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Antiderivatives

Find the general antiderivative of a given function.

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 \(f,\) how can we find a function with derivative \(f?\) If we can find a function \(F\) with derivative \(f,\) we call \(F\) an antiderivative of \(f.\)

Consider the function \(f(x)=2x.\) Knowing the power rule of differentiation, we conclude that \(F(x)={x}^{2}\) is an antiderivative of \(f\) since \({F}^{'}(x)=2x.\) Are there any other antiderivatives of \(f?\) Yes; since the derivative of any constant \(C\) is zero, \({x}^{2}+C\) is also an antiderivative of \(2x.\) Therefore, \({x}^{2}+5\) and \({x}^{2}-\sqrt{2}\) are also antiderivatives. Are there any others that are not of the form \({x}^{2}+C\) for some constant \(C?\) The answer is no. From Corollary \(2\) of the Mean Value Theorem, we know that if \(F\) and \(G\) are differentiable functions such that \({F}^{'}(x)={G}^{'}(x),\) then \(F(x)-G(x)=C\) for some constant \(C.\) 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.

Condensed — the full section is in OpenStax Calculus Volume 1.

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 \(f,\) we use the notation \({f}^{'}(x)\) or \(\frac{df}{dx}\) to denote the derivative of \(f.\) Here we introduce notation for antiderivatives. If \(F\) is an antiderivative of \(f,\) we 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 \(f.\)

Given the terminology introduced in this definition, the act of finding the antiderivatives of a function \(f\) is usually referred to as integrating \(f.\)

For a function \(f\) and an antiderivative \(F,\) the functions \(F(x)+C,\) where \(C\) is any real number, is often 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 form \({x}^{2}+C,\) we write

\[\int 2x\ dx={x}^{2}+C.\]

The collection of all functions of the form \({x}^{2}+C,\) where \(C\) is any real number, is known as the family of antiderivatives of \(2x.\) shows a graph of this family of antiderivatives.

For some functions, evaluating indefinite integrals follows directly from properties of derivatives. For example, for \(n\ne \text{-}1,\)

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

which comes directly from

\[\frac{d}{dx}(\frac{{x}^{n+1}}{n+1})=(n+1)\frac{{x}^{n}}{n+1}={x}^{n}.\]\[\int f(x)dx=F(x)+C\]\[\int f(x)dx=F(x)+C\]\[\frac{d}{dx}(F(x)+G(x))={F}^{'}(x)+{G}^{'}(x)=f(x)+g(x).\]\[\int (f(x)+g(x))dx=F(x)+G(x)+C.\]\[\int (f(x)-g(x))dx=F(x)-G(x)+C.\]\[\frac{d}{dx}(kf(x))=k\frac{d}{dx}F(x)=k{f}^{'}(x)\]\[\int kf(x)dx=kF(x)+C.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

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

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

is a simple example of a differential equation. Solving this equation means finding a function \(y\) with a derivative \(f.\) Therefore, the solutions of 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 problem of finding a function \(y\) that satisfies a differential equation

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

with the additional condition

\[y({x}_{0})={y}_{0}\]

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

\[\frac{dy}{dx}=6{x}^{2}\]\[y(1)=5\]

Example

Try it.

Solve the initial-value problem

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

Solution

First we need to solve the differential equation. If \(\frac{dy}{dx}=\text{sin}\ x,\) then

\[y=\int \text{sin}(x)dx=\text{-}\text{cos}\ x+C.\]

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

\[C=5+\text{cos}(0)=6.\]

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

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

If \(F\) is an antiderivative of \(f,\) then every antiderivative of \(f\) is of the form \(F(x)+C\) for some constant \(C.\)
Solving the initial-value problem
\[\frac{dy}{dx}=f(x),y({x}_{0})={y}_{0}\]
requires us first to find the set of antiderivatives of \(f\) and then to look for the particular antiderivative that also satisfies the initial condition.

Antiderivatives

For the following exercises, show that \(F(x)\) are antiderivatives of \(f(x).\)

For the following exercises, find the general form for the antiderivative of the function.

For the following exercises, find the general form for the antiderivative \(F(x)\) of each function \(f(x).\)

For the following exercises, evaluate the integral.

For the following exercises, solve the initial value problem.

For the following exercises, find two possible functions \(f\) given the second- or third-order derivatives.

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

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For each of the following functions, find all antiderivatives.
1. \(f(x)=3{x}^{2}\)
2. \(f(x)=\frac{1}{x}\)
3. \(f(x)=\text{cos}\ x\)
4. \(f(x)={e}^{x}\)
גלה את התשובה
1. Because
  \[\frac{d}{dx}({x}^{3})=3{x}^{2}\]
  then \(F(x)={x}^{3}\) is an antiderivative of \(3{x}^{2}.\) Therefore, every antiderivative of \(3{x}^{2}\) is of the form \({x}^{3}+C\) for some constant \(C,\) and every function of the form \({x}^{3}+C\) is an antiderivative of \(3{x}^{2}.\)
2. Let \(f(x)=\text{ln}|x|.\) For \(x>0,f(x)=\text{ln}(x)\) and
  \[\frac{d}{dx}(\text{ln}\ x)=\frac{1}{x}.\]
  For \(x<0,f(x)=\text{ln}(\text{-}x)\) and
  \[\frac{d}{dx}(\text{ln}(\text{-}x))=-\frac{1}{\text{-}x}=\frac{1}{x}.\]
  Therefore,
  \[\frac{d}{dx}(\text{ln}|x|)=\frac{1}{x}.\]
  Thus, \(F(x)=\text{ln}|x|\) is an antiderivative of \(\frac{1}{x}.\) Therefore, every antiderivative of \(\frac{1}{x}\) is of the form \(\text{ln}|x|+C\) for some constant \(C\) and every function of the form \(\text{ln}|x|+C\) is an antiderivative of \(\frac{1}{x}.\)
3. We have
  \[\frac{d}{dx}(\text{sin}\ x)=\text{cos}\ x,\]
  so \(F(x)=\text{sin}\ x\) is an antiderivative of \(\text{cos}\ x.\) Therefore, every antiderivative of \(\text{cos}\ x\) is of the form \(\text{sin}\ x+C\) for some constant \(C\) and every function of the form \(\text{sin}\ x+C\) is an antiderivative of \(\text{cos}\ x.\)
4. Since
  \[\frac{d}{dx}({e}^{x})={e}^{x},\]
  then \(F(x)={e}^{x}\) is an antiderivative of \({e}^{x}.\) Therefore, every antiderivative of \({e}^{x}\) is of the form \({e}^{x}+C\) for some constant \(C\) and every function of the form \({e}^{x}+C\) is an antiderivative of \({e}^{x}.\)
Find all antiderivatives of \(f(x)=\text{sin}\ x.\)

גלה את התשובה

\(\text{-}\text{cos}\ x+C\)
Each of the following statements is of the form \(\int f(x)dx=F(x)+C.\) Verify that each statement is correct by showing that \({F}^{'}(x)=f(x).\)
1. \(\int (x+{e}^{x})dx=\frac{{x}^{2}}{2}+{e}^{x}+C\)
2. \(\int x{e}^{x}dx=x{e}^{x}-{e}^{x}+C\)
גלה את התשובה
1. 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.
2. Using the product rule, 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 product of the antiderivatives. Furthermore, the product of antiderivatives, \({x}^{2}{e}^{x}\text{/}2\) is not 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 general, the product of antiderivatives is not an antiderivative of a product.
Verify that \(\int x\ \text{cos}\ x\ dx=x\ \text{sin}\ x+\text{cos}\ x+C.\)

גלה את התשובה

\(\frac{d}{dx}(x\ \text{sin}\ x+\text{cos}\ x+C)=\text{sin}\ x+x\ \text{cos}\ x-\text{sin}\ x=x\ \text{cos}\ x\)
Evaluate each of the following indefinite integrals:
1. \(\int (5{x}^{3}-7{x}^{2}+3x+4)dx\)
2. \(\int \frac{{x}^{2}+4\sqrt[3]{x}}{x}dx\)
3. \(\int \frac{4}{1+{x}^{2}}dx\)
4. \(\int \text{tan}\ x\ \text{cos}\ x\ dx\)
גלה את התשובה
1. Using , we can integrate each of the four terms in the integrand separately. We obtain
  \[\int (5{x}^{3}-7{x}^{2}+3x+4)dx=\int 5{x}^{3}dx-\int 7{x}^{2}dx+\int 3x\ dx+\int 4dx.\]
  From the second part of , each coefficient can be written in front of the integral sign, which gives
  \[\int 5{x}^{3}dx-\int 7{x}^{2}dx+\int 3x\ dx+\int 4dx=5\int {x}^{3}dx-7\int {x}^{2}dx+3\int x\ dx+4\int 1dx.\] Using the power rule for integrals, we conclude that
  \[\int (5{x}^{3}-7{x}^{2}+3x+4)dx=\frac{5}{4}{x}^{4}-\frac{7}{3}{x}^{3}+\frac{3}{2}{x}^{2}+4x+C.\]
2. Rewrite the integrand as
  \[\frac{{x}^{2}+4\sqrt[3]{x}}{x}=\frac{{x}^{2}}{x}+\frac{4\sqrt[3]{x}}{x}.\]
  Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have
  \[\begin{array}{ll}\int (x+\frac{4}{{x}^{2\text{/}3}})dx & =\int x\ dx+4\int {x}^{-2\text{/}3}dx \\ & =\frac{1}{2}{x}^{2}+4\frac{1}{(\frac{-2}{3})+1}{x}^{(-2\text{/}3)+1}+C \\ & =\frac{1}{2}{x}^{2}+12{x}^{1\text{/}3}+C.\end{array}\]
3. Using , write the integral as
  \[4\int \frac{1}{1+{x}^{2}}dx.\]
  Then, use the fact that \({\text{tan}}^{-1}(x)\) is an antiderivative of \(\frac{1}{(1+{x}^{2})}\) to conclude that
  \[\int \frac{4}{1+{x}^{2}}dx=4\ {\text{tan}}^{-1}(x)+C.\]
4. Rewrite the integrand as
  \[\text{tan}\ x\ \text{cos}\ x=\frac{\text{sin}\ x}{\text{cos}\ x}\ \text{cos}\ x=\text{sin}\ x.\]
  Therefore,
  \[\int \text{tan}\ x\ \text{cos}\ x=\int \text{sin}\ x=\text{-}\text{cos}\ x+C.\]
Evaluate \(\int (4{x}^{3}-5{x}^{2}+x-7)dx.\)

גלה את התשובה

\({x}^{4}-\frac{5}{3}{x}^{3}+\frac{1}{2}{x}^{2}-7x+C\)
Solve the initial-value problem
\[\frac{dy}{dx}=\text{sin}\ x,y(0)=5.\]

גלה את התשובה

First we need to solve the differential equation. If \(\frac{dy}{dx}=\text{sin}\ x,\) then
\[y=\int \text{sin}(x)dx=\text{-}\text{cos}\ x+C.\]
Next we need to look for a solution \(y\) that satisfies the initial condition. The initial condition \(y(0)=5\) means we need a constant \(C\) such that \(\text{-}\text{cos}\ x+C=5.\) Therefore,
\[C=5+\text{cos}(0)=6.\]
The solution of the initial-value problem is \(y=\text{-}\text{cos}\ x+6.\)
Solve the initial value problem \(\frac{dy}{dx}=3{x}^{-2},y(1)=2.\)

גלה את התשובה

\(y=-\frac{3}{x}+5\)
A car is traveling at the rate of \(88\) ft/sec \((60\) mph) when the brakes are applied. The car begins decelerating at a constant rate of \(15\) ft/sec².
1. How many seconds elapse before the car stops?
2. How far does the car travel during that time?
גלה את התשובה
1. First we introduce variables for this problem. Let \(t\) be the time (in seconds) after the brakes are first applied. Let \(a(t)\) be the acceleration of the car (in feet per seconds squared) at time \(t.\) Let \(v(t)\) be the velocity of the car (in feet per second) at time \(t.\) Let \(s(t)\) be the car’s position (in feet) beyond the point where the brakes are applied at time \(t.\)
  The car is traveling at a rate of \(88\ \text{ft/sec}.\) Therefore, the initial velocity is \(v(0)=88\) ft/sec. Since the car is decelerating, the acceleration is
  \[a(t)=-15{\ \text{ft/s}}^{2}.\]
  The acceleration is the derivative of the velocity,
  \[{v}^{'}(t)=-15.\]
  Therefore, we have an initial-value problem to solve:
  \[{v}^{'}(t)=-15,v(0)=88.\]
  Integrating, we find that
  \[v(t)=-15t+C.\]
  Since \(v(0)=88,C=88.\) Thus, the velocity function is
  \[v(t)=-15t+88.\]
  To find how long it takes for the car to stop, we need to find the time \(t\) such that the velocity is zero. Solving \(-15t+88=0,\) we obtain \(t=\frac{88}{15}\) sec.
2. To find how far the car travels during this time, we need to find the position of the car 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, we need to solve the initial-value problem
  \[{s}^{'}(t)=-15t+88,s(0)=0.\]
  Integrating, we have
  \[s(t)=-\frac{15}{2}{t}^{2}+88t+C.\]
  Since \(s(0)=0,\) the constant is \(C=0.\) Therefore, the position function 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 it take for the car to stop? How far will the car travel?

גלה את התשובה

\(2.93\ \text{sec},\ 64.5\ \text{ft}\)
\(F(x)=5{x}^{3}+2{x}^{2}+3x+1,f(x)=15{x}^{2}+4x+3\)

גלה את התשובה

\({F}^{'}(x)=15{x}^{2}+4x+3\)
\(F(x)={x}^{2}+4x+1,f(x)=2x+4\)
\(F(x)={x}^{2}{e}^{x},f(x)={e}^{x}({x}^{2}+2x)\)

גלה את התשובה

\({F}^{'}(x)=2x{e}^{x}+{x}^{2}{e}^{x}\)
\(F(x)=\text{cos}\ x,f(x)=\text{-}\text{sin}\ x\)
\(F(x)={e}^{x},f(x)={e}^{x}\)

גלה את התשובה

\({F}^{'}(x)={e}^{x}\)
\(f(x)=\frac{1}{{x}^{2}}+x\)
\(f(x)={e}^{x}-3{x}^{2}+\text{sin}\ x\)

גלה את התשובה

\(F(x)={e}^{x}-{x}^{3}-\text{cos}(x)+C\)
\(f(x)={e}^{x}+3x-{x}^{2}\)
\(f(x)=x-1+4\ \text{sin}(2x)\)

גלה את התשובה

\(F(x)=\frac{{x}^{2}}{2}-x-2\ \text{cos}(2x)+C\)
\(f(x)=5{x}^{4}+4{x}^{5}\)
\(f(x)=x+12{x}^{2}\)

גלה את התשובה

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

גלה את התשובה

\(F(x)=\frac{2}{5}{(\sqrt{x})}^{5}+C\)
\(f(x)={x}^{1\text{/}3}+{(2x)}^{1\text{/}3}\)
\(f(x)=\frac{{x}^{1\text{/}3}}{{x}^{2\text{/}3}}\)

גלה את התשובה

\(F(x)=\frac{3}{2}{x}^{2\text{/}3}+C\)
\(f(x)=2\ \text{sin}(x)+\text{sin}(2x)\)
\(f(x)={\text{sec}}^{2}(x)+1\)

גלה את התשובה

\(F(x)=x+\text{tan}(x)+C\)
\(f(x)=\text{sin}\ x\ \text{cos}\ x\)
\(f(x)={\text{sin}}^{2}(x)\text{cos}(x)\)

גלה את התשובה

\(F(x)=\frac{1}{3}{\text{sin}}^{3}(x)+C\)
\(f(x)=\frac{1}{2}{\text{csc}}^{2}(x)+\frac{1}{{x}^{2}}\)

גלה את התשובה

\(F(x)=-\frac{1}{2}\ \text{cot}(x)-\frac{1}{x}+C\)
\(f(x)=\text{csc}\ x\ \text{cot}\ x+3x\)
\(f(x)=4\ \text{csc}\ x\ \text{cot}\ x-\text{sec}\ x\ \text{tan}\ x\)

גלה את התשובה

\(F(x)=\text{-}\text{sec}\ x-4\ \text{csc}\ x+C\)
\(f(x)=8\ \text{sec}\ x(\text{sec}\ x-4\ \text{tan}\ x)\)
\(f(x)=\frac{1}{2}{e}^{-4x}+\text{sin}\ x\)

גלה את התשובה

\(F(x)=-\frac{1}{8}{e}^{-4x}-\text{cos}\ x+C\)
\(\int (-1)dx\)
\(\int \text{sin}\ x\ dx\)

גלה את התשובה

\(\text{-}\text{cos}\ x+C\)
\(\int (4x+\sqrt{x})dx\)
\(\int \frac{3{x}^{2}+2}{{x}^{2}}dx\)

גלה את התשובה

\(3x-\frac{2}{x}+C\)
\(\int (\text{sec}\ x\ \text{tan}\ x+4x)dx\)
\(\int (4\sqrt{x}+\sqrt[4]{x})dx\)

גלה את התשובה

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

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Antiderivatives

Find the general antiderivative of a given function.
Explain the terms and notation used for an indefinite integral.
State the power rule for integrals.
Use antidifferentiation to solve simple initial-value problems.
for each constant
if
Because
Let

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

נסה את שלך.

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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