Mathematics Advanced • Year 12 • Module 6 • Lesson 6

Areas Between Curves

Build fluency in finding intersection points, deciding which curve is on top, and evaluating the area between two curves.

Build · Skill Drill

1. Quick recall

Answer each in the space provided. 1 mark each

Q1.1 Complete the area-between-curves formula for f(x) ≥ g(x) on [a, b]:

A = ∫_{____}^____ [ ____________ − ____________ ] dx

Q1.2 State the three setup steps in order:

Step 1: ____________________________________________________________

Step 2: ____________________________________________________________

Step 3: ____________________________________________________________

Q1.3 If two curves cross inside [a, b] at x = c, what must you do to the integral, and why?

Stuck? Revisit lesson § Formula Reference and § Setting Up.

2. Worked example — area between y = x and y = x²

This is the lesson's "Worked Example". Follow each line carefully.

Problem. Find the area between y = x and y = x² on the region they enclose.

Step 1 — Find intersections by solving f(x) = g(x).

x = x² ⇒ x − x² = 0 ⇒ x(1 − x) = 0 ⇒ x = 0 or x = 1

Reason: the curves meet where their y-values are equal.

Step 2 — Decide which curve is on top on [0, 1] by testing a point.

At x = 0.5: x = 0.5 vs x² = 0.25 ⇒ x > x²

Reason: y = x lies above y = x² on this interval, so top − bottom = x − x².

Step 3 — Set up and evaluate the integral.

A = ∫₀¹ (x − x²) dx = [ x²/2 − x³/3 ]₀¹

= (1/2 − 1/3) − 0 = 3/6 − 2/6 = 1/6

Reason: evaluate the antiderivative at the upper and lower limits.

Conclusion. Area = 1/6 square units.

3. Faded example — fill in the missing steps

Find the area between y = 2x and y = x² on the region they enclose. Fill in each blank. 4 marks

Step 1 — Intersections:

2x = x² ⇒ x² − 2x = 0 ⇒ x( __________ ) = 0

So x = ______ or x = ______.

Step 2 — Test x = 1 to decide which is on top: y = 2x gives ______, y = x² gives ______, so ____________ is on top.

Step 3 — Integral setup:

A = ∫₀^____ ( __________ − __________ ) dx

Step 4 — Evaluate:

A = [ x² − x³/3 ]₀^____ = ( ______ − ______ ) − 0 = __________

Conclusion. Area = ______________ square units.

Stuck? Revisit lesson § Worked Example and the "Try It Now" box.

4. Graduated practice — find the enclosed area

For each pair, find the area of the region enclosed (or the region described). Show the intersection step, the top/bottom test, and the integral.

Foundation — single setup, clean numbers (4 questions)

Q	Curves & interval	Intersections	Area
4.1 1	y = 4 and y = x²
4.2 1	y = x and y = x³ on [0, 1]
4.3 1	y = 6 − x² and y = 2
4.4 1	y = 3x and y = x²

Standard — HSC-typical difficulty (6 questions)

Show at least the intersection step, the top/bottom test and the evaluated integral.

4.5 Find the area between y = x² and y = 2x + 3. 2 marks

4.6 Find the area between y = x² − 4x and y = x. 2 marks

4.7 Find the area between y = e^x and y = e^−x from x = 0 to x = 1. 2 marks

4.8 Find the area between y = sin x and y = cos x from x = 0 to x = π/4. 2 marks

4.9 Find the area enclosed between y = x² and y = 8 − x². 2 marks

4.10 Find the area between y = √x and y = x/2 on the region they enclose. 2 marks

Extension — curves cross / split required (2 questions)

4.11 Find the total area between y = x³ − x and y = 0 from x = −1 to x = 1. Split the integral and explain why splitting is required. 3 marks

4.12 Find the total area between y = sin x and y = 0 from x = 0 to x = 2π. (Hint: sin x changes sign at x = π.) 3 marks

Stuck on 4.11 / 4.12? Sketch first to see where the "top" changes. Revisit § More Examples.

5. Self-check the easy 3

Tick when you've verified your method on the first three.

For 4.1 I solved x² = 4 to get x = ±2 first, then integrated top − bottom = 4 − x² on [−2, 2].

For 4.2 I tested a point in (0, 1) — e.g. x = 0.5 — to confirm y = x is above y = x³ before integrating.

For 4.3 I subtracted the constant from the parabola, not the other way around, and my answer is positive.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Formula

A = ∫_a^b [ f(x) − g(x) ] dx, where f(x) ≥ g(x) on [a, b].

Q1.2 — Three steps

Step 1: Find the intersection points by solving f(x) = g(x). Step 2: Test a point to decide which function is on top on each subinterval. Step 3: Set up and evaluate the integral ∫ (top − bottom) dx.

Q1.3 — Splitting

You must split the integral at x = c and integrate (top − bottom) on each subinterval separately. Without splitting, the "top" function changes mid-interval and the parts would cancel, giving the wrong (often negative) signed total instead of the geometric area.

Q3 — Faded example (y = 2x, y = x²)

Step 1: 2x = x² ⇒ x² − 2x = 0 ⇒ x(x − 2) = 0, so x = 0 or x = 2.
Step 2: At x = 1, y = 2x gives 2; y = x² gives 1; so y = 2x is on top.
Step 3: A = ∫₀² ( 2x − x² ) dx.
Step 4: A = [ x² − x³/3 ]₀² = ( 4 − 8/3 ) − 0 = 4/3.
Conclusion: Area = 4/3 square units.

Q4.1 — y = 4 and y = x²

x² = 4 ⇒ x = ±2. On [−2, 2], 4 ≥ x². A = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 32/3.

Q4.2 — y = x and y = x³ on [0, 1]

On (0, 1), x > x³ (test x = 0.5: 0.5 > 0.125). A = ∫₀¹ (x − x³) dx = [x²/2 − x&sup4;/4]₀¹ = 1/2 − 1/4 = 1/4.

Q4.3 — y = 6 − x² and y = 2

6 − x² = 2 ⇒ x² = 4 ⇒ x = ±2. On [−2, 2], 6 − x² ≥ 2 (parabola is above the line). A = ∫₋₂² (6 − x² − 2) dx = ∫₋₂² (4 − x²) dx = 32/3.

Q4.4 — y = 3x and y = x²

3x = x² ⇒ x(x − 3) = 0, so x = 0, 3. On (0, 3), 3x > x² (test x = 1: 3 > 1). A = ∫₀³ (3x − x²) dx = [3x²/2 − x³/3]₀³ = 27/2 − 9 = 9/2.

Q4.5 — y = x² and y = 2x + 3

x² = 2x + 3 ⇒ x² − 2x − 3 = (x − 3)(x + 1) = 0, so x = −1, 3. At x = 0: line gives 3, parabola gives 0, so line is on top. A = ∫₋₁³ (2x + 3 − x²) dx = [x² + 3x − x³/3]₋₁³ = (9 + 9 − 9) − (1 − 3 + 1/3) = 9 − (−5/3) = 32/3.

Q4.6 — y = x² − 4x and y = x

x² − 4x = x ⇒ x² − 5x = 0 ⇒ x(x − 5) = 0, so x = 0, 5. At x = 1: line gives 1, parabola gives −3, so line is on top. A = ∫₀⁵ (x − (x² − 4x)) dx = ∫₀⁵ (5x − x²) dx = [5x²/2 − x³/3]₀⁵ = 125/2 − 125/3 = 125/6.

Q4.7 — y = e^x and y = e^−x on [0, 1]

On (0, 1], e^x > e^−x (test x = 1: e > 1/e). A = ∫₀¹ (e^x − e^−x) dx = [e^x + e^−x]₀¹ = (e + e⁻¹) − (1 + 1) = e + 1/e − 2 ≈ 1.086.

Q4.8 — y = sin x and y = cos x on [0, π/4]

On (0, π/4), cos x > sin x (test x = 0: 1 > 0). A = ∫₀^π/4 (cos x − sin x) dx = [sin x + cos x]₀^π/4 = (√2/2 + √2/2) − (0 + 1) = √2 − 1 ≈ 0.414.

Q4.9 — y = x² and y = 8 − x²

x² = 8 − x² ⇒ 2x² = 8 ⇒ x = ±2. On (−2, 2), 8 − x² is on top (test x = 0: 8 > 0). A = ∫₋₂² (8 − x² − x²) dx = ∫₋₂² (8 − 2x²) dx = [8x − 2x³/3]₋₂² = (16 − 16/3) − (−16 + 16/3) = 64/3.

Q4.10 — y = √x and y = x/2

√x = x/2 ⇒ x = x²/4 ⇒ x² − 4x = 0 ⇒ x = 0 or x = 4. On (0, 4), √x > x/2 (test x = 1: 1 > 0.5). A = ∫₀⁴ (√x − x/2) dx = [(2/3)x^3/2 − x²/4]₀⁴ = (16/3 − 4) − 0 = 4/3.

Q4.11 — y = x³ − x and y = 0 on [−1, 1]

Roots at x = −1, 0, 1. On [−1, 0], x³ − x ≥ 0 (test x = −0.5: −0.125 + 0.5 > 0); on [0, 1], x³ − x ≤ 0 (test x = 0.5: 0.125 − 0.5 < 0). The "top" function (the larger of the two curves) changes at x = 0, so we must split or the signed parts cancel.
A = ∫₋₁⁰ (x³ − x) dx + ∫₀¹ −(x³ − x) dx = [x&sup4;/4 − x²/2]₋₁⁰ + [−x&sup4;/4 + x²/2]₀¹ = (0 − (1/4 − 1/2)) + (−1/4 + 1/2 − 0) = 1/4 + 1/4 = 1/2.

Q4.12 — Total area between y = sin x and y = 0 on [0, 2π]

sin x ≥ 0 on [0, π] and sin x ≤ 0 on [π, 2π], so split at π.
A = ∫₀^π sin x dx + ∫_π^2π −sin x dx = [−cos x]₀^π + [cos x]_π^2π = (1 − (−1)) + (1 − (−1)) = 2 + 2 = 4 square units.