RevisionRun — IB Math Practice That Actually Works

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Knowledge Gaps Found

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Skills tracked

528

97 Topics+42 Features+389 Micro-skills

Your Weak Areas — Pinpointed

Topic

Integration by parts

—

Skill

Apply chain rule in trig contexts

—

Feature

3D geometry problems

—

Topic

Areas between curves

—

Skill

Isolate subject with fractions

—

Not just "weak at calculus" — we identify exactly which step or context trips you up.

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Overall Mastery

Based on actual understanding

59%

across all topics

Topic Mastery

This week's change

Number & Algebra

+5%78%

Functions

+12%65%

Geometry & Trig

-3%42%

Statistics & Prob

+8%71%

Calculus

+2%38%

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Recommended For You

Targeting 528 skills across your weak areas

Highest Impact Questions

94

Integration by parts

parts apply formulatrigWeak skill

87

Chain rule in 3D

apply chain rule3dGap detected

82

Areas between curves

area between curves setupDue for review

65

Quadratic equations

complete squarealgebraic inputReinforce

Impact Score

How much this question improves your weakest micro-skills — not just topics.

Exam Mode

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Paper 1 Mock Exam

IB Math AA HL • No calculator

110 marks

Reading Time

5 minutes

Writing Time

2 hours

Reading Time

5:00

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Exam conditions:

No calculator allowed
Formula booklet available
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Solve the equation 2ˣ⁺¹ = 32.

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Find the area enclosed by the curve y = x²sin(x) and the x-axis for 0 ≤ x ≤ π.

hardExam8pts

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement...

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Solve the equation 2ˣ⁺¹ = 32.

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The function f is defined by f(x) = 2x³ − 3x² + 1. Find f'(x).

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In triangle ABC, AB = 8 cm, BC = 6 cm and angle ABC = 120°. Find the area...

hardExam10pts

A particle moves along a straight line. Its velocity v ms⁻¹ at time t seconds...

hardExam9pts

The function f(x) = ax³ + bx² + cx + d has a local maximum at (−1, 5) and...

Detailed Solution & Markscheme

Step-by-step with IB mark allocation

Marks: 4/4(100%)

Rearrange so all y terms (with dy) are on one side and all x terms (with dx) are on the other.

1√1 − y² dy = 4x1 + 2x² dx

Integrate both sides with respect to their variables.

∫ 1√1 − y² dy = arcsin y

Let u = 1 + 2x² so that du = 4x dx.

∫ 4x1 + 2x² dx = ln(1 + 2x²) + C

arcsin y = ln(1 + 2x²) + C

Use y = 0 when x = 0 to find C.

0 = ln(1 + 2 · 0²) + C = ln 1 + C ⟹ C = 0

Take the sine of both sides to get the explicit solution.

y = sin(ln(1 + 2x²))

MMethodAAccuracyRReasoning

Interactive Solutions

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IB Markscheme

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Marks: 4/4(100%)

1√1 − y² dy = 4x1 + 2x² dx

∫ 4x1 + 2x² dx = ln(1 + 2x²) + C

0 = ln 1 + C ⟹ C = 0

y = sin(ln(1 + 2x²))

MMethodAAccuracyRReasoning

Mark Your Mistakes

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Marks: 0/4(0%)

Rearrange so all y terms (with dy) are on one side and all x terms (with dx) are on the other.

1√1 − y² dy = 4x1 + 2x² dx

Integrate both sides with respect to their variables.

∫ 1√1 − y² dy = arcsin y

Let u = 1 + 2x² so that du = 4x dx.

∫ 4x1 + 2x² dx = ln(1 + 2x²) + C

arcsin y = ln(1 + 2x²) + C

Use y = 0 when x = 0 to find C.

0 = ln(1 + 2 · 0²) + C = ln 1 + C ⟹ C = 0

Take the sine of both sides to get the explicit solution.

y = sin(ln(1 + 2x²))

Formula References

Linked to your IB formula booklet

y = ln(1 + 2x²)

y = sin(ln(1 + 2x²))

Integrate both sides of the separated equation:

∫ 1√1 − y² dy = arcsin y

Formula Booklet • IB 5.15g

∫ 1√a² − x² dx = arcsinxa + C

For the RHS, let u = 1 + 2x² so du = 4x dx:

∫ 4x1 + 2x² dx = ln(1 + 2x²) + C

Combined: arcsin y = ln(1 + 2x²) + C

Using y = 0 when x = 0 gives C = 0. Taking sine of both sides:

y = sin(ln(1 + 2x²))

Hover the book icon to see formula booklet references

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