Integration as the Reverse of Differentiation

Introduction

Integration is one of the two fundamental operations of calculus. At this stage of the 9709 course, integration is introduced as the reverse process of differentiation — that is, given a derivative, we work backwards to recover the original function. This is sometimes called antidifferentiation, and the result is called an indefinite integral.

This topic forms the foundation for all later integration work in Pure Mathematics 1, including finding areas under curves and solving problems involving displacement, velocity, and acceleration. In the examination, questions may ask you to integrate expressions directly, find a constant of integration using given conditions, or connect integration explicitly to differentiation.

Core Concept

Recall from differentiation that if $y = x^n$ , then $\dfrac{\mathrm{d}y}{\mathrm{d}x} = nx^{n-1}$ .

Integration reverses this: if $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^{n-1}$ , then $y = \dfrac{x^n}{n} + c$ .

More generally, the indefinite integral of a function $f(x)$ is written

\int f(x)\, \mathrm{d}x

and represents the family of all functions whose derivative is $f(x)$ . The constant of integration $c$ is essential — it accounts for the fact that any constant differentiates to zero, so infinitely many functions share the same derivative.

Integrating $(ax + b)^n$

The syllabus requires integration of expressions of the form $(ax + b)^n$ where $n$ is any rational number except $-1$ , and $a, b$ are constants with $a \neq 0$ .

This result follows directly from the chain rule in reverse. When differentiating $(ax+b)^{n+1}$ , the chain rule produces a factor of $a(n+1)$ . To reverse this, we divide by that factor.

Key Formulae & Definitions

Power rule for integration:

\int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + c, \qquad n \neq -1

Integration of $(ax + b)^n$ :

\int (ax+b)^n \, \mathrm{d}x = \frac{(ax+b)^{n+1}}{a(n+1)} + c, \qquad n \neq -1,\ a \neq 0

Constant multiple rule:

\int k\,f(x)\,\mathrm{d}x = k\int f(x)\,\mathrm{d}x

Sum and difference rule:

\int \bigl[f(x) \pm g(x)\bigr]\,\mathrm{d}x = \int f(x)\,\mathrm{d}x \pm \int g(x)\,\mathrm{d}x

Note: The formula $\dfrac{(ax+b)^{n+1}}{a(n+1)} + c$ is valid for all rational $n$ except $n = -1$ , including negative integers and fractions. Always divide by both $a$ and $(n+1)$ .

Worked Examples

Example 1 — Polynomial and negative power

Find $\displaystyle\int \left(3x^2 - \frac{4}{x^3} + 5\right)\mathrm{d}x$ .

Step 1: Rewrite negative powers using index notation.

\frac{4}{x^3} = 4x^{-3}

So the integral becomes:

\int \left(3x^2 - 4x^{-3} + 5\right)\mathrm{d}x

Step 2: Integrate term by term using the power rule.

= \frac{3x^3}{3} - \frac{4x^{-2}}{-2} + 5x + c

Step 3: Simplify each term.

= x^3 + 2x^{-2} + 5x + c

\boxed{= x^3 + \frac{2}{x^2} + 5x + c}

Justification: Each term uses $\int x^n\,\mathrm{d}x = \dfrac{x^{n+1}}{n+1}+c$ . The constant of integration is added once at the end.

Example 2 — Integrating $(ax+b)^n$ with a fractional power

Find $\displaystyle\int (2x - 3)^{1/2}\,\mathrm{d}x$ .

Step 1: Identify $a$ , $b$ , and $n$ .

a = 2,\quad b = -3,\quad n = \tfrac{1}{2}

Step 2: Apply the formula $\displaystyle\int(ax+b)^n\,\mathrm{d}x = \frac{(ax+b)^{n+1}}{a(n+1)}+c$ .

= \frac{(2x-3)^{\,3/2}}{2 \cdot \frac{3}{2}} + c

Step 3: Simplify the denominator.

2 \cdot \frac{3}{2} = 3

\boxed{= \frac{(2x-3)^{3/2}}{3} + c}

Verification by differentiation: $\dfrac{\mathrm{d}}{\mathrm{d}x}\!\left[\dfrac{(2x-3)^{3/2}}{3}\right] = \dfrac{1}{3} \cdot \dfrac{3}{2}(2x-3)^{1/2} \cdot 2 = (2x-3)^{1/2}$ ✓

Example 3 — Finding a particular function given a condition

A curve has $\dfrac{\mathrm{d}y}{\mathrm{d}x} = (3x+1)^{-2}$ and passes through the point $(1,\, 2)$ . Find the equation of the curve.

Step 1: Integrate $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to find $y$ .

y = \int (3x+1)^{-2}\,\mathrm{d}x = \frac{(3x+1)^{-1}}{3 \cdot (-1)} + c = -\frac{1}{3(3x+1)} + c

Step 2: Use the condition $(1, 2)$ to find $c$ .

2 = -\frac{1}{3(3(1)+1)} + c = -\frac{1}{12} + c

c = 2 + \frac{1}{12} = \frac{25}{12}

Step 3: Write the equation.

\boxed{y = -\frac{1}{3(3x+1)} + \frac{25}{12}}

Common Mistakes & Examiner Pitfalls

Mistake	What goes wrong	How to avoid it
Forgetting to divide by $a$ in $(ax+b)^n$	Writing $\dfrac{(ax+b)^{n+1}}{n+1}$ only	Always divide by both $a$ and $(n+1)$
Omitting the constant of integration $c$	Loses a mark in every indefinite integral	Add $+c$ as a matter of habit
Applying the formula when $n = -1$	Formula breaks down (division by zero)	Recognise $(ax+b)^{-1}$ needs a different approach (ln) — outside this syllabus objective
Not rewriting roots/fractions as indices first	Inability to apply power rule	Always convert: $\sqrt[3]{x} = x^{1/3}$ , $\dfrac{1}{x^2} = x^{-2}$
Adding $c$ to each term separately	Technically valid but messy, and wastes time	One single $+c$ at the end of the whole integral
Integrating $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and forgetting to apply the given point	Equation of curve is wrong	Always substitute coordinates to evaluate $c$

Practice Questions

Q1. Find $\displaystyle\int \left(4x^3 - \frac{3}{\sqrt{x}} + 2\right)\mathrm{d}x$ .

<details><summary>Show answer</summary>

Rewrite: $\dfrac{3}{\sqrt{x}} = 3x^{-1/2}$

\int \left(4x^3 - 3x^{-1/2} + 2\right)\mathrm{d}x

= \frac{4x^4}{4} - \frac{3x^{1/2}}{1/2} + 2x + c

= x^4 - 6x^{1/2} + 2x + c

\boxed{= x^4 - 6\sqrt{x} + 2x + c}

</details>

Q2. Find $\displaystyle\int (5 - 2x)^4\,\mathrm{d}x$ .

<details><summary>Show answer</summary>

Here $a = -2$ , $n = 4$ .

\int (5-2x)^4\,\mathrm{d}x = \frac{(5-2x)^5}{(-2)(5)} + c = \frac{(5-2x)^5}{-10} + c

\boxed{= -\frac{(5-2x)^5}{10} + c}

Check by differentiating: $-\dfrac{1}{10} \cdot 5(5-2x)^4 \cdot (-2) = (5-2x)^4$ ✓

</details>

Q3. Find $\displaystyle\int \frac{6}{(4x-1)^3}\,\mathrm{d}x$ .

<details><summary>Show answer</summary>

Rewrite: $\dfrac{6}{(4x-1)^3} = 6(4x-1)^{-3}$

Here $a = 4$ , $n = -3$ .

\int 6(4x-1)^{-3}\,\mathrm{d}x = 6 \cdot \frac{(4x-1)^{-2}}{4 \cdot (-2)} + c = 6 \cdot \frac{(4x-1)^{-2}}{-8} + c

= -\frac{3}{4}(4x-1)^{-2} + c

\boxed{= -\frac{3}{4(4x-1)^2} + c}

</details>

Q4. A curve is such that $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 4x + 1$ . Given that the curve passes through $(2,\, 5)$ , find $y$ in terms of $x$ .

<details><summary>Show answer</summary>

Integrate:

y = \int (6x^2 - 4x + 1)\,\mathrm{d}x = 2x^3 - 2x^2 + x + c

Apply condition $(2, 5)$ :

5 = 2(8) - 2(4) + 2 + c = 16 - 8 + 2 + c = 10 + c

c = -5

\boxed{y = 2x^3 - 2x^2 + x - 5}

</details>

Q5. Find $\displaystyle\int \left(\sqrt[3]{x^2} + \frac{1}{2(3x+2)^2}\right)\mathrm{d}x$ .

<details><summary>Show answer</summary>

Rewrite: $\sqrt[3]{x^2} = x^{2/3}$ and $\dfrac{1}{2(3x+2)^2} = \dfrac{1}{2}(3x+2)^{-2}$

Integrate the first term:

\int x^{2/3}\,\mathrm{d}x = \frac{x^{5/3}}{5/3} = \frac{3x^{5/3}}{5}

Integrate the second term ( $a=3$ , $n=-2$ ):

\int \tfrac{1}{2}(3x+2)^{-2}\,\mathrm{d}x = \frac{1}{2} \cdot \frac{(3x+2)^{-1}}{3 \cdot (-1)} = -\frac{1}{6}(3x+2)^{-1}

Combine:

\boxed{\frac{3x^{5/3}}{5} - \frac{1}{6(3x+2)} + c}

</details>

Connections

Prerequisite topics (assumed known):

Differentiating Powers — the power rule $\dfrac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n-1}$ is directly reversed here; fluency with it makes integration immediate.
The Chain Rule — differentiating $(ax+b)^n$ via the chain rule provides the exact mechanism that the $(ax+b)^n$ integration formula undoes. Verifying integration results by differentiating back is strongly recommended.

Topics that follow directly:

Definite Integration and Area — using the same integration techniques with limits to find areas under and between curves.
Integration of $\frac{1}{x}$ and exponential/trigonometric functions (Pure Mathematics 2/3) — extends integration to the case $n = -1$ and beyond.
Kinematics using Calculus — applying integration to displacement–velocity–acceleration problems, a key application in Mechanics components.

Introduction

Core Concept

Integrating $(ax + b)^n$

Key Formulae & Definitions

Worked Examples

Example 1 — Polynomial and negative power

Example 2 — Integrating $(ax+b)^n$ with a fractional power

Example 3 — Finding a particular function given a condition

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Introduction

Core Concept

Integrating (ax+b)n(ax + b)^n(ax+b)n

Key Formulae & Definitions

Worked Examples

Example 1 — Polynomial and negative power

Example 2 — Integrating (ax+b)n(ax+b)^n(ax+b)n with a fractional power

Example 3 — Finding a particular function given a condition

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Integrating $(ax + b)^n$

Example 2 — Integrating $(ax+b)^n$ with a fractional power