The Antiderivative of a Function
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.10: “Antiderivatives” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-10-antiderivatives |
| Textbook used in class | Stewart, Calculus, Section 3.9: “Antiderivatives” |
Opening Scenario
You know how to go forward: given a function $f$, differentiate to get $f'$. Now work backward: given $f'$, find $f$.
If someone hands you the formula for the slope of a curve at every point, can you reconstruct the curve? Almost -- you can reconstruct the shape, but not the height. A curve can be shifted up or down without changing any of its slopes. That is why antiderivatives come with a free constant.
Quick Reference
Definition. $F$ is an antiderivative of $f$ on an interval $I$ if $F'(x) = f(x)$ for every $x \in I$.
General antiderivative. If $F$ is one antiderivative of $f$, then every antiderivative of $f$ on $I$ has the form $F(x) + C$ for some constant $C$.
Notation (introduced here; developed further in Chapter 4): The general antiderivative is written using the indefinite integral symbol: $$\int f(x)\,dx = F(x) + C.$$
Key Concepts
1. Antiderivatives Are Families of Functions
A derivative is unique: each function $f$ has exactly one derivative $f'$. But a function $f$ has infinitely many antiderivatives, one for each value of the constant $C$.
Example 1. Find antiderivatives of $f(x) = 2x$.
Notice that $\dfrac{d}{dx}(x^2) = 2x$. So $F(x) = x^2$ is one antiderivative.
But $G(x) = x^2 + 7$ also satisfies $G'(x) = 2x$. So does $H(x) = x^2 - 100$.
The general antiderivative is $x^2 + C$, where $C$ can be any real number. All antiderivatives of $2x$ are vertical shifts of the parabola $y = x^2$.
Boxed answer: $\displaystyle\int 2x\,dx = x^2 + C$.
2. Why All Antiderivatives Differ by a Constant
This is a theorem, not just a pattern. Here is why it is true.
Suppose $F$ and $G$ are both antiderivatives of $f$ on an interval $I$. Then $F'(x) = f(x)$ and $G'(x) = f(x)$, so the function $h(x) = G(x) - F(x)$ satisfies $h'(x) = G'(x) - F'(x) = 0$ for all $x \in I$.
By the zero-derivative corollary of the Mean Value Theorem, if $h'(x) = 0$ on an entire interval, then $h$ is constant on that interval. So $G(x) - F(x) = C$ for some fixed constant $C$, i.e., $G(x) = F(x) + C$.
The key ingredient is the MVT. The statement “all antiderivatives of $f$ differ by a constant” is not obvious -- it is a consequence of the MVT.
“The constant $C$ is always $0$ unless the problem specifies otherwise.” No. The general antiderivative requires $C$ as an unknown constant. Dropping the $C$ gives one specific antiderivative, not the general one, and leads to wrong answers in initial value problems.
3. Verifying an Antiderivative
To check that $F$ is an antiderivative of $f$: differentiate $F$ and confirm that $F'(x) = f(x)$.
Example 2. Verify that $F(x) = \dfrac{x^3}{3} + 5x - 2$ is an antiderivative of $f(x) = x^2 + 5$.
$F'(x) = x^2 + 5 = f(x)$. Confirmed.
Example 3. Is $G(x) = \sin^2 x$ an antiderivative of $g(x) = \sin(2x)$?
$G'(x) = 2\sin x \cos x = \sin(2x) = g(x)$.
Yes, $G$ is an antiderivative of $g$.
4. The Constant C Carries Meaning
In applied problems, the constant $C$ is determined by additional information about the function. Until that information is provided, $C$ is genuinely unknown, and leaving it out changes the answer.
Example 4. The slope of a curve at every point $(x, y)$ is $f'(x) = 3x^2$. Two facts: one curve in this family passes through $(0, 4)$ and another passes through $(0, -1)$.
General antiderivative: $f(x) = x^3 + C$.
Curve through $(0, 4)$: $f(0) = 0 + C = 4$, so $C = 4$ and $f(x) = x^3 + 4$.
Curve through $(0, -1)$: $C = -1$ and $f(x) = x^3 - 1$.
These are two different functions with the same derivative -- two different antiderivatives of the same $f'$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Omitting $+ C$ | Writing $\int 3x^2\,dx = x^3$ | The general antiderivative is $x^3 + C$; omitting $C$ gives one particular antiderivative, not the family |
| Forgetting to verify by differentiating | Guessing $F(x) = \cos x$ is the antiderivative of $f(x) = \sin x$ | Differentiate: $(\cos x)' = -\sin x \neq \sin x$. The antiderivative of $\sin x$ is $-\cos x + C$ |
| Thinking $C$ must be determined | Writing $C = 0$ with no justification | $C$ is only determined when an initial condition (a specific value of $F$ at a specific point) is given |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Verify that $F(x) = \dfrac{x^4}{4} - 3x^2 + 7$ is an antiderivative of $f(x) = x^3 - 6x$.
Show answer
$F'(x) = x^3 - 6x = f(x)$. Confirmed.
Problem 2. Write the general antiderivative of $f(x) = 5$.
Show answer
We need $F'(x) = 5$. The function $F(x) = 5x$ has $F'(x) = 5$.
$\displaystyle\int 5\,dx = 5x + C$.
Level 2 -- Multiple Steps
Problem 3. The slopes of a family of curves at every $x$ are given by $f'(x) = 6x^2 - 2x$. Find the member of the family passing through $(1, 3)$.
Show answer
General antiderivative: $f(x) = 2x^3 - x^2 + C$.
Condition $f(1) = 3$: $2(1) - 1 + C = 1 + C = 3$, so $C = 2$.
Boxed answer: $f(x) = 2x^3 - x^2 + 2$.
Mastery Checklist
Mental Model
An antiderivative is a backward derivative: knowing the rate of change everywhere, reconstruct the quantity. But the rate of change at every point does not determine the starting height. The family of all antiderivatives -- all vertical shifts of one specific antiderivative -- is like a stack of parallel curves, all with the same slopes at every $x$.
The constant $C$ is the missing piece of information. An initial condition $F(a) = b$ selects one specific curve from the stack by pinning down the height at $x = a$.
Connections
Within Chapter 3
- MVT Corollary (Section 3.2): The theorem that all antiderivatives differ by a constant follows directly from the zero-derivative corollary.
Toward Chapter 4
- The indefinite integral: Chapter 4 introduces the notation $\int f(x)\,dx$ systematically and builds a table of antiderivative formulas.
- The Fundamental Theorem of Calculus: The connection between antiderivatives and definite integrals (the area under a curve) is the central theorem of Chapter 4.
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