Anti-Differentiation

In short: Anti-differentiation, or integration, undoes differentiation: it finds a function from its gradient. Add one to the power and divide by the new power, so $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ , and always add the constant $C$ .

Starting point: Years 9 and 10

Inverse operations

Every operation has an opposite that undoes it. Adding undoes subtracting. Square roots undo squaring. If you know the answer and the operation, you can work backwards to the start.

Lines with the same gradient

$y = 2 x$ , $y = 2 x + 5$ and $y = 2 x - 3$ all have gradient 2. They are parallel, just moved up or down. If all you know is "the gradient is 2", the line could be any of them: $y = 2 x + c$ .

To pick the right one, you need a point. If the line passes through $(1, 6)$ , then $6 = 2 + c$ , so $c = 4$ and $y = 2 x + 4$ .

Powers ready to go

As with differentiation, write roots and fractions as powers first: $x = x^{\frac{1}{2}}$ and $\frac{1}{x ^{2}} = x^{- 2}$ .

Anti-differentiation

Idea

Undo the derivative
Primitive F with F' = f

Constant

+ C
Find C from a point

Rules

Add one, divide by new power
(ax + b)ⁿ
eˣ and e^(ax)
1/x gives ln|x|
sin and cos

Notation

∫ f(x) dx

Year 12Primitives and standard integrals

$F (x)$ is a primitive of $f (x)$ if $F^{'} (x) = f (x)$ . We write

$\int f (x) d x = F (x) + C$

The $\int$ sign and $d x$ together mean "anti-differentiate with respect to $x$ ". $C$ is the constant of integration.

$f (x)$	$\int f (x) d x$
$x^{n}$ ( $n \neq = - 1$ )	$\frac{x ^{n + 1}}{n + 1} + C$
$(a x + b)^{n}$	$\frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + C$
$e^{a x + b}$	$\frac{1}{a} e^{a x + b} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$\frac{f ^{'} ( x )}{f ( x )}$	$ln ∣ f (x)∣ + C$
$sin a x$	$- \frac{1}{a} cos a x + C$
$cos a x$	$\frac{1}{a} sin a x + C$
$sec^{2} a x$	$\frac{1}{a} tan a x + C$
$a^{x}$	$\frac{a ^{x}}{ln a} + C$
$f^{'} (x) [f (x)]^{n}$	$\frac{[ f ( x ) ] ^{n + 1}}{n + 1} + C$

Examples.

$\int (6 x^{2} - 4 x + 1) d x = 2 x^{3} - 2 x^{2} + x + C$

$\int \frac{1}{x} d x = \int x^{- \frac{1}{2}} d x = 2 x^{\frac{1}{2}} + C = 2 x + C$

$\int (2 x + 1)^{4} d x = \frac{( 2 x + 1 ) ^{5}}{10} + C$

$\int \frac{2 x}{x ^{2} + 3} d x = ln (x^{2} + 3) + C$

Finding C. If $f^{'} (x) = 6 x - 2$ and $f (1) = 5$ , then $f (x) = 3 x^{2} - 2 x + C$ and $5 = 3 - 2 + C$ , so $C = 4$ .

The reverse chain rule. If the derivative of the inside is sitting outside, the power rule works on the bracket: $\int 3 x^{2} (x^{3} + 1)^{4} d x = \frac{( x ^{3} + 1 ) ^{5}}{5} + C$ . If the number is off, fix it: $\int x (x^{2} + 1)^{3} d x = \frac{1}{2} \int 2 x (x^{2} + 1)^{3} d x = \frac{( x ^{2} + 1 ) ^{4}}{8} + C$ .

Divide first. $\int \frac{x ^{3} + 2 x}{x} d x = \int (x^{2} + 2) d x = \frac{x ^{3}}{3} + 2 x + C$ .

Use an identity. $tan^{2} x = sec^{2} x - 1$ , so $\int tan^{2} x d x = tan x - x + C$ .

Integrating twice. If $f^{''} (x) = 6 x$ and $f$ has a stationary point at $(1, 4)$ : $f^{'} (x) = 3 x^{2} + C$ , and $f^{'} (1) = 0$ gives $C = - 3$ . Then $f (x) = x^{3} - 3 x + D$ , and $f (1) = 4$ gives $D = 6$ .

Differentiate, hence integrate. If a question shows $\frac{d}{d x} (x ln x) = ln x + 1$ , then $\int (ln x + 1) d x = x ln x + C$ , and you can rearrange to integrate $ln x$ itself: $\int ln x d x = x ln x - x + C$ .

Quick summary

Anti-differentiation undoes differentiation. $F$ is a primitive of $f$ if $F^{'} = f$ .
Functions with the same gradient differ by a constant, so always add $+ C$ .
Power rule backwards: add one to the power, divide by the new power.
$\int \frac{1}{x} d x = ln ∣ x ∣ + C$ and $\int e^{a x} d x = \frac{1}{a} e^{a x} + C$ .
Use a given point to find $C$ .

Quiz

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Try 3 questions now

I tripled a number and subtracted 4 to get 20. What was the number?
1. A 5
2. B 8
3. C 6
4. D 72
Show answer
B: 8
Undo: add 4 to get 24, then divide by 3.
Which line is parallel to $y = 5 x + 1$ ?
1. A $y = x + 5$
2. B $y = 5 x - 7$
3. C $y = - 5 x + 1$
4. D $y = \frac{1}{5} x$
Show answer
B: $y = 5 x - 7$
Same gradient, 5.
A line has gradient 2 and passes through $(3, 10)$ . Its equation is:
1. A $y = 2 x + 10$
2. B $y = 2 x + 4$
3. C $y = 2 x - 4$
4. D $y = 3 x + 1$
Show answer
B: $y = 2 x + 4$
$10 = 6 + c$ , so $c = 4$ .