5.2: The Definite Integral
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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
Definition and Notation
The definite integral generalizes the concept of the area under a curve. We lift the requirements that
Definition: Definite Integral
If
provided the limit exists. If this limit exists, the function
The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the
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
We call the function
Previously, we discussed the fact that if
Continuous Functions Are Integrable
If
Functions that are not continuous on
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.
Example 5 . 2 . 1 : Evaluating an Integral Using the Definition
Use the definition of the definite integral to evaluate
Solution
We first want to set up a Riemann sum. Based on the limits of integration, we have
Since we are using a right-endpoint approximation to generate Riemann sums, for each
Thus, the function value at the right endpoint of the interval is
Then the Riemann sum takes the form
Using the summation formula for
Now, to calculate the definite integral, we need to take the limit as
Exercise 5 . 2 . 1
Use the definition of the definite integral to evaluate
Use a right-endpoint approximation to generate the Riemann sum.
- Hint
-
Use the solving strategy from Example
.5 . 2 . 1
- 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
Example 5 . 2 . 2 : 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
![A graph of a semi circle in quadrant one over the interval [0,6] with center at (3,0). The area under the curve over the interval [3,6] is shaded in blue.](https://math.libretexts.org/@api/deki/files/12421/5.2.1.png?revision=1)
Exercise 5 . 2 . 2
Use the formula for the area of a trapezoid to evaluate
- Hint
-
Graph the function
and calculate the area under the function on the interval๐ โก ( ๐ฅ ) [ 2 , 4 ] .
- Answer
-
18 square units
Area and the Definite Integral
When we defined the definite integral, we lifted the requirement that
Net Signed Area
Let us return to the Riemann sum. Consider, for example, the function
![A graph of a downward opening parabola over [-1, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each.](https://math.libretexts.org/@api/deki/files/2586/CNX_Calc_Figure_05_02_003.jpeg?revision=1&size=bestfit&width=325&height=387)
Taking the limit as
The quantity
![A graph of a downward opening parabola over [-2, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). The area in quadrant one under the curve is shaded blue and labeled A1. The area in quadrant four above the curve and to the left of x=2 is shaded blue and labeled A2.](https://math.libretexts.org/@api/deki/files/2587/CNX_Calc_Figure_05_02_002.jpeg?revision=1&size=bestfit&width=325&height=350)
Notice that net signed area can be positive, negative, or zero. If the area above the
Example 5 . 2 . 3 : Finding the Net Signed Area
Find the net signed area between the curve of the function
Solution
The function produces a straight line that forms two triangles: one from
where
where
![A graph of an increasing line over [-6, 6] going through the origin and (-3, -6) and (3,6). The area under the line in quadrant one over [0,3] is shaded blue and labeled A1, and the area above the line in quadrant three over [-3,0] is shaded blue and labeled A2.](https://math.libretexts.org/@api/deki/files/12422/5.2.2.png?revision=1)
Analysis
If
Exercise 5 . 2 . 3
Find the net signed area of
- Hint
-
Use the solving method described in Example
.5 . 2 . 3
- Answer
-
6
Total Area
One application of the definite integral is finding displacement when given a velocity function. If
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
![A graph in quadrant 1 with the x-axis labeled as t (hours) and y-axis labeled as v (mi/hr). The area under the line v(t) = 75 is shaded blue over [0,2].](https://math.libretexts.org/@api/deki/files/2590/CNX_Calc_Figure_05_02_015.jpeg?revision=1&size=bestfit&width=566&height=352)
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 miles 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 (Figure
In this case the displacement is zero.
![A graph in quadrants one and four with the x-axis labeled as t (hours) and the y axis labeled as v (mi/hr). The first part of the graph is the line v(t) = 60 over [0,2], and the area under the line in quadrant one is shaded. The second part of the graph is the line v(t) = -40 over [2,5], and the area above the line in quadrant four is shaded.](https://math.libretexts.org/@api/deki/files/2591/CNX_Calc_Figure_05_02_016.jpeg?revision=1&size=bestfit&width=521&height=497)
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
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.
Definition: Net Signed Area
Let
The total area between
Example 5 . 2 . 4 : Finding the Total Area
Find the total area between
Solution
Calculate the

We have
Then, using the formula for the area of a triangle, we obtain
The total area, then, is
Exercise 5 . 2 . 4
Find the total area between the function
- Hint
-
Review the solving strategy in Example
.5 . 2 . 4
- Answer
-
1 8 u n i t s 2
Properties of the Definite Integral
The properties of indefinite integrals apply to definite integrals as well. Definite integrals also have properties that relate to the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals.
Rule: Properties of the Definite Integral
1.
If the limits of integration are the same, the integral is just a line and contains no area.
2.
If the limits are reversed, then place a negative sign in front of the integral.
3.
The integral of a sum is the sum of the integrals.
4.
The integral of a difference is the difference of the integrals.
5.
for constant
6.
Although this formula normally applies when
Example 5 . 2 . 5 : Using the Properties of the Definite Integral
Use the properties of the definite integral to express the definite integral of
Solution
Using integral notation, we have
Exercise 5 . 2 . 5
Use the properties of the definite integral to express the definite integral of
- Hint
-
Use the solving strategy from Example
and the properties of definite integrals.5 . 2 . 5
- Answer
-
6 โข โซ 3 1 ๐ฅ 3 ๐ ๐ฅ โ 4 โข โซ 3 1 ๐ฅ 2 ๐ ๐ฅ + 2 โข โซ 3 1 ๐ฅ ๐ ๐ฅ โ โซ 3 1 3 ๐ ๐ฅ
Example 5 . 2 . 6 : Using the Properties of the Definite Integral
If it is known that
Solution
By property 6,
Thus,
Exercise 5 . 2 . 6
If it is known that
- Hint
-
Use the solving strategy from Example
and the rule on properties of definite integrals.5 . 2 . 6
- Answer
-
โ 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
Comparison Theorem
i. If
ii. If
iii. If
Example 5 . 2 . 7 : Comparing Two Functions over a Given Interval
Compare
Solution
Graphing these functions is necessary to understand how they compare over the interval
![A graph of the function f(x) = sqrt(1 + x^2) in red and g(x) = sqrt(1 + x) in blue over [-2, 3]. The function f(x) appears above g(x) except over the interval [0,1]. A second, zoomed-in graph shows this interval more clearly.](https://math.libretexts.org/@api/deki/files/12425/5.2.5.png?revision=1)
We can see from the graph that over the interval
![A graph showing the functions f(x) = sqrt(1 + x^2) and g(x) = sqrt(1 + x) over [-3, 3]. The area under g(x) in quadrant one over [0,1] is shaded. The area under g(x) and f(x) is included in this shaded area. The second, zoomed-in graph shows more clearly that equality between the functions only holds at the endpoints.](https://math.libretexts.org/@api/deki/files/12426/5.2.6.png?revision=1)
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 Bโ at most schools.
Suppose, however, that we have a function
Let
which is basically the same expression used to calculate the average of discrete values.
But we know
Following through with the algebra, the numerator is a sum that is represented as
This is a Riemann sum. Then, to get the exact average value, take the limit as
Definition: Average Value of a Function
Let
Example 5 . 2 . 8 : Finding the Average Value of a Linear Function
Find the average value of
Solution
First, graph the function on the stated interval, as shown in Figure
![A graph in quadrant one showing the shaded area under the function f(x) = x + 1 over [0,5].](https://math.libretexts.org/@api/deki/files/12427/5.2.7.png?revision=1)
The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid
Thus the average value of the function is
Exercise 5 . 2 . 7
Find the average value of
- Hint
-
Use the average value formula (Equation
), and use geometry to evaluate the integral.5 . 2 . 2
- Answer
-
3
Key Concepts
- The definite integral can be used to calculate net signed area, which is the area above the
-axis less the area below the๐ฅ -axis. Net signed area can be positive, negative, or zero.๐ฅ - 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
Glossary
- average value of a function
- (or
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 interval๐ ๐ โข ๐ฃ โข ๐ )
- definite integral
- a primary operation of calculus; the area between the curve and the
-axis over a given interval is a definite integral๐ฅ
- integrable function
- a function is integrable if the limit defining the integral exists; in other words, if the limit of the Riemann sums as
goes to infinity exists๐
- integrand
- the function to the right of the integration symbol; the integrand includes the function being integrated
- limits of integration
- these values appear near the top and bottom of the integral sign and define the interval over which the function should be integrated
- net signed area
- the area between a function and the
-axis such that the area below the๐ฅ -axis is subtracted from the area above the๐ฅ -axis; the result is the same as the definite integral of the function๐ฅ
- total area
- total area between a function and the
-axis is calculated by adding the area above the๐ฅ -axis and the area below the๐ฅ -axis; the result is the same as the definite integral of the absolute value of the function๐ฅ
- variable of integration
- indicates which variable you are integrating with respect to; if it is
, then the function in the integrand is followed by๐ฅ ๐ โข ๐ฅ