How much land lies between two rivers? What is the profit region between cost and revenue curves? When does one population overtake another? These are all questions about the area between two curves — one of the most practical applications of integration. Unlike areas under a single curve, areas between curves require finding where curves cross, deciding which is on top, and carefully setting up the integral. The skill is not just computing integrals; it's translating a geometric or real-world question into a precise mathematical setup.
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Two curves $y = x^2$ and $y = x$ intersect at $(0,0)$ and $(1,1)$. Without calculating, which curve is on top between these points? How would you find the area between them?
Core Content
To find the area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ where $f(x) \geq g(x)$:
A = ∫_a^b [f(x) - g(x)] dx
Why subtraction? The area under the top curve minus the area under the bottom curve leaves exactly the region between them. This is the Riemann sum idea: sum up thin vertical strips of width $dx$ and height $[f(x) - g(x)]$.
When curves cross: If $f(x) \geq g(x)$ on $[a, c]$ but $g(x) \geq f(x)$ on $[c, b]$, split the integral:
A = ∫_a^c [f(x) - g(x)] dx + ∫_c^b [g(x) - f(x)] dx
Always sketch first. A sketch reveals which curve is on top, where they intersect, and whether you need to split. This step prevents the most common error: getting a negative area because the subtraction was backwards.
Find the area between $y = x^2$ and $y = x$.
The enclosed area.
Step 1: Intersections: x^2 = x → x(x-1) = 0 → x = 0, 1
Step 2: On [0,1], x ≥ x^2 (test x = 0.5: 0.5 > 0.25)
Step 3: A = ∫_0^1 (x - x^2) dx
= [x^2/2 - x^3/3]_0^1
= (1/2 - 1/3) - 0 = 1/6
$\frac{1}{6}$ square units.
Find the area between $y = x^2$ and $y = 2x$.
Answer:
Intersections: $x^2 = 2x \Rightarrow x = 0, 2$. On $[0, 2]$, $2x \geq x^2$.
$A = \int_0^2 (2x - x^2) \, dx = [x^2 - \frac{x^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$.
Example: Find the area between $y = x^3 - x$ and $y = 0$ (the x-axis).
Step 1: x^3 - x = 0 → x(x^2 - 1) = 0 → x = -1, 0, 1
Step 2: On [-1, 0]: x^3 - x ≥ 0 (test x = -0.5: -0.125 + 0.5 > 0)
On [0, 1]: x^3 - x ≤ 0 (test x = 0.5: 0.125 - 0.5 < 0)
Step 3: A = ∫_{-1}^0 (x^3 - x) dx + ∫_0^1 -(x^3 - x) dx
= [x^4/4 - x^2/2]_{-1}^0 + [-(x^4/4) + x^2/2]_0^1
= (0 - (1/4 - 1/2)) + (-1/4 + 1/2 - 0) = 1/4 + 1/4 = 1/2
$A = \int_a^b [\text{top} - \text{bottom}] \, dx$
Find intersections: $f(x) = g(x)$
Test a point to see which is on top
Split at crossings if needed
Between $x = 0$ and $x = 1$, the line $y = x$ lies above the parabola $y = x^2$ (you can verify by testing $x = 0.5$: $0.5 > 0.25$). The area is found by integrating the difference: $\int_0^1 (x - x^2) \, dx = \frac{1}{6}$. Notice that without the sketch or test point, you might accidentally compute $\int_0^1 (x^2 - x) \, dx = -\frac{1}{6}$, which has the right magnitude but the wrong sign. Area is always positive, so we must ensure we subtract the lower function from the upper function.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Find the area between $y = x^2$ and $y = 2x + 3$. Show all working. 3 MARKS
9. Find the area between $y = x^3 - x$ and $y = 0$ from $x = -1$ to $x = 1$. Show all working and explain why the integral must be split. 4 MARKS
10. A company's revenue is $R(x) = 80x$ and cost is $C(x) = x^2 + 20x + 300$. (a) Find the break-even points. (b) Find the total profit between the break-even points. (c) Explain what would happen to the profit if the cost curve shifted upward. 4 MARKS
1. $x^2 = 4 \Rightarrow x = \pm 2$. $A = \int_{-2}^{2} (4 - x^2) \, dx = [4x - \frac{x^3}{3}]_{-2}^{2} = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{32}{3}$.
2. $A = \int_0^1 (e^x - e^{-x}) \, dx = [e^x + e^{-x}]_0^1 = (e + e^{-1}) - (1 + 1) = e + \frac{1}{e} - 2$.
3. $x^3 = x \Rightarrow x = -1, 0, 1$. $A = \int_{-1}^0 (x^3 - x) \, dx + \int_0^1 (x - x^3) \, dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
1. When curves cross, the "top" function changes. If we don't split, part of the integral will be negative, and the signed areas will cancel instead of adding.
2. $100x = x^2 + 20x + 400 \Rightarrow x^2 - 80x + 400 = 0 \Rightarrow x = 40 \pm \sqrt{1200} \approx 5.4, 74.6$. Profit = $\int_{5.4}^{74.6} (-x^2 + 80x - 400) \, dx$.
3. A sketch reveals intersections, which curve is on top, and whether splitting is needed. Without it, setup errors are common.
Q8 (3 marks): $x^2 = 2x + 3 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0$, so $x = -1, 3$ [1]. On $[-1, 3]$, $2x + 3 \geq x^2$ (test $x = 0$: $3 > 0$) [0.5]. $A = \int_{-1}^{3} (2x + 3 - x^2) \, dx = [x^2 + 3x - \frac{x^3}{3}]_{-1}^{3} = (9 + 9 - 9) - (1 - 3 + \frac{1}{3}) = 9 - (-\frac{5}{3}) = \frac{32}{3}$ [1.5].
Q9 (4 marks): Intersections: $x^3 - x = 0 \Rightarrow x(x^2 - 1) = 0$, so $x = -1, 0, 1$ [1]. On $[-1, 0]$, $x^3 - x \geq 0$ (test $x = -0.5$: $-0.125 + 0.5 > 0$) [0.5]. On $[0, 1]$, $x^3 - x \leq 0$ [0.5]. $A = \int_{-1}^{0} (x^3 - x) \, dx + \int_{0}^{1} (x - x^3) \, dx = [\frac{x^4}{4} - \frac{x^2}{2}]_{-1}^{0} + [\frac{x^2}{2} - \frac{x^4}{4}]_{0}^{1} = (0 - (\frac{1}{4} - \frac{1}{2})) + (\frac{1}{2} - \frac{1}{4} - 0) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ [2].
Q10 (4 marks): (a) $80x = x^2 + 20x + 300 \Rightarrow x^2 - 60x + 300 = 0$ [1]. Using quadratic formula: $x = 30 \pm \sqrt{600} \approx 5.5, 54.5$ [0.5]. (b) Profit = $\int_{5.5}^{54.5} (-x^2 + 60x - 300) \, dx = [-\frac{x^3}{3} + 30x^2 - 300x]_{5.5}^{54.5} \approx 14815$ [1.5]. (c) If cost shifts up, break-even points move closer together, reducing the profit interval and the total profit [1].
Climb platforms by finding areas between curves, identifying top/bottom functions, and evaluating integrals. Pool: lesson 6.
Tick when you've finished all activities and checked your answers.