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 anythingsketch 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 confidencethat 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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