Learning Objectives
 State the definition of the definite integral.
 Explain the terms integrand, limits of integration, and variable of integration.
 Explain when a function is integrable.
 Describe the relationship between the definite integral and net area.
 Use geometry and the properties of definite integrals to evaluate them.
 Calculate the average value of a function.
In the preceding section we defined the area under a curve in terms of Riemann sums:
However, this definition came with restrictions. We required to be continuous and nonnegative. Unfortunately, realworld problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area under the curve to a broader set of functions through the use of the definite integral.
Definition and Notation
The definite integral generalizes the concept of the area under a curve. We lift the requirements that be continuous and nonnegative, and define the definite integral as follows.
Definition
If is a function defined on an interval the definite integral of f from a to b is given by
provided the limit exists. If this limit exists, the function is said to be integrable on or is an integrable function.
Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, who is often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S, suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval, The numbers a and b are xvalues and are called the limits of integration; specifically, a is the lower limit and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as Second, the boundaries of the region are called the limits of integration.
We call the function the integrand, and the dx indicates that is a function with respect to x, called the variable of integration. Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:
Previously, we discussed the fact that if is continuous on then the limit exists and is unique. This leads to the following theorem, which we state without proof.
If is continuous on then f is integrable on
Functions that are not continuous on may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable.
It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.
Evaluating an Integral Using the Definition
Use the definition of the definite integral to evaluate Use a rightendpoint approximation to generate the Riemann sum.
Solution
We first want to set up a Riemann sum. Based on the limits of integration, we have and For let be a regular partition of Then
Since we are using a rightendpoint approximation to generate Riemann sums, for each i, we need to calculate the function value at the right endpoint of the interval The right endpoint of the interval is and since P is a regular partition,
Thus, the function value at the right endpoint of the interval is
Then the Riemann sum takes the form
Using the summation formula for we have
Now, to calculate the definite integral, we need to take the limit as We get
Use the definition of the definite integral to evaluate Use a rightendpoint approximation to generate the Riemann sum.
Answer
6
Evaluating Definite Integrals
Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the xaxis.
Using Geometric Formulas to Calculate Definite Integrals
Use the formula for the area of a circle to evaluate
Solution
The function describes a semicircle with radius 3. To find
we want to find the area under the curve over the interval The formula for the area of a circle is The area of a semicircle is just onehalf the area of a circle, or The shaded area in Figure 1 below covers onehalf of the semicircle, or Thus,
Use the formula for the area of a trapezoid to evaluate
Answer
18 square units
Hint
Graph the function and calculate the area under the function on the interval
Area and the Definite Integral
When we defined the definite integral, we lifted the requirement that be nonnegative. But how do we interpret “the area under the curve” when is negative?
Net Signed Area
Let us return to the Riemann sum. Consider, for example, the function , whose graph is shown in Figure 2 below, on the interval Use and choose as the left endpoint of each interval. Construct a rectangle on each subinterval of height and width Δx. When is positive, the product represents the area of the rectangle, as before. When is negative, however, the product represents the negative of the area of the rectangle. The Riemann sum then becomes
Taking the limit as the Riemann sum approaches the area between the curve above the xaxis and the xaxis, less the area between the curve below the xaxis and the xaxis, as shown in Figure 3 below. Then,
Notice that net signed area can be positive, negative, or zero. If the area above the xaxis is larger, the net signed area is positive. If the area below the xaxis is larger, the net signed area is negative. If the areas above and below the xaxis are equal, the net signed area is zero.
Finding the Net Signed Area
Find the net signed area between the curve of the function and the xaxis over the interval
Solution
The function produces a straight line that forms two triangles: one from to and the other from to , see Figure 4 below . Using the geometric formula for the area of a triangle, the area of triangle A_{1}, above the axis, is
where 3 is the base and is the height. The area of triangle A_{2}, below the axis, is
where 3 is the base and 6 is the height. Thus, the net area is
Analysis
If A_{1} is the area above the xaxis and A_{2} is the area below the xaxis, then the net area is Since the areas of the two triangles are equal, the net area is zero.
Find the net signed area of over the interval illustrated in the following image.
Answer
6
Total Area
One application of the definite integral is finding displacement when given a velocity function. If represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. This is a very important application of the definite integral, and we examine it in more detail later in the chapter. For now, we’re just going to look at some basics to get a feel for how this works by studying constant velocities.
When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car travels away from its starting position in a straight line at a speed of 70 mph for 2 hours, then it is 140 mi away from its original position, see Figure 5 below. Using integral notation, we have
In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at a speed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at a speed of 40 mph for 3 hours, it will be back at it starting position, see Figure 6 below. Again, using integral notation, we have
In this case the displacement is zero.
Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area between the curve and the xaxis, regardless of whether that area is above or below the axis. This is called the total area.
Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis (rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is
Bringing these ideas together formally, we state the following definitions.
Let be an integrable function defined on an interval Let A_{1} represent the area between and the xaxis that lies above the axis and let A_{2} represent the area between and the xaxis that lies below the axis. Then, the net signed area between and the xaxis is given by
The total area between and the xaxis is given by
Finding the Total Area
Find the total area between and the xaxis over the interval
Solution
Calculate the xintercept as (set solve for x). To find the total area, take the area below the xaxis over the subinterval and add it to the area above the xaxis on the subinterval , see Figure 7 below.
We have
Then, using the formula for the area of a triangle, we obtain
The total area, then, is
Find the total area between the function and the xaxis over the interval
Answer
18
Properties of the Definite Integral
Definite integrals follow the natural constant multiple, sum, and difference rules we have for sums and limits. There are also properties of definite integrals that relate to the limits of integration. Together, these properties help us manipulate expressions to evaluate definite integrals.
Properties of the Definite Integral
Suppose that the functions and are integrable over all given intervals.

If the limits of integration are the same, the region is just a line segment that has zero area.

If the limits are reversed, then place a negative sign in front of the integral.

The integral of a sum is the sum of the integrals.

The integral of a difference is the difference of the integrals.

The integral of the product of a constant and a function is equal to the constant multiplied by the integral of the function.

Although this formula normally applies when c is between a and b, the formula holds for all values of a, b, and c, provided is integrable on the largest interval.
Using the Properties of the Definite Integral
Use the properties of the definite integral to express the definite integral of over the interval as the sum of three definite integrals.
Solution
Using integral notation, we have We apply properties 3 and 5 to get
Use the properties of the definite integral to express the definite integral of over the interval as the sum of four definite integrals.
Answer
Using the Properties of the Definite Integral
If it is known that and find the value of
Solution
By property 6,
Thus,
If it is known that and find the value of
Answer
−7
Comparison Properties of Integrals
A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphs as well as by their algebraic expressions can often give new insight into the process of integration. Intuitively, we might say that if a function is above another function then the area between and the xaxis is greater than the area between and the xaxis. This is true depending on the interval over which the comparison is made. The properties of definite integrals are valid whether or The following properties, however, concern only the case and are used when we want to compare the sizes of integrals.
Comparison Theorem
 Suppose that the functions and are integrable over the interval . If for then
 If for then
 If m and M are constants such that for then
Comparing Integrals over a Given Interval
Compare the integrals of the functions and over the interval
Solution
We start by comparing the functions and when Since and for , comparing and is equivalent to comparing the expressions and under the roots on . We consider the difference of these expressions:
Since and on , we have that on . It follows that on , and hence
, .
Since both functions and are continuous on , they are integrable over this interval, and we can apply the comparison theorem to conclude that .
Average Value of a Function
We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the following test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you want to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores. In this case, there are six test scores. Thus,
Therefore, your average test grade is approximately 80.33, which translates to a B− at most schools.
Suppose, however, that we have a function that gives us the speed of an object at any time t, and we want to find the object’s average speed. The function takes on an infinite number of values, so we can’t use the process just described. Fortunately, we can use a definite integral to find the average value of a function such as this.
Let be continuous over the interval and let be divided into n subintervals of width Choose a representative in each subinterval and calculate for In other words, consider each as a sampling of the function over each subinterval. The average value of the function may then be approximated as
which is basically the same expression used to calculate the average of discrete values.
But we know so and we get
Following through with the algebra, the numerator is a sum that is represented as and we are dividing by a fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value of the function is given by
This is a Riemann sum. Then, to get the exact average value, take the limit as n goes to infinity. Thus, the average value of a function is given by
Definition
Let be continuous over the interval Then, the average value of the function (denoted by f_{ave}) on is given by
Finding the Average Value of a Linear Function
Find the average value of over the interval
Solution
First, graph the function on the stated interval, as shown in below.
The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid where h represents height, and a and b represent the two parallel sides. Then,
Thus the average value of the function is
Find the average value of over the interval
Answer
3
Hint
Use the average value formula, and use geometry to evaluate the integral.
Key Concepts
 The definite integral can be used to calculate net signed area, which is the area above the xaxis less the area below the xaxis. Net signed area can be positive, negative, or zero.
 The component parts of the definite integral are the integrand, the variable of integration, and the limits of integration.
 Continuous functions on a closed interval are integrable. Functions that are not continuous may still be integrable, depending on the nature of the discontinuities.
 The properties of definite integrals can be used to evaluate integrals.
 The area under the curve of many functions can be calculated using geometric formulas.
 The average value of a function can be calculated using definite integrals.
Key Equations
 Definite Integral
 Properties of the Definite Integral
for constant c
Exercises
In the following exercises, given L_{n} or R_{n} as indicated, express their limits as as definite integrals. (Note that there exist more than one correct answer.)
1.
Answer
2.
3.
Answer
4.
5.
Answer
6.
7.
Answer
In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the xaxis.
Answer
Answer
Answer
In the following exercises, evaluate the integral using area formulas.
14.
15.
Answer
16.
17.
Answer
The integral is the area of the triangle, 9.
18.
19.
Answer
The integral is the area
20.
21.
Answer
The integral is the area of the “big” triangle less the “missing” triangle,
Suppose that , , and In the following exercises, compute the given integrals.
22.
23.
Answer
24.
25.
Answer
26.
27.
Answer
In the following exercises, given that and compute the integrals.
28.
29.
Answer
30.
31.
Answer
32.
33.
Answer
In the following exercises, use the comparison property of the definite integrals.
34. Show that
35. Show that
Answer
The integrand is negative over
36. Show that
37. Show that
Answer
over so over
38. Show that
39. Show that
Answer
Multiply by the length of the interval to get the inequality.
In the following exercises, use the interpretation of the definite integral as a net area to find the average value f_{ave} of f between a and b.
40.
41.
Answer
42.
43.
Answer
44.
45.
Answer
46. Let and Find the values of and by first showing that and
47. Let and Show that
Answer
Since ,
48. Explain why the graphs of a quadratic function (parabola) and a linear function can intersect in at most two points. Suppose that and and that Explain why whenever
49. Suppose that parabola opens downward and has a vertex with a positive coordinate. For which interval is as large as possible?
Answer
The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, and
50. Suppose can be subdivided into subintervals such that either over or over Set
 Explain why
 Then, explain why
51. Suppose f and g are continuous functions such that for every subinterval of Explain why for all values of x.
Answer
If for some then since is continuous, there is an interval containing t_{0} such that over the interval and then over this interval.
52. Suppose the average value of f over is 1 and the average value of f over is 1 where Show that the average value of f over is also 1.
53. Suppose that is a partition of such that the average value of f over each subinterval is equal to 1, Explain why the average value of f over is also equal to 1.
Answer
The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, Dividing through by gives the desired identity.
54. Suppose that for each i such that one has Show that
55. Suppose that for each i such that one has Show that
Answer
56. If what is
57. Estimate using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value
Answer
The average is which is equal to the integral in this case.
58. Estimate by comparison with the area of a single rectangle with height equal to the value of t at the midpoint How does this midpoint estimate compare with the actual value
59. From the graph of shown:
 Explain why
 Explain why, in general, for any value of a.
Answer
a. The graph is antisymmetric with respect to over so the average value is zero. b. For any value of a, the graph between is a shift of the graph over so the net areas above and below the axis do not change and the average remains zero.
60. If f is 1periodic odd, and integrable over is it always true that
61. If f is 1periodic and is it necessarily true that for all A?
Answer
Yes, the integral over any interval of length 1 is the same.
Glossary
 the average value of a function on an interval can be found by calculating the definite integral of the function and dividing that value by the length of the interval
 a primary operation of calculus; the area between the curve and the xaxis over a given interval is a definite integral
 a function is integrable if the limit defining the integral exists; in other words, if the limit of the Riemann sums as n goes to infinity exists
 the function to the right of the integration symbol; the integrand includes the function being integrated
 these values appear near the top and bottom of the integral sign and define the interval over which the function should be integrated
 the area between a function and the xaxis such that the area below the xaxis is subtracted from the area above the xaxis; the result is the same as the definite integral of the function
 total area between a function and the xaxis is calculated by adding the area above the xaxis and the area below the xaxis; the result is the same as the definite integral of the absolute value of the function
 indicates which variable you are integrating with respect to; if it is x, then the function in the integrand is followed by dx