Free notes · JC 1 and 2
H2 Math notes
Free H2 Math notes on one page, for the A-Level H2 Mathematics syllabus (9758). Full notes for 5 of the 17 syllabus chapters, each with the key idea, the facts and formulas to know, a worked exam question, the common mistakes and exam tips. The other chapters have a checklist of what the syllabus asks for, so the page maps the whole syllabus.
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How to use these notes
- Pick one chapter. Read the key idea and the must-know points.
- Close the page and write down what you remember, then check and mark what you missed.
- Try the worked exam question before reading its steps, and read the common mistakes.
- Do questions on that chapter from a school paper and mark them. Redo the wrong ones a few days later.
Syllabus edition
These notes follow SEAB's H2 Mathematics syllabus 9758 for the A-Level examination in 2027. Source: SEAB, H2 Mathematics syllabus 9758 (2027).
Exam format
| Paper | What it is | Time | Marks | Weighting |
|---|---|---|---|---|
| Paper 1 | 10 to 12 Pure Mathematics questions, including one real-world application question worth at least 12 marks. Answer all. | 3 h | 100 | 50% |
| Paper 2 | Section A, 40 marks: 4 to 5 Pure Mathematics questions. Section B, 60 marks: 6 to 8 Probability and Statistics questions, including one application question worth at least 12 marks. Answer all. | 3 h | 100 | 50% |
- A graphing calculator without a computer algebra system is expected. The MF27 list of formulae and results is given in the exam.
- Answers from the calculator without working are allowed unless a question says otherwise, but a wrong answer with no working gets no marks.
The whole syllabus, chapter by chapter
All 17 chapters of the syllabus in order. 5 have full notes on this page. The rest have a checklist of what the syllabus asks you to do.
| Chapter | What it covers | On this page |
|---|---|---|
| 1. Functions | Domain, range, inverse and composite functions. | Full notes |
| 2. Graphs and Transformations | Key features of graphs, conics, transformations and parametric equations. | Checklist |
| 3. Equations and Inequalities | Forming and solving equations and inequalities. | Checklist |
| 4. Sequences and Series | Arithmetic and geometric series, sums to infinity and sigma notation. | Full notes |
| 5. Vectors | Scalar and vector products, lines and planes in three dimensions. | Full notes |
| 6. Complex Numbers | Cartesian and polar forms, Argand diagrams and complex roots. | Full notes |
| 7. Differentiation | Graphs of derivatives, implicit and parametric differentiation, and applications. | Checklist |
| 8. Maclaurin Series | Standard series, deriving series and small angle approximations. | Checklist |
| 9. Integration Techniques | Standard integrals, substitution and integration by parts. | Full notes |
| 10. Definite Integrals | Areas and volumes of revolution. | Checklist |
| 11. Differential Equations | Solving and interpreting first-order equations. | Checklist |
| 12. Probability | Counting, permutations, combinations and conditional probability. | Checklist |
| 13. Discrete Random Variables | Expectation, variance and the binomial distribution. | Checklist |
| 14. Normal Distribution | Normal probabilities and combinations of variables. | Checklist |
| 15. Sampling | Sample means and the Central Limit Theorem. | Checklist |
| 16. Hypothesis Testing | Tests for a population mean. | Checklist |
| 17. Correlation and Linear Regression | Scatter diagrams, the correlation coefficient and regression lines. | Checklist |
H2 Math at a glance
| Chapter | The one thing to remember |
|---|---|
| Functions | f⁻¹ exists only if f is one-one: every horizontal line meets the graph at most once |
| Sequences and Series | AP: S_n = n/2 [2a + (n - 1)d] = n/2 (a + l) |
| Vectors: Lines and Planes | Line: r = a + λd. Plane: r·n = p, or n₁x + n₂y + n₃z = p with normal n |
| Complex Numbers | i² = -1, so (a + bi)(c + di) = (ac - bd) + (ad + bc)i |
| Integration Techniques | ∫f'(x)/f(x) dx = ln|f(x)| + c |
Notes 1 of 5
Functions
Key idea. A function maps each element of its domain to exactly one value, and its range is the set of outputs. Only a one-one function has an inverse f⁻¹, whose domain is the range of f and whose graph is the reflection of f in the line y = x. The composite fg means g first, then f, and it exists only when the range of g lies inside the domain of f.
Must know
- f⁻¹ exists only if f is one-one: every horizontal line meets the graph at most once
- Domain of f⁻¹ = range of f, and range of f⁻¹ = domain of f
- fg exists only if R_g ⊆ D_f, and the domain of fg is the domain of g
- ff⁻¹(x) = x, and the graphs of f and f⁻¹ are reflections in y = x
What the syllabus asks you to do
- Understand the concepts of function, domain and range
- Find inverse functions and composite functions
- State the conditions for an inverse function and a composite function to exist
- Restrict the domain of a function to obtain an inverse function
- Relate the graph of a one-one function to the graph of its inverse
Domains and ranges in the worked example
| Function | Domain | Range |
|---|---|---|
| f(x) = x² - 6x + 5, x ≥ 3 | [3, ∞) | [-4, ∞) |
| f⁻¹(x) = 3 + √(x + 4) | [-4, ∞) | [3, ∞) |
| g(x) = e^(x) + 1 | ℝ | (1, ∞) |
| gf(x) = e^(x² - 6x + 5) + 1 | [3, ∞) | [1 + e^(-4), ∞) |
Three graphs on equal scales. The curve f starts at (0, 1) and rises steeply to (3, 10). The curve f⁻¹ starts at (1, 0) and rises slowly to (10, 3). The dashed line y = x lies between them, and every point (a, b) on f matches a point (b, a) on f⁻¹.
Worked exam question
The function f is defined by f(x) = x² - 6x + 5 for x ≥ k, and g is defined by g(x) = e^(x) + 1 for x ∈ ℝ. (a) Find the smallest value of k for which f⁻¹ exists. (b) Using this value of k, find f⁻¹(x) and state its domain. (c) Determine whether fg and gf exist, and find the range of any composite that does.
- Complete the square: f(x) = (x - 3)² - 4. The turning point is at x = 3, so f is one-one for x ≥ 3, and the smallest k is 3.
- Let y = (x - 3)² - 4. Then (x - 3)² = y + 4 and x - 3 = √(y + 4), taking the positive root since x ≥ 3.
- So f⁻¹(x) = 3 + √(x + 4). The range of f is [-4, ∞), so the domain of f⁻¹ is x ≥ -4.
- For fg: R_g = (1, ∞) and D_f = [3, ∞). Since (1, ∞) is not a subset of [3, ∞), for example g(0) = 2 is not in D_f, fg does not exist.
- For gf: R_f = [-4, ∞), which lies inside ℝ = D_g, so gf exists and gf(x) = e^(x² - 6x + 5) + 1.
- Range of gf: the inputs to g are [-4, ∞), and g is increasing, so the outputs start at e^(-4) + 1 and increase without bound. So R_gf = [1 + e^(-4), ∞).
Answer: (a) k = 3. (b) f⁻¹(x) = 3 + √(x + 4), for x ≥ -4. (c) fg does not exist, gf exists with range [1 + e^(-4), ∞).
Common mistakes
- Giving the range of fg(x) = ln(x² + 1) as all real numbers.Find R_g first, here [1, ∞), then feed it into f to get R_fg = [0, ∞).
- Writing f⁻¹(x) = 3 ± √(x + 4) for f(x) = (x - 3)² - 4, x ≥ 3.Match the sign to the domain: x ≥ 3 means x - 3 ≥ 0, so take the positive root.
- Stating the domain of f⁻¹(x) = 3 + √(x + 4) from the formula alone.Take the domain of f⁻¹ to be the range of f, found from a sketch of f.
- Working out fg(x) as g(f(x)).Remember fg(x) = f(g(x)): the function written next to x acts first.
- Answering 'not one-one' with a sketch and no reason.Name two inputs with the same output, such as f(1) = f(-1) = 3 for f(x) = x² + 2.
- Writing f⁻¹(x) = 1/f(x).Remember f⁻¹ undoes f: for f(x) = 2x, f⁻¹(x) = x/2, not 1/(2x).
Exam tips
- Always state domains when giving an inverse or a composite function. A correct formula with a missing domain is an incomplete answer.
- For range questions, sketch the graph over the given domain and mark the endpoint values. Include or exclude each endpoint deliberately.
- In 'does fg exist' questions, write R_g and D_f as actual sets, compare them, then conclude in a sentence.
Notes 2 of 5
Sequences and Series
Key idea. An arithmetic progression adds a fixed difference d each time, so u_n = a + (n - 1)d. A geometric progression multiplies by a fixed ratio r, so u_n = ar^(n-1), and its sum to infinity a/(1 - r) exists only when |r| < 1. Sequences can also be defined by a recurrence u_(n+1) = f(u_n), and u_n = S_n - S_(n-1) links each term to the sums.
Must know
- AP: S_n = n/2 [2a + (n - 1)d] = n/2 (a + l)
- GP: S_n = a(1 - r^(n))/(1 - r), and S∞ = a/(1 - r) when |r| < 1
- Split a mixed term such as 3r + 2^(r) into an AP part and a GP part, then sum each part
- For any sequence, u_n = S_n - S_(n-1)
What the syllabus asks you to do
- Understand sequences and series, finite and infinite, and a sequence as a function of a positive integer n
- Use the relationship between the nth term u_n and the sum to n terms S_n
- Work with sequences given by a formula for the nth term or by a relation u_(n+1) = f(u_n), using a graphing calculator to generate terms
- Find the sum and difference of two series
- Use the formulae for the nth term and the sum of a finite arithmetic series and a finite geometric series
- State the condition for an infinite geometric series to converge, and find its sum to infinity
Arithmetic and geometric progressions compared
| Feature | Arithmetic progression | Geometric progression |
|---|---|---|
| Pattern | Add d each time | Multiply by r each time |
| nth term | a + (n - 1)d | ar^(n-1) |
| Sum of n terms | n/2 [2a + (n - 1)d] | a(1 - r^(n))/(1 - r), for r ≠ 1 |
| Sum to infinity | Does not exist unless every term is 0 | a/(1 - r), only when |r| < 1 |
| Example | 5, 8, 11, ... with d = 3 | 24, 12, 6, ... with r = 1/2 |
Points for n from 1 to 10 joined by lines. The terms are 2, 4, 5, 5.5, 5.75 and so on, each closing half of the remaining gap to 6. The graph flattens as it nears the dashed horizontal line at 6, showing that the sequence converges to 6.
Worked exam question
At the start of every month, Mei deposits $500 into a savings account. At the end of every month, interest of 0.5% of the amount in the account is added. (a) Show that the amount in the account at the end of the nth month is 100500(1.005^(n) - 1) dollars. (b) Find the least number of months for the amount to exceed $20 000.
- End of month 1: 500(1.005). End of month 2: (500(1.005) + 500) × 1.005 = 500(1.005² + 1.005).
- Following the pattern, the end of month n gives 500(1.005 + 1.005² + ... + 1.005^(n)).
- The bracket is a GP with first term 1.005, ratio 1.005 and n terms, so its sum is 1.005(1.005^(n) - 1)/(1.005 - 1) = 201(1.005^(n) - 1).
- So the amount is 500 × 201(1.005^(n) - 1) = 100500(1.005^(n) - 1) dollars, as required.
- For the amount to exceed $20 000, 100500(1.005^(n) - 1) > 20000, so 1.005^(n) > 1 + 20000/100500 = 1.19900 to 5 decimal places.
- Take natural logarithms: n ln 1.005 > ln 1.19900. Since ln 1.005 > 0, n > 36.4 to 3 significant figures.
- Check: n = 36 gives about $19 766 and n = 37 gives about $20 368, so the least n is 37.
Answer: 37 months
Common mistakes
- Using S∞ = a/(1 - r) for 3 + 6 + 12 + ..., where r = 2.State |r| < 1 before using S∞, and write 'no sum to infinity' otherwise.
- Counting the terms from r = 3 to r = 10 as 7.Use last minus first plus one: 10 - 3 + 1 = 8.
- Concluding n < 11.4 from (2/3)^(n) < 0.01.Note the sign of ln r before dividing, then check the final n with a table of values.
- Summing the first n terms of 3r + 2 as 3n(n + 1)/2 + 2.A constant k added to every term contributes kn to the sum of n terms.
- Using the arithmetic sum formula on 3r + 2^(r) as a whole.Split it into an arithmetic part and a geometric part, and sum each with its own formula.
- Writing the month 2 amount as 500(1.005)², forgetting the second deposit.Write out the first two months in full before generalising to month n.
Exam tips
- For least n questions, show a table from the graphing calculator with the values either side of the boundary. It gives a clear, checkable conclusion.
- When a term mixes types, such as 3r + 2^(r), write the arithmetic part and the geometric part on separate lines, each with its own a and d or a and r.
- In context questions, write the first two or three stages in full before generalising to n. It fixes the powers and the number of terms.
Notes 3 of 5
Vectors: Lines and Planes
Key idea. A line has vector equation r = a + λd, through the point a with direction d. A plane has equation r·n = p, where n is a normal, often found as the vector product of two directions in the plane. Find intersections by substituting the line into the plane, angles with the scalar product, and the foot of a perpendicular by solving for the parameter.
Must know
- Line: r = a + λd. Plane: r·n = p, or n₁x + n₂y + n₃z = p with normal n
- Lines: cos θ = |d₁·d₂|/(|d₁||d₂|). Line and plane: sin θ = |d·n|/(|d||n|)
- Distance from P to the plane r·n = p is |OP·n - p|/|n|
- A normal to a plane containing directions b and c is n = b × c
What the syllabus asks you to do
- Write vector and cartesian equations of lines and planes
- Use the scalar product and the vector product and their properties
- Find the angle between two lines, between a line and a plane, or between two planes
- Find the foot of the perpendicular and the distance from a point to a line or to a plane
- Determine the relationship between two lines (coplanar or skew), a line and a plane, and two planes
Finding the acute angle between lines and planes
| Angle between | Vectors used | Formula |
|---|---|---|
| Two lines | Directions d₁ and d₂ | cos θ = |d₁·d₂|/(|d₁||d₂|) |
| Two planes | Normals n₁ and n₂ | cos θ = |n₁·n₂|/(|n₁||n₂|) |
| A line and a plane | Direction d and normal n | sin θ = |d·n|/(|d||n|) |
Take the normal as direction
n = (1, 2, 2) for the plane x + 2y + 2z = 12.
Write the perpendicular line
r = (1, 1, 0) + λ(1, 2, 2) passes through P.
Substitute into the plane
(1 + λ) + 2(1 + 2λ) + 2(2λ) = 12 gives λ = 1.
Find the foot N
Put λ = 1 into the line to get N(2, 3, 2).
Distance and reflection
PN = 3 units, and the reflection is 2ON - OP = (3, 5, 4).
Worked exam question
The plane Π has equation x + 2y + 2z = 12 and P is the point (1, 1, 0). Find (a) the foot of the perpendicular from P to Π, (b) the perpendicular distance from P to Π, (c) the reflection of P in Π, and (d) the acute angle between Π and the x-axis.
- A normal to Π is n = (1, 2, 2), with |n| = √(1 + 4 + 4) = 3.
- The line through P perpendicular to Π is r = (1, 1, 0) + λ(1, 2, 2), so a general point on it is (1 + λ, 1 + 2λ, 2λ).
- Substitute into Π: (1 + λ) + 2(1 + 2λ) + 2(2λ) = 12, so 3 + 9λ = 12 and λ = 1. The foot is N(2, 3, 2).
- Distance PN = |λ| × |n| = 1 × 3 = 3 units. Check with the formula: |(1, 1, 0)·(1, 2, 2) - 12|/3 = |3 - 12|/3 = 3.
- N is the midpoint of P and its reflection P', so OP' = 2ON - OP = (4, 6, 4) - (1, 1, 0) = (3, 5, 4).
- The x-axis has direction (1, 0, 0). The angle θ between it and Π satisfies sin θ = |(1, 0, 0)·(1, 2, 2)|/(1 × 3) = 1/3, so θ = 19.5°.
Answer: (a) N(2, 3, 2). (b) 3 units. (c) (3, 5, 4). (d) 19.5°.
Common mistakes
- Using cos θ = |d·n|/(|d||n|) for the angle between a line and a plane.Use sin θ for a line and a plane, and cos θ for two lines or two planes.
- Declaring two lines intersecting after solving only two component equations.Substitute λ and μ into the third component. If it fails and the lines are not parallel, they are skew.
- Using λ as the parameter for both lines when testing for intersection.Give each line its own parameter, λ and μ, before equating position vectors.
- Getting (2, 1, 1) instead of (-2, 1, 1) for (1, 0, 2) × (0, 1, -1).Check the result: its scalar product with each original vector must be zero.
- Giving 120° as the angle between two lines.Take |d₁·d₂| in the formula so the angle is acute, which here gives 60°.
- Measuring the distance from P to a plane to a convenient point on the plane.Use the foot of the perpendicular, or the formula |OP·n - p|/|n|.
Exam tips
- Draw a rough diagram for foot of perpendicular and reflection questions, labelling P, N, the line or plane and the direction vectors. It shows which scalar product should be zero.
- You may use the graphing calculator to solve the simultaneous equations for a point on two planes, but write down the equations you solved.
- Keep exact values such as √14 in intermediate steps, and give angles to 0.1° unless the question says otherwise.
Notes 4 of 5
Complex Numbers
Key idea. A complex number z = x + iy, where i² = -1, has real part x, imaginary part y, conjugate z* = x - iy and modulus |z| = √(x² + y²). Add, subtract and multiply by expanding with i² = -1, and divide by multiplying the top and bottom by the conjugate of the denominator. Non-real roots of polynomial equations with real coefficients come in conjugate pairs.
Must know
- i² = -1, so (a + bi)(c + di) = (ac - bd) + (ad + bc)i
- |z| = √(x² + y²), and zz* = x² + y² = |z|²
- z = w only when their real parts are equal and their imaginary parts are equal
- With real coefficients, if 2 + i is a root then 2 - i is also a root
What the syllabus asks you to do
- Understand the extension of the number system from real numbers to complex numbers
- Find the complex roots of a quadratic equation
- Find the modulus, argument and conjugate of a complex number
- Carry out the four operations on complex numbers
- Use the equality of complex numbers
- Use the result that the non-real roots of a polynomial equation with real coefficients occur in conjugate pairs
- Represent complex numbers on an Argand diagram
- Describe the geometrical effects of conjugation, negation, addition, subtraction and multiplication by i
The effect of each operation on z = 3 + 2i
| Operation | Result | Effect on the Argand diagram |
|---|---|---|
| Conjugate z* | 3 - 2i | Reflection in the real axis |
| Negation -z | -3 - 2i | Rotation of 180° about the origin |
| Multiply by i | -2 + 3i | Rotation of 90° anticlockwise about the origin |
| Add w = 1 - i | 4 + i | Translation by the vector (1, -1) |
| Subtract w = 1 - i | 2 + 3i | Translation by the vector (-1, 1) |
Check the coefficients are real
z³ + z² + pz + 15 = 0 has real coefficients, since p is real.
Write the conjugate root
1 - 2i is a root, so 1 + 2i is also a root.
Form the quadratic factor
Sum 2 and product 5 give z² - 2z + 5.
Compare coefficients
(z² - 2z + 5)(z + 3) matches, giving p = -1.
List all the roots
1 + 2i, 1 - 2i and -3.
Worked exam question
The complex numbers z and w satisfy the simultaneous equations z + 2w = 5 and iz + w = 0. (a) Find z and w in the form x + iy. (b) Find |z - w|, and the argument of z in radians, each to 3 significant figures. (c) Describe the position of the point w relative to the point z on an Argand diagram.
- From the second equation, w = -iz.
- Substitute into the first equation: z - 2iz = 5, so z(1 - 2i) = 5.
- Multiply the top and bottom by the conjugate 1 + 2i: z = 5(1 + 2i)/((1 - 2i)(1 + 2i)) = 5(1 + 2i)/5 = 1 + 2i.
- Then w = -i(1 + 2i) = -i - 2i² = 2 - i. Check: z + 2w = 1 + 2i + 4 - 2i = 5.
- z - w = (1 + 2i) - (2 - i) = -1 + 3i, so |z - w| = √(1 + 9) = √10 = 3.16.
- z = 1 + 2i lies in the first quadrant, so arg z = tan⁻¹(2/1) = 1.11 radians.
- w = -iz is z multiplied by i (a 90° anticlockwise rotation) and then negated (a 180° rotation), which is a 90° clockwise rotation about the origin. So |w| = |z| = √5.
Answer: (a) z = 1 + 2i and w = 2 - i. (b) |z - w| = √10 = 3.16 and arg z = 1.11 radians. (c) w is z rotated through 90° clockwise about the origin, and both points are √5 from the origin.
Common mistakes
- Giving arg(-1 - i) as π/4 from tan⁻¹(1).Plot the point first. -1 - i is in the third quadrant, so its argument is -3π/4.
- Giving an argument of 7π/6 when -π < θ ≤ π is required.Add or subtract 2π until the angle lies in -π < θ ≤ π, so 7π/6 becomes -5π/6.
- Assuming the conjugate of a root of z² + iz + 2 = 0 is also a root.Check that every coefficient is real before using conjugate pairs.
- Writing |3 + 4i| = |3| + |4i| = 7.Use |z| = √(x² + y²), so |3 + 4i| = 5. In general, |z + w| is not |z| + |w|.
- Giving the imaginary part of 3 - 4i as -4i.The imaginary part is the real number multiplying i, here -4.
- Collecting real and imaginary parts of 2z + iz* before expanding iz*.Expand every product fully, then collect real and imaginary parts.
Exam tips
- Sketch every complex number on an Argand diagram before writing its argument. A few seconds of sketching prevents quadrant errors.
- When dividing, multiply the top and bottom by the conjugate of the denominator first, then split the result into real and imaginary parts before comparing it with anything.
- In conjugate root questions, state that the coefficients are real before using the conjugate pair, since that reason is part of the method.
Notes 5 of 5
Integration Techniques
Key idea. Integration techniques find integrals that are not simple powers. First check the formula list and the patterns f'(x)/f(x), f'(x)[f(x)]^(n) and f'(x)e^(f(x)). Otherwise use identities for sin²x, cos²x and tan²x, a substitution when one is given, or integration by parts, ∫u (dv/dx) dx = uv - ∫v (du/dx) dx, for products.
Must know
- ∫f'(x)/f(x) dx = ln|f(x)| + c
- By parts: ∫u (dv/dx) dx = uv - ∫v (du/dx) dx, with u the factor that simplifies when differentiated
- cos²x = ½(1 + cos 2x) and sin²x = ½(1 - cos 2x)
- With a substitution in a definite integral, change the limits and do not substitute back
What the syllabus asks you to do
- Integrate f'(x)[f(x)]^(n), including n = -1, which gives ln|f(x)|, and f'(x)e^(f(x))
- Integrate sin²x, cos²x and tan²x
- Integrate 1/(a² + x²), 1/√(a² - x²), 1/(a² - x²) and 1/(x² - a²) using the standard results
- Integrate by a given substitution
- Integrate by parts
- Evaluate definite integrals, including approximate values from a graphing calculator
Choosing an integration method
| Integrand looks like | Method | Example |
|---|---|---|
| f'(x)/f(x) | Recognise it: ln|f(x)| + c | ∫2x/(x² + 5) dx = ln(x² + 5) + c |
| f'(x)[f(x)]^(n) | Recognise it: [f(x)]^(n+1)/(n + 1) + c | ∫x(x² + 1)³ dx = (x² + 1)⁴/8 + c |
| f'(x)e^(f(x)) | Recognise it: e^(f(x)) + c | ∫2xe^(x²) dx = e^(x²) + c |
| 1/(a² - x²) or 1/(x² - a²) | Standard result with a logarithm | ∫1/(x² - 1) dx = ½ ln|(x - 1)/(x + 1)| + c |
| sin²x, cos²x or tan²x | Trigonometric identity first | ∫tan²x dx = tan x - x + c |
| √(a² - x²) | Substitution x = a sin θ | √(4 - x²) with x = 2 sin θ |
| Product such as xe^(2x) or x² ln x | Integration by parts | ∫ln x dx = x ln x - x + c |
A curve that dips just below the x-axis between x = 0.2 and x = 1, crosses the axis at x = 1, then rises steeply to about 7.39 at x = e, where a dashed vertical line marks the upper limit. The area between the curve, the x-axis and the lines x = 1 and x = e is (2e³ + 1)/9, about 4.57.
Worked exam question
Find the exact value of the integral of x² ln x from x = 1 to x = e.
- Use integration by parts with u = ln x and dv/dx = x², since ln x becomes simpler when differentiated.
- Then du/dx = 1/x and v = x³/3.
- The integral equals [(x³/3) ln x] from 1 to e minus the integral from 1 to e of (x³/3)(1/x) dx.
- Simplify the remaining integrand: (x³/3)(1/x) = x²/3, which integrates to x³/9.
- So the integral equals [(x³/3) ln x - x³/9] from 1 to e.
- At x = e, ln e = 1, giving e³/3 - e³/9 = 2e³/9. At x = 1, ln 1 = 0, giving 0 - 1/9 = -1/9.
- Subtract: 2e³/9 - (-1/9) = (2e³ + 1)/9. The graphing calculator gives 4.57 for the integral, which agrees.
Answer: (2e³ + 1)/9, which is about 4.57
Common mistakes
- Choosing u = x² and dv/dx = ln x for ∫x² ln x dx.Make ln x the u whenever it appears, since it differentiates to the simple 1/x.
- Changing the limits to u-values and then substituting back to x as well.Either change the limits and finish in u, or substitute back and use the x-limits, never both.
- Evaluating the integral of 1/(x - 3) from 0 to 2 as ln(-1) - ln(-3).Write ln|x - 3|, giving ln 1 - ln 3 = -ln 3.
- Writing ∫sin²x dx = sin³x/3 + c.Rewrite with sin²x = ½(1 - cos 2x) first, then integrate term by term.
- Leaving dx unchanged after substituting x = 2 sin θ.Write dx = 2 cos θ dθ on its own line before rewriting the integral.
- Writing ∫1/(9 + x²) dx = tan⁻¹(x/3) + c.Copy the formula list result exactly, (1/a) tan⁻¹(x/a), giving (1/3) tan⁻¹(x/3) + c.
Exam tips
- Before integrating, spend a few seconds choosing the method: standard result, f'(x)/f(x) or f'(x)e^(f(x)), identity, substitution or parts. Starting with the wrong method wastes the most time.
- Check any definite integral with the graphing calculator's numerical integration. If the decimal does not match your exact answer, look for a slip before moving on.
- Copy standard results from the formula list exactly, then adapt the constant a to your question in a separate line.
Syllabus checklist for the rest
What the syllabus asks you to do in the chapters, and parts of chapters, that do not have full notes yet. Each point is the syllabus learning outcome in plain words. Tick each one you can do without looking.
Chapter 2. Graphs and Transformations
3 points to check
- Find symmetry, intercepts, turning points and asymptotes of the graphs named in the syllabus, including ellipses, hyperbolas and rational functions.
- Apply transformations y = af(x), y = f(x) + a, y = f(x + a) and y = f(ax) and their combinations, and relate y = f(x) to y = |f(x)|, y = f(|x|) and y = 1/f(x).
- Sketch simple parametric curves.
Chapter 3. Equations and Inequalities
2 points to check
- Form equations, systems of linear equations and inequalities from a problem, and solve them with a graphing calculator.
- Solve inequalities of the form f(x)/g(x) > 0 with linear or quadratic f and g, inequalities with |x|, and inequalities by graphs.
Chapter 7. Differentiation
1 point to check
- Relate the graph of f'(x) to f(x), differentiate implicit and parametric functions, find and classify stationary points, and solve tangent, normal, maxima and minima and connected rates of change problems.
Chapter 8. Maclaurin Series
1 point to check
- Use the standard series for (1 + x)ⁿ, eˣ, sin x, cos x and ln(1 + x), derive the first terms of a series, state where a standard series converges, and use small angle approximations.
Chapter 10. Definite Integrals
1 point to check
- Find areas under a curve, between a curve and a line or two curves, and below the x-axis, and volumes of revolution about the x- or y-axis.
Chapter 11. Differential Equations
1 point to check
- Solve dy/dx = f(x)g(y) for general and particular solutions, including by a given substitution, and form and interpret a differential equation from a problem.
Chapter 12. Probability
1 point to check
- Use counting principles, permutations and combinations, and calculate probabilities, including conditional probability.
Chapter 13. Discrete Random Variables
1 point to check
- Find expectations and variances, and use the binomial distribution B(n, p) as a model, stating when it is suitable.
Chapter 14. Normal Distribution
1 point to check
- Find normal probabilities and unknown parameters, and use E(aX + b), Var(aX + b), E(aX + bY) and Var(aX + bY) for independent X and Y.
Chapter 15. Sampling
1 point to check
- Use the distribution of the sample mean, the Central Limit Theorem for large samples, and unbiased estimates of population mean and variance.
Chapter 16. Hypothesis Testing
1 point to check
- Set up null and alternative hypotheses, carry out 1-tail and 2-tail tests for a population mean, and interpret the result in context.
Chapter 17. Correlation and Linear Regression
1 point to check
- Interpret the product moment correlation coefficient, find least squares regression lines, choose the right line to predict, and use square, reciprocal or log transformations.
If your child is struggling
- Find the exact chapter and point. The checklists and common mistakes on this page show where an answer breaks down.
- Go back to that chapter's worked question and do it again without the steps.
- Ask the school teacher which chapters the class has covered and which are tested next.
- If several chapters are weak, work on one at a time, with short timed questions, rather than rereading everything.