Integrals · Lesson 1 of 9

Antiderivatives and Indefinite Integrals

“Recover functions from their rates of change and build a reliable first toolkit for integration.”

Learning Objectives

• Explain an antiderivative and the family represented by an indefinite integral. • Interpret integral notation and use the standard elementary integrals. • Apply linearity and algebraic simplification before integrating. • Use an additional condition to determine the constant of integration. • Check antiderivatives by differentiation and respect their domains.

Recovering a function from its derivative

Differentiation answers a forward question: if the position of a moving object is known, how quickly is its position changing? Integration begins with the reverse question. If the velocity is known, what position function could have produced it? A second motivating question is how to measure a region bounded by a curve. Later in the chapter, the Fundamental Theorem of Calculus will connect these two questions.

Suppose a function has derivative $x^{2}$ . Since the derivative of $x^{3} /3$ is $x^{2}$ , we have found one possible original function. But adding $7$ , subtracting $2$ , or adding any fixed real number gives the same derivative. A rate of change alone does not determine a starting position.

Definition

Antiderivative or primitive

A function $F$ is an antiderivative of $f$ on an interval if $F^{'} (x) = f (x)$ at every point of that interval where the derivative is considered.

An entire family of antiderivativesLaTeX

\int f (x) d x = F (x) + C, F' (x) = f (x)

The integrand is

f (x)

x

is the variable of integration, and

C

is an arbitrary real constant called the constant of integration. The integral symbol represents the whole family, not a single selected member.

Why is adding a constant the only freedom? If $G$ and $F$ have the same derivative on an interval, then $(G - F)^{'} (x) = 0$ . A function with zero derivative throughout an interval is constant there, so $G - F = C$ . This interval condition matters: an integrand such as $1/ x$ is not defined across $x = 0$ , and its antiderivatives are considered on intervals that avoid zero.

Notation or term	Meaning
$\int f (x) d x$	The indefinite integral of $f$ with respect to $x$ .
$f (x)$	The integrand: the function being integrated.
$d x$	Indicates the variable of integration.
$F (x)$	One antiderivative satisfying $F^{'} (x) = f (x)$ .
$C$	An arbitrary constant selecting a member of the family.
Integrate	Find the family of antiderivatives.

The variable need not be called $x$ . For example, integrating $y^{4}$ with respect to $y$ gives $y^{5} /5 + C$ . Changing the letter does not change the mathematical relationship. The variable following the integral determines which quantity is allowed to vary.

Standard integrals come from known derivatives

Every derivative formula can be read backwards to obtain an integral formula. This is more reliable than memorising an unconnected list: whenever you are unsure about a sign or coefficient, differentiate the proposed answer. In particular, increasing a power changes both the exponent and the multiplying factor.

Power ruleLaTeX

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C (n \neq = - 1)

Use the rule on intervals where the real-valued power is defined. For

n = 0

, it gives

\int 1 d x = x + C

. The exception

n = - 1

needs the logarithmic rule.

Logarithmic ruleLaTeX

\int \frac{1}{x} d x = ln ∣ x ∣ + C (x \neq = 0)

Here

ln

means the natural logarithm. When

x < 0

, differentiating

ln (- x)

still gives

1/ x

, which explains the absolute value.

Exponential rulesLaTeX

\int e^{x} d x = e^{x} + C, \int a^{x} d x = \frac{a ^{x}}{ln a} + C (a > 0, a \neq = 1)

The derivative of

a^{x}

a^{x} ln a

. Dividing by

ln a

reverses that factor. The base

a = 1

instead gives the constant integrand

1

Integrand	One antiderivative
$cos x$	$sin x$
$sin x$	$- cos x$
$sec^{2} x$	$tan x$
$csc^{2} x$	$- cot x$
$sec x tan x$	$sec x$
$csc x cot x$	$- csc x$
$1/ 1 - x^{2}$	$sin^{- 1} x$ or $- cos^{- 1} x$
$1/ (1 + x^{2})$	$tan^{- 1} x$

The two antiderivatives in the inverse-sine row differ by the constant $π /2$ on their common interval. This is an early example of equivalent answers: different expressions can represent the same family after the arbitrary constant is included. The square-root formula is used on $- 1 < x < 1$ ; trigonometric formulas are used on intervals where their integrands are defined.

Example — Inspection with a chain factor

Problem
Find one antiderivative of $cos 2 x$ , of $3 x^{2} + 4 x^{3}$ , and of $1/ x$ .

1.The derivative of $sin 2 x$ is $2 cos 2 x$ . Divide by $2$ to compensate: one antiderivative of $cos 2 x$ is $\frac{1}{2} sin 2 x$ .
2.Differentiate $x^{3} + x^{4}$ : the result is $3 x^{2} + 4 x^{3}$ . Thus $x^{3} + x^{4}$ is one antiderivative.
3.The derivative of $ln ∣ x ∣$ is $1/ x$ on either interval avoiding zero. Thus $ln ∣ x ∣$ is one antiderivative there. Add $C$ to any of these answers when the full indefinite integral is requested.

Combining integrals without changing their meaning

Differentiation respects addition and multiplication by a constant. Integration inherits the corresponding rules because we can differentiate both sides to check them. This lets us break a complicated sum into familiar pieces, but it does not give a rule for splitting products or quotients into separate integrals.

LinearityLaTeX

\int i = 1 \sum n k_{i} f_{i} (x) d x = i = 1 \sum n k_{i} \int f_{i} (x) d x

The

k_{i}

are constants. The final arbitrary constants can be combined into one

C

. Subtraction is included by using a negative coefficient.

For the addition rule, the derivative of an antiderivative of $f + g$ is $f + g$ . The derivative of the sum of an antiderivative of $f$ and one of $g$ is also $f + g$ . The two expressions therefore differ only by a constant. The same reasoning proves the constant-multiple rule, and repeating the rules gives the finite-sum version.

The inverse relationshipsLaTeX

\frac{d}{d x} (\int f (x) d x) = f (x), \int f' (x) d x = f (x) + C

The first relationship removes the arbitrary constant by differentiating. The second recovers all functions with the given derivative.

Example — Simplify a quotient first

Problem
Evaluate $\int (x^{3} - 1) / x^{2} d x$ .

1.On an interval avoiding $x = 0$ , divide each numerator term by $x^{2}$ : $(x^{3} - 1) / x^{2} = x - x^{- 2}$ .
2.Use linearity and the power rule: the two antiderivatives are $x^{2} /2$ and $- x^{- 1}$ , respectively.
3.Subtracting the second gives $x^{2} /2 + 1/ x + C$ . Differentiation returns $x - 1/ x^{2}$ , confirming the answer.

Example — Fractional powers and an exponential

Problem
Evaluate $\int (x^{2/3} + 1) d x$ and $\int (x^{3/2} + 2 e^{x} - 1/ x) d x$ .

1.For the first integral, increase $2/3$ to $5/3$ and divide by $5/3$ . The answer is $3 x^{5/3} /5 + x + C$ .
2.For the second, integrate each term separately. The power term becomes $2 x^{5/2} /5$ , the exponential term becomes $2 e^{x}$ , and the last term becomes $- ln ∣ x ∣$ .
3.Thus the second answer is $2 x^{5/2} /5 + 2 e^{x} - ln ∣ x ∣ + C$ , on intervals where all terms are real and defined, here $x > 0$ .

Example — Recognising trigonometric derivatives

Problem
Evaluate $\int (sin x + cos x) d x$ , $\int csc x (csc x + cot x) d x$ , and $\int (1 - sin x) / cos^{2} x d x$ .

1.For the sum, integrate term by term to get $- cos x + sin x + C$ .
2.Expand the second integrand as $csc^{2} x + csc x cot x$ . Its integral is $- cot x - csc x + C$ .
3.Rewrite the third integrand as $sec^{2} x - sec x tan x$ . Its integral is $tan x - sec x + C$ . These rewritings expose derivatives already in the standard table.

A family of vertically shifted antiderivatives— At the same horizontal position, all three curves have equal slopes. Their differences are constants.

Selecting one member of the family

An additional value turns the family into a particular function. First integrate to obtain the general expression. Only then substitute the supplied value to determine $C$ . This is exactly how a known starting position supplies information missing from a velocity function.

Example — Using a starting condition

Problem
Find $F$ if $F^{'} (x) = 4 x^{3} - 6$ and $F (0) = 3$ .

1.Integrate the derivative: $F (x) = x^{4} - 6 x + C$ .
2.At $x = 0$ , the equation becomes $3 = C$ .
3.Therefore $F (x) = x^{4} - 6 x + 3$ . Its derivative and its value at zero both satisfy the given conditions.

Common mistake

The power rule cannot be used with $n = - 1$ , because it would require division by zero. Also, $\int f g d x$ is generally not $(\int f d x) (\int g d x)$ . Verify a claimed rule by differentiating it rather than extending linearity to operations it does not cover.

What these methods can express

A continuous function can have an antiderivative that is not expressible using the elementary functions in this chapter. For example, $e^{- x^{2}}$ does not have an elementary antiderivative. Failure to recognise an elementary formula is different from proving that no antiderivative exists.

Quick check

Which expression represents every antiderivative of $2 x$ on the real line?

Quick check

What is $\int x^{- 1} d x$ on an interval avoiding zero?

Quick check

If $F^{'} (x) = 3 x^{2}$ and $F (1) = 5$ , what is $F (x)$ ?

Quick check

Which is an antiderivative of $csc x cot x$ ?

Quick check

Two antiderivatives of the same function on one interval must differ by what?

Practice Problems

Find $\int (4 e^{3 x} + 1) d x$ by inspection.
Find $\int (x - 1/ x)^{2} d x$ after expanding.
Find $\int (x^{3} + 5 x^{2} - 4) / x^{2} d x$ .
Find $\int sec x (sec x + tan x) d x$ .
Find $F$ when $F^{'} (x) = 6 x^{2} - 4$ and $F (1) = 7$ .
Explain why $sin^{- 1} x + C$ and $- cos^{- 1} x + C$ describe the same family on $(- 1, 1)$ .
A particle has velocity $v (t) = 6 t + 2$ and position $s (0) = 5$ . Find $s (t)$ and check the starting value.

Key Takeaways

Integration recovers a function from its derivative. An indefinite integral is a family $F (x) + C$ on an appropriate interval. Standard integrals come from derivative formulas; $1/ x$ needs a logarithm. Linearity handles sums and constant multiples. A supplied function value determines $C$ , and differentiation checks the answer.

Start of chapter

Next · Lesson 2

Integration by Substitution

Recovering a function from its derivative

Standard integrals come from known derivatives

Combining integrals without changing their meaning

Selecting one member of the family

Quiz

Practice Problems

Key Takeaways