Lesson 1: Antiderivatives

Hi everyone! Your Calculus I class focused a lot on taking derivatives of functions. If you took MAT 1475 at CityTech, you saw antiderivatives near the end of the semester. This lesson is a refresher.

Learning goals:

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

Topic: Volume 1, Section 4.10 : Antiderivatives (link to textbook section).

WeBWorK:

Integration – Antiderivatives

WeBWorK Calculus I Review:

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Motivation

If we know the derivative $f'(x)$ of a function $f(x)$ , how can we find $f(x)$ and why might we want to?

Antiderivatives (also known as “going backwards”)

Warmup exercise 1

Differentiate $f(x) = 2x^4$ .

Show answer 1

Since $f(x)=2x^4$ is a constant multiple of a power of $x$ , we can use the power rule for derivatives and the constant multiple rule to differentiate $f(x)$ . So the derivative $f'(x)= 4*2x^{4-1} = 8x^3$ .

Example: In the warmup exercise, we saw that the functions $f(x)=2x^4$ and $f'(x) = 8x^3$ are related: the second function is the derivative of the first function. But we can express this relationship in another way: the function $f(x) = 2x^4$ is an antiderivative of the function $f'(x)=8x^3$ .

Notation: Usually, we’ll use use different notation to indicate this relationship: let $f(x)$ represent a given function and let $F(x)$ represent an antiderivative of $f(x)$ .

Example: In the warmup exercise, if $f(x) = 8x^3$ , then $F(x) = 2x^4$ is an antiderivative of $f(x)$ . This is because $F'(x) = 8x^3$ , which is equal to $f(x)$ .

Definition: An antiderivative of a function $f(x)$ is a function $F(x)$ whose derivative is $f(x)$ . That is, $F'(x) = f(x)$ for all $x$ values in the domain of $f$ .

Indefinite integration (a.k.a. so many antiderivatives!)

Exercise 2

Let $f(x) = 8x^3$ . We saw that $F(x) = 2x^4$ is an antiderivative of $f(x)$ . Can you find another antiderivative of $f(x)$ ?

Answer 2

We saw that $F(x) = 2x^4$ is an antiderivative of $f(x)$ because $F'(x) = f(x)$ . So the question is asking us to find another function $F(x)$ so that its derivative $F'(x)=8x^3$ . If we remember that the derivative of any constant is zero, then it’s easy to make new antiderivatives out of old ones. For example, since $\frac{d}{dx}(1)=0$ , then if we let $F(x) = 2x^4 +1$ , then its derivative $F'(x) = 8x^3$ , so our old $F(x)$ and our new $F(x)$ have the same derivative.

Example: In the answer for exercise 2, we saw how easy it is to make new antiderivatives out of old ones: just add any constant. The graph linked here shows a whole family of antiderivatives of $f(x)=8x^3$ ; drag the slider for the constant $C$ to see how the graphs are related.

We are often interested not in just one antiderivative of a function $f(x)$ , but in all possible antiderivatives of $f(x)$ . Exercise 2 shows us that if we have one antiderivative $F(x)$ of $f(x)$ , then we can obtain another one by adding a constant. In fact, any antiderivative of $f(x)$ can be obtained in this way.

So if $F(x)$ is any antiderivative of $f(x)$ , then the family of all possible antiderivatives of $f(x)$ can be written as $F(x) + C$ where $C$ stands for all possible constants. Sometimes this is called the general antiderivative of $f(x)$ .

Example: In our above examples, if $f(x) = 8x^3$ , then the general antiderivative is written as $2x^4+C$ , where $C$ stands for all possible constants.

Definition: Let $F(x)$ be any antiderivative of $f(x)$ . Then the indefinite integral of $f(x)$ with respect to the variable $x$ is written as $\int f(x) dx$ . The indefinite integral represents the general antiderivative, so $\int f(x) dx = F(x) + C$ where $C$ stands for all possible constants. The act of finding an indefinite integral is called integration or antidifferentiation.

Exercise 3

Evaluate the indefinite integral:

$\int 8x^3 dx.$

Answer 3

Since the indefinite integral is the general antiderivative, and since we know $F(x) = 2x^4$ is an antiderivative of $f(x) = 8x^3$ , that means that $\int 8x^3 dx = 2x^4 + C$ , where $C$ stands for all possible constants.

Video 1 below introduces the main concepts outlined above, outlines basic integration rules, and shows a few more examples. Pay close attention to what happens when the variable of integration is not $x$ . Compare the integration rules to the ones in your textbook (linked here) in table 4.13 and pay special attention to the language used (for example, notice how “integrand” and “variable of integration” are used).

Video 1

Pro tip: you should always check your antiderivatives by differentiating them.

Examples of indefinite integration

The next few short videos show more examples of indefinite integration. Over the next several sessions, you’ll be learning techniques for integrating much more complicated functions than these basic ones.

Video 2 below shows another few examples of indefinite integration of some basic functions.

Video 2

Video 3 below shows an example of indefinite integration with a negative exponent.

Video 3

Video 4 below shows an example of indefinite integration of a product.

Video 4

Video 5 below shows an example with a variety of kinds of terms.

Video 5

Indefinite integration of basic trigonometric functions follows the same rules as the examples above; but you have to remember the derivatives of the six elementary trigonometric functions so that you can recognize them. Video 6 below walks you through some examples.

Video 6

Initial value problems (a.k.a. only one antiderivative)

Recall that when you’re finding an indefinite integral of a function $f(x)$ , you’re finding all possible antiderivatives of $f(x)$ . At times, you may need to find one particular antiderivative that satisfies a certain condition. Usually, that condition is just that the graph of the antiderivative must pass through a certain point; we call this point the initial value.

Exercise 4

Find the antiderivative $F(x)$ of $f(x)=8x^3$ that satisfies $F(1)=7$ .

Answer 4

We saw in exercise 3 that the indefinite integral (which is the general antiderivative) $\int 8x^3 dx = 2x^4 + C$ where $C$ represents all possible constants. That means that any particular antiderivative $F(x)$ has the form $2x^4 + C$ for some particular constant $C$ . So once we find the value of $C$ so that $F(1)=7$ , we’ll have our answer.

Since $F(x) = 2x^4+C$ , $F(1) = 2*1^4 + C=2+C$ . Since we’re given that $F(1)=7$ , we have to solve the equation $7=2+C$ for the variable $C$ . By subtracting $2$ from both sides, we see that $7-2 = C$ , so $C=5$ .

That means that the antiderivative $F(x)$ of $f(x)=8x^3$ that satisfies $F(1)=7$ is $F(x) = 2x^4 + 5$ .

Example: You can see how this initial value problem in exercise 4 works graphically as well. Go back to the graph linked here (this is the same graph you saw before). If you drag the $C$ slider until the red graph goes through the green point at $(1,7)$ , you’ll see that the $C$ value is $5$ .

Examples of initial value problems

Videos 7 and 8 below show how to solve two more initial value problems. The ends of the videos show a picture for how this applies to “slope fields for differential equations.” Slope fields are outside the scope of this course, but understanding that picture will help you see how initial value problems relate to one of the main problems of your Calculus I class: finding the slope of the tangent line.

Video 7

Video 8

Some initial value problems involve the second derivative of a function instead of just the first. An example illustrating an application of this type of problem in real life can be found in Video 8 below.

Applications (a.k.a. no really, this is real life!)

We can find many applications of indefinite integration and initial value problems whenever quantities are changing. Physics is a great place to find meaningful applications. Recall from your Calculus I class that if $s(t)$ is the position of an object at time $t$ , then

the instantaneous rate of change of position is instantaneous velocity $v(t)$ , so $s'(t) = v(t)$ ,
the instantaneous rate of change of velocity is acceleration $a(t)$ , so $v'(t) = a(t)$ and $s''(t) = a(t)$ .

Video 8 below shows an example of an initial value problem with constant acceleration and the initial values are used to find the velocity and position functions. Notice that since you’re given the second derivative of the function you’re interested in, you have to solve two initial value problems.