Indefinite Integral Notation

5 min read

Reference: Stewart §4.4

A Symbol for “Antiderivative”

You’ve already learned that if $F'(x) = f(x)$, then $F$ is an antiderivative of $f$. But writing “the antiderivative of $f$” repeatedly gets tedious. We need shorthand.

The notation $\int f(x)\, dx$ is read “the indefinite integral of $f(x)$ with respect to $x$.” It means any function whose derivative is $f(x)$.

The key insight: there’s not just ONE antiderivative: there’s a whole family. If $F(x)$ works, so does $F(x) + 7$, or $F(x) - \pi$, or $F(x) + C$ for any constant $C$. That’s why we always write:

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

where $C$ is the constant of integration.

Prerequisite Skills

Before You Start

Prerequisite Check: Can you answer these?

From Antiderivatives:

Find a function $F(x)$ such that $F'(x) = 3x^2$.
If $F(x) = x^4$ is one antiderivative of $4x^3$, name two other antiderivatives.

From FTC Part 2: (helpful but not required)

If $F'(x) = f(x)$, what does $\int_a^b f(x)\, dx$ equal in terms of $F$?

Check Your Answers

$F(x) = x^3$ (since $\frac{d}{dx}[x^3] = 3x^2$)
$F(x) = x^4 + 1$ and $F(x) = x^4 - 7$ (or any $x^4 + C$)
$\int_a^b f(x)\, dx = F(b) - F(a)$

If these feel unfamiliar, review the prerequisite pages before continuing.

Quick Reference

Property	Value
Concept	Indefinite Integrals & Net Change
Chapter	Chapter 4, Section 4
Difficulty	Beginner
Time	~15 minutes

Key Concepts

The Indefinite Integral Symbol

$$\int f(x)\, dx = F(x) + C \quad \text{means} \quad F'(x) = f(x)$$

Components:

$\int$: the integral sign (elongated S for “sum”)
$f(x)$: the integrand (the function being integrated)
$dx$: indicates the variable of integration
$F(x) + C$: the general antiderivative (family of functions)

Definite vs. Indefinite: A Critical Distinction

Feature	Definite Integral	Indefinite Integral
Symbol	$\int_a^b f(x)\, dx$	$\int f(x)\, dx$
Result	A number	A function (or family)
Has limits?	Yes ($a$ to $b$)	No
Has $+C$?	No	Yes (always!)
Example	$\int_0^2 x\, dx = 2$	$\int x\, dx = \frac{x^2}{2} + C$

Common error

Forgetting $+C$ on indefinite integrals is the most frequent mistake in calculus. The $+C$ represents the entire family of antiderivatives.

The Connection (FTC2 in Disguise)

The Fundamental Theorem connects these two types:

$$\int_a^b f(x)\, dx = \left[\int f(x)\, dx\right]_a^b = F(b) - F(a)$$

To evaluate a definite integral:

Find the indefinite integral $F(x) + C$
Evaluate $F(b) - F(a)$ (the $C$ cancels!)

Why the Same Symbol?

The integral sign $\int$ is used for both because they’re deeply connected:

The indefinite integral gives the tool (the antiderivative)
The definite integral gives the answer (a number via FTC2)

Think of it like: the indefinite integral is the recipe, the definite integral is the dish you make with that recipe.

Practice Problems

Level 1 Interpreting Notation

Which of the following correctly expresses the relationship between $\int x^2\, dx$ and $\frac{x^3}{3} + C$?

(a) $\int x^2\, dx = \frac{x^3}{3} + C$ because $\frac{d}{dx}\left[\frac{x^3}{3} + C\right] = x^2$

(b) $\int x^2\, dx = \frac{x^3}{3}$ (no $+C$ needed)

Thought Process

The definition: $\int f(x)\, dx = F(x) + C$ means $F'(x) = f(x)$.

Check (a): Does $\frac{d}{dx}\left[\frac{x^3}{3} + C\right] = x^2$?

Using power rule: $\frac{d}{dx}\left[\frac{x^3}{3}\right] = \frac{3x^2}{3} = x^2$ and $\frac{d}{dx}[C] = 0$.

Yes! So (a) is correct.

Show Answer

(a) is correct.

The statement $\int x^2\, dx = \frac{x^3}{3} + C$ is verified by differentiation:

$$\frac{d}{dx}\left[\frac{x^3}{3} + C\right] = x^2 \checkmark$$

(b) is wrong because indefinite integrals always need $+C$
(c) is wrong: integration is not simply “undoing” differentiation in a trivial way; it produces a family of functions

Level 2 Definite vs. Indefinite

Classify each as producing a number or a function:

(a) $\int_0^4 \sqrt{t}\, dt$

(b) $\int \cos\theta\, d\theta$

(d) $\int e^x\, dx$

Thought Process

Look for limits of integration:

If there are limits (numbers on the integral sign) → definite integral → number
If there are no limits → indefinite integral → function

Show Answer

(a) Number: definite integral (limits 0 to 4)

(b) Function: indefinite integral (answer is $\sin\theta + C$)

(d) Function: indefinite integral (answer is $e^x + C$)

Level 3 Verification by Differentiation

Verify that $\int \sec^2 x\, dx = \tan x + C$ by differentiating the right side.

Thought Process

To verify $\int f(x)\, dx = F(x) + C$, we need to show $F'(x) = f(x)$.

Here, we need to show $\frac{d}{dx}[\tan x + C] = \sec^2 x$.

Recall: $\frac{d}{dx}[\tan x] = \sec^2 x$ (a standard derivative)

Show Answer

Verification:

$$\frac{d}{dx}[\tan x + C] = \frac{d}{dx}[\tan x] + \frac{d}{dx}[C]$$

$$= \sec^2 x + 0 = \sec^2 x \checkmark$$

Since differentiating $\tan x + C$ gives $\sec^2 x$, we have verified that:

$$\boxed{\int \sec^2 x\, dx = \tan x + C}$$

Level 4 Finding a Specific Antiderivative

We know $\int 2x\, dx = x^2 + C$ represents a family of functions.

(a) Which member of this family passes through the point $(3, 5)$?

(b) Which member has a $y$-intercept of $-7$?

Thought Process

(a) We need $F(x) = x^2 + C$ where $F(3) = 5$. Solve: $3^2 + C = 5 \Rightarrow 9 + C = 5 \Rightarrow C = -4$

(b) $y$-intercept means the value when $x = 0$. We need $F(0) = -7$. Solve: $0^2 + C = -7 \Rightarrow C = -7$

Show Answer

(a) Finding $C$ using point $(3, 5)$:

If $F(x) = x^2 + C$ and $F(3) = 5$: $$9 + C = 5$$ $$C = -4$$

Answer: $\boxed{F(x) = x^2 - 4}$

(b) Finding $C$ using $y$-intercept $-7$:

$y$-intercept means $F(0) = -7$: $$0 + C = -7$$ $$C = -7$$

Answer: $\boxed{F(x) = x^2 - 7}$

Level 5 Why Can't We Drop the +C?

A student argues: “Since $\frac{d}{dx}[x^2] = 2x$, we have $\int 2x\, dx = x^2$. The $+C$ is just being pedantic.”

Construct a counterexample showing why omitting $+C$ leads to contradictions when solving initial value problems.

Hint: Consider two different functions with the same derivative.

Thought Process

Consider the problem: Find $f(x)$ such that $f'(x) = 2x$ and $f(0) = 5$.

If we claim $\int 2x\, dx = x^2$ (no $+C$), then $f(x) = x^2$.

But then $f(0) = 0 \neq 5$. Contradiction!

The $+C$ is essential for matching initial conditions.

Show Answer

Counterexample:

Consider these two functions:

$f(x) = x^2$
$g(x) = x^2 + 5$

Both have derivative $2x$: $$f'(x) = 2x, \quad g'(x) = 2x$$

Now consider the initial value problem:

Find $F(x)$ such that $F'(x) = 2x$ and $F(0) = 5$.

Without $+C$: If $\int 2x\, dx = x^2$, then $F(x) = x^2$, giving $F(0) = 0 \neq 5$. Contradiction!

With $+C$: $\int 2x\, dx = x^2 + C$, so $F(0) = 0 + C = 5$ gives $C = 5$. Thus $F(x) = x^2 + 5$ solves the problem. $\checkmark$

Conclusion: The $+C$ represents the infinitely many antiderivatives that differ by a vertical shift. Omitting it claims there’s only ONE antiderivative, which is false. $\square$

Common Misconceptions

Common misconception

the indefinite integral $\int f(x)\,dx$ produces a number, just as the definite integral does.

This is the input-output-confusion error. The definite integral $\int_a^b f(x)\,dx$ takes a function and two limits as inputs and outputs a real number. The indefinite integral $\int f(x)\,dx$ takes a function as input and outputs a family of functions, written $F(x) + C$. Writing $\int x^2\,dx = \tfrac{1}{3}$ without a $+C$ and without limits conflates the two objects; $\tfrac{1}{3}$ is the value of $\int_0^1 x^2\,dx$, not of the indefinite integral of $x^2$.

Mastery Checklist

I can state what $\int f(x)\, dx = F(x) + C$ means in terms of derivatives
I always include $+C$ when writing indefinite integrals
I can distinguish between definite (number) and indefinite (function) integrals
I can verify an integral formula by differentiating
I can find a specific antiderivative given an initial condition

Mental Model

The Integral Symbol as a “Family Photo”

Think of $\int f(x)\, dx = F(x) + C$ as a family photo of all functions whose derivative is $f(x)$.

Every family member looks slightly different (different $C$ values)
But they all share the same “genetic trait” (same derivative)
The $+C$ is like saying “and all their siblings”

When you need a specific family member (like for an initial condition), you use additional information to find the right $C$.

Connections

Looking back:

Antiderivatives: the concept this notation represents
FTC Part 2: connects indefinite to definite integrals

Looking ahead:

Computing Indefinite Integrals: using the table of formulas
u-Substitution: technique for harder integrals

Real-world connections:

In physics: Position is the indefinite integral of velocity; the $+C$ represents unknown initial position
In differential equations: Solutions always have arbitrary constants representing initial conditions

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FTC Part 2	Skills Index	Computing Indefinite Integrals

Last updated: 2026-01-22