IB Maths AA SL Topic 5 — Calculus Paper 2 🎯 Skill ~4 min practice

AA SL Area Between Curves skills

A definite integral gives you a signed value, but area is always positive. For curves bounded by the x-axis use absolute value or split at zeros; for two curves, integrate top minus bottom between their intersection points. Sketch first, integrate second.

The Method

Case 1

Curve & x-axis

A = |∫_a^b f(x) dx|

if curve crosses the x-axis, split at the zeros and add absolute values

Case 2

Two Curves

A = ∫_a^b [f(x) − g(x)] dx

f is the top curve, g is the bottom · limits are intersection points

Sketch the curve(s) — figure out which is on top, where they cross, and whether the area dips below the x-axis.
Find the limits of integration — given in the question, or set f(x) = g(x) to find intersections.
Integrate top minus bottom (or |f| if just curve and x-axis), then evaluate.

The setup — top minus bottom

The area between two curves is the integral of f(x) − g(x) where f is on top. The limits a and b are where the curves intersect.

⚠️

Negative areas don’t exist — but negative integrals do

If a curve dips below the x-axis, the integral over that piece is negative — but that’s not the area. To find area, take the absolute value, or split the integral at each x-axis crossing and add the magnitudes. Always sketch first to spot the crossings.

GDC shortcut for area between curves (Paper 2)

TI-84 Plus

Find intersections via 2nd → TRACE → 5: intersect
Press MATH → 9: fnInt(
Enter fnInt(f − g, X, a, b) with the top curve first
Take absolute value if you suspect sign issues

Casio fx-CG50

Use G-Solve → INTSECT for intersections
From RUN-MAT, OPTN → CALC → ∫dx
Enter ∫(f − g, a, b) top curve first
Or use G-Solve → ∫dx on the graph directly

Worked examples

WE 1 EASY

Find the area enclosed by y = 6x − x² and the x-axis.

step 1 — find x-axis intersections 6x − x² = 0 → x(6 − x) = 0 x = 0, x = 6step 2 — sketch — parabola opens down, above x-axis between 0 and 6step 3 — integrate A = ∫₀⁶ (6x − x²) dx = [3x² − x³/3]₀⁶ = (108 − 72) − 0 = 36A = 36 (units²) curve is fully above x-axis here, so positive integral = area directly!

WE 2 MEDIUM

Find the area enclosed between y = x + 2 and y = x².

step 1 — find intersections x² = x + 2 → x² − x − 2 = 0 (x − 2)(x + 1) = 0 x = −1 and x = 2step 2 — identify top curve at x = 0: line gives 2, parabola gives 0 → line is on topstep 3 — integrate top − bottom A = ∫₋₁² [(x + 2) − x²] dx = [x²/2 + 2x − x³/3]₋₁² F(2) = 2 + 4 − 8/3 = 10/3 F(−1) = 1/2 − 2 + 1/3 = −7/6 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2A = 9/2 (units²) always sub a midpoint x-value to confirm which curve is on top!

WE 3 HARD

Find the area enclosed by y = x³ − x and the x-axis between x = −1 and x = 1.

step 1 — find zeros in the interval x³ − x = x(x² − 1) = 0 x = −1, 0, 1step 2 — split at x = 0 on [−1, 0]: try x = −0.5 → −0.125 + 0.5 = 0.375 (above) on [0, 1]: try x = 0.5 → 0.125 − 0.5 = −0.375 (below)step 3 — integrate each piece, take |val| ∫₋₁⁰ (x³ − x) dx = [x⁴/4 − x²/2]₋₁⁰ = 0 − (1/4 − 1/2) = 1/4 ∫₀¹ (x³ − x) dx = (1/4 − 1/2) − 0 = −1/4step 4 — add the magnitudes A = 1/4 + |−1/4| = 1/2A = 1/2 (units²) if you’d integrated straight from −1 to 1, you’d have gotten 0 — symmetric cancellation!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Sketch first — every time.

Q1 EASY Find the area enclosed by y = 4 − x² and the x-axis. Show answer ▼Hide answer ▲

zeros: x = ±2 → curve above between A = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 16 − 16/3 = 32/3 A = 32/3 (units²)

Q2 EASY Find the area between y = x² and the x-axis from x = 1 to x = 3. Show answer ▼Hide answer ▲

x² is positive on [1, 3] A = ∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3 A = 26/3 (units²)

Q3 MEDIUM Find the area enclosed between y = x² and y = 2x. Show answer ▼Hide answer ▲

intersections x² = 2x → x² − 2x = 0 → x = 0, 2 at x = 1: 2x = 2, x² = 1 → 2x is on top A = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3 A = 4/3 (units²)

Q4 MEDIUM Find the total area between y = x² − 4x + 3 and the x-axis from x = 0 to x = 3. Show answer ▼Hide answer ▲

find zeros in [0, 3] (x − 1)(x − 3) = 0 → x = 1, 3 split at x = 1 on [0,1]: at x=0.5, y=1.25 (above) on [1,3]: at x=2, y=−1 (below) ∫₀¹ = [x³/3 − 2x² + 3x]₀¹ = 1/3 − 2 + 3 = 4/3 ∫₁³ = (9 − 18 + 9) − 4/3 = 0 − 4/3 = −4/3 A = 4/3 + |−4/3| = 8/3 A = 8/3 (units²) “total area” → split at zeros and add magnitudes!

Q5 HARD Find the area enclosed between y = √x and y = x²/4, for x ≥ 0. Show answer ▼Hide answer ▲

intersections √x = x²/4 → square: x = x⁴/16 16x = x⁴ → x(x³ − 16) = 0 x = 0 or x³ = 16 → x = ∛16 but check: at x = 4: √4 = 2, 16/4 = 4 — wait, equal at x = 0 and x = ? re-solve: try x = 4: √4 = 2, x²/4 = 4 — not equal x = 0 and the next intersection from x³ = 16 → x = 16^1/3 ≈ 2.52 at x = 1: √1 = 1, 1/4 — √x is on top A = ∫₀^∛16 (x^1/2 − x²/4) dx = [2x^3/2/3 − x³/12]₀^∛16 use GDC: A ≈ 1.68 (3 sf) A ≈ 1.68 (units², 3 sf) when limits are awkward, finish on the GDC — Paper 2 problem!

⚠ Common mistakes

Not sketching first. Without a sketch you can’t tell which curve is on top, where the region is, or whether the curve dips below the x-axis.
Bottom minus top. The area integrand is f(x) − g(x) where f is on top — flipping gives a negative answer.
Forgetting to split at x-axis crossings. If the curve dips below, the integral underestimates the area — split into pieces and add magnitudes.
Reporting a negative integral as area. Area is always positive; take absolute value if needed.
Wrong limits. For “enclosed area between curves”, the limits are the intersection x-values, not whatever the question last mentioned.
Forgetting units squared. Area is always in (units)² — write cm², m², or “square units”.

📖

Want the theory?

Read the full Area Between Curves notes for the link to the fundamental theorem of calculus, why we subtract top minus bottom, and worked exam-style problems.

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