Antiderivative — definition, family of solutions (+C): Question bank

Before you start, recall the three load-bearing facts everything here leans on:

$\frac{d}{d x} (C) = 0$ — a constant has zero slope, so adding it hides no information about the shape.
Zero derivative on ONE interval ⇒ constant — this is the Mean Value Theorem doing the heavy lifting.
$+ C$ is the only freedom; two antiderivatives on a connected interval differ by nothing else.

True or false — justify

True or false: If

F^{'} (x) = G^{'} (x)

for all

x

in an interval, then

F (x) = G (x)

False. Equal slopes force

F

and

G

to differ by a constant, not to be identical —

F - G

has zero derivative, so it is some constant that may be nonzero.

True or false:

\int f (x) d x

names a single function.

False. It names an entire family

F (x) + C

; the

+ C

packages infinitely many parallel curves, so no single curve is "the" answer.

True or false: Since

C

vanishes when you differentiate back, writing it is optional bookkeeping.

False.

C

is invisible after differentiating but it is the whole content of an indefinite integral — dropping it claims a unique answer where infinitely many exist.

True or false: Every function has an antiderivative that you can write with elementary functions.

False. Having an antiderivative (guaranteed for continuous

f

by the Fundamental Theorem of Calculus) is different from being able to express it in elementary form — e.g.

e^{- x^{2}}

has an antiderivative but no elementary formula.

True or false: If two antiderivatives of

f

agree at one point, they agree everywhere on that interval.

True. They differ by a constant

C

; if they agree at one point then

C = 0

there, and a constant is the same everywhere, so they coincide on the whole interval.

True or false: The graphs of

F (x) + C

for different

C

can cross each other.

False. They are pure vertical translates; a fixed vertical shift never lets two parallel curves meet. Crossing would require them to share a height at some

x

, contradicting the constant gap.

True or false:

\int 0 d x = 0

False. The antiderivatives of the zero function are exactly the constants, so

\int 0 d x = C

— the shape is flat, but the height is still unknown.

True or false: On a definite integral

\int_{a}^{b} f d x

, the choice of

C

changes the answer.

False.

[F + C]_{a}^{b} = (F (b) + C) - (F (a) + C) = F (b) - F (a)

; the constant cancels, which is exactly why definite integrals are single numbers (see Definite Integral).

Spot the error

Find the flaw: "

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C

for every real

n

It breaks at

== n = - 1 ==

, where

n + 1 = 0

gives division by zero. That gap is precisely where the logarithm takes over:

\int x^{- 1} d x = ln ∣ x ∣ + C

(see Logarithmic & Exponential Integrals).

Find the flaw: "

\int \frac{1}{x} d x = ln x + C

The domain

x \neq = 0

excludes

x < 0

under

ln x

. The correct antiderivative valid on both signs is

ln ∣ x ∣ + C

, since

\frac{d}{d x} ln ∣ x ∣ = 1/ x

for all

x \neq = 0

Find the flaw: "

\frac{d}{d x} ln ∣ x ∣ = 1/ x

, and

1/ x

is defined nowhere risky, so one constant

C

covers the whole answer."

The domain splits into two disconnected intervals

(- \infty, 0)

and

(0, \infty)

. The "only-a-constant" theorem needs a single connected interval, so each piece gets its own constant

C_{1}, C_{2}

Find the flaw: "

H^{'} (x) = 0

everywhere on the domain

{x \neq = 0}

, therefore

H

is a single constant."

Zero derivative forces constancy only on a connected interval. Across the hole at

x = 0

the two pieces can sit at different heights, so

H

can be one constant on the left and another on the right.

Find the flaw: "

\int cos x d x = sin x

, done."

Missing the

+ C

: the correct family is

sin x + C

. Without a condition like

F (0) = 5

you cannot pin one curve, so leaving off

C

is an unfinished (and technically false) answer.

Find the flaw: "Since differentiation is a function, integration must be its inverse function."

Differentiation is not one-to-one — it maps a whole family

F + C

to the same

f

— so it has no true inverse function. "Antiderivative" recovers the family, not a single input.

Why questions

Why does adding a constant leave the slope unchanged at every point?

Because slope is a rate of change, and a constant never changes; formally

\frac{d}{d x} (F + C) = F^{'} + 0 = f

. Geometrically, sliding a curve straight up moves every point equally, so tangent directions are untouched.

Why must we invoke the Mean Value Theorem to conclude "zero derivative ⇒ constant"?

Intuition says flat slope means flat curve, but proving it requires MVT: on any

[a, b]

H (b) - H (a) = H^{'} (c) (b - a) = 0

, so

H

takes the same value everywhere on the interval.

Why does a single initial condition, not two, suffice to pin down one antiderivative?

The family has exactly one degree of freedom, the constant

C

. One equation

F (x_{0}) = y_{0}

solves for that one unknown, selecting a unique curve (this is the setup of an initial value problem).

Why is

+ C

the only freedom — why not

+ C x

+ C sin x

Because

\frac{d}{d x} (C x) = C \neq = 0

and

\frac{d}{d x} (C sin x) = C cos x \neq = 0

in general; adding those would change the slope. Only a pure constant has derivative zero, so only a constant is invisible to differentiation.

Why can the same

f

describe curves at wildly different heights?

Because

f

only encodes steepness at each

x

, never absolute height. Knowing the shape leaves the "starting height" completely free — that missing height is the

+ C

Edge cases

Edge case: What is the general antiderivative of

f (x) = 0

Any constant:

\int 0 d x = C

. The curve is perfectly flat, but its height is still undetermined, so the family is all horizontal lines.

Edge case: On the domain

x \neq = 0

, how many free constants does

\int \frac{1}{x} d x

truly have?

Two, one per connected interval:

ln (- x) + C_{1}

for

x < 0

and

ln x + C_{2}

for

x > 0

. A textbook "

ln ∣ x ∣ + C

" quietly assumes you stay on one side.

Edge case: If

f

is defined only at isolated points (not on an interval), does "antiderivative" even make sense?

Not in the usual sense — the definition requires

F^{'} = f

on an interval, because slope is a limiting notion needing nearby points. Isolated points give no interval to differentiate over.

Edge case: Can two antiderivatives of

f

on the same interval have graphs that are tangent (touch without crossing) at one point?

No — they differ by a constant, so their vertical gap is fixed. Touching would need the gap to reach zero at one point while being nonzero elsewhere, which a constant cannot do.

Edge case: For a continuous

f

[a, b]

, is an antiderivative guaranteed to exist even if you cannot write a formula?

Yes. The Fundamental Theorem of Calculus builds one as

G (x) = \int_{a}^{x} f (t) d t

; existence is guaranteed by continuity, independent of whether a closed-form expression is available.

Recall One-line summary of every trap here

Every trap on this page is one of three ideas in disguise: (1) slope hides height → always $+ C$ ; (2) zero slope means constant only on ONE connected interval → watch for domain holes; (3) the power rule dies at $n = - 1$ → logarithm appears.

Connections

Mean Value Theorem — the engine behind "zero derivative ⇒ constant."
Power Rule (Integration) — the $n = - 1$ trap.
Logarithmic & Exponential Integrals — where $ln ∣ x ∣$ comes from and why the absolute value.
Differential Equations — Initial Value Problems — how a condition fixes $C$ .
Fundamental Theorem of Calculus — existence of antiderivatives, link to definite integrals.
Definite Integral — where $+ C$ harmlessly cancels.