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.
| Property | Value |
|---|---|
| Concept | Antiderivatives |
| Chapter | 3.9 |
| Difficulty | Beginner |
| Time | ~15 minutes |
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 $f$.
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 $f$.
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 $x^2$.
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 $C$.
y
↑ ╱ F(x) + 2
│ ╱
│ ╱ ╱ F(x) + 1
│ ╱ ╱
│╱ ╱ ╱ F(x)
│ ╱ ╱
│╱ ╱ ╱ F(x) - 1
│ ╱ ╱
│╱ ╱
└──────────→ xThe 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 $f$).
Verify that $F(x) = -\cos x$ is an antiderivative of $f(x) = \sin x$.
Find an antiderivative of $f(x) = 3x^2$.
Hint: What function has derivative $3x^2$?
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) $F_4(x) = 2x^2$
Find the most general antiderivative of $f(x) = \frac{1}{x^2}$ on its natural domain.
Be careful: What is the domain of $f$?
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).
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.
If you know that $f(x) > 0$ for all $x$, what can you conclude about any antiderivative $F(x)$?
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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|---|---|---|
| Newton's Method | Skills Index | Antidifferentiation Rules |
Last updated: 2026-01-22