Areas Between Curves
How much land lies between two rivers? What is the profit region between a cost and revenue curve? When does one population overtake another? All of these are questions about the area between two curves. The skill is not just computing integrals — it's translating a real-world question into a precise mathematical setup: find intersections, decide which curve is on top, integrate the difference.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
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?
Every area-between-curves problem follows the same three steps. Master the steps and no setup will ever trip you up.
Step 1 — Find intersection points by solving $f(x) = g(x)$. These are your limits of integration.
Step 2 — Test a point in each interval to decide which curve is on top.
Step 3 — Integrate $[\text{top} - \text{bottom}]$. If curves cross, split into separate integrals.
Key facts
- Area $= \int_a^b [\text{top} - \text{bottom}]\,dx$
- Find intersections by solving $f(x) = g(x)$
- Split at intersection points if curves cross
Concepts
- Why we subtract the lower curve from the upper
- When curves cross and why we must split
- Geometric meaning of the integral as accumulated strips
Skills
- Find intersection points of polynomial curves
- Set up area integrals correctly with the right subtraction order
- Evaluate areas, splitting when curves cross
To find the area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ where $f(x) \geq g(x)$:
Why subtraction? The area under the top curve minus the area under the bottom curve leaves exactly the region between them. Think of thin vertical strips of width $dx$ and height $[f(x) - g(x)]$ — summing these up gives the total area.
When curves cross: If $f(x) \geq g(x)$ on $[a, c]$ but $g(x) \geq f(x)$ on $[c, b]$, split:
Always sketch first. A sketch reveals which curve is on top, where they intersect, and whether you need to split. This prevents the most common error: a negative area because the subtraction was backwards.
Area between $y = x$ (upper) and $y = x^2$ (lower) from $0$ to $1$ equals $\frac{1}{6}$.
Formula: $A = \int_a^b [f(x) - g(x)]\,dx$ where $f(x) \geq g(x)$; Step 1: Find intersections by solving $f(x) = g(x)$
Pause — copy the area formula $A = \int_a^b [f(x) - g(x)]\,dx$ where $f(x) \geq g(x)$, and the setup method (find intersections by solving $f(x) = g(x)$, always sketch first) into your book.
Did you get this? True or false: the area between $y = x$ and $y = x^2$ from $0$ to $1$ is $\displaystyle\int_0^1 (x - x^2)\,dx$.
We just saw that $A = \int_a^b [f(x) - g(x)]\,dx$ works when $f \geq g$ throughout $[a,b]$. That raises a question: what if the curves cross inside the interval, so that which function is on top changes — making a single unsplit integral give zero or a negative result? This card answers it → split the integral at every crossing point, use a test value to confirm which curve is on top in each sub-interval, and add the pieces.
Example: Find the area between $y = x^3 - x$ and $y = 0$ (the $x$-axis).
Step 1 — Intersections: $x^3 - x = 0 \Rightarrow x(x^2-1) = 0 \Rightarrow x = -1, 0, 1$
Step 2 — Which is on top?
- On $[-1, 0]$: test $x = -0.5$: $(-0.5)^3 - (-0.5) = -0.125 + 0.5 = 0.375 > 0$, so $x^3 - x$ is above the $x$-axis.
- On $[0, 1]$: test $x = 0.5$: $0.125 - 0.5 = -0.375 < 0$, so $x^3 - x$ is below the $x$-axis.
Step 3 — Split and integrate:
Notice: if you simply computed $\int_{-1}^{1}(x^3 - x)\,dx$ without splitting, the signed areas would cancel and you would get $0$ — completely wrong for a geometric area question.
When curves cross inside the interval, split the integral at every crossing point; In each sub-interval, the top function changes — always re-check with a test point
Pause — copy the crossing-curves rule (split integral at every crossing point, confirm top function with a test point in each sub-interval, add absolute values of each piece) into your book.
Quick check: To find the area between $y = x^2$ and $y = 4$, which integral is correct?
Worked examples · 3 in a row, reveal as you go
Find the area between $y = x^2$ and $y = x$.
Find the area between $y = x^3 - x$ and $y = 0$ from $x = -1$ to $x = 1$.
A revenue curve is $R(x) = 100x$ and cost is $C(x) = x^2 + 20x + 400$. Find the break-even points and the total profit between them.
Fill in the blank: The area between $y = x^2$ and $y = 2x$ is $\displaystyle\int_0^2 (2x - x^2)\,dx =$ ___
Common errors · the 3 traps that cost marks
Odd one out: Three of these statements about area between curves are correct. Which one is wrong?
Quick-fire practice · 3 problems
Find the area between $y = x^2$ and $y = 4$
Find the area between $y = e^x$ and $y = e^{-x}$ from $x = 0$ to $x = 1$
Find the area between $y = x^3$ and $y = x$
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: $\int_0^1 (x - x^2)\,dx = \frac{1}{6}$. 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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the area between $y = x^2$ and $y = 2x + 3$. Show all working. (3 marks)
Q2. 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)
Q3. 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)
Comprehensive answers (click to reveal)
Drill 1: Intersections $x = \pm 2$. $A = \int_{-2}^{2}(4-x^2)\,dx = [4x - \frac{x^3}{3}]_{-2}^{2} = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{32}{3}$
Drill 2: $A = \int_0^1(e^x - e^{-x})\,dx = [e^x + e^{-x}]_0^1 = (e + e^{-1}) - 2 = e + \frac{1}{e} - 2$
Drill 3: Intersections $x = -1, 0, 1$. $A = \int_{-1}^{0}(x^3-x)\,dx + \int_{0}^{1}(x-x^3)\,dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Q1 (3 marks): $x^2 = 2x+3 \Rightarrow (x-3)(x+1)=0$, so $x = -1, 3$ [1]. On $[-1,3]$, $2x+3 \geq x^2$ (test $x=0$: $3>0$). $A = \int_{-1}^{3}(2x+3-x^2)\,dx = [x^2+3x-\frac{x^3}{3}]_{-1}^{3} = (9+9-9)-(1-3+\frac{1}{3}) = 9+\frac{5}{3} = \frac{32}{3}$ [2].
Q2 (4 marks): Intersections $x = -1, 0, 1$ [1]. On $[-1,0]$: $x^3-x \geq 0$ (test $-0.5$: positive) [0.5]. On $[0,1]$: $x^3-x \leq 0$ (test $0.5$: negative) [0.5]. $A = \int_{-1}^{0}(x^3-x)\,dx + \int_{0}^{1}(x-x^3)\,dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ [2].
Q3 (4 marks): (a) $80x = x^2+20x+300 \Rightarrow x^2-60x+300=0 \Rightarrow x = 30 \pm \sqrt{600} \approx 5.5, 54.5$ [1]. (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) Shifting the cost curve upward means the profit function $R-C$ decreases everywhere, the break-even points move closer together, and the total profit (area between curves) decreases [1.5].
Five timed questions on areas between curves. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Enter the arenaClimb platforms by finding areas between curves, identifying top/bottom functions, and evaluating integrals correctly.
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