Single-Variable-Calculus · Unit 19 · Video 1 · Interactive Practice

Average Value: Two Heights of a Semicircle

IKey Formulas

Formula	Meaning	What you need
$\displaystyle \lim_{n\to\infty}\frac{y_1 + \cdots + y_n}{n} = \frac{1}{b-a}\int_a^b f(x)\,dx$	The continuous average of $f$ on $[a, b]$	The Riemann sum with $\Delta x = \tfrac{b-a}{n}$ , divided by the length $b - a$
$\displaystyle \frac{1}{b-a}\int_a^b c\,dx = c$	The average of a constant is that constant	The factor $\tfrac{1}{b-a}$ , which is chosen to make this true
$\displaystyle \frac12\int_{-1}^{1}\sqrt{1-x^2}\,dx = \frac{\pi}{4}$	Average height of the unit semicircle with respect to $x$	The integral is the semicircle's area, $\tfrac{\pi}{2}$
$\displaystyle \frac1\pi\int_0^{\pi}\sin\theta\,d\theta = \frac{2}{\pi}$	Average height with respect to arc length (the angle $\theta$ )	Height $\sin\theta$ , antiderivative $-\cos\theta$

Key Insight: An average is an integral divided by the length of the interval, and it is only defined once you name the variable you average over. Over $x$ the semicircle's height averages $\tfrac{\pi}{4} \approx 0.785$ ; over arc length it averages $\tfrac{2}{\pi} \approx 0.637$ , because equal steps of arc crowd toward the low ends.

IIVisualization 1 — Samples Become the Integral

The average of $n$ equally spaced samples of $\sqrt{1-x^2}$ tends to $\tfrac12\int_{-1}^{1}\sqrt{1-x^2}\,dx = \tfrac{\pi}{4}$ .

Samples n 6

IIIVisualization 2 — Equal Steps in x or in θ?

The same semicircle, sampled evenly along the diameter or evenly along the arc, gives two different average heights.

Equal steps in Points n 13

IVVisualization 3 — Guess the Average Height

The average is the height of the rectangle on $[a, b]$ whose area equals the area under the curve.

Function Your guess 0.5

VQuiz Questions

Problem 1 · Average of a Parabola

Given: $f(x) = x^2$ on the interval $0 \le x \le 3$ — find the average value of $f$ .

✅ Correct!

\tfrac13\int_0^3 x^2\,dx = \tfrac13 \cdot 9 = 3

❌ That is the integral, not the average.

\int_0^3 x^2\,dx = 9

; the average divides it by the length of the interval,

3

❌ That averages only the endpoint values.

\tfrac{f(0) + f(3)}{2} = \tfrac92

uses two samples; the continuous average uses all of them,

\tfrac13\int_0^3 x^2\,dx

❌ That is $f$ at the midpoint.

f\!\left(\tfrac32\right) = \tfrac94

, but the average of

f

is not

f

of the average

x

when

f

curves.

❌ Not quite. Compute

\tfrac{1}{b-a}\int_a^b f(x)\,dx

with

a = 0

b = 3

Show solution

The interval has length $b - a = 3 - 0 = 3$ , so

\text{Average} = \frac{1}{3}\int_0^3 x^2\,dx = \frac13\left[\frac{x^3}{3}\right]_0^3 = \frac13 \cdot 9 = 3.

Problem 2 · The Average of √x Is Not √ of the Average

Given: $f(x) = \sqrt{x}$ on $0 \le x \le 4$ — find the average value of $f$ .

✅ Correct!

\tfrac14\int_0^4 \sqrt{x}\,dx = \tfrac14 \cdot \tfrac23 \cdot 4^{3/2} = \tfrac14 \cdot \tfrac{16}{3} = \tfrac43

❌ That is $f$ at the average $x$ .

\sqrt{2} \approx 1.414

f(2)

. Because

\sqrt{x}

bends, its average

\tfrac43 \approx 1.333

is different.

❌ That is the integral alone.

\int_0^4 \sqrt{x}\,dx = \tfrac{16}{3}

; dividing by the length

4

gives the average.

❌ That averages the endpoint values.

\tfrac{f(0) + f(4)}{2} = 1

ignores every value in between.

❌ Not quite. Integrate

x^{1/2}

from

0

4

and divide by

4

Show solution

\frac{1}{4}\int_0^4 x^{1/2}\,dx = \frac14\left[\frac23 x^{3/2}\right]_0^4 = \frac14 \cdot \frac23 \cdot 8 = \frac43.

Compare $f(2) = \sqrt{2} \approx 1.414$ : the average of a function is generally not the function's value at the midpoint.

Problem 3 · A Semicircle of Radius 2, Averaged over Arc Length

Given: the upper semicircle of radius $2$ centred at the origin. A point is chosen on the arc, with pieces of equal arc length equally likely — find the average height of the point.

✅ Correct! Height

2\sin\theta

for

0 \le \theta \le \pi

\tfrac1\pi\int_0^\pi 2\sin\theta\,d\theta = \tfrac{4}{\pi} \approx 1.27

❌ That is the average over $x$ .

\tfrac14\int_{-2}^{2}\sqrt{4-x^2}\,dx = \tfrac14 \cdot 2\pi = \tfrac{\pi}{2}

picks the point evenly along the diameter, a different experiment.

❌ That is the unit semicircle's answer. On radius

2

the height at angle

\theta

2\sin\theta

, so every height doubles.

❌ That is the integral, not the average.

\int_0^\pi 2\sin\theta\,d\theta = 4

; the angle runs over an interval of length

\pi

, so divide by

\pi

❌ Not quite. Arc length is

2\theta

, so equal arc length means equal steps in

\theta

; average

2\sin\theta

over

[0, \pi]

Show solution

A point at angle $\theta$ on the circle of radius $2$ is $(2\cos\theta, 2\sin\theta)$ , and its arc length from the right end is $s = 2\theta$ . Equal arc lengths are equal steps in $\theta$ , so we average over $0 \le \theta \le \pi$ :

\frac{1}{\pi}\int_0^\pi 2\sin\theta\,d\theta = \frac{2}{\pi}\Big[-\cos\theta\Big]_0^\pi = \frac{2}{\pi}(1 + 1) = \frac{4}{\pi}.

Averaging over $s$ directly gives the same: $\tfrac{1}{2\pi}\int_0^{2\pi} 2\sin\tfrac{s}{2}\,ds = \tfrac{4}{\pi}$ . Over $x$ the answer is $\tfrac{\pi}{2} \approx 1.57$ , larger, exactly as for the unit semicircle.

Problem 4 · Average Temperature over a Day

Given: from $t = 0$ to $t = 12$ hours the temperature is $T(t) = 20 + 10\sin\!\left(\tfrac{\pi t}{12}\right)$ degrees — find the average temperature over those 12 hours.

✅ Correct!

\tfrac{1}{12}\int_0^{12} T\,dt = 20 + \tfrac{10}{12} \cdot \tfrac{24}{\pi} = 20 + \tfrac{20}{\pi} \approx 26.4

degrees.

❌ That averages the lowest and highest temperatures.

\tfrac{20 + 30}{2} = 25

would be right for a linear rise; the sine stays near its peak longer than near its ends.

❌ The sine does not average to zero here. Over

0 \le t \le 12

the angle

\tfrac{\pi t}{12}

runs only from

0

\pi

, where

\sin

is never negative.

❌ That is the integral, in degree-hours. Divide

\int_0^{12} T\,dt

by the length of the time interval,

12

❌ Not quite. Average the constant

20

and the sine term separately, then add.

Show solution

The constant $20$ averages to itself. For the sine term, an antiderivative of $\sin\tfrac{\pi t}{12}$ is $-\tfrac{12}{\pi}\cos\tfrac{\pi t}{12}$ :

\int_0^{12} 10\sin\frac{\pi t}{12}\,dt = 10 \cdot \frac{12}{\pi}\Big[-\cos\frac{\pi t}{12}\Big]_0^{12} = \frac{120}{\pi}(1 + 1) = \frac{240}{\pi}.

\text{Average} = 20 + \frac{1}{12} \cdot \frac{240}{\pi} = 20 + \frac{20}{\pi} \approx 26.37.

The sine part averages $\tfrac{2}{\pi}$ of its peak $10$ : the same $\tfrac{2}{\pi}$ as the semicircle's arc average.

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