-axis, but again, to “appreciate the math”, we’ll first learn how to approximate the area using upper, lower, and midpoint sums.
Using Upper and Lower Sums to Approximate Area
Here’s an example of using a lower sum to estimate area; in this case, it’s a left-hand sum, since the upper left part of the rectangle touches the curve. The notation can be scary looking, but it’s not that bad.
Using Midpoint Rule to Approximate Area
The midpoint rule uses sums that touch the function at the center of the rectangles that are under the curve and above the $ x$-axis. We compute the area approximation the same way, but evaluate the function right in between (the midpoint of) each of the rectangles; this will be height of the rectangles (the “$ y$”). We’ll compute the $ \Delta x$ the same way. We’ll see an example below.
Upper, Lower, and Midpoint Sums Problems
Here are examples of upper, lower, and midpoint sum problems:
Trapezoidal Rule
The Trapezoidal Rule, another variation of a Riemann Sum, evaluates areas under a curve (above the $ x$-axis), but uses trapezoids instead of rectangles. A trapezoid is a quadrilateral (four-sided figure) with two sides parallel (the bases) and two sides not parallel. Remember from Geometry how to get the area of a trapezoid; here’s how we’ll use it in Calculus:
We’ll want to create little trapezoids (with bases as $ y$-values) under the curve and above the $ x$-axis between two $ x$ points, and then add up all those areas. Here is the formal definition of the Trapezoidal Rule, as it relates to Definite Integration, which we’ll learn about in the next section:
Here are some Trapezoidal Rule problems. Note that when we have “$ n$” intervals, we will get “$ n+1$” $ f(x)$ values.
You may be given the actual function, or you may be given values at certain points so you’d have to “build” the trapezoids. Let’s do a problem where we don’t have equal intervals:
Area by Limit Definition
The area by limit definition takes the same principals we’ve been using to find the sums of rectangles to find area, but goes one step further. We’ll be finding the area between a function and the $ x$-axis between two $ x$ points, but doing it in a way that we’ll use as many rectangles as we can by taking the limit of the number of rectangles as that limit goes to $ \infty $. This way we’ll get the most accurate area we can, since the more rectangles we have, the closer we get the true area.
Again, this is a tough concept to grasp, but we’ll just use a formula that (hate to say it!) you’ll want to memorize how to use, as shown below. Note that $ \Sigma $ (the sigma sign) means to start with the first value, plug it in and go up to the last value, and taking the sum of all of those terms. Notice from the picture that this formula is closest to the midpoint rule.
Here’s the area by limit formula:
Here are the formulas that will be used:
To simplify and get rid of summation signs, use these summation formulas (usually given, but if not, memorize):
All this looks a little scary, and it’s a bit tedious, but here the steps:
- Use the area formula $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\,\sum\limits_{{i=1}}^{n}{{f\left[ {\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)\cdot \left( {\frac{{b-a}}{n}} \right)} \right]}}$ with the given function and interval, and then simplify as much as you can.
- Separate the summations, if needed, and then leave only the “$ i$’s” in the summation (by moving everything else to the outside).
- Use the above summation formulas to turn “$ i$’s” into “$ n$’s”.
- Simplify and take the limit of what’s left (remember that $ \displaystyle \underset{{n\to \,\infty }}{\mathop{{\lim }}}\,\frac{a}{{{{n}^{r}}}}=0$).
Also, don’t forget these useful summation properties:
$ \displaystyle \sum\limits_{{i=1}}^{n}{{\left( {{{a}_{i}}+{{b}_{i}}} \right)\,=}}\sum\limits_{{i=1}}^{n}{{{{a}_{i}}\,}}+\sum\limits_{{i=1}}^{n}{{{{b}_{i}}\,}}\,\,\,\,\,\,\,\,\,\,\,\sum\limits_{{i=1}}^{n}{{c{{a}_{i}}\,=}}c\sum\limits_{{i=1}}^{n}{{{{a}_{i}}\,}}$
Here’s an example:
Here’s another Area by Limit Definition problem. This can get a little messy, so you have to be careful. And I promise you that we’ll have a much easier way to get these areas soon! They just want you to “appreciate the math” at this point:
Here’s a tricky problem on area by limit definition (sums) that you might see on a test. And (spoiler alert!), you’ll see in the possible answers how we’ll do these problems so much easier using Definite Integration in the next section.
Understand these problems, and practice, practice, practice!
On to Definite Integration – you’re ready!