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 calculatingmake 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 confidencethat 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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