calculus integration, average height of point on semi circle

Question

i was recently watching a single variable calculus video of mit 18.01, lecture 23. in that it is said that average height of a point on semicircle with respect to arc length is 2/pi.I have a hard time to understand that point. i understand why average height of point on semi circle with respect to x is pi/4. but i dont understand with respect to arc length. plz can somebody help me.

Answer 1

The "average value" of any formula $\varphi$ with respect to any increasing variable $\xi$ is defined as

$$ \frac{\int \varphi \, d\xi}{\int d\xi} .$$

For a unit semicircle, arc length is equal to the angle $\theta$, so we can write the average of height $y$ with respect to arc length as

$$ \frac{\int_0^\pi y \, d\theta}{\int_0^\pi d\theta} = \frac{\int_0^\pi \sin\theta\, d\theta}{\int_0^\pi d\theta} .$$

From there it's simple to find the answer you quoted.

Answer 2

The average value with respect to $x$ over interval $[a,b]$ is found by:

Cut the interval into $n$ equal subintervals of length $\Delta x = \frac{b-a}{n}$. Note that $n = \frac{\Delta x}{b-a}$. On each subinterval, use a single value of $x$ (which I'll call $x_i$). Sum the values of $f\left( x_i \right)$ and divide by $n$

And the limit as $n$ increases without bound is the average value of $f$ on $[a,b]$.

$$\lim_{n \to \infty} \sum_{i=1}^n \frac{f(x_i)}{n} = \lim_{n \to \infty} \sum_{i=1}^n \frac{f(x_i)\Delta x}{b-a} = \frac{1}{b-a} \int_a^b f(x) dx$$

For the upper unit semicircle, we have an average height

$$\frac 1 2 \lim_{n\to \infty}\sum_{i=1}^n \sqrt{1-x_i^2} \Delta x = \frac{1}{2} \int_{-1}^1 \sqrt{1-x^2}dx$$

The average with respect to arc length cuts the curve of total length $T$ into $n$ equal subarcs each of length $\Delta t = \frac{T}{n}$ and again uses a representative value, then takes the limit as the number of subarcs of the curve increases without bound.

$$\lim_{n \to \infty} \sum_{i=1}^n \frac{f(t_i)}{n} = \lim_{n \to \infty} \sum_{i=1}^n \frac{f(t_i)\Delta t}{T} = \frac{1}{T} \int_0^T f(t) dt$$

For the upper unit semicircle, we have $t \in [0,\pi]$ we cut the semicircle into $n$ equal arcs, each of length $\Delta t = \frac{\pi}{n}$. Note that $n = \frac{\pi}{\Delta t}$. For each subarc, we use a $t_i$ The height of the point on the semicircle as a function of $t_i$ is $\sin{t_i}$. So the average value of height w.r.t. arc length is

$$\lim_{n \to \infty} \sum_{i=1}^n \frac {\sin{t_i}}{n} = \lim_{n \to \infty}\sum_{i=1}^n \frac {\sin{t_i} \Delta t}{T} = \frac{1}{T} \int_0^\pi \sin{t} dt$$