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? What if the speed is $v(t) = 3t^2$ metres per second? Integration is the tool that turns changing rates into total amounts. It is the inverse of differentiation — and one of the most powerful ideas in mathematics.
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If $f'(x) = 2x$, what could $f(x)$ be? Make a prediction before reading on.
Core Content
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 called the integral sign. It 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: d/dx(x^4) = 4x^3, so ∫4x^3 dx = x^4 + 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.
For any power $n \neq -1$:
∫x^n dx = x^(n+1)/(n+1) + C
Check: Differentiate $\frac{x^{n+1}}{n+1}$: $\frac{d}{dx}(\frac{x^{n+1}}{n+1}) = \frac{(n+1)x^n}{n+1} = x^n$ ✓
Examples:
Critical: This rule does NOT work for $n = -1$. The case $\int \frac{1}{x} \, dx$ is special and equals $\ln|x| + C$.
Find $\int (3x^2 + 4x - 5) \, dx$.
The antiderivative.
Integrate term by term:
∫3x^2 dx = 3 · x^3/3 = x^3
∫4x dx = 4 · x^2/2 = 2x^2
∫(-5) dx = -5x
Combine: x^3 + 2x^2 - 5x + C
$\int (3x^2 + 4x - 5) \, dx = x^3 + 2x^2 - 5x + C$.
Find $\int (x^3 - 2x + 7) \, dx$.
Answer:
$\frac{x^4}{4} - x^2 + 7x + C$.
The best way to verify an integral is to differentiate your answer and check you get back the original function.
Example: Is $\int 5x^4 \, dx = x^5 + C$ correct?
Differentiate: $\frac{d}{dx}(x^5 + C) = 5x^4$ ✓
Always do this check — it catches sign errors, coefficient mistakes, and power errors instantly.
$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$
($n \neq -1$)
$\int k \, dx = kx + C$
Integrate term by term
Differentiate your answer
If $f'(x) = 2x$, then $f(x) = x^2 + C$. 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.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Find $\int (4x^3 - 3x^2 + 2x - 1) \, dx$. Show all working. 3 MARKS
9. If $\frac{dy}{dx} = 3x^2 - 4x + 2$ and the curve passes through $(1, 5)$, find the equation of the curve. 3 MARKS
10. 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
1. $\frac{x^5}{5} + C$
2. $2x^3 + C$
3. $\frac{x^4}{2} - \frac{5x^2}{2} + 3x + C$
4. $\int x^{-3} \, dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$
5. $\int x^{1/3} \, dx = \frac{x^{4/3}}{4/3} + C = \frac{3}{4}x^{4/3} + C$
1. The power rule gives $\frac{x^0}{0}$ which is undefined. The integral of $x^{-1}$ is $\ln|x| + C$.
2. $y = 2x^2 - 3x + C$. At $(1, 5)$: $5 = 2 - 3 + C$, so $C = 6$. $y = 2x^2 - 3x + 6$.
3. $s(t) = t^2 + 3t + C$. At $t = 0$, $s = 0$, so $C = 0$. $s(t) = t^2 + 3t$.
Q8 (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].
Q9 (3 marks): $y = \int (3x^2 - 4x + 2) \, dx = x^3 - 2x^2 + 2x + C$ [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].
Q10 (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: $s(3) = 4.9 \times 9 = 44.1$ metres [0.5]. $C = 0$ because we defined the origin at the release point [0.25]. If thrown upward from 10 m with $v(0) = 5$: $v(t) = 5 - 9.8t$ (upward positive, gravity negative) [0.25], $s(t) = 5t - 4.9t^2 + C$ [0.25]. $s(0) = 10$ gives $C = 10$ [0.25], so $s(t) = 5t - 4.9t^2 + 10$ [0.5]. The initial height and velocity now appear as constants in the displacement function.
Climb platforms using antiderivatives, power rule, and checking answers. Pool: lesson 1.
Tick when you've finished all activities and checked your answers.