# Anti-Differentiation

> Working backwards from a gradient. Primitives, the constant C, and the standard integrals.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/anti-differentiation/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The footprints on the beach
Ella finds a line of footprints in the sand at Bondi. From how far apart they are, she can tell where the walker sped up and slowed down. But can she tell where the walker **started**?

Not from the footprints alone. Knowing how fast something changed tells you the shape of the walk, but not the starting point. You need one extra clue, such as "the walker started at the lifeguard tower." Anti-differentiation is the same: from a gradient you can rebuild a function, up to a starting value we call $+ C$.

**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 $\int x^n\,dx = \frac{x^{n+1}}{n + 1} + C$, 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.

### Worked example: Working backwards
"I doubled a number and added 3 to get 17. What was it?" Undo: subtract 3, then halve. $\dfrac{17 - 3}{2} = 7$.

### Lines with the same gradient

$y = 2x$, $y = 2x + 5$ and $y = 2x - 3$ 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: $y = 2x + c$.

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

### Powers ready to go

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

### Key idea: Many functions, one gradient
Adding a constant moves a graph up or down but does not change its steepness. So many functions share the same gradient function. They differ only by a constant.

### Watch out: Do not forget the constant
A line with gradient 3 is $y = 3x + c$, not just $y = 3x$. The constant matters when a point is given.

**Mind map**

```text
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 11: Undoing the power rule
Differentiating takes one off the power and multiplies by the old power. To undo it, do the opposite in reverse order: **add one to the power, then divide by the new power.**

Which function has derivative $3x^2$? It is $x^3$, or $x^3 + 5$, or $x^3 - 2$. We write $x^3 + C$.

Check: $\dfrac{d}{dx}\left(\dfrac{x^4}{4}\right) = x^3$. So a function with gradient $x^3$ is $\dfrac{x^4}{4} + C$.

### Year 12: Primitives and standard integrals
$F(x)$ is a **primitive** of $f(x)$ if $F'(x) = f(x)$. We write

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

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

| $f(x)$ | $\displaystyle\int f(x)\,dx$ |
|---|---|
| $x^n$ ($n \neq -1$) | $\dfrac{x^{n+1}}{n + 1} + C$ |
| $(ax + b)^n$ | $\dfrac{(ax + b)^{n+1}}{a(n + 1)} + C$ |
| $e^{ax + b}$ | $\dfrac{1}{a}e^{ax + b} + C$ |
| $\dfrac{1}{x}$ | $\ln\lvert x \rvert + C$ |
| $\dfrac{f'(x)}{f(x)}$ | $\ln\lvert f(x) \rvert + C$ |
| $\sin ax$ | $-\dfrac{1}{a}\cos ax + C$ |
| $\cos ax$ | $\dfrac{1}{a}\sin ax + C$ |
| $\sec^2 ax$ | $\dfrac{1}{a}\tan ax + C$ |
| $a^x$ | $\dfrac{a^x}{\ln a} + C$ |
| $f'(x)\big[f(x)\big]^n$ | $\dfrac{\big[f(x)\big]^{n+1}}{n + 1} + C$ |

**Examples.**

$\displaystyle\int (6x^2 - 4x + 1)\,dx = 2x^3 - 2x^2 + x + C$

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

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

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

**Finding C.** If $f'(x) = 6x - 2$ and $f(1) = 5$, then $f(x) = 3x^2 - 2x + C$ 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: $\displaystyle\int 3x^2(x^3 + 1)^4\,dx = \frac{(x^3 + 1)^5}{5} + C$. If the number is off, fix it: $\displaystyle\int x(x^2 + 1)^3\,dx = \frac{1}{2}\int 2x(x^2 + 1)^3\,dx = \frac{(x^2 + 1)^4}{8} + C$.

**Divide first.** $\displaystyle\int \frac{x^3 + 2x}{x}\,dx = \int (x^2 + 2)\,dx = \frac{x^3}{3} + 2x + C$.

**Use an identity.** $\tan^2 x = \sec^2 x - 1$, so $\displaystyle\int \tan^2 x\,dx = \tan x - x + C$.

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

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

## 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.
- $\displaystyle\int \frac{1}{x}\,dx = \ln\lvert x \rvert + C$ and $\displaystyle\int e^{ax}\,dx = \frac{1}{a}e^{ax} + C$.
- Use a given point to find $C$.

## Quiz

Try these questions from the chapter quiz.

1. I tripled a number and subtracted 4 to get 20. What was the number?
   - A. 5
   - B. 8
   - C. 6
   - D. 72
   - Answer: B. 8. Undo: add 4 to get 24, then divide by 3.

2. Which line is parallel to $y = 5x + 1$?
   - A. $y = x + 5$
   - B. $y = 5x - 7$
   - C. $y = -5x + 1$
   - D. $y = \dfrac{1}{5}x$
   - Answer: B. $y = 5x - 7$. Same gradient, 5.

3. A line has gradient 2 and passes through $(3, 10)$. Its equation is:
   - A. $y = 2x + 10$
   - B. $y = 2x + 4$
   - C. $y = 2x - 4$
   - D. $y = 3x + 1$
   - Answer: B. $y = 2x + 4$. $10 = 6 + c$, so $c = 4$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/anti-differentiation/#quiz