4. Applications of Derivatives

# 4.10 Antiderivatives

### Learning Objectives

• Find the general antiderivative of a given function.
• Explain the terms and notation used for an indefinite integral.
• State the power rule for integrals.
• Use antidifferentiation to solve simple initial-value problems.

At this point, we have seen how to calculate derivatives of many functions and have been introduced to a variety of their applications. We now ask a question that turns this process around: Given a function how do we find a function with the derivative and why would we be interested in such a function?

We answer the first part of this question by defining antiderivatives. The antiderivative of a function is a function with a derivative Why are we interested in antiderivatives? The need for antiderivatives arises in many situations, and we look at various examples throughout the remainder of the text. Here we examine one specific example that involves rectilinear motion. In our examination in Derivatives of rectilinear motion, we showed that given a position function of an object, then its velocity function is the derivative of —that is, Furthermore, the acceleration is the derivative of the velocity —that is, Now suppose we are given an acceleration function but not the velocity function or the position function Since determining the velocity function requires us to find an antiderivative of the acceleration function. Then, since determining the position function requires us to find an antiderivative of the velocity function. Rectilinear motion is just one case in which the need for antiderivatives arises. We will see many more examples throughout the remainder of the text. For now, let’s look at the terminology and notation for antiderivatives, and determine the antiderivatives for several types of functions. We examine various techniques for finding antiderivatives of more complicated functions in the second volume of this text (Introduction to Techniques of Integration).

# The Reverse of Differentiation

At this point, we know how to find derivatives of various functions. We now ask the opposite question. Given a function how can we find a function with derivative If we can find a function derivative we call an antiderivative of

### Definition

A function is an antiderivative of the function if

for all in the domain of

Consider the function Knowing the power rule of differentiation, we conclude that is an antiderivative of since Are there any other antiderivatives of Yes; since the derivative of any constant is zero, is also an antiderivative of Therefore, and are also antiderivatives. Are there any others that are not of the form for some constant The answer is no. From Corollary 2 of the Mean Value Theorem, we know that if and are differentiable functions such that then for some constant This fact leads to the following important theorem.

### General Form of an Antiderivative

Let be an antiderivative of over an interval Then,

1. for each constant the function is also an antiderivative of over
2. if is an antiderivative of over there is a constant for which over

In other words, the most general form of the antiderivative of over is

We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions.

### Finding Antiderivatives

For each of the following functions, find all antiderivatives.

then is an antiderivative of Therefore, every antiderivative of is of the form for some constant and every function of the form is an antiderivative of

b. Let For and

For and

Therefore,

Thus, is an antiderivative of Therefore, every antiderivative of is of the form for some constant and every function of the form is an antiderivative of

c. We have

so is an antiderivative of Therefore, every antiderivative of is of the form for some constant and every function of the form is an antiderivative of

d. Since

then is an antiderivative of Therefore, every antiderivative of is of the form for some constant and every function of the form is an antiderivative of

Find all antiderivatives of

#### Hint

What function has a derivative of

# Indefinite Integrals

We now look at the formal notation used to represent antiderivatives and examine some of their properties. These properties allow us to find antiderivatives of more complicated functions. Given a function we use the notation or to denote the derivative of Here we introduce notation for antiderivatives. If is an antiderivative of we say that is the most general antiderivative of and write

The symbol is called an integral sign, and is called the indefinite integral of

### Definition

Given a function the indefinite integral of denoted

is the most general antiderivative of If is an antiderivative of then

The expression is called the integrand and the variable is the variable of integration.

Given the terminology introduced in this definition, the act of finding the antiderivatives of a function is usually referred to as integrating

For a function and an antiderivative the functions where is any real number, is often referred to as the family of antiderivatives of For example, since is an antiderivative of and any antiderivative of is of the form we write

The collection of all functions of the form where is any real number, is known as the family of antiderivatives of (Figure) shows a graph of this family of antiderivatives.

For some functions, evaluating indefinite integrals follows directly from properties of derivatives. For example, for

which comes directly from

This fact is known as the power rule for integrals.

### Power Rule for Integrals

For

Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in Appendix B.

Integration Formulas
Differentiation Formula Indefinite Integral
for

From the definition of indefinite integral of we know

if and only if is an antiderivative of Therefore, when claiming that

it is important to check whether this statement is correct by verifying that

### Verifying an Indefinite Integral

Each of the following statements is of the form Verify that each statement is correct by showing that

#### Solution

1. Since

the statement

is correct.
Note that we are verifying an indefinite integral for a sum. Furthermore, and are antiderivatives of and respectively, and the sum of the antiderivatives is an antiderivative of the sum. We discuss this fact again later in this section.

2. Using the product rule, we see that

Therefore, the statement

is correct.
Note that we are verifying an indefinite integral for a product. The antiderivative is not a product of the antiderivatives. Furthermore, the product of antiderivatives, is not an antiderivative of since

In general, the product of antiderivatives is not an antiderivative of a product.

Verify that

#### Hint

Calculate

In (Figure), we listed the indefinite integrals for many elementary functions. Let’s now turn our attention to evaluating indefinite integrals for more complicated functions. For example, consider finding an antiderivative of a sum In (Figure)a. we showed that an antiderivative of the sum is given by the sum —that is, an antiderivative of a sum is given by a sum of antiderivatives. This result was not specific to this example. In general, if and are antiderivatives of any functions and respectively, then

Therefore, is an antiderivative of and we have

Similarly,

In addition, consider the task of finding an antiderivative of where is any real number. Since

for any real number we conclude that

These properties are summarized next.

### Properties of Indefinite Integrals

Let and be antiderivatives of and respectively, and let be any real number.

Sums and Differences

Constant Multiples

From this theorem, we can evaluate any integral involving a sum, difference, or constant multiple of functions with antiderivatives that are known. Evaluating integrals involving products, quotients, or compositions is more complicated (see (Figure)b. for an example involving an antiderivative of a product.) We look at and address integrals involving these more complicated functions in Introduction to Integration. In the next example, we examine how to use this theorem to calculate the indefinite integrals of several functions.

### Evaluating Indefinite Integrals

Evaluate each of the following indefinite integrals:

#### Solution

1. Using (Figure), we can integrate each of the four terms in the integrand separately. We obtain

From the second part of (Figure), each coefficient can be written in front of the integral sign, which gives

Using the power rule for integrals, we conclude that

2. Rewrite the integrand as

Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have

3. Using (Figure), write the integral as

Then, use the fact that is an antiderivative of to conclude that

4. Rewrite the integrand as

Therefore,

Evaluate

#### Hint

Integrate each term in the integrand separately, making use of the power rule.

# Initial-Value Problems

We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.

A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation

is a simple example of a differential equation. Solving this equation means finding a function with a derivative Therefore, the solutions of (Figure) are the antiderivatives of If is one antiderivative of every function of the form is a solution of that differential equation. For example, the solutions of

are given by

Sometimes we are interested in determining whether a particular solution curve passes through a certain point —that is, The problem of finding a function that satisfies a differential equation

is an example of an initial-value problem. The condition is known as an initial condition. For example, looking for a function that satisfies the differential equation

and the initial condition

is an example of an initial-value problem. Since the solutions of the differential equation are to find a function that also satisfies the initial condition, we need to find such that From this equation, we see that and we conclude that is the solution of this initial-value problem as shown in the following graph.

### Solving an Initial-Value Problem

Solve the initial-value problem

#### Solution

First we need to solve the differential equation. If then

Next we need to look for a solution that satisfies the initial condition. The initial condition means we need a constant such that Therefore,

The solution of the initial-value problem is

Solve the initial value problem

#### Hint

Find all antiderivatives of

Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function is the derivative of a position function and the acceleration is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.

### Decelerating Car

A car is traveling at the rate of 88 ft/sec mph) when the brakes are applied. The car begins decelerating at a constant rate of 15 ft/sec2.

1. How many seconds elapse before the car stops?
2. How far does the car travel during that time?

#### Solution

1. First we introduce variables for this problem. Let be the time (in seconds) after the brakes are first applied. Let be the acceleration of the car (in feet per seconds squared) at time Let be the velocity of the car (in feet per second) at time Let be the car’s position (in feet) beyond the point where the brakes are applied at time
The car is traveling at a rate of Therefore, the initial velocity is ft/sec. Since the car is decelerating, the acceleration is

The acceleration is the derivative of the velocity,

Therefore, we have an initial-value problem to solve:

Integrating, we find that

Since Thus, the velocity function is

To find how long it takes for the car to stop, we need to find the time such that the velocity is zero. Solving we obtain sec.

2. To find how far the car travels during this time, we need to find the position of the car after sec. We know the velocity is the derivative of the position Consider the initial position to be Therefore, we need to solve the initial-value problem

Integrating, we have

Since the constant is Therefore, the position function is

After sec, the position is ft.

Suppose the car is traveling at the rate of 44 ft/sec. How long does it take for the car to stop? How far will the car travel?

### Key Concepts

• If is an antiderivative of then every antiderivative of is of the form for some constant
• Solving the initial-value problem

requires us first to find the set of antiderivatives of and then to look for the particular antiderivative that also satisfies the initial condition.

For the following exercises, show that are antiderivatives of

1.

2.

3.

4.

5.

#### Solution

For the following exercises, find the antiderivative of the function.

6.

7.

8.

9.

#### Solution

For the following exercises, find the antiderivative of each function

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

#### Solution

For the following exercises, evaluate the integral.

26.

27.

28.

29.

30.

31.

32.

33.

#### Solution

34.

For the following exercises, solve the initial value problem.

35.

36.

37.

38.

39.

#### Solution

For the following exercises, find two possible functions given the second- or third-order derivatives.

40.

41.

#### Solution

42.

43.

44.

45. A car is being driven at a rate of 40 mph when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. How long before the car stops?

#### Solution

5.867 sec

46. In the preceding problem, calculate how far the car travels in the time it takes to stop.

47. You are merging onto the freeway, accelerating at a constant rate of 12 ft/sec2. How long does it take you to reach merging speed at 60 mph?

#### Solution

7.333 sec

48. Based on the previous problem, how far does the car travel to reach merging speed?

49. A car company wants to ensure its newest model can stop in 8 sec when traveling at 75 mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

#### Solution

13.75 ft/sec2

50. A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

For the following exercises, find the antiderivative of the function, assuming

51. [T]

52. [T]

53. [T]

54. [T]

55. [T]

#### Solution

56. [T]

For the following exercises, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false.

57. If is the antiderivative of then is the antiderivative of

#### Solution

True

58. If is the antiderivative of then is the antiderivative of

59. If is the antiderivative of then is the antiderivative of

#### Solution

False

60. If is the antiderivative of then is the antiderivative of

## Glossary

antiderivative
a function such that for all in the domain of is an antiderivative of
indefinite integral
the most general antiderivative of is the indefinite integral of we use the notation to denote the indefinite integral of
initial value problem
a problem that requires finding a function that satisfies the differential equation together with the initial condition