In simple terms
A friendly intro before the formal notes — no formulas yet.
Un-doing Derivatives to Find Area
Integration is the reverse process of differentiation, allowing us to find the original function from its rate of change. We use this to calculate the exact area under curves, a fundamental concept in calculus.
Imagine you have a speedometer reading (your speed, the derivative) for your entire car journey. Integration is like using that data to calculate the total distance you've travelled (the original function). A definite integral from 1 pm to 2 pm would tell you exactly how far you travelled in that hour.
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Integration reverses differentiation — find the antiderivative F(x) with F′(x) = f(x).
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A definite integral ∫ₐᵇ f(x) dx equals the signed area under the curve.
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Evaluate F(b) − F(a) after finding F(x); always substitute limits explicitly.
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Integration by parts: ∫ u dv = uv − ∫ v du — choose u using LIATE.
Explore the concept
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Simulations
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1 simulation
- GeoGebraCore9709 3.5
Area between two curves
Drag the red points on the x-axis to set the limits, or type in your own pair of functions.
Try this
- Drag the two red points to the x-values where the curves cross.
- Read the shaded area, then integrate (upper − lower) by hand between the same limits.
- Push one limit past a crossing point and watch the running total start to fall.
Look for Area = ∫(upper − lower) dx between the intersection points; where the curves swap over, the integral has to be split.
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Key formulas
Tap any symbol to reveal exactly what it means and its units.
Tap a symbol — great for exam definitions
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Full topic notes
Formal explanation with the rigour you need for the exam.
Standard Integrals and Key Forms
In P3, we expand our library of standard integrals. The key is to recognise the function's form and apply the correct rule. Pay close attention to the effect of linear transformations of the form inside the function.
The factor of appears when integrating a function of a linear argument . This comes from reversing the chain rule.
Always include the constant of integration, , for indefinite integrals.
The modulus sign in is crucial as the logarithm is only defined for positive inputs.
Integration by Substitution
Integration by substitution is a technique for simplifying integrals that are not in a standard form. It is essentially the reverse of the chain rule for differentiation. The goal is to choose a substitution, typically letting 'u' equal a part of the integrand, which transforms the integral into a simpler one that we know how to solve. When dealing with definite integrals, remember that the limits of integration must also be converted to the new variable.
Integration by Parts
Integration by parts is our method for integrating a product of two functions, derived from the product rule for differentiation. The key to success is choosing which function to label as 'u' (the one to be differentiated) and which to label as (the one to be integrated). A useful mnemonic is LIATE (Logarithmic, Inverse Trig, Algebraic, Trig, Exponential), which suggests the order of preference for choosing 'u'.
The Integration by Parts Formula: \
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
Find the exact area of the region enclosed by the curve , the x-axis, and the lines and .
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The area is given by the definite integral .
Find the exact value of .
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We have a product of an algebraic function () and a logarithmic function (). We use integration by parts.
How it all connects
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Glossary
Key terms for this topic — skim now; the Check step will test them.
- Indefinite integral
The family of functions whose derivative is the integrand. It is written as , where and is the constant of integration.
- integral of
. A common mistake is forgetting the factor.
- integration by parts
. It is used to integrate the product of two functions.
- integral of the form
This is a standard pattern that integrates to . Recognising this saves you from using substitution.
Quick check
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Teach it back
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Teach it back
Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.
Revision flashcards
Guess first, then flip — retrieval beats re-reading.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
The factor of appears when integrating a function of a linear argument . This comes from reversing the chain rule.
Always include the constant of integration, , for indefinite integrals.
The modulus sign in is crucial as the logarithm is only defined for positive inputs.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Find the value of ∫₀² p(x)/(3x+2) dx, giving your answer in the form a+lnb where a and b are integers.
The variables x and y satisfy the differential equation (x²+1) dy/dx = kxe^(2y), where k is a constant. It is given that y = 0 when x = 0 and that y = -½ when x = 1. Solve the differential equation and find the exact value of y when x = √3.
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Checkpoint
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Before you move on: do 9709/22 · Q5(b) on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.
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