2.2: Averages
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Another frequent ^{1} 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.
Let us start with the definition ^{2}
\begin{align*} \text{arithmetic mean} &= \frac{1}{n}\left( y_1 + y_2 + \cdots + y_n \right)\\ \text{geometric mean} &= \left( y_1 \cdot y_2 \cdots y_n \right)^{\frac{1}{n}}\\ \text{harmonic mean} &= \left[\frac{1}{n}\left( \frac{1}{y_1} + \frac{1}{y_2} + \cdots \frac{1}{y_n} \right)\right]^{1} \end{align*}
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 beyondthescopeofthecoursereading) aside and just use the terms average and mean interchangeably for our purposes here. of the average of a finite set of numbers.
The average (mean) of a set of \(n\) numbers \(y_1\text{,}\) \(y_2\text{,}\) \(\cdots\text{,}\) \(y_n\) is
\begin{gather*} y_{\text{ave}} =\bar y = \langle y\rangle =\frac{y_1+y_2+\cdots+y_n}{n} \end{gather*}
The notations \(y_{\text{ave}}\text{,}\) \(\bar y\) and \(\langle y\rangle\) are all commonly used to represent the average.
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 \(b\text{,}\)
 take the average of the values of \(f\) at the selected points,
 and then take the limit as \(n\) tends to infinity.
Unsurprisingly, this process looks very much like how we computed areas and volumes previously. So let's get to it.
 First fix any natural number \(n\text{.}\)
 Subdivide the interval \(a\le x\le b\) into \(n\) equal subintervals, each of width \(\Delta x=\frac{ba}{n}\text{.}\)
 The subinterval number \(i\) runs from \(x_{i1}\) to \(x_i\) with \(x_i=a+i\frac{ba}{n}\text{.}\)
 Select, for each \(1\le i\le n\text{,}\) one value of \(x\) from subinterval number \(i\) and call it \(x_i^*\text{.}\) So \(x_{i1}\le x_i^*\le x_i\text{.}\)
 The average value of \(f\) at the selected points is
\begin{align*} \frac{1}{n}\sum_{i=1}^n f(x_i^*) =&\frac{1}{ba}\sum_{i=1}^n f(x_i^*) \Delta x &\text{since $\Delta x=\frac{ba}{n}$} \end{align*}
giving us a Riemann sum.
Now when we take the limit \(n\rightarrow\infty\) we get exactly \(\frac{1}{ba}\int_a^b f(x)\, d{x}\text{.}\) That's why we define
Let \(f(x)\) be an integrable function defined on the interval \(a\le x\le b\text{.}\) The average value of \(f\) on that interval is
\begin{gather*} f_{\text{ave}}=\bar f=\langle f\rangle =\frac{1}{ba}\int_a^b f(x)\, d{x} \end{gather*}
Consider the case when \(f(x)\) is positive. Then rewriting Definition 2.2.2 as
\(f_{\text{ave}}\ (ba) = \int_a^b f(x)\, d{x}\)
gives us a link between the average value and the area under the curve. The righthand side is the area of the region
\begin{gather*} \big\{(x,y)\ \big\ a\le x\le b,\ 0\le y\le f(x)\ \big\} \end{gather*}
while the lefthand side can be seen as the area of a rectangle of width \(ba\) and height \(f_{\text{ave}}\text{.}\) Since these areas must be the same, we interpret \(f_{\text{ave}}\) 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 us start with a couple of simple examples and then work our way up to harder ones.
Let \(f(x)= x\) and \(g(x)=x^2\) and compute their average values over \(1 \leq x\leq 5\text{.}\)
Solution: We can just plug things into the definition.
\begin{align*} f_{\text{ave}} &= \frac{1}{51}\int_1^5 x \, d{x}\\ &= \frac{1}{4} \bigg[ \frac{x^2}{2} \bigg]_1^5\\ &= \frac{1}{8} (251) = \frac{24}{8}\\ &= 3 \end{align*}
as we might expect. And then
\begin{align*} g_{\text{ave}} &= \frac{1}{51}\int_1^5 x^2 \, d{x}\\ &= \frac{1}{4} \bigg[ \frac{x^3}{3} \bigg]_1^5\\ &= \frac{1}{12} (1251) = \frac{124}{12}\\ &= \frac{31}{3} \end{align*}
Something a little more trigonometric
Find the average value of \(\sin(x)\) over \(0 \leq x \leq \frac{\pi}{2}\text{.}\)
Solution: Again, we just need the definition.
\begin{align*} \text{average} &= \frac{1}{\frac{\pi}{2}  0} \int_0^{\frac{\pi}{2}} \sin(x) \, d{x}\\ &= \frac{2}{\pi} \cdot \bigg[ \cos(x) \bigg]_0^{\frac{\pi}{2}}\\ &= \frac{2}{\pi} (\cos(\frac{\pi}{2})+\cos(0))\\ &= \frac{2}{\pi}. \end{align*}
We could keep going… But better to do some more substantial examples.
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_{\text{ave}} &= \frac{1}{ba}\int_a^b v(t)\, d{t}\\ &=\frac{1}{ba}\int_a^b x'(t)\, d{t}\\ &=\frac{x(b)x(a)}{ba} & \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 very physics example.
When you plug a light bulb into a socket ^{3} and turn it on, it is subjected to a voltage
\begin{align*} V(t) &= V_0\sin(\omega t\delta) \end{align*}
where
 \(V_0=170\) volts,
 \(\omega=2\pi\times 60\) (which corresponds to \(60\) cycles per second^{ 4}) and
 the constant \(\delta\) is an (unimportant) phase. It just shifts the time at which the voltage is zero
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} (or RMSvoltage). In this example we explain the difference, but to simplify the calculations, let us simplify the voltage function and just use
\begin{align*} V(t) &= V_0 \sin(t) \end{align*}
Since the voltage is a sinefunction, it takes both positive and negative values. If we take its simple average over 1 period then we get
\begin{align*} V_{\text{ave}} &= \frac{1}{2\pi0} \int_0^{2\pi} V_0 \sin(t) \, d{t}\\ &= \frac{V_0}{2\pi}\bigg[  \cos(t)\bigg]_0^{2\pi}\\ &= \frac{V_0}{2\pi}\left( \cos(2\pi) + \cos 0\right) = \frac{V_0}{2\pi}(1+1)\\ &= 0 \end{align*}
This is clearly not a good indication of the typical voltage.
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}
\[ \text{quadratic mean} = \sqrt{\frac{1}{n}\left(y_1^2 + y_2^2 + \cdots + y_n^2 \right) } \nonumber \]
of the voltage (so it is always positive) and then take the square root at the end. That is\begin{align*} V_\mathrm{rms} &= \sqrt{\frac{1}{2\pi0} \int_0^{2\pi} V(t)^2 \, d{t}}\\ &= \sqrt{\frac{1}{2\pi} \int_0^{2\pi} V_0^2 \sin^2(t) \, d{t}}\\ &= \sqrt{\frac{V_0^2}{2\pi} \int_0^{2\pi} \sin^2(t) \, d{t}} \end{align*}
This is called the “root mean square” voltage.
Though we do know how to integrate sine and cosine, we don't (yet) know how to integrate their squares. A quick look at doubleangle formulas^{ 7} gives us a way to eliminate the square:
\begin{gather*} \cos(2\theta) =12\sin^2\theta \implies \sin^2\theta=\frac{1\cos(2\theta)}{2} \end{gather*}
Using this we manipulate our integrand a little more:
\begin{align*} V_\mathrm{rms} &= \sqrt{\frac{V_0^2}{2\pi} \int_0^{2\pi} \frac{1}{2}(1\cos(2t)) \, d{t}}\\ &= \sqrt{ \frac{V_0^2}{4\pi} \bigg[t  \frac{1}{2}\sin(2t) \bigg]_0^{2\pi} }\\ &= \sqrt{ \frac{V_0^2}{4\pi} \left(2\pi  \frac{1}{2}\sin(4\pi)  0 + \frac{1}{2}\sin(0) \right) }\\ &= \sqrt{ \frac{V_0^2}{4\pi} \cdot 2\pi }\\ &= \frac{V_0}{\sqrt{2}} \end{align*}
So if the peak voltage is 170 volts then the RMS voltage is \(\frac{170}{\sqrt{2}}\approx 120.2\text{.}\)
Continuing this very physics example:
Let us take our same light bulb with voltage (after it is plugged in) given by
\begin{align*} V(t) &= V_0\sin(\omega t\delta) \end{align*}
where
 \(V_0\) is the peak voltage,
 \(\omega=2\pi\times 60\text{,}\) and
 the constant \(\delta\) is an (unimportant) phase.
If the light bulb is “100 watts”, then what is its resistance?
To answer this question we need the following facts from physics.
 If the light bulb has resistance \(R\) ohms, this causes, by Ohm's law, a current of
\begin{align*} I(t) &= \frac{1}{R} V(t) & \end{align*}
(amps) to flow through the light bulb.  The current \(I\) is the number of units of charge moving through the bulb per unit time.
 The voltage is the energy required to move one unit of charge through the bulb.
 The power is the energy used by the bulb per unit time and is measured in watts.
So the power is the product of the current times the voltage and, so
\[ P(t)=I(t)V(t) =\frac{V(t)^2}{R} =\frac{V_0^2}{R}\sin^2(\omega t\delta) \nonumber \]
The average power used over the time interval \(a\le t\le b\) is
\begin{align*} P_{\text{ave}} &= \frac{1}{ba}\int_a^b P(t)\, d{t} = \frac{V_0^2}{R(ba)}\int_a^b \sin^2(\omega t\delta)\, d{t} \end{align*}
Notice that this is almost exactly the form we had in the previous example when computing the root mean square voltage.
Again we simplify the integrand using the identity
\[ \cos(2\theta) =12\sin^2\theta \implies \sin^2\theta=\frac{1\cos(2\theta)}{2} \nonumber \]
So
\begin{align*} P_{\text{ave}} &= \frac{1}{ba}\int_a^b P(t)\, d{t} = \frac{V_0^2}{2R(ba)}\int_a^b \big[1\cos(2\omega t2\delta)\big]\, d{t}\\ &=\frac{V_0^2}{2R(ba)}\bigg[t\frac{\sin(2\omega t2\delta)}{2\omega}\bigg]_a^b\\ &=\frac{V_0^2}{2R(ba)}\bigg[ba\frac{\sin(2\omega b2\delta)}{2\omega} +\frac{\sin(2\omega a2\delta)}{2\omega}\bigg]\\ &=\frac{V_0^2}{2R} \frac{V_0^2}{4\omega R(ba)}\big[\sin(2\omega b2\delta)\sin(2\omega a2\delta)\big] \end{align*}
In the limit as the length of the time interval \(ba\) tends to infinity, this converges to \(\frac{V_0^2}{2R}\text{.}\) The resistance \(R\) of a “100 watt bulb” obeys
\begin{align*} \frac{V_0^2}{2R} &=100 & \text{so that} && R &= \frac{V_0^2}{200}. \end{align*}
We finish this example off with two side remarks.
 If we translate the peak voltage to the root mean square voltage using
\begin{align*} V_0 &= V_\mathrm{rms} \cdot \sqrt{2} \end{align*}
then we have\begin{align*} P &= \frac{V^2_{\mathrm{rms}}}{R} \end{align*}
 If we were using direct voltage rather than alternating current then the computation is much simpler. The voltage and current are constants, so
\begin{align*} P &= V \cdot I & \text{but $I = V/R$ by Ohm's law}\\ &= \frac{V^2}{R} \end{align*}
So if we have a direct current giving voltage equal to the root mean square voltage, then we would expend the same power.
Optional — 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 \(\frac{x(b)x(a)}{ba}\) 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}.
Let \(f(x)\) be a continuous function on the interval \(a\le x\le b\text{.}\) Then there is some \(c\) obeying \(a\lt c \lt b\) such that
\begin{gather*} \frac{1}{ba}\int_a^b f(x)\, d{x}=f(c) \qquad\text{or}\qquad \int_a^b f(x)\, d{x} = f(c)\,(ba) \end{gather*}

We will apply the mean value theorem (Theorem 2.13.4 in the CLP1 text) to the function
\[ F(x) = \int_a^x f(t)\,\, d{t} \nonumber \]
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
\begin{align*} f(c) &= F'(c) = \frac{F(b)F(a)}{ba} =\frac{1}{ba}\left\{\int_a^b f(t)\,\, d{t}\int_a^a f(t)\,\, d{t}\right\}\\ &=\frac{1}{ba}\int_a^b f(x)\,\, d{x} \end{align*}
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 \(\int_a^b w(x)\,\, d{x}>0\text{.}\) The average value of \(f\) on that interval, weighted by \(w\text{,}\) is
\[ \frac{\int_a^b f(x)\,w(x)\,\, d{x}}{\int_a^b w(x)\,\, d{x}} \nonumber \]
We typically refer to this simply as the weighted average of \(f\text{.}\)
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.
 If the weight function \(w(x)=C\) for some constant \(C>0\) then the weighted average
\[ \frac{\int_a^b f(x)\,w(x)\,\, d{x}}{\int_a^b w(x)\,\, d{x}} =\frac{\int_a^b f(x)\,C\,\, d{x}}{\int_a^b C\,\, d{x}} =\frac{\int_a^b f(x)\,\, d{x}}{ba} \nonumber \]
is just the usual average.  For any function \(w(x)\ge 0\) and any \(a\lt b\text{,}\) we have \(\int_a^b w(x)\, \, d{x}\ge 0\text{.}\) But for the definition of weighted average to make sense, we need to be able to divide by \(\int_a^b w(x)\, \, d{x}\text{.}\) So we need \(\int_a^b w(x)\, \, d{x}\ne 0\text{.}\)
The next theorem says that a continuous function \(f(x)\) must be equal to its weighted average at some point \(x\text{.}\)
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
\begin{gather*} \frac{\int_a^b f(x)\,w(x)\,\, d{x}}{\int_a^b w(x)\,\, d{x}}=f(c) \qquad\text{or}\qquad \int_a^b f(x)\,w(x)\,\, d{x} = f(c)\int_a^b w(x)\,\, d{x} \end{gather*}

We will apply the generalised mean value theorem (Theorem 3.4.38 in the CLP1 text) to
\[ F(x) = \int_a^x f(t)\,w(t)\, d{t}\qquad G(x) = \int_a^x w(t)\, d{t} \nonumber \]
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
\begin{align*} f(c) &= \frac{F'(c)}{G'(c)} = \frac{F(b)F(a)}{G(b)G(a)} =\frac{\int_a^b f(t)\,w(t)\, d{t}\int_a^a f(t)w(t)\,\, d{t}} {\int_a^b w(t)\, d{t}\int_a^a w(t)\, d{t}}\\ &=\frac{\int_a^b f(t)\,w(t)\,\, d{t}}{\int_a^b w(t)\,\, d{t}} \end{align*}
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:
\[ \frac{1}{ba}\int_a^b f(t)\,\, d{t} =\frac{1}{21}\int_1^2 t\,\, d{t} =\frac{3}{2} \nonumber \]
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
\begin{align*} \frac{\int_a^b f(t)\,w_N(t)\,\, d{t}}{\int_a^b w_N(t)\,\, d{t}} &=\frac{\int_1^2 t^{N+1}\,\, d{t}}{\int_1^2 t^N\,\, d{t}} =\frac{\frac{2^{N+2}1}{N+2}}{\frac{2^{N+1}1}{N+1}} =\frac{N+1}{N+2}\ \frac{2^{N+2}1}{2^{N+1}1} \\ &=\begin{cases} \frac{2\times 7}{3\times 3} = 1.555 &\text{if }N=1 \\ \frac{3\times 15}{4\times 7} = 1.607 &\text{if }N=2 \\ \frac{4\times 31}{5\times 15}= 1.653 &\text{if }N=3 \\ \frac{5\times 63}{6\times 31}= 1.694 &\text{if }N=4 \\ 1.889 &\text{if }N=16 \\ 1.992 &\text{if }N=256 \end{cases} \end{align*}
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
\begin{align*} \lim_{N\rightarrow\infty}\frac{N+1}{N+2}\ \frac{2^{N+2}1}{2^{N+1}1} &=\lim_{N\rightarrow\infty}\frac{N+21}{N+2}\ \frac{2^{N+2}2+1}{2^{N+1}1} \\ &=\lim_{N\rightarrow\infty}\left[1\frac{1}{N+2}\right]\ \left[2+\frac{1}{2^{N+1}1}\right] \\ &=2 \end{align*}
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
\begin{align*} w(x)&=\begin{cases} 1&\text{if }x\ge 0 \\ 1&\text{if }x\lt 0 \end{cases} \\ f(x)&=\begin{cases} x&\text{if }x\ge 0 \\ 0&\text{if }x\lt 0 \end{cases} \end{align*}
Then
\begin{align*} \int_a^b f(x)\,w(x)\,\, d{x} &= \int_0^1 x\,\, d{x} =\frac{1}{2} \\ \int_a^b w(x)\,\, d{x} &= \int_0^1 \, d{x} \int_{0.99}^0\, d{x} = 1  0.99 = 0.01 \end{align*}
As \(c\) runs from \(a\) to \(b\text{,}\) \(f(c)\int_a^b w(x)\,\, d{x}=0.01 f(c)\) runs from \(0\) to \(0.01\) and, in particular, never takes a value anywhere near \(\int_a^b f(x)\,w(x)\,\, d{x}=\frac{1}{2}\text{.}\) There is no \(c\) value which works.
Exercises
Recall that we are using \(\log x\) to denote the logarithm of \(x\) with base \(e\text{.}\) In other courses it is often denoted \(\ln x\text{.}\)
Stage 1
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 \(\int_0^5 f(x)\,\, d{x}\text{.}\)
Suppose a car travels for 5 hours in a straight line, with an average velocity of 100 kph. How far did the car travel?
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, \(n=4\text{.}\))
 Using \(n\) samples, what is the distance between two consecutive sample points \(x_i\) and \(x_{i+1}\text{?}\)
 Assuming \(n \geq 4\text{,}\) what is the \(x\)coordinate of the fourth sample?
 Assuming \(n \geq 4\text{,}\) what is the \(y\)value of the fourth sample?
 Write the approximation of the average value of \(f(x)\) over the interval \([a,b]\) using sigma notation.
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?
Suppose \(f\) is an odd function, defined for all real numbers. What is the average of \(f\) on the interval \([10,10]\text{?}\)
Stage 2
For Questions 16 through 18, let the root mean square of \(f(x)\) on \([a,b]\) be \(\displaystyle\sqrt{\frac{1}{ba}\int_a^b f^2(x)\,\, d{x}}\text{.}\) This is the formula used in Example 2.2.6 in the text.
Find the average value of \(f(x) = \sin(5x)+1\) over the interval \(\pi/2 \le x \le \pi/2 \text{.}\)
Find the average value of the function \(y= x^2\log x\) on the interval \(1 \le x\le e\text{.}\)
Find the average value of the function \(f(x) = 3\cos^3x + 2\cos^2x\) on the interval \(0\le x\le\frac\pi2\text{.}\)
Let \(k\) be a positive constant. Find the average value of the function \(f(x) = \sin(kx)\) on the interval \(0\le x\le \pi/k\text{.}\)
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 \(T(x)=\frac{80}{16x^2}\text{.}\) Determine the average temperature in the rod.
What is the average value of the function \(f(x)=\dfrac{\log x}{x}\) on the interval \([1,e]\text{?}\)
Find the average value of \(f(x)=\cos^2(x)\) over \(0\le x\le 2\pi\text{.}\)
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?
 What is the average over the first day?
 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
\[ N(t) = 350 +200\cos\left(\frac{t}{4380}\pi\right)\quad\text{ ppm} \nonumber \]
What is the relative error in the yearly average concentration of carbon dioxide involved in using \(N(t)\text{,}\) instead of \(C(t)\text{?}\)
You may assume a day has exactly 24 hours, and a year has exactly 8760 hours.
Let \(S\) be the solid formed by rotating the parabola \(y=x^2\) from \(x=0\) to \(x=2\) about the \(x\)axis.
 What is the average area of the circular crosssections of \(S\text{?}\) Call this value \(A\text{.}\)
 What is the volume of \(S\text{?}\)
 What is the volume of a cylinder with circular crosssectional area \(A\) and length 2?
Let \(f(x) = x\text{.}\)
 Calculate the average of \(f(x)\) over \([3,3]\text{.}\)
 Calculate the root mean square of \(f(x)\) over \([3,3]\text{.}\)
Calculate the root mean square of \(f(x) = \tan x\) over \(\left[\frac{\pi}{4},\frac{\pi}{4}\right]\text{.}\)
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?
 Find the average of the force exerted by the spring from \(t=0\) to \(t=6\text{.}\)
 Find the root mean square of the force exerted by the spring from \(t=0\) to \(t=6\text{.}\)
Stage 3
A car travels two hours without stopping. The driver records the car's speed every 20 minutes, as indicated in the table below:
time in hours  0  1/3  2/3  1  4/3  5/3  2 
speed in km/hr  50  70  80  55  60  80  40 
 Use the trapezoidal rule to estimate the total distance traveled in the two hours.
 Use the answer to part (a) to estimate the average speed of the car during this period.
Let \(s(t) = e^t\text{.}\)
 Find the average of \(s(t)\) on the interval \([0,1]\text{.}\) Call this quantity \(A\text{.}\)
 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 \([0,1]\text{.}\)
Consider the two functions \(f(x)\) and \(g(x)\) below, both of which have average \(A\) on \([0,4]\text{.}\)
 Which function has a larger average on \([0,4]\text{:}\) \(f(x)A\) or \(g(x)A\text{?}\)
 Which function has a larger average on \([0,4]\text{:}\) \(f(x)A\) or \(g(x)A\text{?}\)
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?
As in Example 2.2.6, let the root mean square of \(f(x)\) on \([a,b]\) be \(\displaystyle\sqrt{\frac{1}{ba}\int_a^b f^2(x)\,\, d{x}}\text{.}\)
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 \(a \neq 0\text{?}\)
That is — does the result of Question 23 generalize?
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 \(f(a+bx)\) over the interval \([a,b]\text{?}\)
Suppose \(f(t)\) is a continuous function, and \(A(x)\) is the average of \(f(t)\) on the interval from 0 to \(x\text{.}\)
 What is the average of \(f(t)\) on \([a,b]\text{,}\) where \(a \lt b\text{?}\) Give your answer in terms of \(A\text{.}\)
 What is \(f(t)\text{?}\) Again, give your answer in terms of \(A\text{.}\)
 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 \([0,x]\text{.}\)
True or false: \(\lim\limits_{x \to \infty} A(x) = 0\text{.}\)
Let \(A(x)\) be the average of the function \(f(t)=e^{t^2}\) on the interval \([0,x]\text{.}\) What is \(\displaystyle\lim_{x \to \infty} A(x)\text{?}\)
 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.
 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:
 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.
 Some countries supply power at 50 cycles per second. Japan actually supplies both — 50 cycles in the east of the country and 60 in the west.
 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!
 For a finite set of numbers one can compute the “quadratic mean” which is another way to generalise the notion of the average:
 A quick glance at Appendix A.14 will refresh your memory.
 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.