Your weak spots

Insights load after your first practice round.

Module 6 · L1 of 15 ~35 min ⚡ +95 XP available

Introduction to Integration

A car's speedometer shows 60 km/h. After 2 hours, how far has it travelled? You multiply: distance = speed × time. But what if the speed keeps changing? Integration is the tool that turns changing rates into total amounts, the inverse of differentiation, and one of the most powerful ideas in mathematics.

Today's hook, If $f'(x) = 2x$, what could $f(x)$ be? It seems like a simple question, but the answer reveals something surprising: differentiation destroys information that integration can only partially recover. Why can't we know $f(x)$ exactly?

0/5QUESTS

Orient to antiderivatives

Work through the visible teaching and complete each embedded check.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall, your gut answer first

+5 XP warm-up

If $f'(x) = 2x$, what could $f(x)$ be? Without calculating make a prediction before reading on. Think about which function, when differentiated, gives $2x$.

auto-saved

What is integration?

+5 XP to read

Differentiation tells us the rate of change at any instant. Integration does the opposite: given a rate of change, it finds the total change.

If $\frac{d}{dx}(x^3) = 3x^2$, then $\int 3x^2 \, dx = x^3 + C$.

The symbol $\int$ is an elongated "S" representing sum because integration adds up infinitely many tiny pieces.

Key idea: If differentiation asks "how fast?", integration asks "how much in total?"

Example: $\frac{d}{dx}(x^4) = 4x^3$, so $\int 4x^3 \, dx = x^4 + C$.

Differentiation gives the rate of change; integration recovers the total amount.

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

The constant of integration $+C$. Since $\frac{d}{dx}(x^3) = 3x^2$ and $\frac{d}{dx}(x^3 + 5) = 3x^2$ and $\frac{d}{dx}(x^3 - 100) = 3x^2$, the antiderivative of $3x^2$ could be $x^3$ plus any constant. We write $+C$ to represent this unknown constant. Differentiation destroys constant information, and integration cannot recover it without a boundary condition.

What you'll master

Know

Key facts

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$
Integration reverses differentiation
The constant of integration $C$

Understand

Concepts

Integration as the inverse of differentiation
Why the constant $+C$ is needed
The relationship between rate and total amount

Can do

Skills

Find antiderivatives of power functions
Apply the sum and constant multiple rules
Check answers by differentiating

Build the integration language

Work through the visible teaching and complete each embedded check.

Key terms

IntegrationThe process of finding the antiderivative of a function; the inverse of differentiation.

AntiderivativeA function $F(x)$ such that $F'(x) = f(x)$; written $\int f(x) \, dx = F(x) + C$.

Constant of integration ($C$)An arbitrary constant added to every indefinite integral because derivatives of constants are zero.

Integral sign ($\int$)Elongated "S" standing for "sum"; instructs us to find the antiderivative.

Power rule (integration)$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for all $n \neq -1$.

Sum rule$\int (f + g) \, dx = \int f \, dx + \int g \, dx$, integrate term by term.

Use the power rule

Work through the visible teaching and complete each embedded check.

The power rule for integration

core concept

For any power $n \neq -1$:

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

Verification: Differentiate $\frac{x^{n+1}}{n+1}$: $\frac{d}{dx}\!\left(\frac{x^{n+1}}{n+1}\right) = \frac{(n+1)x^n}{n+1} = x^n$ ✓

$\int x^2 \, dx$

$= \dfrac{x^3}{3} + C$. Add 1 to power, divide by new power.

$\int \sqrt{x} \, dx$

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

$\int \frac{1}{x^2} \, dx$

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

Critical exception. The power rule does NOT work for $n = -1$. The case $\int \frac{1}{x} \, dx$ is special and equals $\ln|x| + C$. Applying the power rule here would give $\frac{x^0}{0}$, which is undefined.

Key rules to memorise:

Constant rule: $\int k \, dx = kx + C$
Sum rule: $\int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx$
Constant multiple: $\int k \cdot f(x) \, dx = k \int f(x) \, dx$

Velocity and displacement. A car's velocity is $v(t) = 6t$ m/s. To find its displacement after 3 seconds, integrate: $s = \int 6t \, dt = 3t^2 + C$. If $s = 0$ at $t = 0$, then $C = 0$ and $s(3) = 27$ metres. Without integration, we could only find displacement for constant velocity. Integration lets us handle any changing rate.

Power rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (add 1 to power, divide by new power); Constant: $\int k \, dx = kx + C$

Pause, copy the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ (valid for all $n \ne -1$) and the constant rule $\int k\,dx = kx + C$ into your book.

Did you get this? True or false: the reason we write $+C$ in an indefinite integral is that the derivative of any constant is zero, so integration cannot determine the constant without extra information.

Work through antiderivatives

Work through the visible teaching and complete each embedded check.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC POWER RULE

Find $\int (3x^2 + 4x - 5) \, dx$.

$\int 3x^2 \, dx = 3 \cdot \dfrac{x^3}{3} = x^3$

Apply power rule to first term: add 1 to power, divide by new power.

PROBLEM 2 · FINDING A CONSTANT

If $\frac{dy}{dx} = 3x^2 - 4x + 2$ and the curve passes through $(1, 5)$, find the equation of the curve.

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

Integrate term by term to find the general equation.

PROBLEM 3 · CHECKING YOUR ANSWER

Is $\int 5x^4 \, dx = x^5 + C$ correct? Verify your answer.

Apply the power rule: $\int 5x^4 \, dx = 5 \cdot \dfrac{x^5}{5} + C = x^5 + C$

Raise the power by 1 (4 → 5), divide by 5, multiply by coefficient 5.

Quick check: Which is the correct antiderivative of $4x^3 - 3x^2 + 2x - 1$?

Avoid core integration traps

Work through the visible teaching and complete each embedded check.

Common errors · the 3 traps that cost marks

Trap 01

Forgetting the $+C$

Writing $\int x^2 \, dx = \frac{x^3}{3}$ without the $+C$ is wrong. Markers expect it always on indefinite integrals. It represents an entire family of curves.

Trap 02

Using the power rule for $n = -1$

$\int \frac{1}{x} \, dx \neq \frac{x^0}{0}$, that's undefined. The correct answer is $\ln|x| + C$. This exception must be memorised separately.

Trap 03

Not simplifying coefficients

$\int 6x^2 \, dx = 6 \cdot \frac{x^3}{3} = 2x^3 + C$, not $\frac{6x^3}{3} + C$. Simplify the coefficient at each step to avoid ugly expressions.

Copy into your book. Write out the three core integration rules (power rule, constant rule, sum rule) and the $n = -1$ exception. This is your reference card for the rest of Module 6.

auto-saved

Apply and reflect

Work through the visible teaching and complete each embedded check.

Quick-fire practice · 5 antiderivatives

Find $\int x^4 \, dx$

Find $\int 6x^2 \, dx$

Find $\int (2x^3 - 5x + 3) \, dx$

Find $\int \frac{1}{x^3} \, dx$ (rewrite first)

Find $\int \sqrt[3]{x} \, dx$ (rewrite as power)

Fill in the blank: $\int x^5 \, dx = \dfrac{x^{\boxed{?}}}{6} + C$. What is the missing exponent?

Real world: velocity and displacement

A particle has velocity $v(t) = 2t + 3$ m/s. Find its displacement function $s(t)$ given $s(0) = 0$.

$s(t) = \int (2t + 3) \, dt = t^2 + 3t + C$

At $t = 0$: $s(0) = 0 + 0 + C = 0$, so $C = 0$.

Therefore $s(t) = t^2 + 3t$.

Now try this: if $\frac{dy}{dx} = 4x - 3$ and $y = 5$ when $x = 1$, find $y$.

auto-saved

Odd one out. Three of these are antiderivatives of $x^2$. Which one is NOT?

Revisit your thinking

Earlier you were asked: if $f'(x) = 2x$, what could $f(x)$ be? The answer is $f(x) = x^2 + C$. We cannot determine $C$ without additional information (like a point on the curve). This is why the constant of integration is essential, differentiation destroys constant information, and integration cannot recover it without a boundary condition.

auto-saved

Complete the mixed practice

Complete the question-bank round and short-answer practice, then compare your responses with the comprehensive answers.

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

Q1. Find $\int (4x^3 - 3x^2 + 2x - 1) \, dx$. Show all working. (3 marks)

auto-saved

ApplyBand 43 marks

Q2. If $\dfrac{dy}{dx} = 3x^2 - 4x + 2$ and the curve passes through $(1, 5)$, find the equation of the curve. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. A particle moves with velocity $v(t) = 9.8t$ m/s (free fall under gravity). Find the displacement $s(t)$ given $s(0) = 0$. Calculate how far the particle falls in the first 3 seconds. Explain why the constant of integration is zero in this case, and describe what would change if the particle were thrown upward from a height of 10 m with initial velocity 5 m/s. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Drill 1: $\frac{x^5}{5} + C$ 2: $2x^3 + C$ 3: $\frac{x^4}{2} - \frac{5x^2}{2} + 3x + C$ 4: $-\frac{1}{2x^2} + C$ 5: $\frac{3}{4}x^{4/3} + C$

Q1 (3 marks): $\int 4x^3 \, dx = x^4$ [0.5], $\int (-3x^2) \, dx = -x^3$ [0.5], $\int 2x \, dx = x^2$ [0.5], $\int (-1) \, dx = -x$ [0.5]. Answer: $x^4 - x^3 + x^2 - x + C$ [1].

Q2 (3 marks): $y = x^3 - 2x^2 + 2x + C$ [1]. At $(1, 5)$: $5 = 1 - 2 + 2 + C$ [0.5], so $C = 4$ [0.5]. Equation: $y = x^3 - 2x^2 + 2x + 4$ [1].

Q3 (3 marks): $s(t) = \int 9.8t \, dt = 4.9t^2 + C$ [0.5]. $s(0) = 0$ gives $C = 0$ [0.25], so $s(t) = 4.9t^2$ [0.25]. After 3 seconds: $s(3) = 4.9 \times 9 = 44.1$ m [0.5]. $C = 0$ because we defined the origin at the release point [0.25]. If thrown upward from 10 m: $v(t) = 5 - 9.8t$, $s(t) = 5t - 4.9t^2 + C$. $s(0) = 10$ gives $C = 10$, so $s(t) = 5t - 4.9t^2 + 10$ [0.75].

Review and complete

Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.

Retrieve the lesson before you leave

State the power, constant and sum rules for integration.
Explain how a known point determines the constant of integration.
Connect velocity, displacement and antiderivatives.

Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.

Take the full module quiz

quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →

Mark lesson as complete

Tick when you've finished the practice and review.

← Integral calculus overview Lesson 2 · Integrating Power Functions →

Module overview · Maths Advanced · Checkpoint 1