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

Area Between a Curve and a Line

When a region is bounded by both a curve AND a non-vertical line, the answer is sometimes a SUM (area under curve + triangle under line, when shapes sit next to each other) and sometimes a DIFFERENCE (∫(upper − lower) dx, when they overlap). The sketch tells you which. Triangles use ½ × base × height; curve areas use a definite integral.

📘 What you need to know

Find intersections first: solve curve = line for x. These usually give the integration limits.
SUM case: the curve and line bound NEIGHBOURING regions (one beside the other). Add the area under the curve and the area under the line.
DIFFERENCE case: the curve and line bound a region BETWEEN them. Area = ∫(upper − lower) dx on the overlap interval.
Triangle/trapezium shortcut: area under a line = (1/2) × base × height (triangle) or (1/2)(a + b)h (trapezium). Faster than integrating the line.
Identify “upper” and “lower”: pick a test value of x between intersections, plug both in, compare.
Sketch always helps: a quick GDC graph confirms whether you’re in SUM or DIFFERENCE territory.
If GDC available, ∫|curve − line| dx with absolute value handles “upper vs lower” automatically — but you still need correct intersection limits.
Watch for THREE intersections: curves like cubics can cross a line three times, creating TWO separate enclosed regions — handle each separately.

SUM case vs DIFFERENCE case

Left: curve and line sit side by side, so AREA = ∫curve dx + triangle under line. Right: curve and line overlap between two intersections, so AREA = ∫(upper − lower) dx over the overlap.

Method 1 — SUM

A = ∫_curve + (½ base · height)

two adjacent shapes; integrate curve, use triangle formula for line

Method 2 — DIFFERENCE

A = ∫_a^b (upper − lower) dx

curve & line overlap; integrate the gap between them

🧭 Recipe — area bounded by a curve and a line

Sketch the curve and line — by hand or GDC.
Find intersection points: solve curve = line for x.
Identify the region: is it the OVERLAP between curve and line (DIFFERENCE), or two ADJACENT pieces (SUM)?
Compute: integrate (upper − lower) over the overlap, OR add up the pieces (∫curve + triangle).
Add/subtract as the sketch dictates. Always verify final answer is positive.

Worked examples

WE 1

Line above curve — DIFFERENCE case

Find the area of the region bounded by the curve y = x² and the line y = x + 2.

Step 1 — find intersections (curve = line) x² = x + 2 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0 → x = −1 and x = 2 Step 2 — identify upper and lower at x = 0: curve = 0, line = 2 → line is above Step 3 — set up and evaluate ∫(upper − lower) dx A = ∫₋₁² ((x + 2) − x²) dx = ∫₋₁² (x + 2 − x²) dx = [x²/2 + 2x − x³/3]₋₁² F(2) = 2 + 4 − 8/3 = 18/3 − 8/3 = 10/3 F(−1) = 1/2 − 2 + 1/3 = 3/6 + 2/6 − 12/6 = −7/6 A = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 Area = 9/2 square units classic “lens” between a parabola and a line — always integrate (upper − lower)

WE 2

Curve plus triangle — SUM case

A region is bounded by the curve y = √x from x = 0 to x = 4, the line from (4, 2) to (6, 0), and the x-axis. Find the exact total area.

Step 1 — sketch shows the curve from (0,0) to (4,2), then line down to (6,0) two adjacent shapes side-by-side → SUM case Step 2 — find the line’s equation: slope = (0 − 2)/(6 − 4) = −1, through (4, 2) y − 2 = −(x − 4) → y = 6 − x Step 3 — area under curve (integrate) ∫₀⁴ √x dx = [(2/3) x^(3/2)]₀⁴ = (2/3)(8) − 0 = 16/3 Step 4 — area under line (triangle: base × height / 2) triangle vertices: (4, 0), (6, 0), (4, 2) base = 2 (along x-axis), height = 2 → A = (1/2)(2)(2) = 2 Step 5 — add Total = 16/3 + 2 = 16/3 + 6/3 = 22/3 Total area = 22/3 square units two ADJACENT shapes — just add. No need for ∫(upper − lower); the line and curve aren’t overlapping

WE 3

Parabola above line — exact answer involving a surd

Find the exact area enclosed by the curve y = 4 − x² and the line y = 1.

Step 1 — find intersections 4 − x² = 1 → x² = 3 → x = ±√3 Step 2 — identify upper at x = 0: curve = 4, line = 1 → curve is above on (−√3, √3) Step 3 — integrate (curve − line) A = ∫₋√₃^√3 ((4 − x²) − 1) dx = ∫₋√₃^√3 (3 − x²) dx = [3x − x³/3]₋√₃^√3 at x = √3: 3·√3 − (√3)³/3 = 3√3 − 3√3/3 = 3√3 − √3 = 2√3 at x = −√3: −2√3 (by symmetry) A = 2√3 − (−2√3) = 4√3 Area = 4√3 square units (≈ 6.93) (√3)³ = √3 · 3 = 3√3 — careful with cube of a surd

WE 4

Line crosses cubic at THREE points — two separate regions

Find the total area of the regions enclosed between the curve y = x³ − 3x and the line y = x.

Step 1 — find intersections x³ − 3x = x → x³ − 4x = 0 → x(x² − 4) = 0 → x(x − 2)(x + 2) = 0 → x = −2, 0, 2 (three intersections!) Step 2 — identify upper on each piece on (−2, 0): at x = −1, curve = −1 + 3 = 2, line = −1 → curve above on (0, 2): at x = 1, curve = 1 − 3 = −2, line = 1 → line above Step 3 — integrate each region Region A (curve above): ∫₋₂⁰ ((x³ − 3x) − x) dx = ∫₋₂⁰ (x³ − 4x) dx = [x⁴/4 − 2x²]₋₂⁰ = 0 − (4 − 8) = 4 Region B (line above): ∫₀² (x − (x³ − 3x)) dx = ∫₀² (4x − x³) dx = [2x² − x⁴/4]₀² = (8 − 4) − 0 = 4 Step 4 — add the two areas Total = 4 + 4 = 8 Total area = 8 square units when curve and line cross 3+ times, upper/lower SWAPS between regions — never lump them into one integral

WE 5

Downward parabola above line in first quadrant

The region R is bounded by the curve y = −x² + 7x − 6 and the line y = −x + 6, lying entirely in the first quadrant. Find the exact area of R.

Step 1 — find intersections −x² + 7x − 6 = −x + 6 → −x² + 8x − 12 = 0 → x² − 8x + 12 = 0 → (x − 2)(x − 6) = 0 → x = 2, 6 at intersections: (2, 4) and (6, 0) Step 2 — identify upper at x = 4: curve = −16 + 28 − 6 = 6, line = 2 → curve is above Method 1 (direct DIFFERENCE) A = ∫₂⁶ ((−x² + 7x − 6) − (−x + 6)) dx = ∫₂⁶ (−x² + 8x − 12) dx = [−x³/3 + 4x² − 12x]₂⁶ F(6) = −72 + 144 − 72 = 0 F(2) = −8/3 + 16 − 24 = −8/3 − 8 = −32/3 A = 0 − (−32/3) = 32/3 Method 2 (area under curve − area under line) ∫₂⁶ (−x² + 7x − 6) dx = 56/3 (area under curve) ∫₂⁶ (−x + 6) dx = 8 (triangle under line: (½)(4)(4) = 8) A = 56/3 − 8 = 56/3 − 24/3 = 32/3 ✓ Area of R = 32/3 square units both methods give 32/3 — pick whichever is faster (direct ∫(upper − lower) usually wins for clean intersections)

WE 6

Reciprocal curve and line — exact answer involves ln

Find the exact area of the region enclosed by the curve y = 16/x and the line y = 10 − x, where x > 0.

Step 1 — find intersections 16/x = 10 − x → 16 = 10x − x² → x² − 10x + 16 = 0 → x = (10 ± √(100 − 64))/2 = (10 ± 6)/2 = 2 or 8 at intersections: (2, 8) and (8, 2) Step 2 — identify upper at x = 5: curve = 16/5 = 3.2, line = 5 → line is above on (2, 8) Step 3 — integrate (line − curve) A = ∫₂⁸ ((10 − x) − 16/x) dx = [10x − x²/2 − 16 ln|x|]₂⁸ F(8) = 80 − 32 − 16 ln 8 = 48 − 16 ln 8 F(2) = 20 − 2 − 16 ln 2 = 18 − 16 ln 2 A = (48 − 16 ln 8) − (18 − 16 ln 2) = 30 − 16 ln 8 + 16 ln 2 = 30 − 16 (ln 8 − ln 2) = 30 − 16 ln(8/2) = 30 − 16 ln 4 = 30 − 32 ln 2 (using ln 4 = 2 ln 2) Area = 30 − 32 ln 2 ≈ 7.82 square units when the integrand has 1/x, the antiderivative has ln|x| — and log laws (ln 8 − ln 2 = ln 4) simplify the exact form

💡 Top tips

Always sketch first — even a quick GDC graph tells you SUM vs DIFFERENCE and shows you the limits.
Triangle/trapezium shortcut for the line: faster than integrating. Triangle = (1/2) base × height; trapezium = (1/2)(a + b)h.
Identify upper vs lower with a test point: pick any x between intersections, evaluate both, compare.
For DIFFERENCE on calculator paper: just use GDC’s ∫ feature with the integrand (upper − lower) on the overlap interval.
For multiple-region cases (3+ intersections): never combine; handle each region separately with the right “upper” function.

⚠ Common mistakes

Using ∫(curve − line) when the LINE is above — your answer will be negative. Always check the sketch and integrate (upper − lower).
Treating SUM as DIFFERENCE — if the curve and line bound NEIGHBOURING regions (not overlapping), you ADD areas; don’t subtract.
Forgetting one of three intersections on a cubic-and-line problem — make sure you’ve found ALL solutions to curve = line.
Computing the area of the wrong region — when intersections give multiple options, look at the sketch to identify which enclosed area the question wants.
Wrong sign on the line: y = 6 − x has positive y on (−∞, 6), negative beyond. Don’t accidentally use a region where the line is below the x-axis.

Up next: Area Between 2 Curves. Same idea, but with two curves instead of a curve and a line. The formula is identical: A = ∫(upper − lower) dx over the overlap. The tricky bit is that with two curves, the “upper” and “lower” can swap multiple times — meaning multiple regions, each integrated separately. And finding intersections is often a more demanding algebra problem than curve-and-line.

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