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.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
If $f'(x) = 2x$, what could $f(x)$ be? Without calculating — make a prediction before reading on. Think about which function, when differentiated, gives $2x$.
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 $\int 4x^3 \, dx = x^4 + C$.
Key facts
- $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$
- Integration reverses differentiation
- The constant of integration $C$
Concepts
- Integration as the inverse of differentiation
- Why the constant $+C$ is needed
- The relationship between rate and total amount
Skills
- Find antiderivatives of power functions
- Apply the sum and constant multiple rules
- Check answers by differentiating
For any power $n \neq -1$:
Verification: Differentiate $\frac{x^{n+1}}{n+1}$: $\frac{d}{dx}\!\left(\frac{x^{n+1}}{n+1}\right) = \frac{(n+1)x^n}{n+1} = x^n$ ✓
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: $\int k \cdot f(x) \, dx = k \int f(x) \, dx$
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.
Worked examples · 3 in a row, reveal as you go
Find $\int (3x^2 + 4x - 5) \, dx$.
If $\frac{dy}{dx} = 3x^2 - 4x + 2$ and the curve passes through $(1, 5)$, find the equation of the curve.
Is $\int 5x^4 \, dx = x^5 + C$ correct? Verify your answer.
Quick check: Which is the correct antiderivative of $4x^3 - 3x^2 + 2x - 1$?
Common errors · the 3 traps that cost marks
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.
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?
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 $y$.
Odd one out. Three of these are antiderivatives of $x^2$. Which one is NOT?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find $\int (4x^3 - 3x^2 + 2x - 1) \, dx$. Show all working. (3 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)
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)
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].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms using antiderivatives, power rule, and checking answers. Pool: lesson 1.
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