Year 12 Methods — Average Value of a Function (11J)

Today's lesson

You already know how to average a list of numbers: add them up, divide by how many. Today we average a whole continuous function. The trick: add them up becomes an integral.

Learning intentions

Compute the average value of $f$ on $[a,b]$ using $\bar y = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$
Interpret the average value as the height of the rectangle with the same area as the region under the curve
Use a graphical argument to find the average of a vertically-shifted sine curve (no calculus required)
Apply average value to real contexts — temperature in a room, stock prices over a period

Warm-up — averaging four numbers

Find the average of $4, 7, 10, 3$. Stuck? $\bar y = \dfrac{4+7+10+3}{4} = \dfrac{24}{4} = 6$.

Picture the four numbers as towers of heights 4, 7, 10, 3 next to each other. The "average" is the height you'd get by levelling all four towers to the same height: $6$. The total stuff stays the same; you just spread it evenly.

Part 1 — From discrete to continuous

For a continuous function we replace "add up the heights" with "integrate" and "divide by how many" with "divide by the length of the interval":

AVERAGE VALUE OF A FUNCTION The average value of $f$ on $[a,b]$ is $$\bar y \;=\; \frac{1}{b-a}\int_a^b f(x)\,dx.$$

Geometrically, $\bar y$ is the height of the horizontal line that traps exactly the same area over $[a,b]$ as the curve itself. The region above $\bar y$ (where $f>\bar y$) and the region below $\bar y$ (where $f<\bar y$) are equal in area.

📺 Walkthrough: a curve $y=\tfrac{1}{2}x^2+1$ on $[1,4]$, its average value $\bar y = 4.5$, and the gold rectangle that holds the same area.

Part 2 — Worked Example 1 (CAS)

EXAMPLE 1

Sketch $g(x) = \sqrt{x-5}$ and find the average value of $g(x)$ over the interval $[25, 36]$.

Sketch. Square-root translated right by $5$. Defined for $x \ge 5$. At $x=25$, $g = \sqrt{20} = 2\sqrt 5 \approx 4.47$. At $x=36$, $g = \sqrt{31} \approx 5.57$.

$\bar y = \dfrac{1}{36-25}\displaystyle\int_{25}^{36} \sqrt{x-5}\,dx$

$= \dfrac{1}{11}\left[\tfrac{2}{3}(x-5)^{3/2}\right]_{25}^{36}$

$= \dfrac{2}{33}\bigl(31^{3/2} - 20^{3/2}\bigr) = \dfrac{2}{33}\bigl(31\sqrt{31} - 20\sqrt{20}\bigr)$

Simplify $20\sqrt{20} = 20\cdot 2\sqrt 5 = 40\sqrt 5$:

$\bar y = \dfrac{2}{33}\bigl(31\sqrt{31} - 40\sqrt 5\bigr) \approx 5.04$

📺 Walkthrough: Example 1 step-by-step — antiderivative of $\sqrt{x-5}$, evaluation on $[25, 36]$, exact form and decimal value.

⚠️ Watch the domain. Asked for the average of $\sqrt{x-5}$ on $[0, 25]$? The function isn't even defined on $[0, 5)$. Restrict to $[5, 25]$ before integrating — or note that the question is ill-posed.

Part 3 — Worked Example 2 (graphical — no calculus)

A bumpy curve that oscillates symmetrically around some horizontal line $y = k$ has $\bar y = k$ — no integration needed.

EXAMPLE 2 (graphical)

Find the average value of $y = 3\sin(5x) + 4$ over the interval $\left[0, \tfrac{2\pi}{5}\right]$.

Period. $T = \dfrac{2\pi}{5}$ — so the interval is exactly one full period.

$y = 3\sin(5x) + 4$ is a sine wave oscillating between $4 - 3 = 1$ and $4 + 3 = 7$, centred on $y = 4$.

Over one full period the "bumps above $y=4$" and "bumps below $y=4$" are equal in area — they cancel.

$\bar y = 4$ (no calculation needed!)

📺 Walkthrough: $y = 3\sin(5x) + 4$ over one period — green/red bumps cancel, so $\bar y$ = the vertical translation $= 4$.

💡 Trick to memorise. Over a whole number of periods, the average of $a\sin(\omega x + \varphi) + k$ (or $a\cos$) is just $k$ — the vertical shift. The sine bit averages to zero.

Part 4 — Worked Example 3 (CAS context)

EXAMPLE 3 — Air-conditioning temperature

The temperature in degrees in a room $t$ minutes after air-conditioning is turned on is given by $T(t) = t\sin\!\left(\tfrac{\pi t}{4}\right) + 20$ for $0 \le t \le 14$. Find the average temperature over these 14 minutes.

$\bar T = \dfrac{1}{14}\displaystyle\int_0^{14} \left[t\sin\!\left(\tfrac{\pi t}{4}\right) + 20\right] dt$

CAS gives $\displaystyle\int_0^{14} t\sin\!\left(\tfrac{\pi t}{4}\right) dt = -\dfrac{16}{\pi^2}$ (exact).

$\displaystyle\int_0^{14} 20\, dt = 280$. So the total integral $= 280 - \dfrac{16}{\pi^2}$.

$\bar T = \dfrac{1}{14}\left(280 - \dfrac{16}{\pi^2}\right) = 20 - \dfrac{8}{7\pi^2}$

$\bar T \approx 19.88\,°$C

Part 5 — Worked Example 4 (CAS — stocks comparison)

EXAMPLE 4 — Comparing stocks

Over an 8-month period, two stock prices (in $\$$) are modelled by $\text{Orange: } S(t) = 0.4t + 1$ and $\text{Blue: } B(t) = -0.3t^2 + 2.7t - 0.83 + 2\sin(2t)$. Compare the average price of each stock over the 8 months.

Orange: $\bar S = \dfrac{1}{8}\displaystyle\int_0^8 (0.4t + 1)\, dt = \dfrac{1}{8}\bigl[0.2t^2 + t\bigr]_0^8 = \dfrac{1}{8}(12.8 + 8)$

$= \dfrac{20.8}{8}$

$\bar S = \$2.60$

Blue (CAS): $\bar B = \dfrac{1}{8}\displaystyle\int_0^8 \bigl[-0.3t^2 + 2.7t - 0.83 + 2\sin(2t)\bigr] dt$

$= \dfrac{1}{8}\left[-0.1t^3 + 1.35t^2 - 0.83t - \cos(2t)\right]_0^8$ (careful: $\cos$ is exact, not periodic-cancelling)

$=\dfrac{1}{8}\!\left[(-51.2 + 86.4 - 6.64 - \cos 16) - (0 + 0 - 0 - 1)\right]$

$=\dfrac{1}{8}\!\left[28.56 + 1 - \cos 16\right] = \dfrac{29.56 - \cos 16}{8}$

$\bar B \approx \$3.81$ — Blue stock averaged higher than Orange.

Part 6 — Practice

Practice 6.1 — straight averages.

Average value of $f(x) = x^2$ on $[0, 3]$.
Average value of $f(x) = 2x + 1$ on $[0, 4]$.
Average value of $f(x) = \sin x$ on $[0, \pi]$.
Average value of $f(x) = e^x$ on $[0, 1]$.

a) $\bar y = \dfrac{1}{3}\bigl[\tfrac{x^3}{3}\bigr]_0^3 = \dfrac{1}{3}\cdot 9 = 3$.
b) $\bar y = \dfrac{1}{4}\bigl[x^2 + x\bigr]_0^4 = \dfrac{20}{4} = 5$.
c) $\bar y = \dfrac{1}{\pi}\bigl[-\cos x\bigr]_0^{\pi} = \dfrac{1}{\pi}(1 - (-1)) = \dfrac{2}{\pi} \approx 0.637$.
d) $\bar y = \dfrac{1}{1}\bigl[e^x\bigr]_0^1 = e - 1 \approx 1.718$.

Practice 6.2 — graphical shortcut.

Find the average value of $y = 5\cos(3x) - 2$ over the interval $\left[0, \tfrac{2\pi}{3}\right]$ without integrating.

Period $T = \tfrac{2\pi}{3}$ — the interval is exactly one period. Over one full period, $\cos$ averages to $0$, so the only thing left is the constant: $\bar y = -2$.

Practice 6.3 — application.

The depth (m) of water at a jetty at time $t$ hours is $D(t) = 2\sin\!\left(\tfrac{\pi t}{6}\right) + 5$ for $0 \le t \le 12$. Find the average depth over the 12-hour period.

Period of $\sin\!\left(\tfrac{\pi t}{6}\right)$ is $T = \dfrac{2\pi}{\pi/6} = 12$ — exactly one period. So the sine bit averages to $0$ and $\bar D = 5$ m.

Part 7 — Quick quiz (5 min)

Working program — Cambridge §11J

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

Exercise	Set work
11J — Average value of a function	Q1, Q2 acef, Q3, Q5, Q7
Application questions	Q8, Q9 (modelling contexts)

Tomorrow we mix everything together — areas, integrals, average value — in a recap practice set.

Exit ticket — write in your book

Before you pack up, in your exercise book:

State the average-value formula in your own words (one sentence, no symbols).
Find the average value of $f(x) = 6x$ on $[0, 4]$ — try to do it without integrating (linear function: where is the "balance height"?).
Why does $y = a\sin(\omega x) + k$ have average $k$ over any whole number of periods?