Antiderivatives: Learn It 2

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Antiderivatives: Learn It 2

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

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

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

indefinite integral

Given a function [latex]f[/latex], the indefinite integral of [latex]f[/latex], denoted

[latex]\displaystyle\int f(x) dx[/latex],

is the most general antiderivative of [latex]f[/latex]. If [latex]F[/latex] is an antiderivative of [latex]f[/latex], then

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

The expression [latex]f(x)[/latex] is called the integrand and the variable [latex]x[/latex] is the variable of integration.

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].

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

[latex]\displaystyle\int 2x dx=x^2+C[/latex]

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 1 shows a graph of this family of antiderivatives.

The graphs for y = x2 + 2, y = x2 + 1, y = x2, y = x2 − 1, and y = x2 − 2 are shown. — Figure 1. The family of antiderivatives of [latex]2x[/latex] consists of all functions of the form [latex]x^2+C[/latex], where [latex]C[/latex] is any real number.

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

For [latex]n \ne −1[/latex],

[latex]\displaystyle\int x^n dx=\dfrac{x^{n+1}}{n+1}+C[/latex],

which comes directly from

[latex]\frac{d}{dx}\left(\dfrac{x^{n+1}}{n+1}\right)=(n+1)\dfrac{x^n}{n+1}=x^n[/latex]

This fact is known as the power rule for integrals.

power rule for integrals

For [latex]n \ne −1[/latex],

[latex]\displaystyle\int x^n dx=\dfrac{x^{n+1}}{n+1}+C[/latex]

Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions.

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

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

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

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

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

Each of the following statements is of the form [latex]\displaystyle\int f(x) dx=F(x)+C[/latex]. Verify that each statement is correct by showing that [latex]F^{\prime}(x)=f(x)[/latex].

[latex]\displaystyle\int (x+e^x) dx=\dfrac{x^2}{2}+e^x+C[/latex]
[latex]\displaystyle\int xe^xdx=xe^x-e^x+C[/latex]

Earlier, 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 the last example. 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

[latex]\frac{d}{dx}(F(x)+G(x))=F^{\prime}(x)+G^{\prime}(x)=f(x)+g(x)[/latex]

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

[latex]\displaystyle\int (f(x)+g(x)) dx=F(x)+G(x)+C[/latex]

Similarly,

[latex]\displaystyle\int (f(x)-g(x)) dx=F(x)-G(x)+C[/latex]

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

[latex]\frac{d}{dx}(kf(x))=k\frac{d}{dx}F(x)=kF^{\prime}(x)[/latex]

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

[latex]\displaystyle\int kf(x) dx=kF(x)+C[/latex]

These properties are summarized next.

properties of indefinite integrals

Let [latex]F[/latex] and [latex]G[/latex] be antiderivatives of [latex]f[/latex] and [latex]g[/latex], respectively, and let [latex]k[/latex] be any real number.

Sums and Differences

[latex]\displaystyle\int (f(x) \pm g(x)) dx=F(x) \pm G(x)+C[/latex]

Constant Multiples

[latex]\displaystyle\int kf(x) dx=kF(x)+C[/latex]

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. We look at and address integrals involving these more complicated functions later on in the text. In the next example, we examine how to use this theorem to calculate the indefinite integrals of several functions.

Evaluate each of the following indefinite integrals:

[latex]\displaystyle\int (5x^3-7x^2+3x+4) dx[/latex]
[latex]\displaystyle\int \frac{x^2+4\sqrt[3]{x}}{x} dx[/latex]
[latex]\displaystyle\int \frac{4}{1+x^2} dx[/latex]
[latex]\displaystyle\int \tan x \cos x dx[/latex]

Using the properties of indefinite integrals, we can integrate each of the four terms in the integrand separately. We obtain
[latex]\displaystyle\int (5x^3-7x^2+3x+4) dx=\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx[/latex]

From the Constant Multiples property of indefinite integrals, each coefficient can be written in front of the integral sign, which gives

[latex]\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx=5\displaystyle\int x^3 dx-7\displaystyle\int x^2 dx+3\displaystyle\int x dx+4\displaystyle\int 1 dx[/latex]

Using the power rule for integrals, we conclude that

[latex]\displaystyle\int (5x^3-7x^2+3x+4) dx=\frac{5}{4}x^4-\frac{7}{3}x^3+\frac{3}{2}x^2+4x+C[/latex]
Rewrite the integrand as
[latex]\frac{x^2+4\sqrt[3]{x}}{x}=\frac{x^2}{x}+\frac{4\sqrt[3]{x}}{x}[/latex]

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

[latex]\begin{array}{ll} \displaystyle\int (x+\frac{4}{x^{2/3}}) dx & =\displaystyle\int x dx+4\displaystyle\int x^{-2/3} dx \\ & =\frac{1}{2}x^2+4\frac{1}{(\frac{-2}{3})+1}x^{(-2/3)+1}+C \\ & =\frac{1}{2}x^2+12x^{1/3}+C \end{array}[/latex]
Using the properties of indefinite integrals, write the integral as
[latex]4\displaystyle\int \frac{1}{1+x^2} dx[/latex].

Then, use the fact that [latex]\tan^{-1} (x)[/latex] is an antiderivative of [latex]\frac{1}{1+x^2}[/latex] to conclude that

[latex]\displaystyle\int \frac{4}{1+x^2} dx=4 \tan^{-1} (x)+C[/latex]
Rewrite the integrand as
[latex]\tan x \cos x=\frac{ \sin x}{ \cos x} \cos x= \sin x[/latex].

Therefore,

[latex]\displaystyle\int \tan x \cos x dx=\displaystyle\int \sin x dx=− \cos x+C[/latex]

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “4.10 Antiderivatives” here (opens in new window).