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.
Today's hook, Two curves $y = x^2$ and $y = x$ cross at $(0,0)$ and $(1,1)$. Between those points, which one is on top? And why does it matter which way you set up the integral? By the end of this lesson, a negative area will never fool you again.
0/5QUESTS
1
Orient to areas between curves
Work through the visible teaching and complete each embedded check.
Worksheets
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?
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The three-step method
+5 XP to read
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.
A sketch reveals intersections and which curve is higher. Without it, the subtraction order is guesswork and you risk a negative area.
Area is always positive
If your answer is negative, you have the subtraction backwards. Swap $f$ and $g$, or take absolute value, and check your test point.
Split at crossings
If curves cross between $a$ and $b$, split the integral at the crossing point. Each piece integrates top $-$ bottom for that sub-interval.
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What you'll master
Know
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
Understand
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
Can do
Skills
Find intersection points of polynomial curves
Set up area integrals correctly with the right subtraction order
Evaluate areas, splitting when curves cross
2
Build the area-between-curves language
Work through the visible teaching and complete each embedded check.
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Key terms
Area between curves$A = \int_a^b [f(x) - g(x)]\,dx$ where $f(x) \geq g(x)$ on $[a,b]$.
Intersection pointA point $(x_0, y_0)$ where $f(x_0) = g(x_0)$; found by solving $f(x) = g(x)$.
Top functionThe function with larger $y$-values on a given interval; always goes first in the subtraction.
Split integralWhen curves cross, the integral is split at the crossing: $\int_a^c [\ldots] + \int_c^b [\ldots]$.
Signed area$\int_a^b f(x)\,dx$ can be negative; geometric area always uses $|\text{top} - \text{bottom}|$.
Revenue-cost modelProfit between break-even points equals $\int_{x_1}^{x_2}[R(x) - C(x)]\,dx$.
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Set up area integrals
Work through the visible teaching and complete each embedded check.
05
Setting up the integral
core concept
To find the area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ where $f(x) \geq g(x)$:
$$A = \int_a^b [f(x) - g(x)]\,dx$$
Why subtraction? The area under the top curve minus the area under the bottom curve leaves exactly the region between them. 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 $\frac{1}{6}$.
Profit region in economics. A company's revenue is $R(x) = 120x - x^2$ and cost is $C(x) = 60x + 500$. The company breaks even when $R(x) = C(x)$, that is when $-x^2 + 60x - 500 = 0$, so at $x = 10$ or $x = 50$. The total profit generated between the break-even points is $\int_{10}^{50}[R(x) - C(x)]\,dx$, the area between the curves literally represents accumulated profit over that production range. The mathematical structure, find intersections, determine which is higher, integrate the difference, is identical whether you're analysing rivers, profit functions, or population models.
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 $\displaystyle\int_0^1 (x - x^2)\,dx$.
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When curves cross multiple times
core concept
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.
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?
4
Solve area examples
Work through the visible teaching and complete each embedded check.
Worked examples · 3 in a row, reveal as you go
PROBLEM 1 · BASIC AREA BETWEEN CURVES
Find the area between $y = x^2$ and $y = x$.
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$x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, \; x = 1$
Find intersections. These become the limits $a = 0$, $b = 1$.
2
Test $x = 0.5$: $0.5 > 0.25$, so $y = x$ is on top.
Confirm which function is greater on the interval. The top function goes first in the integral.
Integrate top minus bottom ($R - C$) over the profitable interval. The area between the curves is the total accumulated profit.
Fill in the blank: The area between $y = x^2$ and $y = 2x$ is $\displaystyle\int_0^2 (2x - x^2)\,dx =$ ___
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Avoid area-between-curves traps
Work through the visible teaching and complete each embedded check.
Common errors · the 3 traps that cost marks
Trap 01
Subtracting the wrong way
Always integrate $[\text{top function}] - [\text{bottom function}]$. If you get a negative area, you have it backwards. A quick test with a single $x$-value before integrating tells you which is larger, never skip this.
Trap 02
Not splitting at crossings
When curves cross inside the interval, the "top" function changes. Failing to split produces signed areas that cancel, you may get zero or a small number instead of the true geometric area. Identify all crossing points first.
Trap 03
Using the wrong limits
The limits of integration must be the intersection points, not arbitrary values. Solve $f(x) = g(x)$ exactly. If the question gives limits, use those, but check for crossings inside the interval and split if needed.
Odd one out: Three of these statements about area between curves are correct. Which one is wrong?
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Practise and reflect
Work through the visible teaching and complete each embedded check.
Quick-fire practice · 3 problems
1
Find the area between $y = x^2$ and $y = 4$
2
Find the area between $y = e^x$ and $y = e^{-x}$ from $x = 0$ to $x = 1$
3
Find the area between $y = x^3$ and $y = x$
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Revisit your thinking
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.
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Complete the short-answer practice
Apply the lesson methods, then compare each response with its comprehensive answer.
01
Focus-area checkpoint
checkpoint
Use the visible short-answer practice below during this lesson. After this lesson sequence, use Checkpoint 2 for the focus-area checkpoint.
02
Short answer
ApplyBand 43 marks
Q1. Find the area between $y = x^2$ and $y = 2x + 3$. Show all working. (3 marks)
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ApplyBand 54 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)
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AnalyseBand 54 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)
