IB Maths AA HL · Unit 5: Calculus
IB Maths AA HL Advanced Integration Questions
Exam-style IB Maths AA HL advanced integration questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 50 questions
- Paper 1: 44
- Paper 2: 6
- 7 easy
- 21 medium
- 14 hard
- 8 starter
- 3 worked examples
Practise Advanced Integration questions →
AA HL formula booklet
What's examined in AA HL advanced integration
The question bank covers these advanced integration question types (number of questions in brackets):
- Advanced Substitution & Fractions (19)
- Volume and Area Applications (16)
- Integration by Parts (15)
Key formulas
- Standard integrals (HL)
- \(\int \dfrac{1}{\sqrt{a^2-x^2}}\, dx = \arcsin\!\tfrac{x}{a} + C,\ \int \dfrac{1}{a^2 + x^2}\, dx = \tfrac{1}{a}\arctan\!\tfrac{x}{a} + C\)
- Integration by parts
- \(\int u\, dv = uv - \int v\, du\)
- Volume of revolution (about x-axis)
- \(V = \pi \int_a^b y^2\, dx\)
- Volume of revolution (about y-axis)
- \(V = \pi \int_c^d x^2\, dy\)
In the same notation as the IB formula booklet. All AA HL formulas →
Advanced Integration worked examples
Worked example 1: Integration by substitution · easy
Find the exact value of the definite integral $\int_0^1 x e^{x^2} \, dx$.
1. Identify the inner function and set up the substitution: Let $u = x^2$.
2. Differentiate to find the differential relationship: $\frac{du}{dx} = 2x \implies dx = \frac{du}{2x}$.
3. Calculate the new limits of integration in terms of $u$: When $x = 0$, $u = 0^2 = 0$. When $x = 1$, $u = 1^2 = 1$.
4. Substitute $u$, $dx$, and the new limits into the integral: $\int_0^1 x e^u \frac{du}{2x}$.
5. Cancel the $x$ terms and factor out the constant: $\frac{1}{2} \int_0^1 e^u \, du$.
6. Evaluate the definite integral using the anti-derivative: $\frac{1}{2} [e^u]_0^1 = \frac{1}{2}(e^1 - e^0) = \mathbf{\frac{1}{2}(e - 1)}$.
Examiner tip: When using integration by substitution for definite integrals, always convert the upper and lower limits to be in terms of $u$. If you leave them as $x$-values while integrating with respect to $u$, your evaluation will be completely incorrect.
Worked example 2: Integration by parts · medium
Use integration by parts to find the exact value of $\int_1^e x \ln x \, dx$.
1. Select $u$ and $dv$ using the LATE hierarchy (Logarithmic before Algebraic): Let $u = \ln x$ and $dv = x \, dx$.
2. Differentiate $u$ and integrate $dv$: $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$.
3. State the integration by parts formula: $\int u \, dv = uv - \int v \, du$.
4. Substitute the expressions into the formula: $\left[ \frac{x^2}{2} \ln x \right]_1^e - \int_1^e \frac{x^2}{2} \cdot \frac{1}{x} \, dx$.
5. Simplify and evaluate the remaining integral: $\left[ \frac{x^2}{2} \ln x \right]_1^e - \int_1^e \frac{x}{2} \, dx = \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_1^e$.
6. Substitute the limits to find the exact value: $\left(\frac{e^2}{2}(1) - \frac{e^2}{4}\right) - \left(0 - \frac{1}{4}\right) = \frac{e^2}{4} + \frac{1}{4} = \mathbf{\frac{e^2 + 1}{4}}$.
Examiner tip: The "LATE" acronym (Log, Algebra, Trig, Exponential) dictates which function should be $u$. Assigning $dv = \ln x \, dx$ traps you, as you cannot easily integrate it without using parts a second time!
Worked example 3: Volumes of revolution with partial fractions · hard
The region enclosed by the curve $y = \frac{1}{\sqrt{x^2 - 4}}$, the $x$-axis, and the vertical lines $x=3$ and $x=5$ is rotated $360^\circ$ about the $x$-axis. Find the exact volume of the solid generated.
1. State the formula for a volume of revolution about the $x$-axis: $V = \pi \int_a^b y^2 \, dx$.
2. Square the function and set up the integral: $V = \pi \int_3^5 \frac{1}{x^2 - 4} \, dx$.
3. Factorise the denominator and split into partial fractions: $\frac{1}{(x-2)(x+2)} \equiv \frac{A}{x-2} + \frac{B}{x+2}$. Solving gives $A = \frac{1}{4}$ and $B = -\frac{1}{4}$.
4. Rewrite the integral using the partial fractions: $V = \frac{\pi}{4} \int_3^5 \left( \frac{1}{x-2} - \frac{1}{x+2} \right) \, dx$.
5. Integrate to log forms: $\frac{\pi}{4} \big[ \ln|x-2| - \ln|x+2| \big]_3^5$.
6. Evaluate the limits and simplify using log laws: $\frac{\pi}{4} \big( (\ln 3 - \ln 7) - (\ln 1 - \ln 5) \big) = \frac{\pi}{4} \big( \ln \frac{3}{7} + \ln 5 \big) = \mathbf{\frac{\pi}{4} \ln\left(\frac{15}{7}\right)}$.
Examiner tip: Ensure you physically square the function $y$ before attempting to decompose it into partial fractions. Doing partial fractions on $y$ and then squaring the result creates cross-terms that are much harder to integrate!
FAQ
How many IB Maths AA HL advanced integration questions are there?
There are 50 exam-style advanced integration questions in the AA HL question bank (Paper 1: 44 · Paper 2: 6), graded 7 easy, 21 medium, 14 hard, 8 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is advanced integration on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 44 · Paper 2: 6. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA HL Unit 5 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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