How To Draw The Antiderivative Of A Function

4. Applications of Derivatives

iv.10 Antiderivatives

Learning Objectives

Notice the general antiderivative of a given role.
Explain the terms and notation used for an indefinite integral.
State the power rule for integrals.
Apply antidifferentiation to solve simple initial-value issues.

At this point, we have seen how to summate derivatives of many functions and have been introduced to a variety of their applications. We now ask a question that turns this process effectually: Given a function $f,$ how do nosotros notice a part with the derivative $f$ and why would we be interested in such a function?

We respond the first office of this question by defining antiderivatives. The antiderivative of a office $f$ is a function with a derivative $f.$ Why are we interested in antiderivatives? The need for antiderivatives arises in many situations, and we look at various examples throughout the rest of the text. Hither nosotros examine i specific example that involves rectilinear motility. In our exam in Derivatives of rectilinear motion, we showed that given a position function $s(t)$ of an object, so its velocity function $v(t)$ is the derivative of $s(t)$ —that is, $v(t)={s}^{\prime }(t).$ Furthermore, the dispatch $a(t)$ is the derivative of the velocity $v(t)$ —that is, $a(t)={v}^{\prime }(t)=s\text{″}(t).$ At present suppose we are given an acceleration function $a,$ but not the velocity part $v$ or the position office $s.$ Since $a(t)={v}^{\prime }(t),$ determining the velocity function requires u.s.a. to observe an antiderivative of the acceleration function. Then, since $v(t)={s}^{\prime }(t),$ determining the position function requires united states to find an antiderivative of the velocity function. Rectilinear motion is only one case in which the need for antiderivatives arises. We will run into many more examples throughout the remainder of the text. For now, permit's look at the terminology and notation for antiderivatives, and determine the antiderivatives for several types of functions. Nosotros examine various techniques for finding antiderivatives of more complicated functions in the second volume of this text (Introduction to Techniques of Integration).

Indefinite Integrals

We now look at the formal note used to correspond antiderivatives and examine some of their properties. These backdrop allow us to find antiderivatives of more complicated functions. Given a function $f,$ we use the annotation ${f}^{\prime }(x)$ or $\frac{df}{dx}$ to announce the derivative of $f.$ Hither we introduce notation for antiderivatives. If $F$ is an antiderivative of $f,$ nosotros say that $F(x)+C$ is the most general antiderivative of $f$ and write

$\int f(x)dx=F(x)+C.$

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

Definition

Given a function $f,$ the indefinite integral of $f,$ denoted

$\int f(x)dx,$

is the nearly general antiderivative of $f.$ If $F$ is an antiderivative of $f,$ then

$\int f(x)dx=F(x)+C.$

The expression $f(x)$ is called the integrand and the variable $x$ is the variable of integration.

Given the terminology introduced in this definition, the act of finding the antiderivatives of a function $f$ is ordinarily referred to every bit integrating $f.$

For a function $f$ and an antiderivative $F,$ the functions $F(x)+C,$ where $C$ is any real number, is oftentimes referred to as the family of antiderivatives of $f.$ For example, since ${x}^{2}$ is an antiderivative of $2x$ and any antiderivative of $2x$ is of the class ${x}^{2}+C,$ we write

$\int 2xdx={x}^{2}+C.$

The collection of all functions of the class ${x}^{2}+C,$ where $C$ is any real number, is known every bit the family unit of antiderivatives of $2x.$ (Figure) shows a graph of this family of antiderivatives.

For some functions, evaluating indefinite integrals follows directly from properties of derivatives. For instance, for $n\ne \text{−}1,$

$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C,$

which comes directly from

$\frac{d}{dx}(\frac{{x}^{n+1}}{n+1})=(n+1)\frac{{x}^{n}}{n+1}={x}^{n}.$

This fact is known every bit the power dominion for integrals.

Power Dominion for Integrals

For $n\ne \text{−}1,$

$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C.$

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
$\frac{d}{dx}(k)=0$	$\int kdx=\int k{x}^{0}dx=kx+C$
$\frac{d}{dx}({x}^{n})=n{x}^{n-1}$	$\int {x}^{n}dn=\frac{{x}^{n+1}}{n+1}+C$ for $n\ne \text{−}1$
$\frac{d}{dx}(\text{ln}\|x\|)=\frac{1}{x}$	$\int \frac{1}{x}dx=\text{ln}\|x\|+C$
$\frac{d}{dx}({e}^{x})={e}^{x}$	$\int {e}^{x}dx={e}^{x}+C$
$\frac{d}{dx}( \sin x)= \cos x$	$\int \cos xdx= \sin x+C$
$\frac{d}{dx}( \cos x)=\text{−} \sin x$	$\int \sin xdx=\text{−} \cos x+C$
$\frac{d}{dx}( \tan x)={ \sec }^{2}x$	$\int { \sec }^{2}xdx= \tan x+C$
$\frac{d}{dx}( \csc x)=\text{−} \csc x \cot x$	$\int \csc x \cot xdx=\text{−} \csc x+C$
$\frac{d}{dx}( \sec x)= \sec x \tan x$	$\int \sec x \tan xdx= \sec x+C$
$\frac{d}{dx}( \cot x)=\text{−}{ \csc }^{2}x$	$\int { \csc }^{2}xdx=\text{−} \cot x+C$
$\frac{d}{dx}({ \sin }^{-1}x)=\frac{1}{\sqrt{1-{x}^{2}}}$	$\int \frac{1}{\sqrt{1-{x}^{2}}}={ \sin }^{-1}x+C$
$\frac{d}{dx}({ \tan }^{-1}x)=\frac{1}{1+{x}^{2}}$	$\int \frac{1}{1+{x}^{2}}dx={ \tan }^{-1}x+C$
$\frac{d}{dx}({ \sec }^{-1}\|x\|)=\frac{1}{x\sqrt{{x}^{2}-1}}$	$\int \frac{1}{x\sqrt{{x}^{2}-1}}dx={ \sec }^{-1}\|x\|+C$

From the definition of indefinite integral of $f,$ we know

$\int f(x)dx=F(x)+C$

if and merely if $F$ is an antiderivative of $f.$ Therefore, when claiming that

$\int f(x)dx=F(x)+C$

information technology is important to check whether this statement is correct by verifying that ${F}^{\prime }(x)=f(x).$

Verifying an Indefinite Integral

Solution

Since
$\frac{d}{dx}(\frac{{x}^{2}}{2}+{e}^{x}+C)=x+{e}^{x},$

the statement

$\int (x+{e}^{x})dx=\frac{{x}^{2}}{2}+{e}^{x}+C$

is correct.
Note that we are verifying an indefinite integral for a sum. Furthermore, $\frac{{x}^{2}}{2}$ and ${e}^{x}$ are antiderivatives of $x$ and ${e}^{x},$ respectively, and the sum of the antiderivatives is an antiderivative of the sum. We discuss this fact again later in this section.
Using the product dominion, we see that
$\frac{d}{dx}(x{e}^{x}-{e}^{x}+C)={e}^{x}+x{e}^{x}-{e}^{x}=x{e}^{x}.$

Therefore, the statement

$\int x{e}^{x}dx=x{e}^{x}-{e}^{x}+C$

is correct.
Note that we are verifying an indefinite integral for a product. The antiderivative $x{e}^{x}-{e}^{x}$ is not a production of the antiderivatives. Furthermore, the product of antiderivatives, ${x}^{2}{e}^{x}\text{/}2$ is non an antiderivative of $x{e}^{x}$ since

$\frac{d}{dx}(\frac{{x}^{2}{e}^{x}}{2})=x{e}^{x}+\frac{{x}^{2}{e}^{x}}{2}\ne x{e}^{x}.$

In full general, the product of antiderivatives is not an antiderivative of a product.

Verify that $\int x \cos xdx=x \sin x+ \cos x+C.$

Solution

$\frac{d}{dx}(x \sin x+ \cos x+C)= \sin x+x \cos x- \sin x=x \cos x$

In (Figure), we listed the indefinite integrals for many unproblematic functions. Allow's at present turn our attention to evaluating indefinite integrals for more complicated functions. For case, consider finding an antiderivative of a sum $f+g.$ In (Figure)a. we showed that an antiderivative of the sum $x+{e}^{x}$ is given past the sum $(\frac{{x}^{2}}{2})+{e}^{x}$ —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 $F$ and $G$ are antiderivatives of any functions $f$ and $g,$ respectively, then

$\frac{d}{dx}(F(x)+G(x))={F}^{\prime }(x)+{G}^{\prime }(x)=f(x)+g(x).$

Therefore, $F(x)+G(x)$ is an antiderivative of $f(x)+g(x)$ and nosotros accept

$\int (f(x)+g(x))dx=F(x)+G(x)+C.$

Similarly,

$\int (f(x)-g(x))dx=F(x)-G(x)+C.$

In addition, consider the job of finding an antiderivative of $kf(x),$ where $k$ is any real number. Since

$\frac{d}{dx}(kf(x))=k\frac{d}{dx}F(x)=k{F}^{\prime }(x)$

for whatever existent number $k,$ nosotros conclude that

$\int kf(x)dx=kF(x)+C.$

These backdrop are summarized adjacent.

Properties of Indefinite Integrals

Let $F$ and $G$ be antiderivatives of $f$ and $g,$ respectively, and permit $k$ be any real number.

Sums and Differences

$\int (f(x)\text{±}g(x))dx=F(x)\text{±}G(x)+C$

Abiding Multiples

$\int kf(x)dx=kF(x)+C$

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 employ this theorem to summate the indefinite integrals of several functions.

Evaluating Indefinite Integrals

Solution

Using (Effigy), we tin integrate each of the four terms in the integrand separately. We obtain
$\int (5{x}^{3}-7{x}^{2}+3x+4)dx=\int 5{x}^{3}dx-\int 7{x}^{2}dx+\int 3xdx+\int 4dx.$

From the second role of (Effigy), each coefficient can exist written in front end of the integral sign, which gives

$\int 5{x}^{3}dx-\int 7{x}^{2}dx+\int 3xdx+\int 4dx=5\int {x}^{3}dx-7\int {x}^{2}dx+3\int xdx+4\int 1dx.$

Using the power rule for integrals, we conclude that

$\int (5{x}^{3}-7{x}^{2}+3x+4)dx=\frac{5}{4}{x}^{4}-\frac{7}{3}{x}^{3}+\frac{3}{2}{x}^{2}+4x+C.$
Rewrite the integrand equally
$\frac{{x}^{2}+4\sqrt[3]{x}}{x}=\frac{{x}^{2}}{x}+\frac{4\sqrt[3]{x}}{x}=0.$

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

$\begin{array}{cc}\hfill \int (x+\frac{4}{{x}^{2\text{/}3}})dx& =\int xdx+4\int {x}^{-2\text{/}3}dx\hfill \\ & =\frac{1}{2}{x}^{2}+4\frac{1}{(\frac{-2}{3})+1}{x}^{(-2\text{/}3)+1}+C\hfill \\ & =\frac{1}{2}{x}^{2}+12{x}^{1\text{/}3}+C.\hfill \end{array}$
Using (Figure), write the integral as
$4\int \frac{1}{1+{x}^{2}}dx.$

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

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

Therefore,

$\int \tan x \cos x=\int \sin x=\text{−} \cos x+C.$

Evaluate $\int (4{x}^{3}-5{x}^{2}+x-7)dx.$

Solution

${x}^{4}-\frac{5}{3}{x}^{3}+\frac{1}{2}{x}^{2}-7x+C$

Initial-Value Problems

We await 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

$\frac{dy}{dx}=f(x)$

is a unproblematic example of a differential equation. Solving this equation means finding a part $y$ with a derivative $f.$ Therefore, the solutions of (Figure) are the antiderivatives of $f.$ If $F$ is one antiderivative of $f,$ every function of the form $y=F(x)+C$ is a solution of that differential equation. For example, the solutions of

$\frac{dy}{dx}=6{x}^{2}$

are given by

$y=\int 6{x}^{2}dx=2{x}^{3}+C.$

Sometimes we are interested in determining whether a particular solution curve passes through a certain point $({x}_{0},{y}_{0})$ —that is, $y({x}_{0})={y}_{0}.$ The trouble of finding a function $y$ that satisfies a differential equation

$\frac{dy}{dx}=f(x)$

with the additional status

$y({x}_{0})={y}_{0}$

is an instance of an initial-value problem. The condition $y({x}_{0})={y}_{0}$ is known as an initial status. For example, looking for a part $y$ that satisfies the differential equation

$\frac{dy}{dx}=6{x}^{2}$

and the initial condition

$y(1)=5$

is an example of an initial-value trouble. Since the solutions of the differential equation are $y=2{x}^{3}+C,$ to find a function $y$ that likewise satisfies the initial status, we need to observe $C$ such that $y(1)=2{(1)}^{3}+C=5.$ From this equation, we see that $C=3,$ and nosotros conclude that $y=2{x}^{3}+3$ is the solution of this initial-value problem as shown in the post-obit graph.

Solving an Initial-Value Trouble

Solve the initial-value problem

$\frac{dy}{dx}= \sin x,y(0)=5.$

Solution

Beginning we need to solve the differential equation. If $\frac{dy}{dx}= \sin x,$ then

$y=\int \sin (x)dx=\text{−} \cos x+C.$

Next we need to look for a solution $y$ that satisfies the initial condition. The initial status $y(0)=5$ ways nosotros demand a constant $C$ such that $\text{−} \cos x+C=5.$ Therefore,

$C=5+ \cos (0)=6.$

The solution of the initial-value problem is $y=\text{−} \cos x+6.$

Solve the initial value trouble $\frac{dy}{dx}=3{x}^{-2},y(1)=2.$

Solution

$y=-\frac{3}{x}+5$

Initial-value problems arise in many applications. Next nosotros consider a problem in which a driver applies the brakes in a car. Nosotros are interested in how long it takes for the car to stop. Call up that the velocity role $v(t)$ is the derivative of a position function $s(t),$ and the acceleration $a(t)$ is the derivative of the velocity role. In earlier examples in the text, nosotros could summate the velocity from the position and so compute the acceleration from the velocity. In the next example nosotros piece of work the other fashion effectually. Given an acceleration function, we calculate the velocity function. We then employ the velocity function to determine the position function.

Decelerating Car

A car is traveling at the rate of 88 ft/sec $(60$ mph) when the brakes are practical. The motorcar begins decelerating at a constant rate of 15 ft/secⁱⁱ.

How many seconds expire before the automobile stops?
How far does the motorcar travel during that time?

Solution

First we introduce variables for this problem. Let $t$ exist the time (in seconds) subsequently the brakes are start applied. Permit $a(t)$ exist the dispatch of the car (in anxiety per seconds squared) at time $t.$ Allow $v(t)$ be the velocity of the car (in feet per second) at time $t.$ Let $s(t)$ be the car's position (in feet) across the point where the brakes are applied at time $t.$
The auto is traveling at a rate of $88\text{ft/sec}.$ Therefore, the initial velocity is $v(0)=88$ ft/sec. Since the auto is decelerating, the acceleration is
$a(t)=-15{\text{ft/s}}^{2}.$

The acceleration is the derivative of the velocity,

${v}^{\prime }(t)=15.$

Therefore, we have an initial-value problem to solve:

${v}^{\prime }(t)=-15,v(0)=88.$

Integrating, we find that

$v(t)=-15t+C.$

Since $v(0)=88,C=88.$ Thus, the velocity role is

$v(t)=-15t+88.$

To find how long it takes for the car to stop, we demand to find the fourth dimension $t$ such that the velocity is zero. Solving $-15t+88=0,$ we obtain $t=\frac{88}{15}$ sec.
To find how far the car travels during this fourth dimension, nosotros need to find the position of the machine after $\frac{88}{15}$ sec. We know the velocity $v(t)$ is the derivative of the position $s(t).$ Consider the initial position to be $s(0)=0.$ Therefore, nosotros need to solve the initial-value problem
${s}^{\prime }(t)=-15t+88,s(0)=0.$

Integrating, nosotros have

$s(t)=-\frac{15}{2}{t}^{2}+88t+C.$

Since $s(0)=0,$ the constant is $C=0.$ Therefore, the position office is

$s(t)=-\frac{15}{2}{t}^{2}+88t.$

After $t=\frac{88}{15}$ sec, the position is $s(\frac{88}{15})\approx 258.133$ ft.

Suppose the car is traveling at the rate of 44 ft/sec. How long does information technology take for the car to end? How far will the car travel?

[reveal-answer q="923849″]Bear witness Answer[/reveal-answer]
[subconscious-answer a="923849″] $2.93 \sec ,64.5\text{ft}$

Key Concepts

For the following exercises, bear witness that $F(x)$ are antiderivatives of $f(x).$

one. $F(x)=5{x}^{3}+2{x}^{2}+3x+1,f(x)=15{x}^{2}+4x+3$

Solution

${F}^{\prime }(x)=15{x}^{2}+4x+3$

2. $F(x)={x}^{2}+4x+1,f(x)=2x+4$

3. $F(x)={x}^{2}{e}^{x},f(x)={e}^{x}({x}^{2}+2x)$

Solution

${F}^{\prime }(x)=2x{e}^{x}+{x}^{2}{e}^{x}$

4. $F(x)= \cos x,f(x)=\text{−} \sin x$

5. $F(x)={e}^{x},f(x)={e}^{x}$

Solution

${F}^{\prime }(x)={e}^{x}$

For the following exercises, find the antiderivative of the part.

6. $f(x)=\frac{1}{{x}^{2}}+x$

7. $f(x)={e}^{x}-3{x}^{2}+ \sin x$

Solution

$F(x)={e}^{x}-{x}^{3}- \cos (x)+C$

8. $f(x)={e}^{x}+3x-{x}^{2}$

9. $f(x)=x-1+4 \sin (2x)$

Solution

$F(x)=\frac{{x}^{2}}{2}-x-2 \cos (2x)+C$

For the post-obit exercises, notice the antiderivative $F(x)$ of each role $f(x).$

10. $f(x)=5{x}^{4}+4{x}^{5}$

11. $f(x)=x+12{x}^{2}$

Solution

$F(x)=\frac{1}{2}{x}^{2}+4{x}^{3}+C$

12. $f(x)=\frac{1}{\sqrt{x}}$

13. $f(x)={(\sqrt{x})}^{3}$

Solution

$F(x)=\frac{2}{5}{(\sqrt{x})}^{5}+C$

fourteen. $f(x)={x}^{1\text{/}3}+{(2x)}^{1\text{/}3}$

15. $f(x)=\frac{{x}^{1\text{/}3}}{{x}^{2\text{/}3}}$

Solution

$F(x)=\frac{3}{2}{x}^{2\text{/}3}+C$

16. $f(x)=2 \sin (x)+ \sin (2x)$

17. $f(x)={ \sec }^{2}(x)+1$

Solution

$F(x)=x+ \tan (x)+C$

eighteen. $f(x)= \sin x \cos x$

nineteen. $f(x)={ \sin }^{2}(x) \cos (x)$

Solution

$F(x)=\frac{1}{3}{ \sin }^{3}(x)+C$

20. $f(x)=0$

21. $f(x)=\frac{1}{2}{ \csc }^{2}(x)+\frac{1}{{x}^{2}}$

Solution

$F(x)=-\frac{1}{2} \cot (x)-\frac{1}{x}+C$

22. $f(x)= \csc x \cot x+3x$

23. $f(x)=4 \csc x \cot x- \sec x \tan x$

[reveal-reply q="93930″]Show Reply[/reveal-answer]
[hidden-reply a="93930″] $F(x)=\text{−} \sec x-4 \csc x+C$

24. $f(x)=8 \sec x( \sec x-4 \tan x)$

25. $f(x)=\frac{1}{2}{e}^{-4x}+ \sin x$

Solution

$F(x)=-\frac{1}{8}{e}^{-4x}- \cos x+C$

For the following exercises, evaluate the integral.

26. $\int (-1)dx$

27. $\int \sin xdx$

[reveal-answer q="436260″]Show Answer[/reveal-answer]
[subconscious-reply a="436260″] $\text{−} \cos x+C$

28. $\int (4x+\sqrt{x})dx$

29. $\int \frac{3{x}^{2}+2}{{x}^{2}}dx$

Solution

$3x-\frac{2}{x}+C$

xxx. $\int ( \sec x \tan x+4x)dx$

31. $\int (4\sqrt{x}+\sqrt[4]{x})dx$

Solution

$\frac{8}{3}{x}^{3\text{/}2}+\frac{4}{5}{x}^{5\text{/}4}+C$

32. $\int ({x}^{-1\text{/}3}-{x}^{2\text{/}3})dx$

33. $\int \frac{14{x}^{3}+2x+1}{{x}^{3}}dx$

Solution

$14x-\frac{2}{x}-\frac{1}{2{x}^{2}}+C$

34. $\int ({e}^{x}+{e}^{\text{−}x})dx$

For the following exercises, solve the initial value problem.

35. ${f}^{\prime }(x)={x}^{-3},f(1)=1$

Solution

$f(x)=-\frac{1}{2{x}^{2}}+\frac{3}{2}$

36. ${f}^{\prime }(x)=\sqrt{x}+{x}^{2},f(0)=2$

37. ${f}^{\prime }(x)= \cos x+{ \sec }^{2}(x),f(\frac{\pi }{4})=2+\frac{\sqrt{2}}{2}$

Solution

$f(x)= \sin x+ \tan x+1$

38. ${f}^{\prime }(x)={x}^{3}-8{x}^{2}+16x+1,f(0)=0$

39. ${f}^{\prime }(x)=\frac{2}{{x}^{2}}-\frac{{x}^{2}}{2},f(1)=0$

Solution

$f(x)=-\frac{1}{6}{x}^{3}-\frac{2}{x}+\frac{13}{6}$

For the post-obit exercises, observe 2 possible functions $f$ given the 2d- or third-lodge derivatives.

40. $f\text{″}(x)={x}^{2}+2$

41. $f\text{″}(x)={e}^{\text{−}x}$

Solution

Answers may vary; one possible respond is $f(x)={e}^{\text{−}x}$

42. $f\text{″}(x)=1+x$

43. $f\text{‴}(x)= \cos x$

Answers may vary; one possible answer is $f(x)=\text{−} \sin x$

44. $f\text{‴}(x)=8{e}^{-2x}- \sin x$

45.A machine is being driven at a rate of 40 mph when the brakes are applied. The car decelerates at a constant charge per unit of 10 ft/sec^two. How long earlier the car stops?

46.In the preceding problem, calculate how far the motorcar travels in the fourth dimension it takes to finish.

47.You are merging onto the state highway, accelerating at a constant rate of 12 ft/sec². How long does it take you to reach merging speed at 60 mph?

48.Based on the previous problem, how far does the car travel to reach merging speed?

49.A automobile company wants to ensure its newest model can stop in 8 sec when traveling at 75 mph. If we presume constant deceleration, find the value of deceleration that accomplishes this.

fifty.A automobile company wants to ensure its newest model tin cease in less than 450 ft when traveling at threescore mph. If we assume constant deceleration, discover the value of deceleration that accomplishes this.

For the following exercises, find the antiderivative of the function, assuming $F(0)=0.$