Integration (AA HL)
Integration is the reverse process of differentiation - given a derivative, it finds the function it came from. This topic covers the standard integrals, the reverse chain rule for composite functions, definite integrals and the areas they represent, and the HL techniques of integration by substitution, integration by parts, and volumes of revolution.
What the syllabus says
This topic maps onto four points in the official IB Analysis & Approaches syllabus, spanning SL foundations and the HL extension.
| Code | Syllabus content |
|---|---|
| SL5.10 | Indefinite integral of \(x^n\) (\(n\neq-1\)), \(\sin x\), \(\cos x\), \(e^x\) and \(\dfrac1x\). The composites of any of these with the linear function \(ax+b\). Integration by inspection (reverse chain rule) or by substitution for expressions of the form \(\int kg'(x)f(g(x))\,dx.\) |
| SL5.11 | Definite integrals, including the analytical approach \(\int_a^b g'(x)\,dx = g(b)-g(a).\) Areas of a region enclosed by a curve and the \(x\)-axis, where \(f(x)\) can be positive or negative, found without the use of technology. Areas between curves. |
| AHL5.16 | Integration by substitution; on examination papers the substitution will be provided unless the integral is already of the reverse chain rule form \(\int k\,g'(x)f(g(x))\,dx\). Integration by parts, including repeated integration by parts. |
| AHL5.17 | Area of the region enclosed by a curve and the \(y\)-axis in a given interval. Volumes of revolution about the \(x\)-axis or \(y\)-axis. |
SL5.10-5.11 are core AA content examinable at both levels; AHL5.16-5.17 are HL-only.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is an indefinite integral?
An indefinite integral is the general antiderivative of a function - a whole family of curves that all share the same gradient function, differing only by a constant. It's written with an integral sign and no limits, and always needs a "+C" at the end.
e.g. \(\displaystyle\int 2x\,dx = x^2 + C\), since \(\dfrac{d}{dx}(x^2+C) = 2x\) for any constant \(C\).
What is a definite integral?
A definite integral has an upper and lower limit and evaluates to a single number, found using the fundamental theorem of calculus: \(\int_a^b g'(x)\,dx = g(b)-g(a).\) No constant of integration is needed since it cancels out.
e.g. \(\displaystyle\int_0^2 2x\,dx = \left[x^2\right]_0^2 = 4 - 0 = 4.\)
What is the reverse chain rule?
The reverse chain rule integrates a composite function with a linear inner function \(ax+b\) by integrating as normal, then dividing by \(a\) to correct for the chain rule that would apply on the way back to differentiating.
e.g. \(\displaystyle\int \cos(4x+1)\,dx = \tfrac14\sin(4x+1)+C\), since differentiating gives back \(\cos(4x+1)\) exactly.
What is integration by parts?
Integration by parts breaks down the integral of a product into a simpler integral, using the formula \(\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx.\) Choose \(u\) so that it becomes simpler when differentiated.
e.g. \(\displaystyle\int xe^{2x}\,dx = \tfrac12xe^{2x} - \tfrac14e^{2x}+C\), taking \(u=x,\ dv=e^{2x}dx.\)
What is a volume of revolution?
A volume of revolution is the solid formed when a region is rotated fully around an axis. Rotating about the \(x\)-axis uses \(V = \pi\int_a^b y^2\,dx\), summing up the volumes of infinitely many thin circular discs.
e.g. Rotating \(y=\ln x\) between \(x=1\) and \(x=e\) about the \(x\)-axis gives \(V = \pi\int_1^e(\ln x)^2\,dx \approx 2.26.\)
Key formulas
A handful of standard results and two named techniques cover the whole topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The standard integrals, the fundamental theorem of calculus and integration by parts are all on the official formula booklet.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C,\ n\neq-1\) | Power rule for integration | ✓ Yes |
| \(\displaystyle\int \tfrac1x\,dx = \ln|x|+C\) | Reciprocal integral | ✓ Yes |
| \(\displaystyle\int \sin x\,dx=-\cos x+C,\ \int\cos x\,dx=\sin x+C,\ \int e^x\,dx=e^x+C\) | Standard integrals | ✓ Yes |
| \(\displaystyle\int_a^b g'(x)\,dx = g(b)-g(a)\) | Fundamental theorem of calculus | ✓ Yes |
| \(\displaystyle\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx\) | Integration by parts | ✓ Yes |
| Composite with \(ax+b\): integrate, then divide by \(a\) | Reverse chain rule | Not a listed formula - pattern to recognise |
Indefinite vs definite integrals
The two forms use the same antiderivative but produce very different kinds of answer.
| Feature | Indefinite integral | Definite integral |
|---|---|---|
| Notation | \(\int f(x)\,dx\) | \(\int_a^b f(x)\,dx\) |
| Result | A function (family of curves) | A number |
| Constant of integration | Always needed: \(+C\) | Not needed - it cancels |
| Typical use | Finding an antiderivative or a curve's equation | Finding an area or accumulated change |
Standard integrals and the reverse chain rule
Every indefinite integral on this course is either a standard result, a linear composite of one, or found by inspection using the reverse chain rule.
Power rule
\[\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C,\ n\neq-1\]
Raise the power by one and divide by the new power.
✓ In the formula bookletStandard functions
\[\int\sin x\,dx=-\cos x+C,\quad \int\tfrac1x\,dx=\ln|x|+C\]
Learn these alongside the corresponding derivatives - each is the reverse of a standard derivative.
✓ In the formula bookletReverse chain rule
\[\int \cos(4x+1)\,dx = \tfrac14\sin(4x+1)+C\]
Integrate as if the inner function were just \(x\), then divide by the coefficient of \(x\) inside.
Not in the formula booklet - pattern to recogniseDefinite integrals and area
A definite integral gives a signed value - you need to think about the sign before calling the result an "area".
Fundamental theorem of calculus
\[\int_a^b g'(x)\,dx = g(b)-g(a)\]
Find the antiderivative, then substitute the upper limit and subtract the result at the lower limit.
✓ In the formula bookletArea below the axis
Where a curve dips below the \(x\)-axis, the definite integral over that interval comes out negative. Take the modulus of that piece to get a genuine area.
Not in the formula booklet - reasoning stepAreas between curves
Integrate the difference of the two functions, top minus bottom, between their points of intersection.
Not in the formula booklet - techniqueHL techniques
These extend integration to products and compositions that the reverse chain rule alone can't handle, and to 3D volumes.
Integration by substitution
Replace part of the integrand with a new variable \(u\) to turn an awkward integral into a standard one. On the exam, a substitution is provided unless the integral is already of the reverse-chain-rule form.
Not a listed formula - techniqueIntegration by parts
\[\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx\]
Used for products where one factor simplifies on differentiating, such as \(x\ln x\) or \(xe^{2x}\). Sometimes needs repeating.
✓ In the formula bookletVolume of revolution
\[V = \pi\int_a^b y^2\,dx\]
Rotating the region under a curve fully around the \(x\)-axis; swap \(x\) and \(y\) roles for a rotation about the \(y\)-axis.
✓ In the formula bookletWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Consider the indefinite integral \(\displaystyle\int (6x^2 - 4x + 3)\,dx.\)
(a) Find its value.
(b) Given that the curve passes through \((0,5),\) find \(C.\)
Worked solution
(a) \(\int (6x^2-4x+3)\,dx\) M1
\(= 2x^3 - 2x^2 + 3x + C.\) A1 A1
(b) At \(x=0:\ C=5.\) A1
Consider the indefinite integral \(\displaystyle\int (2x-1)^5\,dx.\)
(a) Find its value.
(b) Verify your answer by differentiating it.
Worked solution
(a) Raise the power and divide by the new power and the inner coefficient: M1
\(\int (2x-1)^5\,dx\) A1
\(= \dfrac{(2x-1)^6}{12} + C.\) A1
(b) \(\dfrac{d}{dx}\left(\dfrac{(2x-1)^6}{12}\right)=\dfrac{6(2x-1)^5\cdot2}{12}=(2x-1)^5,\) matching the integrand. R1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting "+C" on an indefinite integral. Every indefinite integral represents a family of curves, not a single one - dropping the constant loses an accuracy mark even if the rest of the working is correct.
- Not taking the modulus when a curve dips below the axis. A definite integral over a region where \(f(x)<0\) comes out negative - if the question asks for an area, you need to take the modulus of that piece before adding it to the rest.
- Missing the \(\tfrac1a\) factor on a linear composite. Integrating \(\cos(4x+1)\) as \(\sin(4x+1)+C\) instead of \(\tfrac14\sin(4x+1)+C\) forgets to correct for the inner function's coefficient.
- Choosing \(u\) and \(dv\) poorly in integration by parts. Pick \(u\) to be the factor that gets simpler when differentiated (like \(x\) or \(\ln x\)) - choosing the other way round usually makes the resulting integral harder, not easier.
Using your GDC
Every major GDC model - TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50 - has a built-in numerical integration function that evaluates a definite integral directly from the function, without you needing to find an antiderivative by hand. It's most useful for checking a Paper 1 answer on the calculator paper, or for evaluating an integral the syllabus doesn't expect you to do analytically. Remember that a numerical result for area below the \(x\)-axis will come out negative, so take the modulus if the question asks for area rather than a signed integral.
See the full GDC guide for model-specific button sequences across every topic.
Ready to practise properly?
Integration questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
Do I need to know integration by parts for AA HL?
Yes, integration by parts is HL-only content, including repeated integration by parts for cases like the integral of \(x^2e^x.\) It uses the formula \(\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx.\)
How do I find the area between a curve and the x-axis when the curve dips below?
Write the definite integral first, then check the sign of the region. Where the curve is below the \(x\)-axis the integral comes out negative, so take the modulus of that part before adding it to any part above the axis.
Can I use my GDC to evaluate integrals I can't do by hand?
Yes - every major model has a numerical integration function that evaluates a definite integral directly. It's especially useful for checking an analytic answer or for integrals the syllabus doesn't expect you to do by hand. See the GDC guide for model-specific instructions.
What's the difference between an indefinite and a definite integral?
An indefinite integral gives a family of antiderivatives and always needs a \(+C\), since any constant disappears on differentiation. A definite integral has limits and evaluates to a single number - typically an area or an accumulated quantity - with no constant needed.
Sub-topics
Integration broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AA HL syllabus unit, in case you want to keep going.