IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~10 min read

Area Between 2 Curves

Same idea as curve-and-line: A = ∫(upper − lower) dx. The two trickier features here are that both boundaries are curves (no triangle shortcut) and the “upper” curve can SWAP between regions. If the curves cross three times, you get TWO regions — compute each separately with the correct upper, then add.

📘 What you need to know

Core formula: A = ∫_a^b (y₁ − y₂) dx, where y₁ is the upper curve and y₂ is the lower curve.
Find intersections by solving f(x) = g(x) — these are the integration limits.
Identify upper/lower with a test point between intersections.
Two or more intersections → look for SWAPS: the upper curve can change between regions. Each region needs its own integral with the correct upper.
Below the x-axis is NOT a problem — (upper − lower) is always positive in a correctly-identified region, even if both curves are negative.
No modulus needed if you set up (upper − lower) correctly; the integrand is already positive.
GDC shortcut: ∫|f(x) − g(x)| dx handles upper/lower automatically — but you still need the intersection limits.
Trig and other transcendental cases often give exact surd answers like 2√2 — recognise these from the standard trig values.

The core formula

Area between two curves (between intersections) A = ∫_a^b (y₁ − y₂) dx where y₁ = upper, y₂ = lower, and a, b = intersection x-values

Two curves crossing three times create TWO regions: in R₁ the blue curve is above; in R₂ the red curve is above. Total area = ∫(f − g) on R₁ + ∫(g − f) on R₂ — the upper/lower SWAPS between regions, so you can never combine into one integral.

One region vs multiple regions

One region (2 intersections)

A = ∫_a^b (y₁ − y₂) dx

curves cross twice → single enclosed region; one integral

Two regions (3 intersections)

A = ∫_R₁ + ∫_R₂
(upper SWAPS)

curves cross thrice → upper changes between regions; two integrals

The most common HL trap: a cubic and a parabola often cross 3 times, but students treat it as a single region and write one integral. Result: the two regions’ signed values partially cancel, giving the wrong total. ALWAYS check for upper/lower swap between consecutive intersections.

🧭 Recipe — area between two curves

Sketch both curves on the same axes — GDC graphing helps.
Solve f(x) = g(x) for ALL intersections. There may be 2, 3, or more.
Identify upper vs lower on each piece using a test point between consecutive intersections.
Write a separate integral for each region using ∫(upper − lower) dx.
Evaluate each integral; sum them for the total area. (Or use GDC’s ∫|f − g| dx shortcut.)

Worked examples

WE 1

Two parabolas — single region, surd answer

Find the exact area enclosed by the curves y = x² + 1 and y = 5 − x².

Step 1 — find intersections x² + 1 = 5 − x² → 2x² = 4 → x² = 2 → x = ±√2 Step 2 — identify upper at x = 0: y₁ = 1, y₂ = 5 → y₂ = 5 − x² is the upper curve Step 3 — integrate (upper − lower) A = ∫₋√₂^√2 ((5 − x²) − (x² + 1)) dx = ∫₋√₂^√2 (4 − 2x²) dx = [4x − 2x³/3]₋√₂^√2 at √2: 4√2 − 2(2√2)/3 = 4√2 − 4√2/3 = 12√2/3 − 4√2/3 = 8√2/3 at −√2: −8√2/3 (by symmetry) A = 8√2/3 − (−8√2/3) = 16√2/3 Area = 16√2/3 square units (≈ 7.54) careful with (√2)³ = √2 · 2 = 2√2 — surds powered up need watching

WE 2

Two simple quadratics — clean rational answer

Find the exact area enclosed by the curves y = x² and y = 2x − x².

Step 1 — find intersections x² = 2x − x² → 2x² − 2x = 0 → 2x(x − 1) = 0 → x = 0, 1 Step 2 — identify upper at x = 1/2: y₁ = 1/4, y₂ = 1 − 1/4 = 3/4 → y₂ is upper Step 3 — integrate A = ∫₀¹ ((2x − x²) − x²) dx = ∫₀¹ (2x − 2x²) dx = [x² − 2x³/3]₀¹ = (1 − 2/3) − 0 = 1/3 Area = 1/3 square units when both curves are parabolas and meet at clean integers, the answer is usually a simple fraction

WE 3

Parabola and cubic — single region on [0, 1]

Find the exact area enclosed by the curves y = x² and y = x³.

Step 1 — find intersections x² = x³ → x²(1 − x) = 0 → x = 0 (tangent) or x = 1 Step 2 — identify upper at x = 1/2: x² = 1/4, x³ = 1/8 → x² is upper Step 3 — integrate A = ∫₀¹ (x² − x³) dx = [x³/3 − x⁴/4]₀¹ = (1/3 − 1/4) − 0 = 4/12 − 3/12 = 1/12 Area = 1/12 square units x = 0 is a TANGENCY (curves touch there but don’t cross) — even so, they enclose the region [0, 1] where x² > x³

WE 4

Two curves crossing 3 times — upper/lower SWAPS

Find the total area of the regions enclosed by the curves y = x³ and y = x⁵.

Step 1 — find intersections x³ = x⁵ → x³(1 − x²) = 0 → x = 0, ±1 (three intersections) Step 2 — identify upper on each region on (−1, 0): at x = −1/2, x³ = −1/8, x⁵ = −1/32 → x⁵ is upper (less negative) on (0, 1): at x = 1/2, x³ = 1/8, x⁵ = 1/32 → x³ is upper Step 3 — integrate each region with correct upper Region A: ∫₋₁⁰ (x⁵ − x³) dx = [x⁶/6 − x⁴/4]₋₁⁰ = 0 − (1/6 − 1/4) = −(2/12 − 3/12) = 1/12 Region B: ∫₀¹ (x³ − x⁵) dx = [x⁴/4 − x⁶/6]₀¹ = (1/4 − 1/6) − 0 = 3/12 − 2/12 = 1/12 Step 4 — add Total = 1/12 + 1/12 = 2/12 = 1/6 Total area = 1/6 square units by symmetry, the two regions have equal area. But you MUST integrate them separately — using (x³ − x⁵) over [−1, 1] would give 0 because the two regions cancel

WE 5

Cubic and quadratic — three intersections, exam-style

The curves y = f(x) and y = g(x) are defined by f(x) = x³ and g(x) = 3x² − 2x. Find the exact total area of the regions enclosed between the two curves.

Step 1 — find intersections: solve f(x) = g(x) x³ = 3x² − 2x → x³ − 3x² + 2x = 0 → x(x² − 3x + 2) = 0 → x(x − 1)(x − 2) = 0 → x = 0, 1, 2 (three intersections) Step 2 — identify upper on each region on (0, 1): at x = 1/2, f = 1/8, g = 3/4 − 1 = −1/4 → f is upper on (1, 2): at x = 3/2, f = 27/8, g = 27/4 − 3 = 15/4 → g is upper (since 15/4 = 30/8 > 27/8) Step 3 — integrate each region Region 1 (0 to 1, f upper): ∫₀¹ (f − g) dx = ∫₀¹ (x³ − 3x² + 2x) dx = [x⁴/4 − x³ + x²]₀¹ = 1/4 − 1 + 1 = 1/4 Region 2 (1 to 2, g upper): ∫₁² (g − f) dx = ∫₁² (3x² − 2x − x³) dx = [x³ − x² − x⁴/4]₁² at 2: 8 − 4 − 4 = 0 at 1: 1 − 1 − 1/4 = −1/4 Region 2 area = 0 − (−1/4) = 1/4 Step 4 — total Total = 1/4 + 1/4 = 1/2 Total area = 1/2 square units classic exam trap: if you integrate (f − g) over [0, 2] directly you get 1/4 − 1/4 = 0. The regions cancel because upper SWAPPED at x = 1

WE 6

Trig — exact area between sin and cos

Find the exact area enclosed by y = sin x and y = cos x between their first two intersection points in the interval 0 ≤ x ≤ 2π.

Step 1 — find intersections in [0, 2π] sin x = cos x → tan x = 1 → x = π/4, 5π/4 Step 2 — identify upper on (π/4, 5π/4) at x = π/2: sin = 1, cos = 0 → sin is upper Step 3 — integrate (sin − cos) A = ∫(π/4)^(5π/4) (sin x − cos x) dx = [−cos x − sin x](π/4)^(5π/4) Evaluate at limits at 5π/4: −cos(5π/4) − sin(5π/4) = −(−√2/2) − (−√2/2) = √2/2 + √2/2 = √2 at π/4: −cos(π/4) − sin(π/4) = −√2/2 − √2/2 = −√2 Subtract A = √2 − (−√2) = 2√2 Area = 2√2 square units (≈ 2.83) sin and cos repeatedly cross — between EACH consecutive pair of intersections, they enclose an area of exactly 2√2 by symmetry

💡 Top tips

Sketch ALWAYS — even a rough GDC graph shows how many regions there are and which curve is upper where.
Test point between intersections — substitute any x-value strictly between two consecutive intersections to identify upper/lower.
Three+ intersections = MULTIPLE regions — never combine; always split into separate integrals.
(upper − lower) is ALWAYS positive in a correctly-identified region — if your integral comes out negative, you’ve swapped them.
GDC shortcut: ∫_a^b|f(x) − g(x)| dx gives total area between curves between intersections a and b, handling upper/lower automatically.

⚠ Common mistakes

Combining multiple regions into one integral — when curves cross 3+ times, upper SWAPS, and a single integral has the two regions partially cancel.
Missing an intersection — make sure you find ALL solutions to f(x) = g(x) when factoring; don’t stop after one.
Getting upper/lower backwards — always test with a value between intersections. Don’t guess.
Taking |∫(f − g) dx| as the area — wrong if the regions partially cancel. Modulus must be INSIDE the integral, not OUTSIDE.
Trying to use the modulus formula for the line (which works for curve-and-line) when both are curves — just stick with ∫(upper − lower) dx.

🎉 That closes the Techniques & Applications of Integration chapter! You can now handle every flavour of integration problem: trig and exponential standards, reverse chain rule, substitution, definite integrals with their six properties, negative integrals and modulus, area between a curve and a line, and area between two curves. Next up in Topic 5 are the more advanced chapters: Optimisation, Kinematics, Basic Limits & Continuity, and then the HL-only material — Further Differentiation, Further Integration, Differential Equations, and Maclaurin Series.

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