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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.
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A["Antiderivatives<br/>(§3.9)"]
B["FTC Part 2<br/>(§4.3)"]
end
subgraph "This Skill"
C["Indefinite Integral<br/>Notation"]
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subgraph "Unlocks"
D["Computing<br/>Indefinite Integrals"]
E["u-Substitution<br/>(§4.5)"]
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A --> C
B -.-> C
C --> D
C --> E
click C "indefinite-integral-notation.html"
click D "computing-indefinite-integrals.html"
click A "../ch3-sec9/rectilinear-motion.html"
click B "../ch4-sec3/ftc-part2.html"
click E "../ch4-sec5/u-substitution-definite.html"
From Antiderivatives:
From FTC Part 2: (helpful but not required)
If these feel unfamiliar, review the prerequisite pages before continuing.
| Property | Value |
|---|---|
| Concept | Indefinite Integrals & Net Change |
| Chapter | Chapter 4, Section 4 |
| Difficulty | Beginner |
| Time | ~15 minutes |
$$\int f(x)\, dx = F(x) + C \quad \text{means} \quad F'(x) = f(x)$$
Components:
| 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 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:
The integral sign $\int$ is used for both because they're deeply connected:
Think of it like: the indefinite integral is the recipe, the definite integral is the dish you make with that recipe.
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)
(c) $\int x^2\, dx = x^2$ because the integral and derivative cancel
Classify each as producing a number or a function:
(a) $\int_0^4 \sqrt{t}\, dt$
(b) $\int \cos\theta\, d\theta$
(c) $\int_{-1}^{1} (x^3 + x)\, dx$
(d) $\int e^x\, dx$
Verify that $\int \sec^2 x\, dx = \tan x + C$ by differentiating the right side.
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 $-7$?
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.
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 $f(x)$.
When you need a specific family member (like for an initial condition), you use additional information to find the right $C$.
Looking back:
Looking ahead:
Real-world connections:
| Previous | Up | Next |
|---|---|---|
| FTC Part 2 | Skills Index | Computing Indefinite Integrals |
Last updated: 2026-01-22