In this section we will find connections between differential calculus (derivatives and antiderivatives) and integral calculus (definite integrals). These connections between the major ideas of calculus are important enough to be called the Fundamental Theorem of Calculus. These connections will also explain why we use the term indefinite integral for the set of all antiderivatives, and why we use such similar notations for antiderivatives and definite integrals.
Let be a continuous function defined on . The definite integral is the “area under ” on . We can turn this concept into a function by letting the upper (or lower) bound vary.
Let . It computes the area under on as illustrated in Figure 5.4.1. We can study this function using our knowledge of the definite integral. For instance, since .
The first part of the Fundamental Theorem of Calculus tells us how to find derivatives of these kinds of functions.
Let be continuous on and let . Then is a continuous function on , differentiable on , and
In order to see why this is true, we must compute . Suppose and are in . Theorem 5.2.1 implies that
which we can rewrite as
This allows us to simplify the numerator of the difference quotient in our limit as follows:
so we see that
Assume for the moment that . Since and are both in and is continuous on , is also continuous on . Applying the Extreme Value Theorem (Theorem 3.1.1), we know that must have an absolute minimum value and an absolute maximum value on this interval. In other words, whenever . Using Theorem 5.3.3, we can now say that
Computing the outer integrals, this becomes
Since , we may divide by to obtain
Now suppose that . Preceding as before, we know that has an absolute minimum value and an absolute maximum value on the interval . We know that whenever , so we have
Once again we compute to obtain
Since , we can divide by to obtain:
We are now ready to compute the desired limit,
Whether or , we know that
where and are both between and . Note that
so the Squeeze Theorem (Theorem 1.3.5) says that
Since is continuous at , we know that
Finally, we know that
so applying the Squeeze Theorem again tells us that
Therefore as desired. Because the limit exists, Exercise 35. in Section 2.1 implies that is continuous on as well. All that remains is to show that is continuous at and . But repeating the preceding argument shows that , and similarly for near .∎
Initially this seems simple, as demonstrated in the following example.
Let . What is ?
SolutionUsing the Fundamental Theorem of Calculus, we have
.
This simple example reveals something incredible: is an antiderivative of . Therefore, for some value of . (We can find , but generally we do not care. We know that , which allows us to compute . In this case, .)
We have done more than found a complicated way of computing an antiderivative. Consider a function defined on an open interval containing , and . Suppose we want to compute . First, let . Using the properties of the definite integral found in Theorem 5.2.1, we know
We now see how indefinite integrals and definite integrals are related: we can evaluate a definite integral using antiderivatives. Furthermore, Theorem 5.1.1 told us that any other antiderivative differs from by a constant: . This means that , and the formula we’ve just found holds for any antiderivative. Consequently, it does not matter what value of we use, and we might as well let . This proves the second part of the Fundamental Theorem of Calculus.
Let be continuous on and let be any antiderivative of . Then
We spent a great deal of time in the previous section studying . Using the Fundamental Theorem of Calculus, evaluate this definite integral.
SolutionWe need an antiderivative of . All antiderivatives of have the form ; for simplicity, choose .
The Fundamental Theorem of Calculus states
This is the same answer we obtained using limits in the previous section, just with much less work.
A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus. Instead of explicitly writing , the notation is used. Thus the solution to Example 5.4.2 would be written as:
Evaluate the following definite integrals.
Solution
(This is interesting; it says that the area under one “hump” of a sine curve is 2.)
This integral is interesting; the integrand is a constant function, hence we are finding the area of a rectangle with width and height 2. Notice how the evaluation of the definite integral led to .
In general, if is a constant, then .
Part 1 of the Fundamental Theorem of Calculus (FTC) states that given , . Using other notation, . While we have just practiced evaluating definite integrals, sometimes finding antiderivatives is impossible and we need to rely on other techniques to approximate the value of a definite integral. Functions written as are useful in such situations.
It may be of further use to compose such a function with another. As an example, we may compose with to get
What is the derivative of such a function? The Chain Rule can be employed to state
An example will help us understand this.
Find the derivative of .
SolutionWe can view as being the function composed with ; that is, . The Fundamental Theorem of Calculus states that . The Chain Rule gives us
Normally, the steps defining and are skipped.
Practice this once more.
Find the derivative of .
SolutionNote that . Viewed this way, the derivative of is straightforward:
We established, starting with Key Idea 2.2.1, that the derivative of a position function is a velocity function, and the derivative of a velocity function is an acceleration function. Now consider definite integrals of velocity and acceleration functions. Specifically, if is a velocity function, what does mean?
The Fundamental Theorem of Calculus states that
where is any antiderivative of . Since is a velocity function, must be a position function, and measures a change in position, or displacement.
How would we measure total distance traveled? We have to consider the intervals when and when . Therefore,
A ball is thrown straight up with velocity given by ft/s, where is measured in seconds. Find, and interpret,
Solution
Using the Fundamental Theorem of Calculus, we have
Thus if a ball is thrown straight up into the air with velocity , the height of the ball, 1 second later, will be 4 feet above the initial height. We will see in part 2. that the distance traveled is much farther. It has gone up to its peak and is falling down, but the difference between its height at and is 4 ft.
Here we are trying to find the total distance traveled by the ball. We must first consider where and .
This means for and for so we have
Integrating a rate of change function gives total change. Velocity is the rate of position change; integrating velocity gives the total change of position, i.e., displacement.
Integrating a speed function gives a similar, though different, result. Speed is also the rate of position change, but does not account for direction. So integrating a speed function gives total change of position, without the possibility of “negative position change.” Hence the integral of a speed function gives distance traveled.
As acceleration is the rate of velocity change, integrating an acceleration function gives total change in velocity. We do not have a simple term for this analogous to displacement. If miles/h and is measured in hours, then
means the velocity has increased by 15m/h from to .
Consider the graph of a function in Figure 5.4.2(a) and the area defined by . Three rectangles are then drawn; in (b), the height of the rectangle is greater than on , hence the area of this rectangle is is greater than .
In (c), the height of the rectangle is smaller than on , hence the area of this rectangle is less than .
Finally, in (d) the height of the rectangle is such that the area of the rectangle is exactly that of . Since rectangles that are “too big”, as in (b), and rectangles that are “too little,” as in (c), give areas greater/lesser than , it makes sense that there is a rectangle, whose top intersects somewhere on , whose area is exactly that of the definite integral.
We state this idea formally in a theorem.
Let be continuous on . There exists a value in such that
This is an existential statement; exists, but we do not provide a method of finding it. Theorem 5.4.3 is directly connected to the Mean Value Theorem of Differentiation, given as Theorem 3.2.1.
If , then . Otherwise, we define the following for in :
Applying Theorem 5.4.1, we know is differentiable on and that for any in . We may now apply the Mean Value Theorem for Differentiation (Theorem 3.2.1) to see that there is a value in such that
Note that and that by Theorem 5.4.2. Therefore we can rewrite our equation as:
We demonstrate the principles involved in this version of the Mean Value Theorem in the following example.
Consider . Find a value guaranteed by the Mean Value Theorem.
SolutionWe first need to evaluate . (This was previously done in Example 5.4.3.)
Thus we seek a value in such that .
In Figure 5.4.3 is sketched along with a rectangle with height . The area of the rectangle is the same as the area under on .
Let be a function on with such that . Consider :
When is shifted by , the amount of area under above the -axis on is the same as the amount of area below the -axis above ; see Figure 5.4.4 for an illustration of this. In this sense, we can say that is the average value of on .
The value is the average value in another sense. First, recognize that the Mean Value Theorem can be rewritten as
for some value of in . Next, partition the interval into equally spaced subintervals, and choose any in . The average of the numbers , , …, is:
Multiply this last expression by 1 in the form of :
Now take the limit as :
This tells us this: when we evaluate at (somewhat) equally spaced points in , the average value of these samples is as .
This leads us to a definition.
Let be continuous on . The average value of on is , where is a value in guaranteed by the Mean Value Theorem. I.e.,
An application of this definition is given in the following example.
An object moves back and forth along a straight line with a velocity given by on , where is measured in seconds and is measured in ft/s.
What is the average velocity of the object?
SolutionBy our definition, the average velocity is:
We can understand the above example through a simpler situation. Suppose you drove 100 miles in 2 hours. What was your average speed? The answer is simple: displacement/time = 100 miles/2 hours = 50 mph.
What was the displacement of the object in Example 5.4.8? We calculate this by integrating its velocity function: ft. Its final position was 3 feet from its initial position after 3 seconds: its average velocity was 1 ft/s.
This section has laid the groundwork for a lot of great mathematics to follow. The most important lesson is this: definite integrals can be evaluated using antiderivatives. Since the previous section established that definite integrals are the limit of Riemann sums, we can later create Riemann sums to approximate values other than “area under the curve,” convert the sums to definite integrals, then evaluate these using the Fundamental Theorem of Calculus. This will allow us to compute the work done by a variable force, the volume of certain solids, the arc length of curves, and more.
The downside is this: generally speaking, computing antiderivatives is much more difficult than computing derivatives. Much of our time in Chapter 8 will be devoted to techniques of finding antiderivatives so that a wide variety of definite integrals can be evaluated.
How are definite and indefinite integrals related?
What constant of integration is most commonly used when evaluating definite integrals?
T/F: If is a continuous function, then is also a continuous function.
The definite integral can be used to find “the area under a curve.” Give two other uses for definite integrals.
In Exercises 5–34., use the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral.
Explain why for all values of .
In Exercises 37–40., find a value guaranteed by the Mean Value Theorem.
In Exercises 41–46., find the average value of the function on the given interval.
on
on
on
on
on
on
In Exercises 47–50., a velocity function of an object moving along a straight line is given. Find (a) the displacement of the object over the given time interval and (b) the total distance traveled by the object over the given time interval.
ft/s on
ft/s on
ft/s on
ft/s on
In Exercises 51–54., an acceleration function of an object moving along a straight line is given. Find the change of the object’s velocity over the given time interval.
ft/s on
ft/s on
ft/s on
ft/s on
In Exercises 55–62., use the Fundamental Theorem of Calculus Part 1 to find .