Antiderivatives and Indefinite Integrals

Part 1 · Hook

Why this matters

Tell someone a car's speed at every moment and ask where it is. They cannot say, not without knowing where it started. Knowing a derivative pins a function down only up to a starting value, and that missing piece is the

+ C

that students lose points for forgetting.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the derivative of $\tan x$ ?

$\sec^2 x$
$\sec x \tan x$
$-\csc^2 x$
$\cot x$

Show the answer

$\frac{d}{dx}\tan x = \sec^2 x$ .

Correct: $\sec^2 x$ :
$\sec x \tan x$ :
$-\csc^2 x$ :
$\cot x$ :

2. Write $\frac{1}{\sqrt{x}}$ as a power of $x$ .

$x^{-1/2}$
$x^{1/2}$
$x^{-2}$
$-x^{1/2}$

Show the answer

$\sqrt{x} = x^{1/2}$ , and $\frac{1}{x^a} = x^{-a}$ .

Correct: $x^{-1/2}$ :
$x^{1/2}$ :
$x^{-2}$ :
$-x^{1/2}$ :

3. If $F' = f$ , what is $\int_1^4 f(x)\,dx$ ?

$F(4) - F(1)$
$F(1) - F(4)$
$f(4) - f(1)$
$F(4)$

Show the answer

The Fundamental Theorem, part 2.

Correct: $F(4) - F(1)$ :
$F(1) - F(4)$ :
$f(4) - f(1)$ :
$F(4)$ :

Part 4 · See it

See it first

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. — The curves y = x³/3 - x + C: vertical shifts of one another, with the same slope at every x. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A constant has derivative 0if $F' = f$ , then $(F + C)' = f$ for every constant $C$
Two antiderivatives of f on an intervaldiffer by a constant, so the whole family is F(x) + C
Write the family as $\int f(x)\,dx$ $\int f(x)\,dx = F(x) + C$ : an indefinite integral always ends in $+ C$
You also know one point on the graphsubstitute it to find $C$ and pick out the one function

Part 6 · Key ideas

Key ideas

Indefinite integral: $\int f(x)\,dx = F(x) + C$ , the family of all antiderivatives. No limits, a family, never a number.
Power rule backward: $\int x^n\,dx$ $= \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$ ; and $\int \frac{1}{x}\,dx = \ln|x| + C$ .
Rewrite before integrating: roots as powers, products expanded, a fraction over a single term split up.
Check every answer by differentiating it. Find $C$ from a known point by substituting it.

Part 7 · Misconception

A common mistake

The wrong idea: $\int \sin x\,dx = \cos x + C$ , because sine and cosine go together.

What actually happens: $\frac{d}{dx}\cos x = -\sin x$ , so $\int \sin x\,dx = -\cos x + C$ . Differentiate your answer to check the sign.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A function known from its derivative and one value

A function

f

has derivative

f'(x) = 3x^2 - 4x + 1

, and its graph passes through

(2, 5)

.

What is known about f
Given	Value
$f'(x)$	$3x^2 - 4x + 1$
$f(2)$	5

1. Find $\int (3x^2 - 4x + 1)\,dx$ . Include the constant: write $+ C$ .

Type an expression for $\int (3x^2 - 4x + 1)\,dx =$ .

Show the answer

Term by term: $\int 3x^2\,dx = x^3$ , $\int 4x\,dx = 2x^2$ , $\int 1\,dx = x$ . So $\int (3x^2 - 4x + 1)\,dx$ $= x^3 - 2x^2$ $+ x$ $+ C$ . Check: the derivative is $3x^2 - 4x + 1$ .

Answer: $\int (3x^2 - 4x + 1)\,dx =$ $x^{3} - 2x^{2} + x + C$

2. $f(x) = x^3 - 2x^2 + x + C$ . Find the value of $C$ that makes the graph pass through $(2, 5)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f(2)$ $= 2^3 - 2(2)^2$ $+ 2$ $+ C$ $= 2$ $+ C$ $= 5$ , so $C = 3$ .

Answer: 3

3. Write $f(x)$ .

Type an expression for $f(x) =$ .

Show the answer

$f(x) = x^3 - 2x^2 + x + 3$ . Check: $f'(x) = 3x^2 - 4x + 1$ and $f(2) = 8 - 8 + 2 + 3 = 5$ .

Answer: $f(x) =$ $x^{3} - 2x^{2} + x + 3$

4. Using the function you found, evaluate it at the origin: what is $f(0)$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f(0) = 0 - 0 + 0 + 3 = 3$ .

Answer: 3

5. Rewrite as a power of $x$ , then antidifferentiate: $\int \frac{1}{x^3}\,dx$ . Include the constant: write $+ C$ .

Type an expression for $\int \frac{1}{x^3}\,dx =$ .

Show the answer

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

Answer: $\int \frac{1}{x^3}\,dx =$ $-\frac{1}{2x^2} + C$

6. Which is $\int \frac{1}{x}\,dx$ ?

$\ln|x| + C$
$\frac{x^0}{0} + C$
$-\frac{1}{x^2} + C$
$\frac{1}{\ln x} + C$

Show the answer

For $x > 0$ , $(\ln x)' = \frac{1}{x}$ ; for $x < 0$ , $(\ln(-x))' = \frac{1}{x}$ too. So $\int \frac{1}{x}\,dx = \ln|x| + C$ .

Correct: $\ln|x| + C$ : Correct: $\frac{d}{dx}\ln|x| = \frac{1}{x}$ for $x \ne 0$ .
$\frac{x^0}{0} + C$ : The power rule needs $n \ne -1$ ; for $x^{-1}$ it would divide by 0. The antiderivative is $\ln|x|$ .
$-\frac{1}{x^2} + C$ : That is the derivative of $\frac{1}{x}$ , the wrong direction.
$\frac{1}{\ln x} + C$ : The derivative of $\frac{1}{\ln x}$ is $-\frac{1}{x(\ln x)^2}$ , not $\frac{1}{x}$ .

Part 9 · Summary

Summary

Every derivative rule read backward gives an antidifferentiation rule. The indefinite integral

\int f(x)\,dx = F(x) + C

is the whole family of antiderivatives, so it always carries

+ C

. Rewrite the integrand into basic forms first, check by differentiating, and use a known point to find

C

.

Part 10 · Up next

What comes next

Part 11 · Connections