Area Between Curves — Introduction
Two curves enclose a region of the plane — your job is to measure it exactly. The key idea is deceptively simple: integrate the difference of the upper and lower functions over the interval. But getting limits right, handling sign correctly, and knowing when to split the region separates a confident solver from a lost one. This lesson builds every piece from scratch.
The curves $y = x^2$ and $y = x$ both pass through the origin. Without computing anything — sketch the region they enclose and estimate its area in square units. Write your reasoning below.
Every area-between-curves problem reduces to one subtraction inside an integral: integrate the top curve minus the bottom curve over the correct interval.
If $f(x) \geq g(x)$ on $[a,b]$, the enclosed area is:
$A = \displaystyle\int_a^b [f(x) - g(x)]\,dx$
The limits $a$ and $b$ are the x-coordinates of the intersection points unless the problem specifies otherwise.
Key facts
- $A = \displaystyle\int_a^b [f(x) - g(x)]\,dx$ when $f(x) \geq g(x)$
- Limits come from solving $f(x) = g(x)$
- The integrand is always (upper) $-$ (lower), never reversed
Concepts
- Why the formula measures the gap between curves, not the gap from the $x$-axis
- Why area is always positive even if curves dip below the $x$-axis
- The connection between Riemann sums and the integral formula
Skills
- Sketch two curves and identify the enclosed region
- Find intersection points algebraically
- Set up and evaluate $\int_a^b [f(x)-g(x)]\,dx$
Suppose two curves $y = f(x)$ and $y = g(x)$ intersect at $x = a$ and $x = b$, with $f(x) \geq g(x)$ throughout $[a,b]$. The area of the region between them is:
Why does subtracting work? Think of thin vertical strips of width $\Delta x$. Each strip has height $f(x) - g(x)$ (the gap between the curves) and area $[f(x)-g(x)]\,\Delta x$. Adding infinitely many strips gives the integral. The key insight is that both curves can be anywhere relative to the $x$-axis — all that matters is the difference between them.
Finding the limits: Set $f(x) = g(x)$ and solve for $x$. In the simplest case there are two solutions $x = a$ and $x = b$ which become the limits.
Suppose two curves $y = f(x)$ and $y = g(x)$ intersect at $x = a$ and $x = b$, with $f(x) \geq g(x)$ throughout $[a,b]$. The area of the region between them is:
Pause — copy the area formula $A=\int_a^b[f(x)-g(x)]\,dx$ and the condition $f(x)\geq g(x)$ on $[a,b]$ into your book.
Quick check: The curves $y = 4$ and $y = x^2$ enclose a region. Which integral correctly gives its area?
We just saw that the area between two curves where $f(x)\geq g(x)$ on $[a,b]$ is $A=\int_a^b[f(x)-g(x)]\,dx$. That raises a question: before you can apply this formula you need the limits $a$ and $b$ — how do you find the intersection points, and what do you do when the equation $f(x)=g(x)$ has more than two solutions? This card answers it → set $f(x)=g(x)$, solve, and check that $f\geq g$ on each sub-interval.
The limits of integration come from the intersection points of the two curves. To find them:
- Set the two equations equal: $f(x) = g(x)$.
- Rearrange to $f(x) - g(x) = 0$.
- Factorise or use the quadratic formula.
- The real solutions are the $x$-coordinates of the intersection points.
Then substitute back to find the $y$-coordinates if needed (e.g., for sketching).
Example: Find where $y = 6x - x^2$ and $y = x$ intersect.
$6x - x^2 = x \Rightarrow 5x - x^2 = 0 \Rightarrow x(5-x) = 0$
So $x = 0$ or $x = 5$. The region runs from $x=0$ to $x=5$.
The limits of integration come from the intersection points of the two curves. To find them:
Pause — copy the intersection-point method: set $f(x)=g(x)$, solve for $x$, verify which curve is upper on each interval, then apply the formula into your book.
Did you get this? True or false: the curves $y = x^2$ and $y = 2x$ intersect at $x = 0$ and $x = 2$.
Worked examples · 3 in a row, reveal as you go
Find the area enclosed by $y = x$ and $y = x^2$.
Find the area enclosed by $y = 6x - x^2$ and $y = x$.
Find the area between $y = x^2 - 4$ and $y = 0$ (the $x$-axis) on the interval where the parabola is below the axis.
Fill the gap: The area between $y=x$ and $y=x^2$ on $[0,1]$ equals $\displaystyle\int_0^1(x-x^2)\,dx = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when $f(x) \leq 0$ and $g(x) \leq 0$ on $[a,b]$, the area between the curves can still be found using $\int_a^b [f(x)-g(x)]\,dx$ (where $f$ is the upper curve).
Activities · practice with the ideas
Find the area enclosed by $y = x^2$ and $y = 2x$. (Sketch first.)
Find the area between $y = 4$ and $y = x^2$.
Set up (but do not evaluate) the integral for the area between $y = \sqrt{x}$ and $y = x^2$ on $[0,1]$.
The region between $y = x$ and $y = x^3$ from $x=0$ to $x=1$. Find the area.
Explain in one sentence why the area between $y = f(x)$ and $y = g(x)$ is always positive when $f(x) \geq g(x)$, even if both functions are negative.
Odd one out: Three of these are correct steps in a standard area-between-curves problem. Which step is NOT correct?
Earlier you estimated the area between $y=x$ and $y=x^2$. The exact answer is $\dfrac{1}{6} \approx 0.167$ square units.
The region is a thin sliver between two curves that start and end at the same points $(0,0)$ and $(1,1)$. The key insight: even though the integral formula looks like it "measures from the $x$-axis", subtracting the two functions measures only the gap between them, regardless of where they sit on the plane.
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. State the integral that gives the area enclosed by $y = 2x$ and $y = x^2$. (1 mark)
Q2. Find the exact area enclosed by $y = 4$ and $y = x^2$. (3 marks)
Q3. Find the area of the region bounded by $y = x^2$ and $y = x + 2$. Show all working including finding the intersection points. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. Intersections: $x=0,2$; $A=\int_0^2(2x-x^2)\,dx=\frac{4}{3}$. · 2. Intersections: $x=\pm2$; $A=\int_{-2}^{2}(4-x^2)\,dx=\frac{32}{3}$. · 3. On $[0,1]$: $\sqrt{x}\geq x^2$; $A=\int_0^1(\sqrt{x}-x^2)\,dx$. · 4. $A=\int_0^1(x-x^3)\,dx=\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1=\frac{1}{4}$. · 5. Because $f(x)-g(x)\geq0$ for all $x\in[a,b]$, the integral of a non-negative function is non-negative.
Q1 (1 mark): Intersections: $x^2=2x\Rightarrow x=0,2$. $A=\int_0^2(2x-x^2)\,dx$ [1].
Q2 (3 marks): $x^2=4\Rightarrow x=\pm2$ [1]. $A=\int_{-2}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-2}^{2}$ [1]. $=\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)=16-\frac{16}{3}=\dfrac{32}{3}$ [1].
Q3 (3 marks): $x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow(x-2)(x+1)=0$, so $x=-1,2$ [1]. On $[-1,2]$: $y=x+2$ is above $y=x^2$. $A=\int_{-1}^{2}(x+2-x^2)\,dx=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}$ [1]. $=\left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{20}{6}+\frac{7}{6}=\dfrac{27}{6}=\dfrac{9}{2}$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering area-between-curves questions. Lighter alternative to the boss.
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