Approximating Area
In This Chapter
- Using rectangles to approximate area
- Right, left, and midpoint sums
- Trapezoidal approximations
- Parabolic approximations with Simpson’s Rule
Have you ever seen the movie Speed with Keanu Reeves and Sandra Bullock? If not, here’s a recap. Everyone’s trapped in this city bus, which will explode if the speedometer goes below 50 mph. So, you’ve got this killer, runaway bus that’s flying around the city and can’t stop—the perfect breeding ground for destruction, disaster, high drama, mayhem, and a budding romance between the movie’s two stars. (Darn that Keanu. Talk about being in the right place at the right time ….)
By now, you probably feel like you’re on that bus. Calculus is tearing all over the place, never slowing down, never stopping, and (unfortunately) never inhabited by such attractive movie stars. The more you learn about derivatives, the more you have to remember about the things that preceded them. Just when you understand something, another (seemingly unrelated) topic pops up to confound your understanding. When will this bus slow down? Actually, the bus slows down right now.
You may feel a slight lurching in the pit of your stomach as we slow to a complete stop, and start discussing something completely and utterly different for a while. Until now, we’ve spent a ton of time talking about rates of change and tangent slopes. That’s pretty much over. Instead, we’re going to start talking about finding the area under curves. I know that’s a big change, but it’ll all come together in the end. For now, take a deep breath, and enjoy a much slower pace for a few chapters as we talk about something different. And if you see Sandra Bullock, tell her I said hi.
Riemann Sums
Let me begin by saying something deeply philosophical. Curves are really, really curvy. It is this inherent curviness that makes it hard to find the area beneath them. For example, take a look at the graph of y = x2 + 1 in Figure 14.1 (only the interval [0,3] is pictured).
Figure 14.1
If only the shaded region between y = x2 + 1 and the x-axis were a square or rectangle—that would make finding the area so much easier.
We want to try and figure out exactly how much area is represented by the shaded space. We don’t have any formulas from geometry to help us find the area of such a curved figure, so we’re going to need to come up with some new techniques. To start with, we’re going to approximate that area using figures for which we already have area formulas. Even though it seems kind of lame, we’re going to approximate the shaded area using rectangles. The process of using rectangles to approximate area is called Riemann sums.
Definition
A Riemann sum is an approximation of an area calculated using rectangles.
We will be exploring simple Riemann sums. Some calculus courses will explore very complicated sums, which involve crazy formulas containing sigma signs (Σ). In my opinion, these won’t help you understand the underlying calculus concepts, so I omit them.
Critical Point
When I say we are looking for the area beneath the curve, I actually mean the area between the curve and the x-axis; otherwise, the area beneath a curve would almost always be infinite. You can always assume that you are finding the area between the curve and the x-axis unless the problem states otherwise.
Right and Left Sums
I’m going to approximate that shaded area beneath y = x2 + 1 using three rectangles. Because I’m only finding the area on the x-interval [0,3], that means I’ll be using three rectangles, each of width 1. (If I had been using six rectangles on an interval of length 3, each rectangle would have width 
.) How high should I make each rectangle? Well, I choose to use a right sum, which means that the rectangles will have the height reached by the function at the right side of each interval, as pictured in Figure 14.2.
Figure 14.2
I’m using three rectangles to approximate the area on [0,1], [1,2], and [2,3].
The rectangle on [0,1] will have the height reached at the far right side of the interval (i.e., x = 1), which is 2. Similarly, the second rectangle is 5 units tall, because that is the height of the function at x = 2, the right side of its interval. Therefore, the heights of the rectangles are 2, 5, and 10, from left to right. The width of each rectangle is 1.
We can approximate the area beneath the curve by adding the areas of the three rectangles together. Because the area of a rectangle is equal to its length times its width, the total area captured by the rectangles is 1 ∙ 2 + 1 ∙ 5 + 1 ∙ 10 = 17. Therefore, the right Riemann sum approximation with n = 3 rectangles is 17.
Critical Point
It was easy to see that the width of every rectangle in our right sum was 1. If the width of the rectangles is not so obvious, use the width formula 
to calculate the width. In this formula, the interval [a,b] is split up into n different rectangles, and each will have width Δx. In our right sum example, 
, because we are splitting up the interval [0,3] into n = 3 rectangles.
Clearly, the area covered by the rectangles is much more than is beneath the curve. In fact, it looks like a lot more. This should tell you that we have got to come up with better methods later (and indeed we will). For now, let’s have a go at the same area problem, but this time use four rectangles and left sums.
Definition
The kind of sum you’re calculating depends on how high you make the rectangles. If you use the height at each rectangle’s left boundary, you’re finding left sums. If you use the height at the right boundary of each rectangle, the result is right sums. Obviously, midpoint sums use the height reached by the function in the middle of each interval.
Example 1: Approximate the area beneath the curve f(x) = x2 + 1 on the interval [0,3] using a left Riemann sum with four rectangles.
Solution: To find how wide each of the four rectangles will be, use the formula 
:
If each of the four intervals is 
wide, and the rectangles start at 0, then the rectangles will be defined by the intervals 
. (This is because 
, 
, etc.) You will be using the heights reached by the function at the left boundary of each interval. Therefore, the heights will be 
, as illustrated in Figure 14.3.
Figure 14.3
Each of the four rectangles is 
wide, and they are as high as the function f(x) = x2 + 1 at the left edge of each rectangle, hence left sums is the result.
The area of each rectangle is its width times its height, so the total area is:
This number underestimates the actual area beneath the curve, because there are large pieces of that area missed by our rectangles.
Midpoint Sums
Calculating midpoint sums is similar to calculating right and left sums. The only difference is (you guessed it) how you define the heights of the rectangles. In our ongoing example of f(x) = x2 + 1 on the x-interval [0,3], let’s say we wanted to calculate midpoint sums using (to make it easy) n = 3 rectangles. As before, the intervals defining the rectangles’ boundaries will be [0,1], [1,2], and [2,3], and each rectangle will have a width of 1. What about the heights?
Look at the interval [0,1]. If we were using left sums, the height of the rectangle would be f(0). Using right sums, it’d be f(1). However, we’re using midpoint sums, so you use the function value at the midpoint of the interval, which in this case is 
. Therefore, the height of the rectangle is 
. If you apply this to all three intervals, the midpoint Riemann approximation of the area would be:
You’ve got problems
Problem 1: Approximate the area beneath the curve g(x) = –cos x on the interval 
using n = 4 rectangles and (1) left sums, (2) right sums, and (3) midpoint sums.
The Trapezoidal Rule
Unfortunately, unless you use a ton of rectangles, Riemann sums are just not all that accurate. The Trapezoidal Rule, however, is often a more accurate way to approximate area beneath a curve. Instead of constructing rectangles, this method uses small trapezoids. In effect, these trapezoids look the same as their predecessor rectangles near their bases, but completely different at the top. To construct the trapezoids, you mark the height of the function at the beginning and end of the width interval (which is still calculated by the formula 
) and connect those two points. Figure 14.4 shows how the Trapezoidal Rule approximates the area beneath our favorite function in the whole world, y = x2 + 1.
Critical Point
This is going to freak you out. Remember how the left and right sums offset one another when we approximated the area beneath y = x2 + 1—one too big and the other too small? Well, the Trapezoidal Rule (with n trapezoids) is exactly the average of the left and right sums (with n rectangles). We already know that the right sum of y = x2 + 1 (with n = 3) is 17. You can find the corresponding left sum to be 8. If you calculate the Trapezoidal Rule approximation (with n = 3 trapezoids), you get 12.5, which is the average of 8 and 17.
Figure 14.4
The “tops” of our approximating shapes are no longer parallel to the x-axis. Instead they connect the function’s heights at the interval endpoints.
There’s a lot less room for error with this rule, and it’s actually just as easy to use as Riemann sums were. One difference—this one requires that you memorize a formula.
Critical Point
If you’re dying to know the actual area beneath y = x2 + 1 on the interval [0,3], it is exactly 12. Of our approximations so far, the midpoint sum came the closest (even though we used only three rectangles with this method but four with left sums).
The Trapezoidal Rule: The approximate area beneath a continuous curve f(x) on the interval [a,b] using n trapezoids equals:
In practice, you pop the correct numbers into the fraction at the beginning and then evaluate the function at every interval boundary. Except for the endpoints, you’ll multiply all the values by 2.
The area of any trapezoid is one-half the height times the sum of the bases (the bases are the parallel sides). For the trapezoid in Figure 14.5, the area is 
. You may not be used to seeing trapezoids tipped on their side like this—in geometry, the bases are usually horizontal, not vertical. The reason you see all those 2’s in the Trapezoidal Rule is that every base is used twice for consecutive trapezoids except for the bases at the endpoints.
Figure 14.5
Our approximate trapezoids are simply right trapezoids shoved onto their sides, with bases b1 and b2, and height h.
Critical Point
There’s another way to get better approximations using Riemann sums. If you increase the number of rectangles you use, the amount of error decreases. However, the amount of calculating you have to do increases. Eventually, we’ll find a way to obtain the exact area without much work at all. The way is rooted in Riemann sums, but uses an infinite number of rectangles in order to eliminate any error completely.
Let’s go straight into an example, and you’ll see that the Trapezoidal Rule is not very hard at all. Just for grins, let’s use f(x) = x2 + 1 yet again to see if the Trapezoidal Rule can beat out our current best estimate of 11.75 given by the midpoint sum.
Example 2: Approximate the area beneath f(x) = x2 + 1 on the interval [0,3] using the Trapezoidal Rule with n = 5 trapezoids.
Solution: Because you are using five trapezoids, you need to determine how wide each will be, so apply the Δx formula:
Kelley’s Cautions
Although the Trapezoidal Rule’s formula contains the expression 
, you still use the formula 
to find the width of the trapezoids. Don’t get them confused—they are separate formulas.
Therefore, the boundaries of the intervals will start at x = 0 and progress in steps of 
and 3. These numbers belong in the formula as a, x1, x2, x3, x4, and b. So according to the Trapezoidal Rule, the area is approximately:
This is actually the closest approximation yet, although it is a bit too big. Had this curve been concave down instead of up, the result would have underestimated the area. Can you see why? The teeny bit of error would have been outside, rather than inside, the curve.
You’ve got problems
Problem 2: Approximate the area beneath y = sin x on the interval [0,π] using the Trapezoidal Rule with n = 4 trapezoids.
Simpson’s Rule
Our final area-approximating tool is Simpson’s Rule. Geometrically, it creates tiny little parabolas (rather than the slanted trapezoidal interval roofs) to wrap even closer around the function we’re approximating. The formula is astonishingly similar to the Trapezoidal Rule, but here’s the catch: you can only use an even number of subintervals.
Simpson’s Rule: The approximate area under the continuous curve f(x) on the closed interval [a,b] using an even number of subintervals, n, is:
In this formula, the outermost terms get multiplied by nothing. However, beginning with the second term, you multiply consecutive terms by 4, then 2, then 4, then 2, etc. Make sure you always start with 4, though. Back to Old Faithful, f(x) = x2 + 1, for an example.
Example 3: Approximate the area beneath that confounded function f(x) = x2 + 1 on the closed interval [0,3], this time using Simpson’s Rule and n = 6 subintervals.
Solution: Some quick calculating tells us that our subintervals will have the width of 
. Now, to the formula we go:
Remember to multiply 
by 4, the next term by 2, etc. However, the first and last terms get no additional coefficient:
Whoa! Because Simpson’s Rule uses quadratic approximations, and this is a quadratic function, you get the exact answer. This only happens for areas beneath quadratic equations, though.
You’ve got problems
Problem 3: Approximate the area beneath 
on the interval [1,5] using Simpson’s Rule with n = 4 subintervals.
The Least You Need to Know
- Riemann sums use rectangles to approximate the area beneath a curve; the heights of these rectangles are based on the height of the function at the left end, right end, or midpoint of each subinterval.
- The width of each subinterval in all the approximating techniques is
. - The Trapezoidal Rule is the average of the left and right sums, and usually gives a better approximation than either does individually.
- Simpson’s Rule uses intervals topped with parabolas to approximate area; therefore, it gives the exact area beneath quadratic functions.