Antiderivative: Definition & Examples

An antiderivative of a function

f

is a function

F

whose derivative is

f

F'(x)=f(x).

The task runs differentiation backwards: recover a level from a known rate of change. For example, a model of marginal trading cost can tell an execution desk how total cost changes with order size. An antiderivative recovers the cost curve, but the marginal rate alone cannot reveal its starting level.

One rate, many totals

Suppose a hypothetical execution desk models marginal cost as

c(q)=0.04+0.002q, \qquad q\geq0,

where $q$ is shares traded and $c(q)$ is USD per share. Quantity is treated as continuous. This is a smooth teaching model of execution costs, excluding the purchase price of the shares, rather than a quoted fee schedule or a fitted market-impact model.

Reading the polynomial differentiation rule backwards gives

C(q)=0.04q+0.001q^2+K.

Differentiate to check: $C'(q)=0.04+2(0.001)q=c(q)$ . The constant $K$ disappears. Thus every choice of $K$ gives an antiderivative with exactly the same marginal costs.

Keep the units. The coefficient

0.04

has units USD/share and

0.002

has units USD/share². Multiplication by

q

q^2

leaves each term in

C(q)

in USD. The constant

K

must also be in USD; it is not another per-share charge.

Suppose the desk allocates USD 10 of setup overhead before trading, so that $C(0)=10$ . Substitution gives $K=10$ , selecting

C(q)=0.04q+0.001q^2+10.

At 100 shares,

C(100)=4+10+10=24\text{ USD}.

The increase from trading those shares is only $C(100)-C(0)=14$ USD. The remaining USD 10 was already present at the starting point. Setting the constant to zero without evidence would recover the variable-cost curve, but miss this model's total cost.

Synthetic cost curves with starting constants USD 0, USD 10 and USD 20 reach USD 14, USD 24 and USD 34 at 100 shares. All share marginal cost 0.04 plus 0.002q USD per share and rise by USD 14.

Original hypothetical continuous-quantity cost model. Changing the starting cost shifts the whole curve vertically, leaving its slopes and interval changes unchanged.

Why the family has just one free constant

On an interval, if

F

is one antiderivative of

f

, every antiderivative has the form $F+K$ , where

K

is constant. If

G'=F'

, then

(G-F)'=0

; the mean value theorem implies that

G-F

is constant on the interval. This is the constant-family theorem. OpenStax §4.10 states both directions.

The interval matters. For $f(x)=1/x$ , the domain excludes zero and splits into two intervals. Its antiderivatives may be

F(x)=\begin{cases} \ln x+K_+, & x>0,\\ \ln(-x)+K_-, & x<0. \end{cases}

There is no requirement that $K_+=K_-$ . A known value on the positive side cannot determine the negative side's constant. One starting value fixes one member of the family on a connected interval.

A function, a family, or an accumulated change?

These closely related objects answer different questions:

Object	What it tells you in the cost model
An antiderivative $0.04q+0.001q^2$	One function whose derivative is the specified marginal cost
The indefinite integral $\int c(q)\,dq$	The complete family $0.04q+0.001q^2+K$
The definite integral $\int_0^{100}c(q)\,dq$	The accumulated increase, USD 14
The initial-value condition $C(0)=10$	The extra information selecting the actual total-cost curve in this model

Because

c

is continuous, the fundamental theorem of calculus connects reconstruction to accumulation:

C(q)=C(0)+\int_0^q c(u)\,du.

Here $u$ is the integration variable. The integral supplies the change from zero shares to $q$ shares; $C(0)$ supplies the starting level. Any arbitrary constant cancels when subtracting endpoints. Adding a further arbitrary constant to a definite integral would change its answer.

The useful habit is to differentiate the proposed antiderivative and check the stated starting value separately. Passing the derivative check proves the rate is correct; it cannot establish that the starting cost is correct. MIT's introduction emphasizes this lost constant when reversing differentiation.

For a fuller treatment of reconstruction and standard rules, see Antiderivatives and Initial-Value Problems. The public page provides a preview; the complete lesson requires membership.

Sources

James Stewart, Single Variable Calculus: Early Transcendentals, 6th edition (Thomson Brooks/Cole, 2008), §4.9, “Antiderivatives,” and §§5.3–5.4 on the fundamental theorem, indefinite integrals and net change.
OpenStax, Calculus Volume 1, §4.10, “Antiderivatives”, especially Theorem 4.14 and initial-value problems.
MIT, Calculus for Beginners, §12.1, “The Anti-derivative”, on reversing differentiation, constants and verification.

One rate, many totals

Why the family has just one free constant

A function, a family, or an accumulated change?

Sources

Related terms

Definite integral

Derivative (calculus)

Fundamental theorem of calculus

Indefinite integral

Interval