4.9 Antiderivatives

In addition to finding derivatives, there is an important “inverse” problem: Given the derivative, find the function itself. For example, in physics we may know the velocity \(v(t)\) (the derivative) and wish to compute the position \(s(t)\) of an object. Since \(s'(t) = v(t)\), this amounts to finding a function whose derivative is \(v(t)\). A function \(F(x)\) whose derivative is \(f(x)\) is called an antiderivative of \(f(x)\).

DEFINITION Antiderivatives

A function \(F(x)\) is an antiderivative of \(f(x)\) on \((a, b)\) if \(F'(x) = f(x)\) for all \(x \in (a, b)\).

Examples:

\[ F'(x) = \frac{d}{dx}(-\cos x) = \sin x = f(x) \]

\[ F'(x) = \frac{d}{dx}\left(\frac{1}{3}x^3\right) = x^2=f(x) \]

One critical observation is that antiderivatives are not unique. We are free to add a constant \(C\) because the derivative of a constant is zero, and so, if \(F'(x) = f(x)\), then \((F(x) + C)' = f(x)\). For example, each of the following is an antiderivative of \(x^{2}\):

\[ \frac{1}{3}x^3,\quad \frac{1}{3}x^3+5,\quad \frac{1}{3}x^3-4 \]

Are there any antiderivatives of \(f(x)\) other than those obtained by adding a constant to a given antiderivative \(F(x)\)? Our next theorem says that the answer is no if \(f(x)\) is defined on an interval \((a, b)\).

THEOREM 1 The General Antiderivative

Let \(F(x)\) be an antiderivative of \(f(x)\) on \((a, b)\). Then every other antiderivative on \((a, b)\) is of the form \(F(x) + C\) for some constant \(C\).

Proof

If \(G(x)\) is a second antiderivative of \(f(x)\), set \(H(x) = G(x) - F(x)\). Then \[ H'(x) = G'(x) - F'(x) = f(x) - f(x) = 0. \] By the Corollary in Section 4.7, \(H(x)\) must be a constant—say, \(H(x) = C\) —and therefore \(G(x) = F(x) + C\).

GRAPHICAL INSIGHT

The graph of \(F(x) + C\) is obtained by shifting the graph of \(F(x)\) vertically by \(C\) units. Since vertical shifting moves the tangent lines without changing their slopes, it makes sense that all of the functions \(F(x) + C\) have the same derivative (Figure 4.84). Theorem 1 tells us that conversely, if two graphs have parallel tangent lines, then one graph is obtained from the other by a vertical shift.

Figure 4.84: The tangent lines to the graphs of \(y = F(x)\) and \(y = F(x) + C\) are parallel.

We often describe the general antiderivative of a function in terms of an arbitrary constant \(C\), as in the following example.

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Example 1

Find two antiderivatives of \(f(x) = \cos x\). Then determine the general antiderivative.

Solution The functions \(F(x) = \sin x\) and \(G(x) = \sin x + 2\) are both antiderivatives of \(f(x)\). The general antiderivative is \(F(x) = \sin x + C\), where \(C\) is any constant.

The process of finding an antiderivative is called integration. We will see why in Chapter 5, when we discuss the connection between antiderivatives and areas under curves given by the Fundamental Theorem of Calculus. Anticipating this result, we begin using the integral sign \(\int\), the standard notation for antiderivatives.

NOTATION Indefinite Integral

The terms “antiderivative” and “indefinite integral” are used interchangeably. In some textbooks, an antiderivative is called a “primitive function.”

The notation

\[ \int f(x)dx = F(x) + C\quad\text{means that}\quad F'(x) = f(x) \]

We say that \(F(x) + C\) is the general antiderivative or indefinite integral of \(f(x)\).

The function \(f(x)\) appearing in the integral sign is called the integrand. The symbol \(dx\) is a differential. It is part of the integral notation and serves to indicate the independent variable. The constant \(C\) is called the constant of integration.

There are no Product, Quotient, or Chain Rules for integrals. However, we will see that the Product Rule for derivatives leads to an important technique called Integration by Parts (Section 7.1) and the Chain Rule leads to the Substitution Method (Section 5.6).

Some indefinite integrals can be evaluated by reversing the familiar derivative formulas. For example, we obtain the indefinite integral of \(x^{n}\) by reversing the Power Rule for derivatives.

THEOREM 2 Power Rule for Integrals

\[ \boxed{\displaystyle\int x^ndx = \frac{x^{n+1}}{n+1}+C\quad\text{for }n\neq1} \]

Proof

We just need to verify that \(F(x)=\frac{x^{n+1}}{n+1}\) is an antiderivative of \(f(x) = x^{n}\):

\[ F'(x)=\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) = \frac{1}{n+1}\left((n+1)x^n\right) = x^n \]

In words, the Power Rule for Integrals says that to integrate a power of \(x\), “add one to the exponent and then divide by the new exponent.” Here are some examples:

\[ \int x^5dx = \frac{1}{6}x^6 + C,\quad \int x^{-9}dx = -\frac{1}{8}x^{-8} + C,\quad \int x^{\frac{3}{5}}dx = \frac{5}{8}x^{\frac{8}{5}} + C \]

The Power Rule is not valid for \(n = -1\). In fact, for \(n = -1\), we obtain the meaningless result

\[ \int x^{-1}dx = \frac{x^0}{0} + C\quad\text{(meaningless)} \]

Notice that in integral notation, we treat dx as a movable variable, and thus we write \(\int\frac{1}{x}dx\) as \(\int\frac{dx}{x}\).

Recall, however, that the derivative of the natural logarithm is \(\frac{d}{dx}\ln x =\frac{1}{x}\). This shows that \(F(x) = \ln x\) is an antiderivative of \(y=\frac{1}{x}\). Thus, for \(n = -1\), instead of the Power Rule we have

\[ \int \frac{dx}{x} = \ln x + C \]

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This formula is valid for \(x \gt 0\), where \(\ln x\) is defined. We would like to have an antiderivative of \(y=\frac{1}{x}\) on its full domain, namely on \(\{x : x \neq 0\}\). To achieve this end, we extend \(F(x)\) to an even function by setting \(F(x) = \ln |x|\) (Figure 4.85). Then \(F(x) = F(-x)\), and by the Chain Rule, \(F'(x) = -F'(-x)\). For \(x \lt 0\), we obtain

\[ \frac{d}{dx}\ln|x| = F'(x) = -F'(-x) = -\frac{1}{-x} = \frac{1}{x} \]

This proves that \(\frac{d}{dx}\ln|x|=\frac{1}{x}\) for all \(x \neq 0\).

Figure 4.85: ---

THEOREM 3 Antiderivative of \(y=\frac{1}{x}\)

The function \(F(x) = \ln |x|\) is an antiderivative of \(y=\frac{1}{x}\) in the domain \(\{x : x \neq 0\}\); that is,

\[ \boxed{\displaystyle\int\frac{dx}{x} = \ln|x|+C}\tag{1} \]

The indefinite integral obeys the usual linearity rules that allow us to integrate “term by term.” These rules follow from the linearity rules for the derivative (see Exercise 79.)

THEOREM 4 Linearity of the Indefinite Integral

  • Sum Rule: \(\displaystyle\int(f(x)+g(x))dx = \int f(x)dx + \int g(x)dx\)
  • Scalar Multiples Rule: \(\displaystyle\int cf(x) dx = c\int f(x)dx\)

Example 2

Evaluate \(\displaystyle\int(3x^4-5x^{\frac{2}{3}}+x^{-3})dx\).

Solution We integrate term by term and use the Power Rule:

When we break up an indefinite integral into a sum of several integrals as in Example 2, it is not necessary to include a separate constant of integration for each integral.

\begin{align*} \int(3x^4-5x^{\frac{2}{3}}+x^{-3})dx &= \int 3x^4dx- \int 5x^{\frac{2}{3}}dx+ \int x^{-3}dx &&\text{(Sum Rule)}\\ &=3\int x^4dx- 5\int x^{\frac{2}{3}}dx+ \int x^{-3}dx &&\text{(Multiple Rule)}\\ &=3\left(\frac{x^5}{5}\right) - 5\left(\frac{x^{\frac{5}{3}}}{5/3}\right) + \frac{x^{-2}}{-2} + C &&\text{(Power Rule)}\\ &=\frac{3}{5}x^5 - 3x^{\frac{5}{3}} - \frac{1}{2}x^{-2} + C&& \end{align*}

To check the answer, we verify that the derivative is equal to the integrand:

\[ \frac{d}{dx}\left(\frac{3}{5}x^5 - 3x^{\frac{5}{3}} - \frac{1}{2}x^{-2} + C\right) = 3x^4-5x^{\frac{2}{3}}+x^{-3} \]

Example 3

Evaluate \(\displaystyle\int\left(\frac{5}{x}-3x^{-10}\right)\).

Solution Apply Eq. (1) and the Power Rule:

\begin{align*} \int\left(\frac{5}{x}-3x^{-10}\right)& = 5\int\frac{dx}{x} - 3\int x^{-10}dx\\ &=5\ln|x|-3\left(\frac{x^{-9}}{-9}\right) + C = 5\ln|x| + \frac{1}{3}x^{-9} +C \end{align*}

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The differentiation formulas for the trigonometric functions give us the following integration formulas. Each formula can be checked by differentiation.

Basic Trigonometric Integrals

\begin{align*} \int\sin xdx &= -\cos x + C &\int\cos xdx &=\sin x+C\\ \int\sec^2xdx &= \tan x + C &\int\csc^2xdx&=-\cot x+C\\ \int\sec x\tan xdx&=\sec x+C &\int\csc x\cot x&=-\csc x+C\\ \end{align*}

Similarly, for any constants \(b\) and \(k\) with \(k \neq 0\), the formulas

\[ \frac{d}{dx}\sin(kx+b) = k\cos(kx+b)\quad \frac{d}{dx}\cos(kx+b) = -k\sin(kx+b) \]

translate to the following indefinite integral formulas:

\[\boxed{ \begin{aligned} \displaystyle\int\cos(kx+b)dx&=\dfrac{1}{k}\sin(kx+b) + C\\ \displaystyle\int\sin(kx+b)dx&=-\frac{1}{k}\cos(kx+b) + C \end{aligned} }\]

Example 4

Evaluate \(\displaystyle\int(\sin(8t-3)+20\cos9t)dt\).

Solution

\begin{align*} \int(\sin(8t-3)+20\cos9t)dt &= \int\sin(8t-3)dt+20\int\cos9tdt\\ &=\frac{1}{b}\cos(8t-3) + \frac{20}{9}\sin 9t + C \end{align*}

4.9.1 Integrals Involving the Exponential Function

The formula \((e^{x})' = e^{x}\) says that \(f(x) = e^{x}\) is its own derivative. But this means that \(f(x) = e^{x}\) is also its own antiderivative. In other words,

\[ \boxed{\displaystyle\int e^x dx = e^x + C} \]

More generally, for any constants \(b\) and \(k \neq 0\),

\[ \boxed{\displaystyle\int e^{kx+b} dx = \frac{1}{k}e^{kx+b} + C} \]

Example 5

Evaluate

(a) \(\displaystyle\int(3e^x - 4)dx\) and

(b) \(\displaystyle\int 12e^{7-3x}dx\).

Solution

(a) \(\displaystyle\int(3e^x - 4)dx = 3\displaystyle\int e^xdx - \displaystyle\int4dx = 3e^x -4x+C\)

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(b) \(\displaystyle\int 12e^{7-3x}dx = 12\displaystyle\int e^{7-3x}=12\left(\frac{1}{-3}e^{7-3x}\right)=-4e^{7-3x}+C\)

4.9.2 Initial Conditions

We can think of an antiderivative as a solution to the differential equation

\[ \frac{dy}{dx} = f(x) \tag{2} \]

An initial condition is like the \(y\)-intercept of a line, which determines one particular line among all lines with the same slope. The graphs of the antiderivatives of \(f(x)\) are all parallel (Figure 4.84 in Section 4.9), and the initial condition determines one of them.

In general, a differential equation is an equation relating an unknown function and its derivatives. The unknown in Eq. (2) is a function \(y = F(x)\) whose derivative is \(f(x)\); that is, \(F(x)\) is an antiderivative of \(f(x)\).

Eq. (2) has infinitely many solutions (because the antiderivative is not unique), but we can specify a particular solution by imposing an initial condition—that is, by requiring that the solution satisfy \(y(x_{0}) = y_{0}\) for some fixed values \(x_{0}\) and \(y_{0}\). A differential equation with an initial condition is called an initial value problem.

Example 6

Solve \(\frac{dy}{dx}=4x^7\) subject to the initial condition \(y(0) = 4\).

Solution First, find the general antiderivative:

\[ y(x) = \int 4x^7dx = \frac{1}{2}x^8 + C \]

Then choose \(C\) so that the initial condition is satisfied: \(y(0) = 0 + C = 4\). This yields \(C = 4\), and our solution is \(y = \frac{1}{2}x^8+4\).

Example 7

Solve the initial value problem \(\frac{dy}{dt}=\sin(\pi t)\), \(y(2) = 2\).

Solution First find the general antiderivative:

\[ y(t) = \int\sin(\pi t)dt = -\frac{1}{\pi}\cos(\pi t) + C \]

Then solve for \(C\) by evaluating at \(t = 2\):

\[ y(2) = -\frac{1}{\pi}\cos(2\pi) + C=2\implies C = 2 +\frac{1}{\pi} \]

The solution of the initial value problem is \(y(t)=-\frac{1}{\pi}\cos(\pi t) + 2 +\frac{1}{\pi}\).

Example 8

A car traveling with velocity \(24\text{ m/s}\) begins to slow down at time \(t = 0\) with a constant deceleration of \(a = -6\text{ m/s}^{2}\). Find

(a) the velocity \(v(t)\) at time \(t\), and

(b) the distance traveled before the car comes to a halt.

Solution

(a) The derivative of velocity is acceleration, so velocity is the antiderivative of acceleration:

Relation between position, velocity, and acceleration:

\begin{align*} s'(t)&=v(t)&s(t)&=\int v(t)dt\\ v'(t)&=a(t)&v(t)&=\int a(t)dt \end{align*}

\[ v(t) = \int adt = \int(-6)dt=-6t+C \]

The initial condition \(v(0) = C = 24\) gives us \(v(t) = -6t + 24\).

(b) Position is the antiderivative of velocity, so the car’s position is

\[ s(t) = \int v(t)dt = \int(-6t+24)dt = -3t^2+24t+C_1 \]

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where \(C_{1}\) is a constant. We are not told where the car is at \(t = 0\), so let us set \(s(0) = 0\) for convenience, getting \(C_{1} = 0\). With this choice, \(s(t) = -3t^{2} + 24t\). This is the distance traveled from time \(t = 0\).

The car comes to a halt when its velocity is zero, so we solve

\[ v(t) = -6t+24 = 0\implies t=4\text{ s} \]

The distance traveled before coming to a halt is \(s(4) = -3(4^{2}) + 24(4) = 48\text{ m}\).

4.9.3 Section 4.9 Summary

\[ \int f(x)dx = F(x) + C \]

\begin{align*} \int x^ndx &= \frac{x^{n+1}}{n+1}+C&&(n\neq-1)\\ \int\sin(kx+b)dx&=-\frac{1}{k}\cos(kx+b)+C&&(k\neq0)\\ \int\cos(kx+b)dx&= \frac{1}{k}\sin(kx+b)+C&&(k\neq0)\\ \int e^{kx+b}dx&=\frac{1}{k}e^{kx+b} + C&&(k\neq0)\\ \int\frac{dx}{x}&=\ln|x|+C \end{align*}