At this point, we have seen how to calculate derivatives of many functions and have been introduced to a variety of their applications. We now ask a question that turns this process around: Given a function [latex]f[/latex], how do we find a function with the derivative [latex]f[/latex] and why would we be interested in such a function?
We answer the first part of this question by defining antiderivatives. The antiderivative of a function [latex]f[/latex] is a function with a derivative [latex]f[/latex]. Why are we interested in antiderivatives? The need for antiderivatives arises in many situations, and we look at various examples throughout the remainder of the text. Here we examine one specific example that involves rectilinear motion. In our examination in Derivatives of rectilinear motion, we showed that given a position function [latex]s(t)[/latex] of an object, then its velocity function [latex]v(t)[/latex] is the derivative of [latex]s(t)[/latex]—that is, [latex]v(t)=s^<\prime>(t)[/latex]. Furthermore, the acceleration [latex]a(t)[/latex] is the derivative of the velocity [latex]v(t)[/latex]—that is, [latex]a(t)=v^<\prime>(t)=s^<\prime \prime>(t)[/latex]. Now suppose we are given an acceleration function [latex]a[/latex], but not the velocity function [latex]v[/latex] or the position function [latex]s[/latex]. Since [latex]a(t)=v^<\prime>(t)[/latex], determining the velocity function requires us to find an antiderivative of the acceleration function. Then, since [latex]v(t)=s^<\prime>(t)[/latex], determining the position function requires us to find an antiderivative of the velocity function. Rectilinear motion is just one case in which the need for antiderivatives arises. We will see many more examples throughout the remainder of the text. For now, let’s look at the terminology and notation for antiderivatives, and determine the antiderivatives for several types of functions. We examine various techniques for finding antiderivatives of more complicated functions in the second volume of this text (Introduction to Techniques of Integration).
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].
A function [latex]F[/latex] is an antiderivative of the function [latex]f[/latex] if
[latex]F^<\prime>(x)=f(x)[/latex]for all [latex]x[/latex] in the domain 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^-\sqrt[/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.
Let [latex]F[/latex] be an antiderivative of [latex]f[/latex] over an interval [latex]I[/latex]. Then,
In other words, the most general form of the antiderivative of [latex]f[/latex] over [latex]I[/latex] is [latex]F(x)+C[/latex].
We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions.
Answer: a. Because [latex-display]\frac(x^3)=3x^2[/latex-display] then [latex]F(x)=x^3[/latex] is an antiderivative of [latex]3x^2[/latex]. Therefore, every antiderivative of [latex]3x^2[/latex] is of the form [latex]x^3+C[/latex] for some constant [latex]C[/latex], and every function of the form [latex]x^3+C[/latex] is an antiderivative of [latex]3x^2[/latex]. b. Let [latex]f(x)=\ln |x|[/latex]. For [latex]x>0, \, f(x)=\ln (x)[/latex] and [latex]\frac(\ln x)=\frac[/latex]. For [latex]x<0, \, f(x)=\ln (−x)[/latex] and [latex]\frac(\ln (−x))=-\frac=\frac[/latex]. Therefore, [latex]\frac(\ln |x|)=\frac[/latex]. Thus, [latex]F(x)=\ln |x|[/latex] is an antiderivative of [latex]\frac[/latex]. Therefore, every antiderivative of [latex]\frac[/latex] is of the form [latex]\ln |x|+C[/latex] for some constant [latex]C[/latex] and every function of the form [latex]\ln |x|+C[/latex] is an antiderivative of [latex]\frac[/latex]. c. We have [latex]\frac(\sin x)= \cos x[/latex], so [latex]F(x)= \sin x[/latex] is an antiderivative of [latex] \cos x[/latex]. Therefore, every antiderivative of [latex] \cos x[/latex] is of the form [latex] \sin x+C[/latex] for some constant [latex]C[/latex] and every function of the form [latex] \sin x+C[/latex] is an antiderivative of [latex] \cos x[/latex]. d. Since [latex]\frac(e^x)=e^x[/latex], then [latex]F(x)=e^x[/latex] is an antiderivative of [latex]e^x[/latex]. Therefore, every antiderivative of [latex]e^x[/latex] is of the form [latex]e^x+C[/latex] for some constant [latex]C[/latex] and every function of the form [latex]e^x+C[/latex] is an antiderivative of [latex]e^x[/latex].
Find all antiderivatives of [latex]f(x)= \sin x[/latex].