An indefinite integral, written ∫ f(x) dx = F(x) + C, is the family of all antiderivatives of f, meaning every function F whose derivative is f, where C is an arbitrary constant (AP Calc Topic 6.8, FUN-6.C.1).
An indefinite integral is differentiation run in reverse. Instead of asking "what is the derivative of F?" you ask "which functions have f as their derivative?" The CED states it precisely in FUN-6.C.1: ∫ f(x) dx = F(x) + C, where F'(x) = f(x) and C is any constant. That "+ C" isn't decoration. Since the derivative of any constant is zero, x², x² + 5, and x² − 17 all have the same derivative, so the answer to ∫ 2x dx has to be the whole family x² + C, not one single function.
Notice what's missing compared to a definite integral. There are no bounds on the integral sign, and the answer is a function (plus C), not a number. Every antiderivative rule you know is just a derivative rule read backwards (FUN-6.C.2). And per FUN-6.C.3, some functions simply have no closed-form antiderivative, which is why the AP exam sometimes hands you an integral you're meant to evaluate with a calculator or leave in integral form rather than solve by hand.
Indefinite integrals live in Topic 6.8 of Unit 6 (Integration and Accumulation of Change), supporting learning objective 6.8.A, which asks you to determine antiderivatives using your knowledge of derivatives. They're the engine behind Topic 6.7, because the Fundamental Theorem of Calculus (LO 6.7.A) says you evaluate a definite integral ∫(a to b) f(x) dx by finding an antiderivative F and computing F(b) − F(a). In other words, every analytic definite-integral problem starts with an indefinite-integral skill. On the BC side, Topic 9.5 (LO 9.5.A) extends the same idea to vector-valued functions, where you integrate each component and use initial conditions to pin down the constants. From Unit 6 through Unit 9, if you can't antidifferentiate, you can't move.
Keep studying AP Calculus Unit 9
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view galleryAntiderivative (Unit 6)
These two terms are nearly the same idea wearing different outfits. An antiderivative is one specific function F with F' = f, while the indefinite integral ∫ f(x) dx is the notation for the entire family of antiderivatives, which is exactly why the + C is required.
Fundamental Theorem of Calculus (Unit 6)
The FTC is the bridge between indefinite and definite integrals. It says the area-style definite integral ∫(a to b) f(x) dx equals F(b) − F(a) for any antiderivative F, so finding an indefinite integral is step one of evaluating almost any definite integral analytically.
Velocity Function, v(t) (Units 4 and 8)
Particle motion is indefinite integration with a story attached. Integrating v(t) gives position x(t) + C, and an initial condition like x(0) = 3 lets you solve for C. This turns a family of functions into one particular solution.
Integrating Vector-Valued Functions (Unit 9, BC only)
Topic 9.5 is the BC remix of the same skill. You integrate a rate vector component by component, picking up a constant of integration in each component, then use initial conditions to find the particular solution (LO 9.5.A).
Indefinite integrals show up most directly as "evaluate ∫ ... dx" multiple-choice items, where the answer choices differ by a sign, a coefficient, or the presence of + C. Practice questions in this style ask things like ∫(5cos(x) + 2x) dx or ∫(2eˣ − 3sin(x)) dx, so you need trig, exponential, and power-rule antiderivatives cold, including the sign flips (the antiderivative of sin is −cos). You should also recognize vocabulary questions, since "an integral with no specified bounds" is the definition of an indefinite integral. On FRQs, the term rarely appears verbatim, but the skill is everywhere. Any time you apply the FTC, recover position from velocity, or solve a differential equation with an initial condition, you're antidifferentiating and handling the + C. Forgetting the + C on an FRQ where it matters (like a general solution to a differential equation) costs you the point.
An indefinite integral has no bounds and its answer is a family of functions, F(x) + C. A definite integral ∫(a to b) f(x) dx has bounds and its answer is a single number, often interpreted as net accumulation or area under the curve. The FTC connects them. You use the indefinite-integral skill (find F) to compute the definite integral's number, F(b) − F(a), and notice the C cancels in that subtraction, which is why definite integrals never need a + C.
An indefinite integral ∫ f(x) dx equals F(x) + C, where F is any function with F'(x) = f(x) and C is an arbitrary constant (FUN-6.C.1).
The + C is required because infinitely many functions share the same derivative, so the answer is a family of functions, not one function.
Every antiderivative rule is a differentiation rule run in reverse, so knowing your derivatives is the foundation (FUN-6.C.2).
Definite integrals output a number while indefinite integrals output a function family, and the FTC links them through F(b) − F(a).
An initial condition, like a starting position or x(0) = 3, lets you solve for C and turn the general family into one particular solution.
Some functions have no closed-form antiderivative (FUN-6.C.3), so don't panic if an exam integral can't be solved by hand.
It's the family of all antiderivatives of a function, written ∫ f(x) dx = F(x) + C where F'(x) = f(x). It appears in Topic 6.8 of Unit 6 and is the reverse process of differentiation.
An indefinite integral has no bounds and gives a function plus C, while a definite integral has bounds a and b and gives a number, computed via the FTC as F(b) − F(a). The C cancels in that subtraction, so definite integrals never include + C.
Yes, for indefinite integrals and general solutions to differential equations. Omitting it can cost you a point on FRQs. You don't write + C for definite integrals, because the constant cancels when you compute F(b) − F(a).
Almost, but not exactly. An antiderivative is one specific function F with F' = f, while the indefinite integral is notation for the entire set of antiderivatives, F(x) + C. That's why the constant is built into the definition.
No. Per the CED (FUN-6.C.3), many functions don't have closed-form antiderivatives. Every continuous function has an antiderivative in theory (you can define one as F(x) = ∫ from a to x of f(t) dt), but you can't always express it with elementary functions.