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## Section5.1Antiderivatives from Formulas

###### Motivating Questions
• What is an antiderivative of a function?

• If we know the equation of a function, can we find the equation of an antiderivative?

• What information do we need in order to find a single antiderivative of a function?

Up to this point, we have been focused on finding the derivative of a function. We will now examine how to find the function whose derivative is the function we are given. Our goal is the following: given a function $f(x)\text{,}$ can we find a function $F(x)$ such that $\displaystyle \frac{dF}{dx}=f(x)\text{?}$

###### The Antiderivative of a Function

An antiderivative of a function $f(x)$ is any function $F(x)$ such that

\begin{equation*} \frac{dF}{dx}=f(x). \end{equation*}

That is, $f(x)$ is the derivative of $F(x)\text{.}$

###### Example5.1

Consider the function $f(x)=2x\text{,}$ can we think of a possible antiderivative?

Solution

A possible antiderivative of this is $F(x)=x^2\text{,}$ since

\begin{equation*} F'(x)=2x=f(x). \end{equation*}

However, the antiderivative is not unique, and there are infinitely many antiderivatives of $f \text{.}$ Another possible antiderivative could be $F(x)=x^2+4$ since again $F'(x)=2x=f(x)\text{.}$ Note that any constant can be added to $x^2$ since the derivative of a constant is 0! Thus the most general antiderivative is

\begin{equation*} F(x)=x^2+C \end{equation*}

for a general constant C.

###### The Integral of a Function

If $F(x)$ is any antiderivative of a function $f(x)\text{,}$ then the indefinite integral of $f(x)$ is

\begin{equation*} \int f(x)dx=F(x)+C, \end{equation*}

where $C$ is called the constant of integration.

The general problem of finding an antiderivative is difficult. In part, this is due to the fact that we are trying to undo the process of differentiating, and the undoing is much more difficult than the doing. For example, while it is evident that an antiderivative of $f(x) = 2x$ is $F(x) = x^2$ and that an antiderivative of $g(x) = 3x^2$ is $G(x) = x^3\text{,}$ combinations of $f$ and $g$ can be far more complicated. This warrants the following question: in general, how do we find an antiderivative of a function given by a formula? We will begin to answer that question in this chapter.

### SubsectionBasic Antiderivatives

What is involved in trying to find an antiderivative for each function? From our experience with derivative rules, we know that derivatives of sums and constant multiples of basic functions are simple to execute, but derivatives involving products, quotients, and composites of familiar functions are more complicated. Therefore, it stands to reason that antidifferentiating products, quotients, and composites of basic functions may be even more challenging. We defer our study of all but the most elementary antiderivatives to later in the text.

###### Constant Rule

For any constant $k\text{:}$

\begin{equation*} \int k dx=kx+C \end{equation*}

###### Example5.2

When evaluating the integral, the variable of integration is indicated by the $dx\text{,}$ or $dy\text{,}$ or $dt$ at the end of the integral.

\begin{equation*} \int 4dt=4t+C \end{equation*}
\begin{equation*} \int \pi dy=\pi y+C \end{equation*}
\begin{equation*} \int \frac{1}{4}dz=\frac{1}{4}z+C \end{equation*}

###### Power Rule

For any $n\neq -1\text{:}$

\begin{equation*} \int x^n dx=\frac{x^{n+1}}{n+1}+C \end{equation*}

###### Example5.3

Evaluate each of the following integrals:

\begin{equation*} \int x^2dx=\frac{x^3}{3}+C \end{equation*}
\begin{equation*} \int y^{3.4} dy=\frac{y^{4.4}}{4.4}+C \end{equation*}

If the function is in the form $x^n$ but is equivalent to this form, you may have to rewrite the function as we did when taking the derivative:

\begin{equation*} \int \frac{1}{z^2}dz=\int z^{-2}dz=\frac{z^{-1}}{-1}+C=\frac{-1}{z}+C \end{equation*}
\begin{equation*} \int \sqrt{x}dx=\int x^{1/2}dx=\frac{z^{3/2}}{3/2}+C=\frac{2}{3}z^{3/2}+C \end{equation*}

###### Natural Log and Exponential Rules
\begin{equation*} \int \frac{1}{x} dx=\ln|x|+C \end{equation*}

It is important to include the absolute values in the natural log since the domain of the natural log is $x\gt0\text{.}$

Exponential Rule

\begin{equation*} \int e^{ax}dx=\frac{e^{ax}}{a}+C \end{equation*}

###### Example5.4
1. \begin{equation*} \int e^{5x}dx=\frac{e^{5x}}{5}+C \end{equation*}
2. \begin{equation*} \int e^{y/3} dy=\frac{e^{y/3}}{1/3}+C=3e^{y/3}+C \end{equation*}

Before proceeding any further it will be useful to state two properties.

###### Properties of Antiderivatives: Sums and Constant Multiples

Let $f$ and $g$ be functions that have an antiderivative and $c$ a constant. Then

\begin{equation*} \int\left(f(x) \pm g(x)\right) dx= \int f(x)dx \pm \int g(x)dx \end{equation*}

In words, the antiderivative of a sum or difference is the sum or difference of the antiderivatives. Or put another way, you can integrate term by term.

\begin{equation*} \int cf(x) dx = c\int f(x)dx \end{equation*}

In words, a multiplied constant stays.

We can now put all these properties together, along with the rules we have stated and evaluate some more difficult examples.

###### Example5.5

We will start by showing every step

\begin{align*} \int (4x^2+5x+6)dx\mathstrut \amp =\int 4x^2dx+\int 5xdx+\int 6dx \\ \mathstrut \amp =4\int x^2dx+5\int x dx+\int 6dx \\ \mathstrut \amp =4\frac{x^3}{3}+5\frac{x^2}{2}+6x+C \end{align*}

In general, we do not need to show the steps of splitting the integral into separate integrals, rather we will simply evaluate term by term as we did when finding the derivative. Similarly, we will not show the step of factoring out a constant, rather the multiplied constant will simply stay along. Unlike taking the derivative, we cannot often simplify the constant in one step.

###### Example5.6

Evaluate each of the following integrals using the rules we have developed.

1. \begin{equation*} \int (5x^3+6x-3)dx \end{equation*}
2. \begin{equation*} \int \left(4x^{2.3}+5\sqrt{x}+\frac{3}{x^2}\right)dx \end{equation*}
3. \begin{equation*} \int \left(\frac{4}{x}+4e^{8x}+6\right)dx \end{equation*}
Solution
1. \begin{equation*} \int (5x^3+6x-3)dx=\frac{5}{4}x^4+3x^2-3x+C \end{equation*}
2. First rewrite the terms to apply the power rule

\begin{equation*} \int \left(4x^{2.3}+5\sqrt{x}+\frac{3}{x^2}\right)dx=\int \left(4x^{2.3}+5x^{1/2}+3x^{-2}\right)dx \end{equation*}

Then apply the rules of integration

\begin{equation*} \int \left(4x^{2.3}+5x^{1/2}+3x^{-2}\right)dx=4\frac{x^{3.3}}{3.3}+5\frac{x^{3/2}}{3/2}-3x^{-1}+C \end{equation*}
3. \begin{equation*} \int \left(\frac{4}{x}+4e^{8x}+6\right)dx=4\ln|x|+\frac{e^{8x}}{2}+6x+C \end{equation*}

### SubsectionApplications of Integration

We will now examine how we can find a specific function $f(x)$ given its derivative $f'(x)\text{.}$ Suppose you know $f'(x)=x^2+4\text{.}$ To find $f(x) \text{,}$ we can integrate:

\begin{equation*} f(x)=\int f'(x) dx=\int (x^2+4)dx=\frac{x^3}{3}+4x+C. \end{equation*}

The issue of course is the $+C\text{.}$ We want to find a single function $f(x)\text{;}$ however, we have found many possible functions, one for each possible value of the C. In order to find a specific function, we need more information. In particular, we need to know the value of the function at a point. If this is given, this is called an initial condition.

###### Example5.7

Find the function $f(x)$ given that $f'(x)=x^2+4$ and $f(0)=3\text{.}$

First we integrate

\begin{equation*} f(x)=\int f'(x)dx=\int (x^2+4) dx=\frac{x^3}{3}+4x+C. \end{equation*}

Then we use the initial condition, $f(0)=3\text{,}$ to find the value of $C \text{:}$

\begin{equation*} f(0)=\frac{0^3}{3}+4(0)+C=3 \end{equation*}

We use this to solve for $C \text{,}$ for which we get $C=3\text{.}$ Thus we have our solution

\begin{equation*} f(x)=\frac{x^3}{3}+4x+3. \end{equation*}

From this example, note that the initial condition does not always have to be at $x=0\text{.}$ Furthermore, note that to find the value of $C$ we must solve an equation.

###### Example5.8

Find the function $f(x)$ given that $f'(x)=x^2+4$ and $f(3)=3\text{.}$

Solution

The first step of evaluating the integral would remain the same as in the previous example. The difference here is in finding the value of $C\text{.}$

First we integrate

\begin{equation*} f(x)=\int f'(x)dx=\int (x^2+4) dx=\frac{x^3}{3}+4x+C. \end{equation*}

Then we use the initial condition, $f(3) = 3 \text{,}$ to find the value of $C \text{:}$

\begin{equation*} f(3)=\frac{3^3}{3}+4(3)+C=3. \end{equation*}

Solving, we get $C=-21\text{,}$ so we now have:

\begin{equation*} f(x)=\frac{x^3}{3}+4x-21. \end{equation*}
###### Example5.9

Find the function $g(t)$ given that $g'(x)=6e^{3t}+\frac{2}{t}$ and $g(1)=5\text{.}$

Solution

First we integrate

\begin{equation*} g(t)=\int g'(t)dt=\int (6e^{3t}+\frac{2}{t}) dt=\frac{6e^{3t}}{3}+2\ln|t|+C, \end{equation*}

then we use the initial condition: $g(1)=5\text{,}$ to find the value of C:

\begin{equation*} g(1)=2e^{3(1)}+2\ln|1|+C=5 \end{equation*}

to solve for $C=5-2e^3$ and thus we have our solution

\begin{equation*} g(t)=2e^{3(t)}+2\ln|t|+5-2e^3 \end{equation*}

Now let's examine some word problems in which we must find the function given the rate of change (derivative).

###### Example5.10

The marginal cost of producing the next unit, after $x$ units of a product have already been produced, is given by

\begin{equation*} C'(x)=x^3-6x, \text{ (in dollars per unit). } \end{equation*}

Find the total cost function, $C(x)\text{,}$ given that the fixed costs are $\ 1200\text{.}$

\begin{equation*} C(x)=\frac{x^4}{4}-3x+1200 \end{equation*}
Solution

We are given the marginal cost, $C'(x) \text{,}$ so to find the cost function we must first integrate.

\begin{equation*} C(x)=\int\left(x^3-6x\right)dx=\frac{x^4}{4}+\frac{6x^2}{2}+C \end{equation*}

Then to find the value of $C$ we will use the fact that the fixed costs are $\ 1200\text{.}$ Since this is fixed, this is the same as saying $C(0)=1200\text{,}$ then we plug in $x=0$ to find the value of $C\text{.}$

\begin{equation*} C(0)=\frac{0^4}{4}+\frac{6(0)^2}{2}+C=1200 \implies C=1200 \end{equation*}

Thus we find the total cost function

\begin{equation*} C(x)=\frac{x^4}{4}-3x+1200. \end{equation*}
###### Example5.11

A company finds that the rate at which the quantity of their item that consumers demand changes with respect to the price and is given by the marginal demand function

\begin{equation*} D'(p)=-\frac{4000}{p^2}, \end{equation*}

where $p$ is the price per item, in dollars. Find the demand function if $1015$ items are demanded by consumers when the price is $\8$ per item.

\begin{equation*} D(p)=\frac{4000}{p}+515 \end{equation*}
Solution

We are given the marginal demand function, $D'(p) \text{,}$ so to find the demand function we must first integrate.

\begin{equation*} D(p)=\int\left(\frac{-4000}{p^2}\right)dp=\int\left(-4000p^{-2}\right)dp=4000p^{-1}+C=\frac{4000}{p}+C. \end{equation*}

Then to find the value of $C$ we will use the fact that $1015$ items are demanded by consumers when the price is $p=8\text{,}$ thus $D(8)=1015\text{.}$

\begin{equation*} D(8)=\frac{4000}{8}+C=1015 \implies C=1015-500=515 \end{equation*}

Thus the demand function is

\begin{equation*} D(p)=\frac{4000}{p}+515. \end{equation*}
###### Example5.12

The rate of change in Cheryl's pulse (in beats per minute per minute) $t$ minutes after she stops exercising is given by

\begin{equation*} R'(t)=-46.964e^{-0.796t}. \end{equation*}

Find $R(t)$ if Cheryl's pulse is $78$ beats per minute 2 minutes after she stopped exercising.

\begin{equation*} R(t)=59e^{-0.796t}+66 \end{equation*}
Solution

We are given the rate of change, $R'(t)\text{,}$ so to find $R(t)$ we must first integrate.

\begin{equation*} R(t)=\int\left(-46.964e^{-0.796t}\right)dt=-46.964\frac{e^{-0.796t}}{-0.796}+C=59e^{-0.796t}+C. \end{equation*}

Then to find the value of $C$ we will use the fact that Cheryl's pulse is $78$ beats per minute 2 minutes after she stopped exercising; that is, $R(2)=78\text{.}$

\begin{equation*} R(2)=59e^{-0.796(2)}+C=78 \implies C=66 \end{equation*}

So we find that

\begin{equation*} R(t)=59e^{-0.796t}+66 \end{equation*}

### SubsectionSummary

• An antiderivative of a function $f(x)$ is any function $F(x)$ such that $\frac{dF}{dx} = f(x) \text{.}$ In other words, it is a function whose derivative is $f(x) \text{.}$ If $F(x)$ is any antiderivative of a function $f(x) \text{,}$ then the indefinite integral of $f(x)$ is

\begin{equation*} \int f(x)dx = F(x) + C, \end{equation*}

where $C$ is called the constant of integration. For a function $f(x) \text{,}$ there can be many antiderivatives of $f(x) \text{.}$

• As with derivatives, there are antiderivative rules which tell us how to integrate familiar functions, such as constant functions, power functions, logs, and exponents. There are also rules on how to integrate sums and constant multiples of functions. Integrating products, quotients, and compositions of functions are a little more involved, and we will explore these in later sections.