In simple terms
A friendly intro before the formal notes — no formulas yet.
Rewinding the Function Clock
If differentiation is finding a function's rate of change, integration is working backwards to find the original function from that rate of change. It's the mathematical equivalent of un-baking a cake to find the recipe.
Imagine you're a food critic. Differentiation is like tasting a finished brownie and describing its flavour profile: 'intensely chocolatey with a hint of salt'. Integration is like being given that flavour profile and having to write the original recipe. You can figure out the main ingredients (flour, sugar, chocolate), but you might not know the exact amount of salt, or if there was a drop of vanilla extract. These small unknown quantities, which don't change the main flavour profile, are like the 'constant of integration'. There are a whole family of similar recipes that could produce that same taste.
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Identify the function to integrate, which is often given as a derivative like or .
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Apply the power rule to each term: increase the power by 1, then divide by the new power — or quote the standard integral for , , or .
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Add the constant of integration at the end. This represents all possible antiderivatives.
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If a point on the curve is given, substitute the and values into your integrated function and solve for to get the particular solution.
Explore the concept
Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.
Step 1
Identify the function to integrate, which is often given as a derivative like or .
Key formulas
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Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Full topic notes
Formal explanation with the rigour you need for the exam.
Antidifferentiation and the indefinite integral
If differentiating gives , then is an antiderivative of , and finding it is integration. We write this using the indefinite integral.
The symbol is the integral sign, is the integrand, and tells us we are integrating with respect to . The result is the most general antiderivative, where and is the constant of integration. The word 'indefinite' signals that we are not yet pinning down a single function — we are describing a whole family at once.
The power rule for integration
Just as differentiation has a power rule, so does integration — and it is the differentiation rule run in reverse. To differentiate we multiply by the power and subtract one from it. To integrate we do the opposite, in the opposite order: add one to the power, then divide by the new power.
For any rational : and with a constant multiplier:
Step 1: Increase the index (power) of by 1.
Step 2: Divide the term by this new index.
Step 3: Add the constant of integration .
Condition: the rule needs , since would divide by zero — that case is handled by below.
A three-second check every time: differentiate your answer in your head. If it returns to the integrand, you integrated correctly. The two commonest slips — dividing by the old index instead of the new one, and accidentally differentiating (multiplying down) instead of integrating — are both caught instantly by this reverse check.
Integrals of the standard functions
Not everything is a power of . Four standard integrals appear constantly and are simply the reverse of derivatives you already know. Learn them as a set — they are quoted, not derived, in an exam.
is its own integral, exactly as it is its own derivative.
integrates to — the modulus is required, and this fills the gap left by the power rule at .
Watch the sign on the trig pair: integrating gives , but integrating gives . The minus sign is the classic dropped mark.
Constant multiples and sums combine as usual: .
Finding the constant: from a family to a particular solution
An indefinite integral gives a family of functions, not one. For instance , and all differentiate to ; the constant is what distinguishes them, shifting the curve vertically. To select the single curve we want, we need one extra piece of information — a point the curve passes through, called a boundary condition or initial condition. Substituting that point lets us solve for and obtain the particular solution.
In Paper 1, 'find the equation of the curve' always means find a specific . Forgetting at the integration step makes the rest impossible — there is no constant left to solve for. Write immediately after integrating, then substitute the given point.
Common mistakes examiners penalise
Forgetting the constant of integration — every indefinite integral needs , and without it a 'find the curve' question cannot be finished. This is the single most penalised slip in the topic.
Dividing by the old index instead of the new one — the power rule divides by , not . , not .
Differentiating by mistake — writing multiplies down instead of building up. Always add one to the index. Reverse-check by differentiating your answer.
Dropping the minus sign on — it is , while .
Trying to use the power rule on — the rule fails at ; the answer is , and the modulus must be there.
Solving for but never writing the final equation — after finding , substitute it back and state . A value of alone is not the equation the question asked for.
Substituting the coordinates the wrong way round — put the -value where appears and the -value equal to the whole expression; swapping them corrupts the value of .
Model answer — marked the way our engine marks it
In Paper 1 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and follow-through (FT) means a wrong number early on need not cost you the marks that depend on it, provided your later steps are correct on your own figure. But that protection only exists if your method is written down. Study how each mark below is earned by a specific line: the integration, the constant, the substitution of the boundary condition, and the final equation.
Where this leads
Antidifferentiation is the engine of the rest of calculus. Once you can reverse the power rule and quote the standard integrals, definite integrals attach limits to find exact areas under curves, and boundary conditions reappear whenever a real quantity is reconstructed from its rate of change — displacement from velocity, or volume from a flow rate. Master the habit here — add one and divide, quote the standard integrals, never forget , and use the given point to pin it down — and the integration that follows becomes variations on a method you already own.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
Find .
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Integrate term by term, treating the constant as .
Find .
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Integrate term by term, pulling constant multipliers outside each integral.
The gradient of a curve is given by . The curve passes through the point . Find the equation of the curve. [4]
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The curve is the integral of its gradient function.
A curve has gradient and passes through the point . Find the equation of the curve. [5]
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Model answer — full working.
How it all connects
The big idea sits in the middle — tap a linked idea to explore the link.
Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Try to recall each definition before you reveal it.
Quick check
Answer in your head first — then tap to check. No pressure.
Revision flashcards
Flip the card. Test yourself before the exam.
What is antidifferentiation?
The process of finding an original function from its derivative . It is the reverse of differentiation: if then is an antiderivative of .
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
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Step 1: Increase the index (power) of by 1.
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Step 2: Divide the term by this new index.
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Step 3: Add the constant of integration .
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Condition: the rule needs , since would divide by zero — that case is handled by below.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Get a Paper 1 calculation marked: find the equation of a curve from its gradient, with full working
Get a Paper 1 calculation marked: find the equation of a curve from its gradient, with full working
Extra simulations & links
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Frequently asked
Checkpoint
One marked question is worth ten re-reads — close the loop before you move on.
Reading it isn’t knowing it — prove it.
Before you move on: do Get a Paper 1 calculation marked: find the equation of a curve from its gradient, with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.