Awful pun. The two main approaches to statistics are frequentism and Bayesianism; the latter named after Bayes’ Theorem which is, in turn, named for Reverend Thomas Bayes. While this (both the approaches to statistics and their history and naming) is a very interesting and quite philosophical topic, it is beyond the scope of this course. The interested reader has plenty of interesting reading here to interest them.
application of integration is computing averages and other statistical quantities. We will not spend too much time on this topic — that is best left to a proper course in statistics — however, we will demonstrate the application of integration to the problem of computing averages.
We are being a little loose here with the distinction between mean and average. To be much more pedantic — the average is the arithmetic mean. Other interesting “means” are the geometric and harmonic means:
All of these quantities, along with the median and mode, are ways to measure the typical value of a set of numbers. They all have advantages and disadvantages — another interesting topic beyond the scope of this course, but plenty of fodder for the interested reader and their favourite search engine. But let us put pedantry (and beyond-the-scope-of-the-course-reading) aside and just use the terms average and mean interchangeably for our purposes here.
Now suppose that we want to take the average of a function \(f(x)\) with \(x\) running continuously from \(a\) to \(b\text{.}\) How do we even define what that means? A natural approach is to
select, for each natural number \(n\text{,}\) a sample of \(n\text{,}\) more or less uniformly distributed, values of \(x\) between \(a\) and \(b\text{,}\)
Select, for each \(1\le i\le n\text{,}\) one value of \(x\) from subinterval number \(i\) and call it \(x_i^*\text{.}\) So \(x_{i-1}\le x_i^*\le x_i\text{.}\)
while the left-hand side can be seen as the area of a rectangle of width \(b-a\) and height \(f_\ave\text{.}\) Since these areas must be the same, we interpret \(f_\ave\) as the height of the rectangle which has the same width and the same area as \(\big\{(x,y)\ \big|\ a\le x\le b,\ 0\le y\le f(x)\ \big\}\text{.}\)
Let \(x(t)\) be the position at time \(t\) of a car moving along the \(x\)-axis. The velocity of the car at time \(t\) is the derivative \(v(t)=x'(t)\text{.}\) The average velocity of the car over the time interval \(a\le t\le b\) is
\begin{align*}
v_\ave &= \frac{1}{b-a}\int_a^b v(t)\dee{t}\\
&=\frac{1}{b-a}\int_a^b x'(t)\dee{t}\\
&=\frac{x(b)-x(a)}{b-a} & \text{by the fundamental theorem of calculus.}
\end{align*}
The numerator in this formula is just the displacement (net distance travelled — if \(x'(t)\ge 0\text{,}\) it’s the distance travelled) between time \(a\) and time \(b\) and the denominator is just the time it took.
Notice that this is exactly the formula we used way back at the start of your differential calculus class to help introduce the idea of the derivative. Of course this is a very circuitous way to get to this formula — but it is reassuring that we get the same answer.
A normal household socket delivers alternating current, rather than the direct current USB supplies. At the risk of yet another “the interested reader” suggestion — the how and why household plugs supply AC current is another worthwhile and interesting digression from studying integration. The interested reader should look up the “War of Currents”. The diligent and interested reader should bookmark this, finish the section and come back to it later.
The voltage \(V_0\) is the “peak voltage” — the maximum value the voltage takes over time. More typically we quote the “root mean square” voltage 5
This example was written in North America where the standard voltage supplied to homes is 120 volts. Most of the rest of the world supplies homes with 240 volts. The main reason for this difference is the development of the light bulb. The USA electrified earlier when the best voltage for bulb technology was 110 volts. As time went on, bulb technology improved and countries that electrified later took advantage of this (and the cheaper transmission costs that come with higher voltage) and standardised at 240 volts. So many digressions in this section!
(or RMS-voltage). In this example we explain the difference, but to simplify the calculations, let us simplify the voltage function and just use
What we actually want here is a measure of how far the voltage is from zero. Now we could do this by taking the average of \(|V(t)|\text{,}\) but this is a little harder to work with. Instead we take the average of the square 6
For a finite set of numbers one can compute the “quadratic mean” which is another way to generalise the notion of the average:
In the limit as the length of the time interval \(b-a\) tends to infinity, this converges to \(\frac{V_0^2}{2R}\text{.}\) The resistance \(R\) of a “100 watt bulb” obeys
Subsection2.2.1Optional — Return to the mean value theorem
Here is another application of the Definition 2.2.2 of the average value of a function on an interval. The following theorem can be thought of as an analogue of the mean value theorem (which was covered in your differential calculus class) but for integrals. The theorem says that a continuous function \(f(x)\) must be exactly equal to its average value for some \(x\text{.}\) For example, if you went for a drive along the \(x\)-axis and you were at \(x(a)\) at time \(a\) and at \(x(b)\) at time \(b\text{,}\) then your velocity \(x'(t)\) had to be exactly your average velocity \(\frac{x(b)-x(a)}{b-a}\) at some time \(t\) between \(a\) and \(b\text{.}\) In particular, if your average velocity was greater than the speed limit, you were definitely speeding at some point during the trip. This is, of course, no great surprise 8
There are many unsurprising things that are true, but there are also many unsurprising things that surprisingly turn out to be false. Mathematicians like to prove things - surprising or not.
By the part 1 of the fundamental theorem of calculus (Theorem 1.3.1), \(F'(x)=f(x)\text{,}\) so the mean value theorem says that there is a \(a\lt c\lt b\) with
In the next section, we will encounter an application in which we want to take the average value of a function \(f(x)\text{,}\) but in doing so we want some values of \(x\) to count more than other values of \(x\text{.}\) That is, we want to weight some \(x\)’s more than other \(x\)’s. To do so, we choose a “weight function” \(w(x)\ge 0\) with \(w(x)\) larger for more important \(x\)’s. Then we define the weighted average of \(f\) as follows.
Let \(f(x)\) and \(w(x)\) be integrable functions defined on the interval \(a\le x\le b\) with \(w(x)\ge 0\) for all \(a\le x\le b\) and with \(\int_a^b w(x)\,\dee{x}>0\text{.}\) The average value of \(f\) on that interval, weighted by \(w\text{,}\) is
Here are a few remarks concerning this definition.
The definition has been rigged so that, if \(f(x)=1\) for all \(x\text{,}\) then the weighted average of \(f\) is \(1\text{,}\) no matter what weight function \(w(x)\) is used.
For any function \(w(x)\ge 0\) and any \(a\lt b\text{,}\) we have \(\int_a^b w(x)\, \dee{x}\ge 0\text{.}\) But for the definition of weighted average to make sense, we need to be able to divide by \(\int_a^b w(x)\, \dee{x}\text{.}\) So we need \(\int_a^b w(x)\, \dee{x}\ne 0\text{.}\)
Theorem2.2.10.Mean Value Theorem for Weighted Integrals.
Let \(f(x)\) and \(w(x)\) be continuous functions on the interval \(a\le x\le b\text{.}\) Assume that \(w(x)\gt 0\) for all \(a\lt x\lt b\text{.}\) Then there is some \(c\) obeying \(a\lt c \lt b\) such that
By the part 1 of the fundamental theorem of calculus (Theorem 1.3.1), \(F'(x)=f(x)w(x)\) and \(G'(x)=w(x)\text{,}\) so the generalised mean value theorem says that there is a \(a\lt c\lt b\) with
In this example, we will take a number of weighted averages of the simple function \(f(x)=x\) over the simple interval \(a=1\le x\le 2=b\text{.}\) As \(x\) increases from \(1\) to \(2\text{,}\) the function \(f(x)\) increases linearly from \(1\) to \(2\text{.}\) So it is no shock that the ordinary average of \(f\) is exactly its middle value:
Pick any natural number \(N\ge 1\) and consider the weight function \(w_N(x)=x^N\text{.}\) Note that \(w_N(x)\) increases as \(x\) increases. So \(w_N(x)\) weights bigger \(x\)’s more than it weights smaller \(x\)’s. In particular \(w_N\) weights the point \(x=2\) by a factor of \(2^N\) (which is greater than \(1\) and grows to infinity as \(N\) grows to infinity) more than it weights the point \(x=1\text{.}\) The weighted average of \(f\) is
As we would expect, the \(w_N\)-weighted average is between \(1.5\) (which is the ordinary, unweighted, average) and \(2\) (which is the biggest value of \(f\) in the interval) and grows as \(N\) grows. The limit as \(N\rightarrow\infty\) of the \(w_N\)-weighted average is
Here is an example which shows what can go wrong with Theorem 2.2.10 if we allow the weight function \(w(x)\) to change sign. Let \(a=-0.99\) and \(b=1\text{.}\) Let
As \(c\) runs from \(a\) to \(b\text{,}\)\(f(c)\int_a^b w(x)\,\dee{x}=0.01 f(c)\) runs from \(0\) to \(0.01\) and, in particular, never takes a value anywhere near \(\int_a^b f(x)\,w(x)\,\dee{x}=\frac{1}{2}\text{.}\) There is no \(c\) value which works.
Below is the graph of a function \(y=f(x)\text{.}\) Its average value on the interval \([0,5]\) is \(A\text{.}\) Draw a rectangle on the graph with area \(\int_0^5 f(x)\,\dee{x}\text{.}\)
A force \(F(x)\) acts on an object from position \(x=a\) metres to position \(x=b\) metres, for a total of \(W\) joules of work. What was the average force on the object?
Suppose we want to approximate the average value of the function \(f(x)\) on the interval \([a,b]\text{.}\) To do this, we cut the interval \([a,b]\) into \(n\) pieces, then take \(n\) samples by finding the function’s output at the left endpoint of each piece, starting with \(a\text{.}\) Then, we average those \(n\) samples. (In the example below, \(n=4\text{.}\))
Suppose \(f(x)\) and \(g(x)\) are functions that are defined for all numbers in the interval \([0,10]\text{.}\)
If \(f(x) \leq g(x)\) for all \(x\) in \([0,10]\text{,}\) then is the average value of \(f(x)\) is less than or equal to the average value of \(g(x)\) on the interval \([0,10]\text{,}\) or is there not enough information to tell?
Suppose \(f(x) \leq g(x)\) for all \(x\) in \([0.01,10]\text{.}\) Is the average value of \(f(x)\) less than or equal to the average value of \(g(x)\) over the interval \([0,10]\text{,}\) or is there not enough information to tell?
For Questions 16 through 18, let the root mean square of \(f(x)\) on \([a,b]\) be \(\displaystyle\sqrt{\frac{1}{b-a}\int_a^b f^2(x)\,\dee{x}}\text{.}\) This is the formula used in Example 2.2.6 in the text.
The temperature in Celsius in a 3 m long rod at a point \(x\) metres from the left end of the rod is given by the function \(T(x)=\frac{80}{16-x^2}\text{.}\) Determine the average temperature in the rod.
The carbon dioxide concentration in the air at a particular location over one year is approximated by \(C(t) = 400+50\cos\left(\frac{t}{12}\pi\right)+200\cos\left(\frac{t}{4380}\pi\right)\) parts per million, where \(t\) is measured in hours.
What is the average carbon dioxide concentration for that location for that year?
Suppose measurements were only made at noon every day: that is, when \(t=12+24n\text{,}\) where \(n\) is any whole number between 0 and 364. Then the daily variation would cease: \(50\cos\left(\frac{(12+24n)}{12}\pi\right) = 50\cos\left(\pi+2\pi n\right) = 50\cos\pi=-50\text{.}\) So, the approximation for the concentration of carbon dioxide in the atmosphere might be given as
A force acts on a spring, and the spring stretches and contracts. The distance beyond its natural length at time \(t\) is \(f(t) = \sin\left(t\pi\right)\) cm, where \(t\) is measured in seconds. The spring constant is 3 N/cm.
What is the force exerted by the spring at time \(t\text{,}\) if it obeys Hooke’s law?
For any point \(t\text{,}\) the difference between \(s(t)\) and \(A\) is \(s(t)-A\text{.}\) Find the average value of \(s(t)-A\) on the interval \([0,1]\text{.}\)
For any point \(t\text{,}\) the absolute difference between \(s(t)\) and \(A\) is \(|s(t)-A|\text{.}\) Find the average value of \(|s(t)-A|\) on the interval \([0,1]\text{.}\)
Suppose the root mean square of a function \(f(x)\) on the interval \([a,b]\) is \(R\text{.}\) What is the volume of the solid formed by rotating the portion of \(f(x)\) from \(a\) to \(b\) about the \(x\)-axis?
Suppose \(f(x)=ax^2+bx+c\text{,}\) and the average value of \(f(x)\) on the interval \([0,1]\) is the same as the average of \(f(0)\) and \(f(1)\text{.}\) What is \(a\text{?}\)
Suppose \(f(x)=ax^2+bx+c\text{,}\) and the average value of \(f(x)\) on the interval \([s,t]\) is the same as the average of \(f(s)\) and \(f(t)\text{.}\) Is it possible that \(a \neq 0\text{?}\)
Let \(f(x)\) be a function defined for all numbers in the interval \([a,b]\text{,}\) with average value \(A\) over that interval. What is the average of \(f(a+b-x)\) over the interval \([a,b]\text{?}\)
Find a function \(f(x)\) with average \(0\) over \([-1,1]\) but \(f(x) \neq 0\) for all \(x\) in \([-1,1]\text{,}\) or show that no such function exists.
Find a continuous function \(f(x)\) with average \(0\) over \([-1,1]\) but \(f(x) \neq 0\) for all \(x\) in \([-1,1]\text{,}\) or show that no such function exists.
Suppose \(f(x)\) is a positive, continuous function with \(\lim\limits_{x \to \infty} f(x)=0\text{,}\) and let \(A(x)\) be the average of \(f(x)\) on \([0,x]\text{.}\)
Let \(A(x)\) be the average of the function \(f(t)=e^{-t^2}\) on the interval \([0,x]\text{.}\) What is \(\displaystyle\lim_{x \to \infty} A(x)\text{?}\)
