Single-Variable-Calculus · Unit 19 · Video 2 · Interactive Practice

Weighted Averages: The Cauldron's Average Temperature

IKey Formulas

Formula	Meaning	What you need
$\displaystyle \frac{\int_a^b f(x)\,w(x)\,dx}{\int_a^b w(x)\,dx}$	Weighted average of $f$ with weight $w$	Two integrals: $f$ times the weight on top, the total weight below
$\displaystyle \frac{\int_a^b c\,w(x)\,dx}{\int_a^b w(x)\,dx} = c$	A constant averages to itself, for any weight	The constant factors out and the two integrals cancel
$\displaystyle \int_0^1 (100 - 30y)\,\pi y\,dy = 40\pi$	Energy to heat the cauldron, in $^\circ\text{C} \cdot \text{m}^3$	Horizontal disks: volume $\pi x^2\,dy = \pi y\,dy$ , temperature $T(y)$
$\displaystyle \frac{40\pi}{\int_0^1 \pi y\,dy} = \frac{40\pi}{\pi/2} = 80$	Volume-weighted average temperature, $80^\circ\text{C}$	Weight $w(y) = \pi y$ ; the plain average over height is $85^\circ\text{C}$

Key Insight: The weight says how much each part counts. The cauldron is wide at the cool top and narrow at the hot bottom, so weighting each level by its volume pulls the average below the $85^\circ\text{C}$ that counts every level equally.

IIVisualization 1 — Average Price of Shares

Shares bought at $10, $20 and $30 average to the price where the share counts balance.

Shares at 10 3 Shares at 20 2 Shares at 30 5

IIIVisualization 2 — Slicing the Cauldron into Disks

Weighting each level by its disk volume $\pi y\,dy$ gives $80^\circ\text{C}$ ; counting every level equally gives $85^\circ\text{C}$ .

Disks n 5 Weight

IVVisualization 3 — Where the Two Averages Fall

For any linear temperature in this cauldron, the volume-weighted average is $\tfrac13 T_{\text{bottom}} + \tfrac23 T_{\text{top}}$ .

T at bottom 100 T at top 70

VQuiz Questions

Problem 1 · A Weight That Grows with x

Given: $f(x) = x$ on $0 \le x \le 1$ with weight $w(x) = x$ — find the weighted average of $f$ .

✅ Correct!

\dfrac{\int_0^1 x \cdot x\,dx}{\int_0^1 x\,dx} = \dfrac{1/3}{1/2} = \dfrac23

: large

x

counts more, so the average moves above

\tfrac12

❌ That is the numerator alone.

\int_0^1 x^2\,dx = \tfrac13

still has to be divided by the total weight

\int_0^1 x\,dx = \tfrac12

❌ That is the plain average, with $w = 1$ . The weight

w(x) = x

makes values near

x = 1

count more than values near

0

❌ The fraction is upside down. The integral of

f w

goes on top and the total weight below; an average of values in

[0, 1]

cannot exceed

1

❌ Not quite. Compute

\int_0^1 f w\,dx

and

\int_0^1 w\,dx

, then divide.

Show solution

\frac{\int_0^1 x \cdot x\,dx}{\int_0^1 x\,dx} = \frac{\left[\tfrac{x^3}{3}\right]_0^1}{\left[\tfrac{x^2}{2}\right]_0^1} = \frac{1/3}{1/2} = \frac23.

Problem 2 · Two Purchases

Given: you buy $100$ shares at $10 and later $300$ shares at $30 — find your average price per share.

✅ Correct! Total spent

1000 + 9000 = 10{,}000

dollars over

400

shares: $25 per share.

❌ That treats both purchases as equal. The plain average of the two prices ignores that three times as many shares were bought at the higher price.

❌ The weights are swapped.

\tfrac{300 \cdot 10 + 100 \cdot 30}{400} = 15

puts the large count on the low price.

❌ That is the total spent. Divide by the total weight, the

400

shares.

❌ Not quite. Weighted average = total spent divided by the total number of shares.

Show solution

The prices are the values and the share counts are the weights:

\frac{10 \cdot 100 + 30 \cdot 300}{100 + 300} = \frac{10{,}000}{400} = 25.

The average is $25 per share, three quarters of the way from $10 to $30, because three quarters of the shares were bought at $30.

Problem 3 · A Colder Cauldron

Given: the same cauldron, the region between $y = x^2$ and $y = 1$ revolved about the $y$ -axis, now with temperature $T(y) = 90 - 60y$ — find the average temperature of the water, weighted by volume.

✅ Correct!

\dfrac{\int_0^1 (90 - 60y)\,\pi y\,dy}{\int_0^1 \pi y\,dy} = \dfrac{25\pi}{\pi/2} = 50

degrees.

❌ That counts every level equally.

\tfrac{90 + 30}{2} = 60

is the average over height; the wide, cool top holds more water.

❌ That is the energy integral alone.

\int_0^1 (90 - 60y)\,\pi y\,dy = 25\pi

; divide by the volume

\tfrac{\pi}{2}

❌ The denominator is the volume $\tfrac{\pi}{2}$ , not $\pi$ . Dividing

25\pi

\pi

leaves out the

\tfrac12

from

\int_0^1 y\,dy

❌ Not quite. Use the disk weight

w(y) = \pi y

for both integrals.

Show solution

The disk at height $y$ has radius $x = \sqrt{y}$ , so its volume is $\pi y\,dy$ and the weight is $w(y) = \pi y$ .

\int_0^1 (90 - 60y)\,\pi y\,dy = \pi\left[45y^2 - 20y^3\right]_0^1 = 25\pi, \qquad \int_0^1 \pi y\,dy = \frac{\pi}{2}.

\text{Average} = \frac{25\pi}{\pi/2} = 50.

Check with $\tfrac13 T_{\text{bottom}} + \tfrac23 T_{\text{top}} = \tfrac13 \cdot 90 + \tfrac23 \cdot 30 = 50$ .

Problem 4 · Temperature That Depends on the Radius

Given: the same cauldron, but now heated from the side, so its temperature depends only on the distance $x$ from the axis — which slicing puts each slice at a single temperature, and what is its weight?

✅ Correct! A shell is the set of points at one distance

x

from the axis, so it has one temperature; its volume is

2\pi x(1 - x^2)\,dx

❌ A disk spans every radius from $0$ to its edge. If

T

depends on

x

, the temperature changes across each disk, so a disk is no longer at a single temperature.

❌ The weight must be the shell's volume. A shell at radius

x

runs from

y = x^2

up to

y = 1

, so its height

1 - x^2

belongs in the weight.

❌ The slicing follows the temperature. Slice along the level sets of

T

: here they are cylinders around the axis, so only shells work without a second averaging.

❌ Not quite. Ask which slices keep the distance from the axis fixed.

Show solution

Choose slices on which $T$ is constant. If $T$ depends only on $x$ , those are cylinders of radius $x$ : the shells. The shell at radius $x$ has circumference $2\pi x$ , height $1 - x^2$ and thickness $dx$ , so

w(x) = 2\pi x(1 - x^2), \qquad \text{Average} = \frac{\int_0^1 T(x)\,2\pi x(1 - x^2)\,dx}{\int_0^1 2\pi x(1 - x^2)\,dx}.

The denominator is the cauldron's volume again, $\tfrac{\pi}{2}$ .

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