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.

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 $-2$ft/s.) How far away from the starting point is the object — what is its displacement?

Here we use “Displacement $=$ Rate$_1$ × Time$_1$ + Rate$_2$ × Time$_2\text{,}$” which is \begin{equation*} \text{ Displacement } \ = 5\cdot10 + (-2)\cdot 4 = 42\text{ ft. } \end{equation*}

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 $[10,14]\text{,}$ 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.

Now consider a more difficult problem.

##### Example5.2.3Finding position using velocity

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

1. What was the initial velocity of the object?

2. What was the maximum height of the object?

3. What was the height of the object at time $t=2\text{?}$

Solution

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

Given a graph of a function $y=f(x)\text{,}$ 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.

##### Definition5.2.5The Definite Integral, Total Signed Area

Let $y=f(x)$ be defined on a closed interval $[a,b]\text{.}$ The total signed area from $x=a$ to $x=b$ under $f$ is:

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

The definite integral of $f$ on $[a,b]$ is the total signed area of $f$ on $[a,b]\text{,}$ denoted \begin{equation*} \int_a^b f(x)\ dx, \end{equation*} where $a$ and $b$ are the bounds of integration.

By our definition, the definite integral gives the “signed area under $f\text{.}$” 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 related to antiderivatives. We have now defined the definite integral, which relates to areas under a function. 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 “$\int$” 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 $f\text{.}$

We practice using this notation.

##### Example5.2.6Evaluating definite integrals

Consider the function $f$ given in Figure 5.2.7.

Find:

1. $\ds \int_0^3 f(x)\ dx$

2. $\ds \int_3^5 f(x)\ dx$

3. $\ds \int_0^5 f(x)\ dx$

4. $\ds \int_0^3 5f(x)\ dx$

5. $\ds \int_1^1 f(x) \ dx$

Solution

This example illustrates some of the properties of the definite integral, given here.

We give a brief justification of Theorem 5.2.9 here.

1. As demonstrated in Example 5.2.6, there is no “area under the curve” when the region has no width; hence this definite integral is 0.

2. This states that total area is the sum of the areas of subregions. It is easily considered when we let $a\lt b\lt c\text{.}$ We can break the interval $[a,c]$ into two subintervals, $[a,b]$ and $[b,c]\text{.}$ The total area over $[a,c]$ is the area over $[a,b]$ plus the area over $[b,c]\text{.}$ It is important to note that this still holds true even if $a\lt b\lt c$ is not true. We discuss this in the next point.

3. This property can be viewed a merely a convention to make other properties work well. (Later we will see how this property has a justification all its own, not necessarily in support of other properties.) Suppose $b\lt a\lt c\text{.}$ The discussion from the previous point clearly justifies \begin{equation} \int_b^a f(x)\ dx + \int_a^c f(x)\ dx = \int_b^c f(x)\ dx. \label{eq_defint1}\tag{5.2.1} \end{equation} However, we still claim that, as originally stated, \begin{equation} \int_a^b f(x)\ dx + \int_b^c f(x)\ dx = \int_a^c f(x)\ dx. \label{eq_defint2}\tag{5.2.2} \end{equation} How do Equations (5.2.1) and (5.2.2) relate? Start with Equation (5.2.1): \begin{align*} \int_b^a f(x)\ dx + \int_a^c f(x)\ dx \amp = \int_b^c f(x)\ dx\\ \int_a^c f(x)\ dx \amp = -\int_b^a f(x)\ dx + \int_b^c f(x)\ dx \end{align*} Property $(3)$ justifies changing the sign and switching the bounds of integration on the $\ds -\int_b^a f(x)\ dx$ term; when this is done, Equations (5.2.1) and (5.2.2) are equivalent. The conclusion is this: by adopting the convention of Property (3), Property (2) holds no matter the order of $a\text{,}$ $b$ and $c\text{.}$ Again, in the next section we will see another justification for this property.

4. Each of these may be non–intuitive. Property (5) states that when one scales a function by, for instance, 7, the area of the enclosed region also is scaled by a factor of 7. Both Properties (4) and (5) can be proved using geometry. The details are not complicated but are not discussed here.

##### Example5.2.10Evaluating definite integrals using Theorem 5.2.9.

Consider the graph of a function $f(x)$ shown in Figure 5.2.11.

1. Which value is greater: $\ds \int_a^b f(x)\ dx$ or $\ds \int_b^c f(x)\ dx\text{?}$

2. Is $\ds \int_a^c f(x)\ dx$ greater or less than 0?

3. Which value is greater: $\ds \int_a^b f(x)\ dx$ or $\ds \int_c^b f(x)\ dx\text{?}$

Solution

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

##### Example5.2.12Evaluating definite integrals using geometry

Evaluate the following definite integrals: \begin{equation*} 1. \ \int_{-2}^5 (2x-4)\ dx \qquad 2.\ \int_{-3}^3 \sqrt{9-x^2}\ dx. \end{equation*}

Solution
##### Example5.2.14Understanding motion given velocity

Consider the graph of a velocity function of an object moving in a straight line, given in Figure 5.2.15, 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.

Solution

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 5.2.16, where a region below $y=x^2$ is shaded. What is its area? The function $y=x^2$ is relatively simple, yet the shape it defines has an area that is not simple to find geometrically.

In Section 5.3 we will explore how to find the areas of such regions.

# Subsection5.2.1Exercises

Terms and Concepts

In the following exercises, a graph of a function $f(x)$ is given. Using the geometry of the graph, evaluate the definite integrals.

Review