Antiderivatives and Indefinite… — AP® Calculus BC Notes

To use the Fundamental Theorem you need antiderivatives, and lots of them. This page collects the basic rules, introduces the notation for "all the antiderivatives at once," and shows how a single known point picks out the function you want.

A family, not a single function

Five curves y = x cubed over 3 - x + C for C = -2, -1, 0, 1 and 2, each a vertical shift of the others. At x = 1.5 a dashed line crosses all five, and a short tangent segment at each crossing shows the same slope, 1.25, because every curve has derivative x squared - 1. — **Figure 1.** y = x³/3 - x + C for five values of C. Same slope at every x; different heights. LevlPrep original diagram.

$x^3/3 - x$ has derivative $x^2 - 1$ . So does $x^3/3 - x + 5$ , and $x^3/3 - x - 2$ : a constant has derivative 0. On an interval, these are the only antiderivatives; two functions with the same derivative differ by a constant. Graphically they are vertical shifts of one curve, as the figure shows.

The indefinite integral names the whole family:

$\int (x^2 - 1)\,dx = \frac{x^3}{3} - x + C.$

$C$ is the constant of integration. Leaving it off says "this one function," which is wrong: the scoring guidelines take a point for a missing $+ C$ .

Three objects with the same sign

Written	Name	What it is	Example with $2x$
$\int_a^b f(x)\,dx$	definite integral	a number	$\int_0^3 2x\,dx$ $= 9$
$\int f(x)\,dx$	indefinite integral	a family of functions	$\int 2x\,dx = x^2 + C$
$\int_a^x f(t)\,dt$	accumulation function	one function of $x$ , 0 at $x = a$	$\int_1^x 2t\,dt$ $= x^2 - 1$

The basic rules

Each rule is a derivative rule read backward. Differentiate the right side and you get the integrand.

Integrand	Antiderivative (add $+ C$ )	Because
$x^n$ , $n \ne -1$	$\frac{x^{n+1}}{n+1}$	power rule
$\frac{1}{x}$	$\ln\|x\|$	$\frac{d}{dx}\ln\|x\| = \frac{1}{x}$
$e^x$	$e^x$	$e^x$ is its own derivative
$\cos x$	$\sin x$	$\frac{d}{dx}\sin x = \cos x$
$\sin x$	$-\cos x$	$\frac{d}{dx}\cos x = -\sin x$
$\sec^2 x$	$\tan x$	$\frac{d}{dx}\tan x = \sec^2 x$
$\sec x \tan x$	$\sec x$	$\frac{d}{dx}\sec x = \sec x\tan x$
$\csc^2 x$	$-\cot x$	$\frac{d}{dx}\cot x = -\csc^2 x$
$\frac{1}{1 + x^2}$	$\arctan x$	inverse tangent derivative
$\frac{1}{\sqrt{1 - x^2}}$	$\arcsin x$	inverse sine derivative

Constant factors come out and sums split: $\int (kf + g)\,dx$ $= k\int f\,dx + \int g\,dx$ . Products and quotients do not split, so rewrite them first.

Worked example 1. Find $\int \left(\sqrt{x} + \frac{4}{x^3} - 2\cos x\right)dx$ .

Rewrite as powers: $\sqrt{x} = x^{1/2}$ and $\frac{4}{x^3} = 4x^{-3}$ .

$\int x^{1/2}\,dx$ $= \frac{x^{3/2}}{3/2}$ $= \frac{2}{3}x^{3/2}$ . $\int 4x^{-3}\,dx$ $= 4\cdot\frac{x^{-2}}{-2}$ $= -\frac{2}{x^2}$ . $\int 2\cos x\,dx = 2\sin x$ .

Answer: $\frac{2}{3}x^{3/2} - \frac{2}{x^2} - 2\sin x + C$ . Check: differentiating gives back $x^{1/2} + 4x^{-3} - 2\cos x$ .

Two more rewrites: expand a power of a sum, $(x^2 + 1)^2 = x^4 + 2x^2 + 1$ , and split a fraction over a single term, $\frac{x^3 - 2x}{x} = x^2 - 2$ . Never integrate the top and bottom of a fraction separately.

Picking one function: use a known point

Worked example 2. $f'(x) = 3x^2 - 4x + 1$ and $f(2) = 5$ . Find $f(x)$ .

Antidifferentiate: $f(x) = x^3 - 2x^2 + x + C$ .

Substitute the point right away: $f(2)$ $= 8 - 8 + 2 + C$ $= 2 + C$ $= 5$ , so $C = 3$ .

$f(x) = x^3 - 2x^2 + x + 3$ .

With a second derivative, antidifferentiate twice; each step brings its own constant, and each needs its own piece of information, such as $f'(0)$ and then $f(0)$ .

Some functions have no formula for an antiderivative

Every continuous function has antiderivatives (the accumulation function $\int_a^x f(t)\,dt$ is one), but some cannot be written with familiar functions: $e^{-x^2}$ and $\sin(x^2)$ are examples. On the exam, integrals of such functions appear on calculator questions, and the calculator evaluates them numerically.