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 [latex]f[/latex], how can we find a function with derivative [latex]f[/latex]? If we can find a function [latex]F[/latex] with derivative [latex]f[/latex], we call [latex]F[/latex] an antiderivative of [latex]f[/latex].

Consider the function [latex]f(x)=2x[/latex]. Knowing the power rule of differentiation, we conclude that [latex]F(x)=x^2[/latex] is an antiderivative of [latex]f[/latex] since [latex]F^{\prime}(x)=2x[/latex]. Are there any other antiderivatives of [latex]f[/latex]? Yes; since the derivative of any constant [latex]C[/latex] is zero, [latex]x^2+C[/latex] is also an antiderivative of [latex]2x[/latex]. Therefore, [latex]x^2+5[/latex] and [latex]x^{2}-\sqrt{2}[/latex] are also antiderivatives. Are there any others that are not of the form [latex]x^2+C[/latex] for some constant [latex]C[/latex]? The answer is no. From Corollary 2 of the Mean Value Theorem, we know that if [latex]F[/latex] and [latex]G[/latex] are differentiable functions such that [latex]F^{\prime}(x)=G^{\prime}(x)[/latex], then [latex]F(x)-G(x)=C[/latex] for some constant [latex]C[/latex]. This fact leads to the following important theorem. We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions.

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 [latex]f[/latex], we use the notation [latex]f^{\prime}(x)[/latex] or [latex]\frac{df}{dx}[/latex] to denote the derivative of [latex]f[/latex]. Here we introduce notation for antiderivatives. If [latex]F[/latex] is an antiderivative of [latex]f[/latex], we say that [latex]F(x)+C[/latex] is the most general antiderivative of [latex]f[/latex] and write

The symbol [latex]\int [/latex] is called an integral sign, and [latex]\int f(x) dx[/latex] is called the indefinite integral of [latex]f[/latex].

Given the terminology introduced in this definition, the act of finding the antiderivatives of a function [latex]f[/latex] is usually referred to as integrating [latex]f[/latex].

For a function [latex]f[/latex] and an antiderivative [latex]F[/latex], the functions [latex]F(x)+C[/latex], where [latex]C[/latex] is any real number, is often referred to as the family of antiderivatives of [latex]f[/latex]. For example, since [latex]x^2[/latex] is an antiderivative of [latex]2x[/latex] and any antiderivative of [latex]2x[/latex] is of the form [latex]x^2+C[/latex], we write

The collection of all functions of the form [latex]x^2+C[/latex], where [latex]C[/latex] is any real number, is known as the family of antiderivatives of [latex]2x[/latex]. (Figure) shows a graph of this family of antiderivatives.

For some functions, evaluating indefinite integrals follows directly from properties of derivatives. For example, for [latex]n \ne −1[/latex],

which comes directly from

This fact is known as the power rule for integrals.

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
[latex]\frac{d}{dx}(k)=0[/latex]	[latex]\int kdx=\int kx^0 dx=kx+C[/latex]
[latex]\frac{d}{dx}(x^n)=nx^{n-1}[/latex]	[latex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/latex] for [latex]n\ne −1[/latex]
[latex]\frac{d}{dx}(\ln |x|)=\frac{1}{x}[/latex]	[latex]\int \frac{1}{x}dx=\ln |x|+C[/latex]
[latex]\frac{d}{dx}(e^x)=e^x[/latex]	[latex]\int e^x dx=e^x+C[/latex]
[latex]\frac{d}{dx}(\sin x)= \cos x[/latex]	[latex]\int \cos x dx= \sin x+C[/latex]
[latex]\frac{d}{dx}(\cos x)=− \sin x[/latex]	[latex]\int \sin x dx=− \cos x+C[/latex]
[latex]\frac{d}{dx}(\tan x)= \sec^2 x[/latex]	[latex]\int \sec^2 x dx= \tan x+C[/latex]
[latex]\frac{d}{dx}(\csc x)=−\csc x \cot x[/latex]	[latex]\int \csc x \cot x dx=−\csc x+C[/latex]
[latex]\frac{d}{dx}(\sec x)= \sec x \tan x[/latex]	[latex]\int \sec x \tan x dx= \sec x+C[/latex]
[latex]\frac{d}{dx}(\cot x)=−\csc^2 x[/latex]	[latex]\int \csc^2 x dx=−\cot x+C[/latex]
[latex]\frac{d}{dx}( \sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}[/latex]	[latex]\int \frac{1}{\sqrt{1-x^2}} dx= \sin^{-1} x+C[/latex]
[latex]\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}[/latex]	[latex]\int \frac{1}{1+x^2} dx= \tan^{-1} x+C[/latex]
[latex]\frac{d}{dx}(\sec^{-1} |x|)=\frac{1}{x\sqrt{x^2-1}}[/latex]	[latex]\int \frac{1}{x\sqrt{x^2-1}} dx= \sec^{-1} |x|+C[/latex]

From the definition of indefinite integral of [latex]f[/latex], we know

if and only if [latex]F[/latex] is an antiderivative of [latex]f[/latex]. Therefore, when claiming that

it is important to check whether this statement is correct by verifying that [latex]F^{\prime}(x)=f(x)[/latex].

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 [latex]f+g[/latex]. In (Figure)a. we showed that an antiderivative of the sum [latex]x+e^x[/latex] is given by the sum [latex](\frac{x^2}{2})+e^x[/latex]—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 [latex]F[/latex] and [latex]G[/latex] are antiderivatives of any functions [latex]f[/latex] and [latex]g[/latex], respectively, then

Therefore, [latex]F(x)+G(x)[/latex] is an antiderivative of [latex]f(x)+g(x)[/latex] and we have

Similarly,

In addition, consider the task of finding an antiderivative of [latex]kf(x)[/latex], where [latex]k[/latex] is any real number. Since

for any real number [latex]k[/latex], we conclude that

These properties are summarized next.

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.

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 [latex]y[/latex] with a derivative [latex]f[/latex]. Therefore, the solutions of (Figure) are the antiderivatives of [latex]f[/latex]. If [latex]F[/latex] is one antiderivative of [latex]f[/latex], every function of the form [latex]y=F(x)+C[/latex] 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 [latex](x_0,y_0)[/latex]—that is, [latex]y(x_0)=y_0[/latex]. The problem of finding a function [latex]y[/latex] that satisfies a differential equation

with the additional condition

is an example of an initial-value problem. The condition [latex]y(x_0)=y_0[/latex] is known as an initial condition. For example, looking for a function [latex]y[/latex] 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 [latex]y=2x^3+C[/latex], to find a function [latex]y[/latex] that also satisfies the initial condition, we need to find [latex]C[/latex] such that [latex]y(1)=2(1)^3+C=5[/latex]. From this equation, we see that [latex]C=3[/latex], and we conclude that [latex]y=2x^3+3[/latex] is the solution of this initial-value problem as shown in the following graph.

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 [latex]v(t)[/latex] is the derivative of a position function [latex]s(t)[/latex], and the acceleration [latex]a(t)[/latex] 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.