Antiderivative Definition

6 min read

Reference: Stewart §3.9

Reversing the Derivative

A physicist measures an object’s velocity and wants to find its position. A biologist knows how fast a population is growing and wants to predict its future size. In both cases, the question is the same: given a derivative, what was the original function?

This “reverse differentiation” process is called finding an antiderivative. If differentiation asks “what is the slope?”, antidifferentiation asks “what curve has this slope?”

The surprising twist: there’s never just one answer. Every antiderivative comes with an arbitrary constant, because the derivative “loses information” about vertical shifts.

Prerequisite Map

Quick Reference

Property	Value
Concept	Antiderivatives
Chapter	3.9
Difficulty	Beginner
Time	~15 minutes

Key Concepts

Definition of Antiderivative

A function $F$ is called an antiderivative of $f$ on an interval $I$ if:

$$\boxed{F'(x) = f(x) \text{ for all } x \text{ in } I}$$

In words: $F$ is an antiderivative of $f$ if the derivative of $F$ gives back $f$.

Example: Recognizing Antiderivatives

Is $F(x) = \frac{1}{3}x^3$ an antiderivative of $f(x) = x^2$?

Check: $F'(x) = \frac{d}{dx}\left(\frac{1}{3}x^3\right) = \frac{1}{3} \cdot 3x^2 = x^2 = f(x)$ ✓

Yes! Differentiating $F$ gives back $f$.

The “+C” Mystery: Why Infinitely Many Antiderivatives?

Here’s the key insight: if $F(x)$ is an antiderivative of $f(x)$, so is $F(x) + 5$, or $F(x) - 17$, or $F(x) + C$ for any constant $C$.

Why? Because the derivative of a constant is zero:

$$\frac{d}{dx}[F(x) + C] = F'(x) + 0 = f(x)$$

So $x^3/3$, $x^3/3 + 1$, $x^3/3 - 100$, and $x^3/3 + \pi$ are ALL antiderivatives of $x^2$.

The General Antiderivative Theorem

But here’s the amazing fact: these are the only antiderivatives!

Theorem: If $F$ is an antiderivative of $f$ on an interval $I$, then the most general antiderivative of $f$ on $I$ is:

$$\boxed{F(x) + C}$$

where $C$ is an arbitrary constant.

Why? If $G$ is another antiderivative of $f$, then $G'(x) = f(x) = F'(x)$, so $(G - F)'(x) = 0$. By a consequence of the Mean Value Theorem, a function with zero derivative must be constant. Thus $G(x) - F(x) = C$ for some constant $C$.

Visualizing the Family of Antiderivatives

y
↑        ╱ F(x) + 2
│      ╱
│    ╱   ╱ F(x) + 1
│  ╱   ╱
│╱   ╱   ╱ F(x)
│  ╱   ╱
│╱   ╱   ╱ F(x) - 1
│  ╱   ╱
│╱   ╱
└──────────→ x

The curves are vertical translates of each other. At any $x$-value, they all have the same slope (because they’re all antiderivatives of the same $f$).

Practice Problems

Level 1 Verifying an Antiderivative

Verify that $F(x) = -\cos x$ is an antiderivative of $f(x) = \sin x$.

Thought Process

To verify, differentiate $F(x)$ and check if you get $f(x)$.

Show Answer

Differentiate $F(x) = -\cos x$:

$$F'(x) = -(-\sin x) = \sin x = f(x)$$

Since $F'(x) = f(x)$, $F(x) = -\cos x$ is indeed an antiderivative of $\sin x$. ✓

Level 2 Finding an Antiderivative

Find an antiderivative of $f(x) = 3x^2$.

Hint: What function has derivative $3x^2$?

Thought Process

Think backwards from the Power Rule. If the derivative is $3x^2$, the original had power 3. Work out what coefficient gives 3 after differentiation.

Show Answer

We need $F(x)$ such that $F'(x) = 3x^2$.

By the Power Rule in reverse, $\frac{d}{dx}(x^3) = 3x^2$.

So $F(x) = x^3$ is an antiderivative.

The general antiderivative is $F(x) = x^3 + C$.

Level 3 Distinguishing Antiderivatives

Let $f(x) = 2x$. Which of the following are antiderivatives of $f$?

(a) $F_1(x) = x^2$ (b) $F_2(x) = x^2 + 7$ (c) $F_3(x) = x^2 - 3$ (d) $F_4(x) = 2x^2$

Thought Process

Differentiate each candidate and check if you get $f(x) = 2x$.

Show Answer

Check each by differentiation:

(a) $F_1'(x) = 2x = f(x)$ ✓ (b) $F_2'(x) = 2x = f(x)$ ✓ (c) $F_3'(x) = 2x = f(x)$ ✓ (d) $F_4'(x) = 4x \neq f(x)$ ✗

Answer: (a), (b), and (c) are all antiderivatives. They differ by constants, as expected.

(d) is NOT an antiderivative because $4x \neq 2x$.

Level 4 Discontinuous Domains

Find the most general antiderivative of $f(x) = \frac{1}{x^2}$ on its natural domain.

Be careful: What is the domain of $f$?

Thought Process

$f(x) = x^{-2}$ is undefined at $x = 0$, so its domain is two separate intervals: $(-\infty, 0)$ and $(0, \infty)$. The theorem about “+C” applies on each interval separately, so there could be different constants on each piece.

Show Answer

First, $f(x) = x^{-2}$, and $\frac{d}{dx}(x^{-1}) = -x^{-2}$, so $\frac{d}{dx}(-x^{-1}) = x^{-2}$.

A particular antiderivative is $-\frac{1}{x}$.

However, the domain of $f$ is $(-\infty, 0) \cup (0, \infty)$: two separate intervals!

The theorem guarantees antiderivatives differ by a constant on each interval, but the constants can be different. So:

$$F(x) = \begin{cases} -\frac{1}{x} + C_1 & \text{if } x < 0 \\ -\frac{1}{x} + C_2 & \text{if } x > 0 \end{cases}$$

where $C_1$ and $C_2$ can be different constants.

Level 5 Uniqueness from the Mean Value Theorem

Prove that if $F$ and $G$ are both antiderivatives of $f$ on an interval $I$, then $G(x) = F(x) + C$ for some constant $C$.

Hint: Consider $H(x) = G(x) - F(x)$ and use the fact that a function with zero derivative on an interval must be constant (a consequence of MVT).

Thought Process

Let $H = G - F$. Show $H'(x) = 0$ everywhere on $I$. Then appeal to the MVT corollary: if $H'(x) = 0$ for all $x$ in an interval, then $H$ must be constant.

Show Answer

Proof:

Let $H(x) = G(x) - F(x)$.

Since $F$ and $G$ are both antiderivatives of $f$: $$H'(x) = G'(x) - F'(x) = f(x) - f(x) = 0$$

for all $x$ in $I$.

By a consequence of the Mean Value Theorem (Corollary 3.2.7 in Stewart): if $H'(x) = 0$ for all $x$ in an interval $I$, then $H$ is constant on $I$.

Therefore $H(x) = C$ for some constant $C$, which means: $$G(x) - F(x) = C$$ $$G(x) = F(x) + C$$

This proves that any two antiderivatives of the same function differ by a constant. ∎

CCI-Style Conceptual Questions

Conceptual Why the Constant?

A student says: “The antiderivative of $2x$ is $x^2$, period. There’s no need for +C.”

Explain why this is incomplete and give a specific counterexample.

Thought Process

Find another function whose derivative is also $2x$. If there’s more than one, then “$x^2$” isn’t the complete answer.

Show Answer

The student is incomplete. While $x^2$ is an antiderivative of $2x$, it’s not the only one.

Counterexample: $x^2 + 5$ also has derivative $2x$: $$\frac{d}{dx}(x^2 + 5) = 2x$$

In fact, $x^2 + 100$, $x^2 - \pi$, and $x^2 + C$ for any constant $C$ are all antiderivatives of $2x$.

The complete answer is: the general antiderivative of $2x$ is $x^2 + C$.

Conceptual Graphical Interpretation

If you know that $f(x) > 0$ for all $x$, what can you conclude about any antiderivative $F(x)$?

Thought Process

$F'(x) = f(x) > 0$ means something specific about the behavior of $F$.

Show Answer

Since $F'(x) = f(x) > 0$ for all $x$, the derivative of $F$ is always positive.

This means $F$ is strictly increasing everywhere.

The graph of any antiderivative of $f$ rises from left to right, never decreasing. Different choices of $C$ shift the curve up or down, but they all have the same increasing behavior.

Common Misconceptions

Common misconception

the antiderivative of $f$ is a single function, not a family.

This is the action-view-of-function error. Antidifferentiation does not produce one answer; it produces a family $F(x) + C$ parameterized by an arbitrary constant $C$, because differentiation destroys information about vertical position. Omitting $+C$ and writing, for example, $x^3$ as the antiderivative of $3x^2$ implies a single curve when infinitely many curves share the same derivative. The constant is not a formality; it represents genuinely distinct functions that all satisfy the equation $F'(x) = f(x)$.

Common misconception

differentiating then antidifferentiating always returns the original function.

This is the action-view-of-function error about the direction of the inverse relationship. Antidifferentiating $f$ and then differentiating the result does return $f$. The reverse order, differentiating $F$ and then antidifferentiating $F'$, returns $F + C$, not necessarily $F$ itself, because the constant is lost during differentiation. These two operations are inverses of each other in only one direction without additional information (such as an initial condition) to pin down $C$.

Mastery Checklist

I can state the definition of an antiderivative
I can verify whether a function is an antiderivative by differentiating
I understand why antiderivatives always include “+C”
I can explain why two antiderivatives of the same function differ by a constant
I understand how the MVT justifies the uniqueness of antiderivatives (up to +C)
I can handle functions with disconnected domains (different constants on each piece)

Mental Model

Think of differentiation as “forgetting” information:

When you differentiate $x^2 + 5$, you get $2x$. The “+5” disappears. Differentiation “forgets” the vertical position of the curve: it only remembers the slopes.

Antidifferentiation is like trying to remember something forgotten. You can recover the shape of the curve (the slopes), but you can’t know exactly where it sat vertically. That’s why you need “+C”: it represents the forgotten information.

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Last updated: 2026-01-22