19. Anti-Differentiation | Mathematics Advanced

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.

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

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 $C = 4$ .

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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