Differentiation tells us the rate of change of a function. But what if we know the rate of change and want to recover the original function? That process is called integration — or anti-differentiation — and it is the mirror image of everything you have learned so far.
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You know that the derivative of $x^2$ is $2x$. So if someone told you that the derivative of an unknown function was $2x$, you might guess that the original function is $x^2$. But what about $x^2 + 3$? Or $x^2 - 7$? All of these also have derivative $2x$. If you only know the derivative, what does that tell you about how many possible original functions there might be?
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Wrong: √(a + b) = √a + √b.
Right: The square root of a sum is not the sum of square roots; √(a+b) cannot be simplified this way.
📚 Core Content
If differentiation tells us the gradient of a function, then integration (or anti-differentiation) lets us recover the original function from its gradient. The notation for integration is:
$$\int f(x) \, dx = F(x) + C$$
where $F'(x) = f(x)$. The symbol $\int$ is called the integral sign, and $dx$ tells us we are integrating with respect to $x$.
Because the derivative of any constant is zero, if $F(x)$ is an anti-derivative of $f(x)$, then so is $F(x) + 5$, $F(x) - 100$, or $F(x) + C$ for any constant $C$. We always include $+C$ when finding an indefinite integral.
For any real number $n \neq -1$:
$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$
Notice that this is the reverse of the differentiation power rule: we increase the power by 1, then divide by the new power.
Just like differentiation, integration is linear:
$$\int kf(x) \, dx = k\int f(x) \, dx$$
$$\int \bigl(f(x) + g(x)\bigr) \, dx = \int f(x) \, dx + \int g(x) \, dx$$
If we know a point that the original function passes through, we can find the exact value of $C$. This is called an initial condition or boundary condition.
For example, if $f'(x) = 2x$ and the curve passes through $(1, 5)$:
🧮 Worked Examples
🧪 Activities
1 $\int (4x^3 - 2x + 7) \, dx$
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2 $\int \frac{5}{x^2} \, dx$
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3 $\int \sqrt{x} \, dx$
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4 Find $f(x)$ given $f'(x) = 2x - 3$ and $f(2) = 5$.
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1 Explain why $\int 2x \, dx = x^2 + C$ and not simply $x^2$.
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2 A particle has velocity $v(t) = 3t^2 + 4t$. Explain how you would find its displacement function $s(t)$ if you know that $s(0) = 2$.
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Earlier you thought about recovering a function from its derivative $2x$.
Because differentiation eliminates constant terms, knowing only the derivative tells us the original function up to an unknown constant. Every function of the form $x^2 + C$ has derivative $2x$. To pinpoint the exact function, we need an additional piece of information — a point the curve passes through. This is why the constant of integration $C$ is essential, and why initial conditions are so powerful in physics and engineering.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Find the following indefinite integrals. Show all working. 4 MARKS
(a) $\int (6x^2 - 4x + 3) \, dx$
(b) $\int \frac{3}{x^4} \, dx$
(c) $\int (x^2 + \sqrt{x}) \, dx$
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7. Find $f(x)$ given that $f'(x) = x^2 + 2x - 1$ and $f(1) = 3$. 3 MARKS
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8. A particle moves in a straight line with acceleration $a(t) = 6t - 4$, where $t$ is in seconds. Initially, the particle has velocity $2$ m/s and is at the origin. 4 MARKS
(a) Find the velocity function $v(t)$.
(b) Find the displacement function $s(t)$.
(c) Find the position of the particle after $2$ seconds.
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1. $x^4 - x^2 + 7x + C$
2. $\int 5x^{-2} \, dx = -5x^{-1} + C = -\frac{5}{x} + C$
3. $\int x^{1/2} \, dx = \frac{2}{3}x^{3/2} + C$
4. $f(x) = x^2 - 3x + C$. Using $f(2) = 5$: $4 - 6 + C = 5 \Rightarrow C = 7$. So $f(x) = x^2 - 3x + 7$.
1. The derivative of $x^2 + C$ is $2x$ for any constant $C$. Since we do not know which constant was in the original function, we must include $+C$ in the general anti-derivative.
2. Integrate $v(t)$ to get $s(t) = t^3 + 2t^2 + C$. Then use $s(0) = 2$ to find $C = 2$. So $s(t) = t^3 + 2t^2 + 2$.
1. B — Using the power rule: $\\int x^3 , dx = \frac{x^4}{4} + C$.
2. B — $\\int 2x , dx = 2 \\cdot \frac{x^2}{2} + C = x^2 + C$.
3. B — Integration (anti-differentiation) reverses the process of differentiation.
4. A — $f(x) = x^2 - 3x + C$. Using $f(1) = 2$: $1 - 3 + C = 2 \\Rightarrow C = 4$. So $f(x) = x^2 - 3x + 4$.
5. B — $\\int x^{-2} , dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C$.
6. B — $\\int x^{1/2} , dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$.
7. A — $v(t) = \\int (6t - 4) , dt = 3t^2 - 4t + C$. Using $v(0) = 2$: $C = 2$. So $v(t) = 3t^2 - 4t + 2$.
Q6 (4 marks): (a) $2x^3 - 2x^2 + 3x + C$ [1]. (b) $-\frac{1}{x^3} + C$ [1]. (c) $\frac{x^3}{3} + \frac{2x^{3/2}}{3} + C$ [2].
Q7 (3 marks): $f(x) = \frac{x^3}{3} + x^2 - x + C$ [1]. $f(1) = \frac{1}{3} + 1 - 1 + C = 3 \Rightarrow C = 3 - \frac{1}{3} = \frac{8}{3}$ [1]. So $f(x) = \frac{x^3}{3} + x^2 - x + \frac{8}{3}$ [1].
Q8 (4 marks): (a) $v(t) = 3t^2 - 4t + 2$ [1]. (b) $s(t) = t^3 - 2t^2 + 2t + C$. $s(0) = 0 \Rightarrow C = 0$ [1]. So $s(t) = t^3 - 2t^2 + 2t$ [1]. $s(2) = 8 - 8 + 4 = 4$ m [1].
Integration as Anti-Differentiation
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