5 Integration

5.2 The Definite Integral

We start with an easy problem. An object travels in a straight line at a constant velocity of 5 ft/s for 10 seconds. How far away from its starting point is the object?

We approach this problem with the familiar “Distance = Rate × Time” equation. In this case, Distance = 5ft/s × 10s = 50 feet.

It is interesting to note that this solution of 50 feet can be represented graphically. Consider Figure 5.2.1, where the constant velocity of 5ft/s is graphed on the axes. Shading the area under the line from t=0 to t=10 gives a rectangle with an area of 50 square units; when one considers the units of the axes, we can say this area represents 50 ft.

margin: 5105t (s)v (ft/s) Figure 5.2.1: The area under a constant velocity function corresponds to distance traveled. Λ

Now consider a slightly harder situation (and not particularly realistic): an object travels in a straight line with a constant velocity of 5ft/s for 10 seconds, then instantly reverses course at a rate of 2ft/s for 4 seconds. (Since the object is traveling in the opposite direction when reversing course, we say the velocity is a constant -2ft/s.) How far away from the starting point is the object — what is its displacement?

Here we use “Distance = Rate1 × Time1 + Rate2 × Time2,” which is

Distance =510+(-2)4=42 ft.

Hence the object is 42 feet from its starting location.

We can again depict this situation graphically. In Figure 5.2.2 we have the velocities graphed as straight lines on [0,10] and [10,14], respectively. The displacement of the object is

“Area above the t-axis  - Area below the t-axis,”

which is easy to calculate as 50-8=42 feet.

margin: 51015-25t (s)v (ft/s) Figure 5.2.2: The total displacement is the area above the t-axis minus the area below the t-axis. Λ

Now consider a more difficult problem.

Example 5.2.1 Finding position using velocity

The velocity of an object moving straight up/down under the acceleration of gravity is given as v(t)=-32t+48, where time t is given in seconds and velocity is in ft/s. When t=0, the object had a height of 0 ft.

  1. (a)

    What was the initial velocity of the object?

  2. (b)

    What was the maximum height of the object?

  3. (c)

    What was the height of the object at time t=2?

SolutionIt is straightforward to find the initial velocity; at time t=0, v(0)=-320+48=48 ft/s.

To answer questions about the height of the object, we need to find the object’s position function s(t). This is an initial value problem, which we studied in the previous section. We are told the initial height is 0, i.e., s(0)=0. We know s(t)=v(t)=-32t+48. To find s, we find the indefinite integral of v(t):


Since s(0)=0, we conclude that C=0 and s(t)=-16t2+48t.

To find the maximum height of the object, we need to find the maximum of s. Recalling our work finding extreme values, we find the critical points of s by setting its derivative equal to 0 and solving for t:


(Notice how we ended up just finding when the velocity was 0ft/s.) The first derivative test shows this is a maximum, so the maximum height of the object is found at


The height at time t=2 is now straightforward to compute: it is s(2)=32ft.

While we have answered all three questions, let’s look at them again graphically, using the concepts of area that we explored earlier.

Figure 5.2.3 shows a graph of v(t) on axes from t=0 to t=3. It is again straightforward to find v(0). How can we use the graph to find the maximum height of the object?

margin: 123-5050t (s)v (ft/s) Figure 5.2.3: A graph of v(t)=-32t+48; the shaded areas help determine displacement. Λ

Recall how in our previous work that the displacement of the object (in this case, its height) was found as the area under the velocity curve, as shaded in the figure. Moreover, the area between the curve and the t-axis that is below the t-axis counted as “negative” area. That is, it represents the object coming back toward its starting position. So to find the maximum distance from the starting point — the maximum height — we find the area under the velocity line that is above the t-axis, i.e., from t=0 to t=1.5. This region is a triangle; its area is

Area =12Base×Height=12×1.5s×48ft/s=36ft,

which matches our previous calculation of the maximum height.

Finally, to find the height of the object at time t=2, we calculate the total signed area under the velocity function from t=0 to t=2. This signed area is equal to s(2), the displacement (i.e., signed distance) from the starting position at t=0 to the position at time t=2. That is,

Displacement = Area above the t-axis - Area below t-axis.

The regions are triangles, and we find


This also matches our previous calculation of the height at t=2.

Notice how we answered each question in this example in two ways. Our first method was to manipulate equations using our understanding of antiderivatives and derivatives. Our second method was geometric: we answered questions looking at a graph and finding the areas of certain regions of this graph.

The above example does not prove a relationship between area under a velocity function and displacement, but it does indicate that there may be a relationship. Section 5.4 will fully establish fact that the area under a velocity function is displacement.

While Figure 5.2.2 shows the graph of a function that has a jump discontinuity, we will now focus on functions that are continuous. Given a graph of a continuous function y=f(x), we will find that there is great use in computing the area between the curve y=f(x) and the x-axis. Because of this, we need to define some terms. The total signed area from x=a to x=b under a continuous function f is

(area under f and above the x-axis on [a,b]) -

(area above f and under the x-axis on [a,b]).

Definition 5.2.1 The Definite Integral

Let y=f(x) be continuous on a closed interval [a,b]. The definite integral of f on [a,b] is the total signed area of f on [a,b], denoted


where a and b are the bounds of integration.

By our definition, the definite integral gives the “signed area under f.” We usually drop the word “signed” when talking about the definite integral, and simply say the definite integral gives “the area under f ” or, more commonly, “the area under the curve.”

The previous section introduced the indefinite integral, which is related to antiderivatives. We have now defined the definite integral, which relates to areas under a curve. The two are very much related, as we’ll see when we learn the Fundamental Theorem of Calculus in Section 5.4. Recall that earlier we said that the “” symbol was an “elongated S” that represented finding a “sum.” In the context of the definite integral, this notation makes a bit more sense, as we are adding up areas under the function f.

We practice using this notation.

Example 5.2.2 Evaluating definite integrals

Consider the function f given in Figure 5.2.4. Find: margin: 12345-11xy Figure 5.2.4: A graph of f(x) in Example 5.2.2. Λ

  1. (a)


  2. (b)


  3. (c)


  4. (d)


  5. (e)



  1. (a)

    03f(x)dx is the area under f on the interval [0,3]. This region is a triangle, so the area is 03f(x)dx=12(3)(1)=1.5.

  2. (b)

    35f(x)dx represents the area of the triangle found under the x-axis on [3,5]. The area is 12(2)(1)=1; since it is found under the x-axis, this is “negative area.” Therefore 35f(x)dx=-1.

  3. (c)

    05f(x)dx is the total signed area under f on [0,5]. This is 1.5+(-1)=0.5.

  4. (d)

    035f(x)dx is the area under 5f on [0,3]. This is sketched in Figure 5.2.5. Again, the region is a triangle, with height 5 times that of the height of the original triangle. Thus the area is 035f(x)dx=15/2=7.5. margin: 12345-55xy Figure 5.2.5: A graph of 5f in Example 5.2.2. (Yes, it looks just like the graph of f in Figure 5.2.4, just with a different y-scale.) Λ

  5. (e)

    11f(x)dx is the area under f on the “interval” [1,1]. This describes a line segment, not a region; it has no width. Therefore the area is 0.

This example illustrates some of the properties of the definite integral, given in Theorem 5.2.1.

So far, when we have computed a definite integral abf(x)dx, we have required that ab. In practice, it is sometimes convenient to be able to compute abf(x)dx for a>b. To do so, we introduce the convention that for any a and b, abf(x)dx=-baf(x)dx. It will be clear why this makes sense after we introduce Riemann sums.

Theorem 5.2.1 Properties of the Definite Integral

Let f and g be continuous on a closed interval I that contains the values a, b, and c, and let k be a constant. The following hold:

  1. (a)


  2. (b)


  3. (c)


  4. (d)


  5. (e)


We will justify these properties after introducing Riemann sums. For now, we note that properties 1 and 5 are illustrated in Example 5.2.2 and property 2 is our convention from above. To see why property 3 makes sense geometrically, consider the figure below:


Property 3 says that the total area under this curve should be the sum of the area under the curve from a to b and the area under the curve from b to c.

What if the picture were like the following?


Then we have


and we can apply property 2.

acf(x)dx =abf(x)dx-cbf(x)dx,so property 2 yields
acf(x)dx =abf(x)dx+bcf(x)dx
Example 5.2.3 Evaluating definite integrals using Theorem 5.2.1.

Consider the graph of a function f(x) shown in Figure 5.2.6. Answer the following: margin: abcxy Figure 5.2.6: A graph of a function in Example 5.2.3. Λ

  1. (a)

    Which value is greater: abf(x)dx or bcf(x)dx?

  2. (b)

    Is acf(x)dx greater or less than 0?

  3. (c)

    Which value is greater: abf(x)dx or cbf(x)dx?


  1. (a)

    abf(x)dx has a positive value (since the area is above the x-axis) whereas bcf(x)dx has a negative value. Hence abf(x)dx is bigger.

  2. (b)

    acf(x)dx is the total signed area under f between x=a and x=c. Since the region below the x-axis looks to be larger than the region above, we conclude that the definite integral has a value less than 0.

  3. (c)

    Notice how the second integral has the bounds “reversed.” Therefore cbf(x)dx represents a positive number, greater than the area described by the first definite integral. Hence cbf(x)dx is greater.

The area definition of the definite integral allows us to use geometry to compute the definite integral of some simple functions.

Example 5.2.4 Evaluating definite integrals using geometry
margin: (-2,-8)(5,6)R1R2-225-10-5510xy (a)-335xy (b) Figure 5.2.7: A graph of f(x)=2x-4 in (a) and f(x)=9-x2 in (b), from Example 5.2.4. Λ

Evaluate the following definite integrals:

1.-25(2x-4)dx    2.-339-x2dx.


  1. (a)

    It is always useful to sketch the function in the integrand, as shown in Figure 5.2.7(a). We see we need to compute the areas of two regions, which we have labeled R1 and R2. Both are triangles, so the area computation is straightforward:

    R1:12(4)(8)=16  R2:12(3)6=9.

    Region R1 lies under the x-axis, hence it is counted as negative area (we can think of the triangle’s height as being “-8”), so

  2. (b)

    Recognize that the integrand of this definite integral describes a half circle, as sketched in Figure 5.2.7(b), with radius 3. Thus the area is:

Example 5.2.5 Understanding motion given velocity
margin: 111138-551015abct (s)y (ft/s) Figure 5.2.8: A graph of a velocity in Example 5.2.5. Λ

Consider the graph of a velocity function of an object moving in a straight line, given in Figure 5.2.8, where the numbers in the given regions gives the area of that region. Assume that the definite integral of a velocity function gives displacement. Find the maximum speed of the object and its maximum displacement from its starting position.

SolutionSince the graph gives velocity, finding the maximum speed is simple: it looks to be 15ft/s.

margin: Note: The displacement of the object is different from the distance traveled since the object moves backwards and forwards at different times in this example. The displacement measures how far the object is from where it started, without regard for how far it actually traveled to get there. Λ

At time t=0, the displacement is 0; the object is at its starting position. At time t=a, the object has moved backward 11 feet. Between times t=a and t=b, the object moves forward 38 feet, bringing it into a position 27 feet forward of its starting position. From t=b to t=c the object is moving backwards again, hence its maximum displacement is 27 feet from its starting position.

We can also analyze the displacement by drawing the path of the particle’s location as time varies, as in Figure 5.2.9.
016-1127 Figure 5.2.9: Number line for Example 5.2.5.
The object starts at the origin, and moves to the left with a negative velocity 11 units. It then reverses direction and moves to the right with a positive velocity 38 units, arriving at 27. Finally, it reverses direction again and moves to the left with a negative velocity 11 units, ending at 26.

margin: 123510xy Figure 5.2.10: What is the area below y=x2 on [0,3]? The region is not a usual geometric shape. Λ

In our examples, we have either found the areas of regions that have nice geometric shapes (such as rectangles, triangles and circles) or the areas were given to us. Consider Figure 5.2.10, where a region below y=x2 is shaded. What is its area? The function y=x2 is relatively simple, yet the shape it defines has an area that is not simple to find geometrically.

In the next section we will explore how to find the areas of such regions.

Exercises 5.2


Terms and Concepts

  1. 1.

    What is “total signed area”?

  2. 2.

    What is “displacement”?

  3. 3.

    What is 33sinxdx?

  4. 4.
    Give a single definite integral that has the same value as 01(2x+3)dx+12(2x+3)dx.


In Exercises 5–10., a graph of a function f(x) is given. Using the geometry of the graph, evaluate the definite integrals.

  1. 5.
    y=-2x+424-4-224xy (a) 01(-2x+4)dx (b) 02(-2x+4)dx (c) 03(-2x+4)dx (d) 13(-2x+4)dx (e) 24(-2x+4)dx (f) 01(-6x+12)dx
  2. 6.
    y=f(x)12345-2-112xy (a) 02f(x)dx (b) 03f(x)dx (c) 05f(x)dx (d) 25f(x)dx (e) 53f(x)dx (f) 03-2f(x)dx
  3. 7.
    y=f(x)123424xy (a) 02f(x)dx (b) 24f(x)dx (c) 242f(x)dx (d) 014xdx (e) 23(2x-4)dx (f) 23(4x-8)dx
  4. 8.
    y=x-11234-1123xy (a) 01(x-1)dx (b) 02(x-1)dx (c) 03(x-1)dx (d) 23(x-1)dx (e) 14(x-1)dx (f) 14((x-1)+1)dx
  5. 9.
    f(x)=4-(x-2)21234123xy (a) 02f(x)dx (b) 24f(x)dx (c) 04f(x)dx (d) 045f(x)dx
  6. 10.
    f(x)=3510123xy (a) 05f(x)dx (b) 37f(x)dx 0 (c) 00f(x)dx (d) abf(x)dx, where 0ab10

In Exercises 11–14., a graph of a function f(x) is given; the numbers inside the shaded regions give the area of that region. Evaluate the definite integrals using this area information.

  1. 11.
    y=f(x)591121123-100-5050xy (a) 01f(x)dx (b) 02f(x)dx (c) 03f(x)dx (a) 12-3f(x)dx (b) 02|f(x)|dx (c) 03|f(x)|dx
  2. 12.
    f(x)=sin(πx/2)4/π4/π1234-11xy (a) 02f(x)dx (b) 24f(x)dx (c) 04f(x)dx (a) 01f(x)dx (b) 02|f(x)|dx (c) 04|f(x)|dx
  3. 13.
    f(x)=3x2-3444-2-112-5510xy (a) -2-1f(x)dx (b) 12f(x)dx (c) -11f(x)dx (a) 01f(x)dx (b) 02|f(x)|dx (c) 01|f(x)|dx
  4. 14.
    f(x)=x21/37/3121234xy (a) 025x2dx (b) 02(x2+3)dx (a) 13(x-1)2dx (b) 24((x-2)2+5)dx

In Exercises 15–16., a graph of the velocity function of an object moving in a straight line is given. Answer the questions based on that graph.

  1. 15.
    123-112t (s)y (ft/s) (a) What is the object’s maximum velocity? (b) What is the object’s maximum displacement? (c) What is the object’s total displacement on [0,3]?
  2. 16.
    12345123t (s)y (ft/s) (a) What is the object’s maximum velocity? (b) What is the object’s maximum displacement? (c) What is the object’s total displacement on [0,5]?
  1. 17.
    An object is thrown straight up with a velocity, in ft/s, given by v(t)=-32t+64, where t is in seconds, from a height of 48 feet. (a) What is the object’s maximum velocity? (b) What is the object’s maximum displacement? (c) When does the maximum displacement occur? (d) When will the object reach a height of 0? (Hint: find when the displacement is -48ft.)
  2. 18.
    An object is thrown straight up with a velocity, in ft/s, given by v(t)=-32t+96, where t is in seconds, from a height of 64 feet. (a) What is the object’s initial velocity? (b) When is the object’s displacement 0? (c) How long does it take for the object to return to its initial height? (d) When will the object reach a height of 210 feet?

In Exercises 19–22., let

  • 02f(x)dx=5,

  • 03f(x)dx=7,

  • 02g(x)dx=-3, and

  • 23g(x)dx=5.

Use these values to evaluate the given definite integrals.

  1. 19.


  2. 20.


  3. 21.


  4. 22.

    Find non-zero values for a and b such that


In Exercises 23–26., let

  • 03s(t)dt=10,

  • 35s(t)dt=8,

  • 35r(t)dt=-1, and

  • 05r(t)dt=11.

Use these values to evaluate the given definite integrals.

  1. 23.


  2. 24.


  3. 25.


  4. 26.
    Find non-zero values for a and b such that 05(as(t)+br(t))dt=0
  1. 27.
    Show that for any real m and b>0 we have 0bmxdx=12mb2.
  2. 28.

    Look at signed areas to show that for any continuous function f and any a and b where a<b, we have that |abf(x)dx|ab|f(x)|dx.


In Exercises 29–32., evaluate the given indefinite integral.

  1. 29.


  2. 30.


  3. 31.


  4. 32.


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