CHAPTER
16

Applications of the Fundamental Theorem

In This Chapter

• Finding yet more curvy area
• Integration’s Mean Value Theorem
• Position equations and distance traveled
• Functions defined by definite integrals
• Pretending you’re Noah: finding arc length

Once you learned how to find the slope of a tangent line (a seemingly meaningless skill), it probably seemed as though the applications for the derivative would never stop. You were finding velocity and rates of change (both instantaneous and average), calculating related variable rates, optimizing functions, determining extrema, and, all in all, bringing peace and prosperity to the universe.

If you think that it’s about time for applications of definite integrals to start pouring in, you must be psychic. (Either that or you read the table of contents.) For now, we’ll look at some of the most popular definite integral-related calculus topics. We’ll start by finding area bounded by two curves (rather than one curve and the x-axis). We’ll briefly backtrack to topics we’ve already discussed, but we’ll spice them up a little with what we now know of integrals. Finally, we’ll look at definite integral functions, also called accumulation functions.

Calculating Area Between Two Curves

This’ll blow your mind. In fact, after you read it, you may question your very sanity. The thin ribbons of consciousness tying you to this mortal world may stretch and break, catapulting you into madness, or at least making you lose your appetite. Perhaps you should sit down before you continue.

You’ve been calculating the area between curves all along without even knowing it. There, I said it. I hope you’re okay.

If you want to calculate the area between two continuous curves, we’ll call them f(x) and g(x), on the same x-interval [a,b], here’s what you do. Set up a definite integral as you did last chapter, with a and b as the lower and upper limits of integration, respectively. You’ll stick either f(x) – g(x) or g(x) – f(x) inside the integral. To decide which one to use, you have to graph the functions—you should subtract the lower graph from the higher graph. For example, in Figure 16.1, g(x) is below f(x) on the interval [a,b].

Figure 16.1

At least on the interval [a, b], the graph of f(x) is always higher than the graph of g(x).

Watch out! If you subtract the functions in the wrong order you’ll get a negative answer, and you should never get a negative answer when finding the area between curves, even if some of that area falls below the x-axis.

What if the curves switch places? For example, look at the graph in Figure 16.2. To the left of x = c, f(x) is above g(x), but when x > c, the functions switch places and g(x) is on top.

Figure 16.2

The graphs take turns on the top bunk—neither is above the other on the entire interval.

To find the shaded area, you’ll have to use two separate definite integrals, one for the interval [a,c], when f(x) is on top, and one for [c,b], when g(x) is:

Example 1: Calculate the area between the functions f(x) = sin 2x and g(x) = cos x on the interval .

Solution: These graphs play leapfrog all along the x-axis, but on the interval , g(x) is definitely above f(x) (see Figure 16.3).

Figure 16.3

On the interval , g(x) = cos x rises above f(x) = sin 2x.

Therefore, the integral will contain g(x) – f(x):

Split this up into separate integrals:

Calculating the first is fairly easy:

Use u-substitution to integrate sin 2x, setting u = 2x:

Don’t forget to change your x boundaries into u boundaries when you u-substitute. For example, to get the new u boundary of 4π, plug the old x boundary of 2π into the u equation: u = 2(4π) = 4π. The final answer is the first integral minus the second:

The Mean Value Theorem for Integration

Think back to the original Mean Value Theorem from Chapter 13. It said that somewhere on an interval, the derivative was equal to the average rate of change for the whole interval. It turns out that integration has its own version of a Mean Value Theorem, but because integration involves area instead of rates of change, it’s a bit different.

A Geometric Interpretation

In essence, the Mean Value Theorem for Integration states that at some point along an interval [a,b], there exists a certain point (c, f(c)) between a and b (see Figure 16.4). If you draw a rectangle whose base is the interval [a,b] and whose height is f(c), the area of that rectangle will be exactly the area beneath the function on [a,b].

Figure 16.4

A visual representation of the Mean Value Theorem for Integration. The area of the shaded rectangle, whose height is f(c), is exactly equal to .

The Mean Value Theorem for Integration: If a function f(x) is continuous on the interval [a,b], then there exists c, a ≤ c ≤ b, such that .

This Mean Value Theorem, like its predecessor, is only an existence theorem. It guarantees that the value x = c and the corresponding key height f(c) exist. You may wonder why it’s so important that a curvy graph and a plain old rectangle must always share the same area. We’ll get to that after the next example.

Example 2: Find the value f(c) guaranteed by the Mean Value Theorem for Integration for the function f(x) = x³ – 4x² + 3x + 4 on the interval [1,4].

Solution: The Mean Value Theorem for Integration states that there is a c between a and b so that . You know everything except what c is, but that’s okay.

Plug in everything you know:

After the quick subtraction problem on the left (and the slightly lengthier definite integral on the right), you should get this:

This means that the area beneath the curve f(x) = x³ – 4x² + 3x + 4 on the interval [1,4] (which is ) is equal to the area of the rectangle whose length is the same as the interval’s length (3) and whose height is (c, ).

The Average Value Theorem

The value f(c) that you found in both Example 2 and Problem 2 has a special name. It is called the average value of the function. If you take the Mean Value Theorem for Integration and divide both sides of it by (b – a), you’ll get the equation for average value:

This is simply a different form of our previous equation, so it doesn’t warrant much more attention. However, some textbooks completely skip over the Mean Value Theorem for Integration and go right to this, which they call the Average Value Theorem. They might word Problem 2 earlier as follows: “Find the average value of on the interval [1,100].” You’d solve the problem the exact same way (see Figure 16.5).

Figure 16.5

Here’s the diagram for the Mean Value theorem for Integration once more. The height of the denoted line is the function’s average value. Although the function dips below and shoots above f(x), that’s how high f(x) is on average.

Here’s one way to think of the relationship between the theorems. Most functions twist and turn throughout their domains. If you could “level out” a function by filling in its valleys and flattening out its peaks until the function was a horizontal line, the height of that line (i.e., its y-value) would be the average value for that function.

Finding Distance Traveled

Definite integrals also play well with position and velocity functions. Remember that derivatives measure a rate of change. Well, it turns out that definite integrals of rate of change functions measure accumulated change. For example, if you are given a function that represents the rate of sales of the new must-have toy, the Super Fantastic Hula Hoop, then the definite integral gives you the actual number of hula hoops sold.

Most often, however, math teachers like to explore this property of integrals as it applies to motion. Specifically, the definite integral of the velocity function of an object gives you the total displacement of the object. A word of caution: you will most often be asked to find the total distance traveled by the object—not the total displacement. To calculate the total distance, you’ll first have to determine where the object changes direction (using a wiggle graph) and then integrate the velocity separately on every interval that direction changes.

Here’s the difference between total distance traveled and total displacement. Let’s say at any hour t, I want to know (in miles) how far I am away from my favorite bright orange 1970s-style easy chair that my wife hates. My initial position (i.e., t = 0 hours) is in the chair, so s(0) = 0. Two hours later, I am at work, 50 miles away from my chair, so s(2) = 50 miles. Once my workday and commute home are complete, I am back home in the chair, and s(12) = 0. I have traveled a total distance of 100 miles, counting my travel away from the chair and back again. However, my displacement is 0. Displacement is the total change in position counting only the beginning and ending position; if the object in question changes direction any time during that interval of time, displacement does not correctly reflect the total distance traveled.

Example 3: In the book The Fellowship of the Ring by J.R.R. Tolkien, a young hobbit named Frodo embarks on an epic, exciting, and hairy-footed adventure to destroy the One Ring in the fires of Mount Doom. Based on a little estimation and the book Journeys of Frodo: An Atlas of J.R.R. Tolkien’s The Lord of the Rings by Barbara Strachey, I have designed an equation modeling Frodo’s journey. During the first four days of his journey (from Hobbiton to the home of Tom Bombadil), his velocity (in miles per day) is given by this equation:

v(t) = – 15.5t³ + 86.25t² – 117.25t + 48.75

For example, v(2) gives his approximate velocity at the exact end of the second day. Find the total distance Frodo travels from t = 0 to t = 4.

Solution: Because you want to find the total distance traveled, you need to determine if Frodo changed direction at any point, and actually started to wander toward Hobbiton rather than away. This is not necessarily caused by poor hobbit navigation, but perhaps by hindrances such as the old forest, getting caught in trees, etc. To see if his direction changed, create a wiggle graph for velocity (see Figure 16.6).

Figure 16.6

The hobbits have a pretty good sense of direction; in fact, they are heading farther and farther away from Hobbiton until just before the end of the fourth day.

Integrate the velocity equation separately, on both of these intervals. Because they are heading slightly backward (i.e., toward their beginning point) on the interval (3.78586,4), that definite integral will be negative. However, because it still represents the distance the hobbits are traveling, you don’t want it to be subtracted from your answer, so turn it into its opposite by multiplying it by –1. You should do this for any negative pieces of your wiggle graph in this type of problem. Therefore, the distance traveled is:

Although the numbers are darn ugly, the premise is very simple. I’ll leave the figuring up to you. You should get 108.298 for the first interval (distance away from Hobbiton) and –(–3.298) for the second, which is the small distance back toward Hobbiton; the sum is 111.596 miles.

Right about now, you’re seeing that the numbers in this problem are not easy whole numbers. They rarely turn out to be so in real-world (or Middle Earth) examples, so calculators are a necessary tool. There are those who would have me burned at the mathematical stake for suggesting such a thing. In fact, I was once yelled at fiercely by the lunch ladies in the high school cafeteria where I worked for suggesting that you should use a calculator to check your answers when converting fractions to decimals. I’ve never received fewer tater tots than I did that day. Lunch ladies can be so bitter.

Accumulation Functions

Before we close out this chapter and make it a fond memory, let’s talk about accumulation functions. You’ll probably see a few of them lurking around contemporary calculus classes, as they are now “in” because of the advent of calculus reform. An accumulation function is a definite integral with a variable expression in one or more of its limits of integration. They are called accumulation functions because they get their value by accumulating area beneath curves, as do all definite integrals.

The most famous accumulation function is the natural logarithmic function: <. The natural log function gets its value by accumulating area under the simple curve ! For example, the value of ln 5, which always seemed so alien to me (where the heck do you get 1.60944?) is equal to the area beneath on the interval [1,5].

Practically speaking, you should be able to evaluate and differentiate accumulation functions, so let’s get to it. Believe it or not, evaluating accumulation functions is just as easy as evaluating any other function—just plug in the correct x-value. Once you plug in the value, you’ll apply the Fundamental Theorem to the resulting definite integral.

Example 4: Given the accumulation function , complete the following tasks:

(a) Calculate f(4).

(b) Determine f ′(x).

Solution:

(a) Plug 4 into x, not t, because f(x) is a function of x. In other words, x becomes the upper limit of integration:

Now you can integrate like normal:

(b) To find the derivative of an accumulation function, look no further than part two of the Fundamental Theorem. In this case, f ′(x) = x – 4. Just plug the top bound into t and multiply by its derivative (which is 1 in this case). Pretty easy, eh? You already knew how to tackle these problems, even before they showed up. Kudos!

Arc Length

At this point, you can do all kinds of crazy math calculating. Geometry told you how to find weird areas, and calculus took that one step further. The kinds of areas you can calculate now would have boggled your mind back in your days of geometric innocence. However, it remains a math skill that has visible and understandable application, even to those who don’t know the difference between calculus and a tuna sandwich. Now, let’s add to your list of skills the ability to find lengths of curves. By the time you’re done, you’ll even be able to prove (finally) that the circumference of a circle really is 2π.

Rectangular Equations

The term “rectangular equations” really means “plain old functions.” Mathematicians use the term for the obvious reason that it takes less time to say (mathematicians are busy people). It turns out that finding the length of a curve (on some x interval) is as easy as calculating a definite integral. In fact, the length of a continuous function f(x) on the interval [a,b] is equal to . In other words, find the derivative of the function, square it, add 1, and integrate the square root of the result over the correct interval.

Example 5: Find the length of the function between points (1,1) and (16,4) on its graph.

Solution: Use the Power Rule to find the derivative of g(x) = x^1/2, and you get . All you do now is plug into the arc length formula:

The integration problem that results is not simple at all. For our purposes, it is enough to know and apply the formula, not to struggle through the integral itself. You’ll find that many (if not most) arc length integrals will end up complicated and require somewhat advanced methods to integrate. We will, however, satisfy ourselves with a computer- or calculator-assisted solution—they have no problem with complex definite integrals. The final answer is approximately 15.3397.

Don’t feel like you’re cheating by using a calculating tool rather than solving this problem by hand. You’d have to know just about every integration technique there is to find the arc lengths of even very simple functions.

The arc length of a curve defined parametrically is found with this definite integral:

In other words, find the derivatives of the x and y equations, square them both, add them together, take the square root, and integrate the whole mess. Note that a and b are limiting values of the parameter t this time—not x boundaries.

Example 6: The parametric representation of a circle with radius 1 (centered at the origin) is x = cosθ, y = sinθ. Prove that the circumference of a circle really is 2π by calculating the arc length of the parametric curve on .

Solution: Start by finding the derivatives of the x and y equations with respect to θ (it’s easy):

Now, plug those values into the parametric arc-length formula and simplify using the Mama theorem (review Chapter 4 if you don’t know what the heck I mean by that):

The Least You Need to Know

• To find the area between two curves, integrate the curve on top subtracted by the curve below it on the proper interval.
• The average value of a function, f(x), over an interval, [a,b], is found by dividing the definite integral of that function, , by the length of the interval itself, b – a.
• To calculate the distance traveled by an object, calculate the definite integral of its velocity function separately for each period of time it changes direction.
• Accumulation functions get their value by gathering area under a curve; they are defined by definite integrals with variables in one or more of their limits of integration.
• You can find the arc length of rectangular or parametric curves via similar definite integral formulas.