Year 12 Methods — Area Between Curves (11I)

Today's lesson

Last lesson you used a definite integral to find the area between one curve and the $x$-axis. Now we'll do the area between two curves — the region trapped between them.

Learning intentions

Find the area enclosed between two curves $y=f(x)$ and $y=g(x)$ by integrating top minus bottom
Find the limits of integration by solving $f(x)=g(x)$ for the intersection points
Handle regions where the upper curve changes — split at the cross-over point

Part 1 — The "top minus bottom" rule

AREA BETWEEN TWO CURVES If $f(x) \ge g(x)$ for all $x$ in $[a,b]$, then the area enclosed between $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is $$A \;=\; \int_a^b \bigl[\,f(x) - g(x)\,\bigr]\,dx.$$

The "top minus bottom" automatically handles any negative regions — if the whole picture is below the $x$-axis, the answer is still positive because $f$ (the higher one) sits above $g$ (the lower one).

⚠️ Which one is on top? Test one $x$-value in the interval — e.g. the midpoint — and see which function is bigger there. If $f$ is bigger at that test point, $f$ is the "top" function on that interval.

📺 Walkthrough: $y=x+1$ above $y=x^2-x-2$ between their intersection points. The "top minus bottom" gives area $= \tfrac{32}{3}$.

Part 2 — Setting up the integral (4-step recipe)

RECIPE

Sketch the two curves to see the enclosed region.
Find intersection points — solve $f(x)=g(x)$. Those $x$-values are your limits.
Identify the top function — substitute a test value between the intersections.
Integrate $\displaystyle\int_a^b (\text{top} - \text{bottom})\,dx$.

EXAMPLE 1 — Line and parabola

Find the area of the region enclosed by $y = x + 1$ and $y = x^2 - x - 2$.

Step 1 (intersections). Set equal: $x+1 = x^2 - x - 2$ $\Rightarrow$ $x^2 - 2x - 3 = 0$ $\Rightarrow$ $(x-3)(x+1) = 0$, so $x = -1$ or $x = 3$.

Step 2 (top function). Test $x=0$: line gives $0+1=1$, parabola gives $0-0-2=-2$. The line is on top.

Step 3 (set up). $A = \displaystyle\int_{-1}^{3} \bigl[(x+1) - (x^2-x-2)\bigr]\,dx = \int_{-1}^{3} (-x^2 + 2x + 3)\,dx$

Step 4 (evaluate). $=\left[-\tfrac{x^3}{3} + x^2 + 3x\right]_{-1}^{3}$

$= \!\left(-9 + 9 + 9\right) - \!\left(\tfrac{1}{3} + 1 - 3\right) = 9 - \!\left(-\tfrac{5}{3}\right) = 9 + \tfrac{5}{3}$

$A = \dfrac{32}{3}$ square units

EXAMPLE 2 — Curve and a line

Find the area enclosed by $y = x^2$ and $y = 2x$.

Intersections: $x^2 = 2x \Rightarrow x(x-2)=0 \Rightarrow x=0,2$.

Test $x=1$: $y=x^2$ gives $1$, $y=2x$ gives $2$. Line is on top.

$A = \displaystyle\int_0^2 (2x - x^2)\,dx = \left[x^2 - \tfrac{x^3}{3}\right]_0^2$

$= \!\left(4 - \tfrac{8}{3}\right) - 0 = \tfrac{12-8}{3}$

$A = \dfrac{4}{3}$ sq units

EXAMPLE 3 — Two parabolas

Find the area enclosed by $y = 4 - x^2$ and $y = x + 2$.

Intersections: $4 - x^2 = x + 2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1) = 0$, so $x=-2,1$.

Test $x=0$: parabola gives $4$, line gives $2$. Parabola is on top.

$A = \displaystyle\int_{-2}^{1} \bigl[(4-x^2) - (x+2)\bigr]\,dx = \int_{-2}^{1} (2 - x - x^2)\,dx$

$= \left[2x - \tfrac{x^2}{2} - \tfrac{x^3}{3}\right]_{-2}^{1}$

$= \!\left(2 - \tfrac{1}{2} - \tfrac{1}{3}\right) - \!\left(-4 - 2 + \tfrac{8}{3}\right) = \tfrac{7}{6} - \!\left(-\tfrac{10}{3}\right) = \tfrac{7}{6} + \tfrac{20}{6}$

$A = \dfrac{27}{6} = \dfrac{9}{2}$ sq units

Part 3 — When the curves cross inside the region

If the upper curve swaps partway through (i.e. the curves cross at three or more points), you must split at every cross-over and treat each piece separately.

📺 Walkthrough: $y=x^3$ vs $y=x$ on $[-1,1]$. The cubic is above on $[-1,0]$ and below on $[0,1]$ — split at $x=0$ and add absolute values to get total area $=\tfrac{1}{2}$.

EXAMPLE 4 — Curves that swap

Find the total area enclosed between $y = x^3$ and $y = x$ on $-1 \le x \le 1$.

Intersections. $x^3 = x \Rightarrow x(x^2-1) = 0 \Rightarrow x = -1, 0, 1$.

Which is on top in each piece? Test $x=-\tfrac{1}{2}$: $x^3 = -\tfrac{1}{8}$, $x = -\tfrac{1}{2}$ — so the cubic is above (less negative). Test $x=\tfrac{1}{2}$: $x^3 = \tfrac{1}{8}$, $x = \tfrac{1}{2}$ — the line is above.

Left piece: $\displaystyle\int_{-1}^{0} (x^3 - x)\,dx = \left[\tfrac{x^4}{4} - \tfrac{x^2}{2}\right]_{-1}^{0} = 0 - \!\left(\tfrac{1}{4} - \tfrac{1}{2}\right) = \tfrac{1}{4}$

Right piece: $\displaystyle\int_{0}^{1} (x - x^3)\,dx = \left[\tfrac{x^2}{2} - \tfrac{x^4}{4}\right]_{0}^{1} = \tfrac{1}{2} - \tfrac{1}{4} = \tfrac{1}{4}$

Total area $= \tfrac{1}{4} + \tfrac{1}{4} = \dfrac{1}{2}$ sq units (symmetry confirms!)

💡 Symmetry shortcut. If a region is symmetric about the origin (or the $y$-axis), you can integrate over half the region and double it. Always sanity-check the answer.

Part 4 — A transcendental example

EXAMPLE 5 — Exponential vs line

Find the exact area enclosed by $y = e^x$, the line $y = x$, the $y$-axis and the line $x = 1$.

Test $x = 0.5$: $e^{0.5} \approx 1.65$, $x = 0.5$. $e^x$ is on top.

(The region is bounded by $x=0$, $x=1$ — no need to solve for intersection; the limits are given.)

$A = \displaystyle\int_0^1 (e^x - x)\,dx = \left[e^x - \tfrac{x^2}{2}\right]_0^1$

$= \!\left(e - \tfrac{1}{2}\right) - (1 - 0) = e - \tfrac{3}{2}$

$A = e - \dfrac{3}{2}$ sq units ($\approx 1.218$)

Practice 4.1 — set up and evaluate.

Area enclosed by $y = x^2$ and $y = x + 2$.
Area enclosed by the two parabolas $y = x^2$ and $y = 8 - x^2$.

a) Intersections: $x^2 = x+2 \Rightarrow (x-2)(x+1)=0$, so $x=-1, 2$. Line on top. $\int_{-1}^{2} [(x+2) - x^2]\,dx = [\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}]_{-1}^{2} = (\tfrac{10}{3}) - (-\tfrac{7}{6}) = \dfrac{9}{2}$ sq units.
b) Intersections: $x^2 = 8 - x^2 \Rightarrow 2x^2 = 8 \Rightarrow x = \pm 2$. The "upside-down" parabola is on top. $\int_{-2}^{2} [(8-x^2) - x^2]\,dx = \int_{-2}^{2} (8 - 2x^2)\,dx = [8x - \tfrac{2x^3}{3}]_{-2}^{2} = \dfrac{64}{3}$ sq units.

Practice 4.2 — split at a crossing.

Find the total area enclosed by $y = \sin x$ and $y = \cos x$ on $\tfrac{\pi}{4} \le x \le \tfrac{5\pi}{4}$.

On $[\tfrac{\pi}{4}, \tfrac{5\pi}{4}]$, $\sin x \ge \cos x$ throughout (they only meet at the endpoints).
$A = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x)\,dx = [-\cos x - \sin x]_{\pi/4}^{5\pi/4}$
$= (-\cos\tfrac{5\pi}{4} - \sin\tfrac{5\pi}{4}) - (-\cos\tfrac{\pi}{4} - \sin\tfrac{\pi}{4})$
$= (\tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2}) - (-\tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{2}}{2}) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$ sq units.

Part 5 — Quick quiz (5 min)

Working program — Cambridge §11I

After the quiz, open Chapter 11 and work through these:

Exercise	Set work
11I — Areas between two curves	Q1, Q2 ace, Q3, Q5, Q7
Chapter 11 review	Multiple-choice Q1–6, Short-answer Q1, Q3 (definite integrals + areas)

Exit ticket — write in your book

Before you pack up, in your exercise book:

State the "top minus bottom" rule for the area between two curves (one line).
Sketch the region enclosed by $y = x$ and $y = x^2$ and write down the integral (do not evaluate).
Why must you split at any intersection point that lies inside the integration interval?