CalculusWithJuliaNotes.jl/quarto/integrals/mean_value_theorem.qmd

# Mean value theorem for integrals


```{julia}
#| echo: false

import Logging
Logging.disable_logging(Logging.Info) # or e.g. Logging.Info
Logging.disable_logging(Logging.Warn)

import SymPy
function Base.show(io::IO, ::MIME"text/html", x::T) where {T <: SymPy.SymbolicObject}
    println(io, "<span class=\"math-left-align\" style=\"padding-left: 4px; width:0; float:left;\"> ")
    println(io, "\\[")
    println(io, sympy.latex(x))
    println(io, "\\]")
    println(io, "</span>")
end

# hack to work around issue
import Markdown
import CalculusWithJulia
function CalculusWithJulia.WeaveSupport.ImageFile(d::Symbol, f::AbstractString, caption; kwargs...)
    nm = joinpath("..", string(d), f)
    u = "![$caption]($nm)"
    Markdown.parse(u)
end

nothing
```

This section uses these add-on packages:


```{julia}
using CalculusWithJulia
using Plots
using QuadGK
```

```{julia}
#| echo: false
#| results: "hidden"
using CalculusWithJulia.WeaveSupport

const frontmatter = (
        title = "Mean value theorem for integrals",
        description = "Calculus with Julia: Mean value theorem for integrals",
        tags = ["CalculusWithJulia", "integrals", "mean value theorem for integrals"],
);
nothing
```

---


## Average value of a function


Let $f(x)$ be a continuous function over the interval $[a,b]$ with $a < b$.


The average value of $f$ over $[a,b]$ is defined by:


$$
\frac{1}{b-a} \int_a^b f(x) dx.
$$

If $f$ is a constant, this is just the contant value, as would be expected. If $f$ is *piecewise* linear, then this is the weighted average of these constants.


#### Examples


##### Example: average velocity


The average velocity between times $a < b$, is simply the change in position during the time interval divided by the change in time. In notation, this would be $(x(b) - x(a)) / (b-a)$. If $v(t) = x'(t)$ is the velocity, then by the second part of the fundamental theorem of calculus, we have, in agreement with the definition above, that:


$$
\text{average velocity} = \frac{x(b) - x(a)}{b-a} = \frac{1}{b-a} \int_a^b v(t) dt.
$$

The average speed is the change in *total* distance over time, which is given by


$$
\text{average speed} =  \frac{1}{b-a} \int_a^b \lvert v(t)\rvert dt.
$$

Let $\bar{v}$ be the average velocity. Then we have $\bar{v} \cdot(b-a) = x(b) - x(a)$, or the change in position can be written as a constant ($\bar{v}$) times the time, as though we had a constant velocity.  This is an old intuition. [Bressoud](http://www.math.harvard.edu/~knill/teaching/math1a_2011/exhibits/bressoud/) comments on the special case known to scholars at Merton College around $1350$ that the distance traveled by an object under uniformly increasing velocity starting at $v_0$ and ending at $v_t$ is equal to the distance traveled by an object with constant velocity of $(v_0 + v_t)/2$.


##### Example


What is the average value of $f(x)=\sin(x)$ over $[0, \pi]$?


$$
\text{average} = \frac{1}{\pi-0} \int_0^\pi \sin(x) dx = \frac{1}{\pi} (-\cos(x)) \big|_0^\pi = \frac{2}{\pi}
$$

Visually, we have:


```{julia}
plot(sin, 0, pi)
plot!(x -> 2/pi)
```

##### Example


What is the average value of the function $f$ which is $1$ between $[0,3]$, $2$ between $(3,5]$ and $1$ between $(5,6]$?


Though not continuous, $f(x)$ is integrable as it contains only jumps. The integral from $[0,6]$ can be computed with geometry: $3\cdot 3 + 2 \cdot 2 + 1 \cdot 1 = 14$. The average then is $14/(6-0) = 7/3$.


##### Example


What is the average value of the function $e^{-x}$ between $0$ and $\log(2)$?


$$
\begin{align*}
\text{average} = \frac{1}{\log(2) - 0} \int_0^{\log(2)} e^{-x} dx\\
&= \frac{1}{\log(2)} (-e^{-x}) \big|_0^{\log(2)}\\
&= -\frac{1}{\log(2)} (\frac{1}{2} - 1)\\
&= \frac{1}{2\log(2)}.
\end{align*}
$$

Visualizing, we have


```{julia}
plot(x -> exp(-x), 0, log(2))
plot!(x -> 1/(2*log(2)))
```

## The mean value theorem for integrals


If $f(x)$ is assumed integrable, the average value of $f(x)$ is defined, as above. Re-expressing gives that there exists a $K$ with


$$
K \cdot (b-a) = \int_a^b f(x) dx.
$$

When we assume that $f(x)$ is continuous, we can describe $K$ as a value in the range of $f$:


> **The mean value theorem for integrals**: Let $f(x)$ be a continuous function on $[a,b]$ with $a < b$. Then there exists $c$ with $a \leq c \leq b$ with
>
> $f(c) \cdot (b-a) = \int_a^b f(x) dx.$`



The proof comes from the intermediate value theorem and the extreme value theorem. Since $f$ is continuous on a closed interval, there exists values $m$ and $M$ with $f(c_m) = m \leq f(x) \leq M=f(c_M)$, for some $c_m$ and $c_M$ in the interval $[a,b]$. Since $m \leq f(x) \leq M$, we must have:


$$
m \cdot (b-a) \leq K\cdot(b-a) \leq M\cdot(b-a).
$$

So in particular $K$ is in $[m, M]$. But $m$ and $M$ correspond to values of $f(x)$, so by the intermediate value theorem, $K=f(c)$ for some $c$ that must lie in between $c_m$ and $c_M$, which means as well that it must be in $[a,b]$.


##### Proof of second part of Fundamental Theorem of Calculus


The mean value theorem is exactly what is needed to prove formally the second part of the Fundamental Theorem of Calculus. Again, suppose $f(x)$ is continuous on $[a,b]$ with $a < b$. For any $a < x < b$, we define $F(x) = \int_a^x f(u) du$. Then the derivative of $F$ exists and is $f$.


Let $h>0$. Then consider the forward difference $(F(x+h) - F(x))/h$. Rewriting gives:


$$
\frac{\int_a^{x+h} f(u) du - \int_a^x f(u) du}{h}  =\frac{\int_x^{x+h} f(u) du}{h} = f(\xi(h)).
$$

The value $\xi(h)$ is just the $c$ corresponding to a given value in $[x, x+h]$ guaranteed by the mean value theorem.  We only know that $x \leq \xi(h) \leq x+h$. But this is plenty - it says that $\lim_{h \rightarrow 0+} \xi(h) = x$. Using the fact that $f$ is continuous and the known properties of limits of compositions of functions this gives $\lim_{h \rightarrow 0+} f(\xi(h)) = f(x)$. But this means that the (right) limit of the secant line expression exists and is equal to $f(x)$, which is what we want to prove. Repeating a similar argument when $h < 0$, finishes the proof.


The basic notion used is simply that for small $h$, this expression is well approximated by the left Riemann sum taken over $[x, x+h]$:


$$
f(\xi(h)) \cdot h = \int_x^{x+h} f(u) du.
$$

## Questions


###### Question


Between $0$ and $1$ a function is constantly $1$. Between $1$ and $2$ the function is constantly $2$. What is the average value of the function over the interval $[0,2]$?


```{julia}
#| hold: true
#| echo: false
f(x) = x < 1 ? 1.0 : 2.0
a,b = 0, 2
val, _ = quadgk(f, a, b)
numericq(val/(b-a))
```

###### Question


Between $0$ and $2$ a function is constantly $1$. Between $2$ and $3$ the function is constantly $2$. What is the average value of the function over the interval $[0,3]$?


```{julia}
#| hold: true
#| echo: false
f(x) = x < 2 ? 1.0 : 2.0
a, b= 0, 2
val, _ = quadgk(f, a, b)
numericq(val/(b-a))
```

###### Question


What integral will show the intuition of the Merton College scholars that the distance traveled by an object under uniformly increasing velocity starting at $v_0$ and ending at $v_t$ is equal to the distance traveled by an object with constant velocity of $(v_0 + v_t)/2$.


```{julia}
#| hold: true
#| echo: false
choices = [
"``\\int_0^t v(u) du = v^2/2 \\big|_0^t``",
"``\\int_0^t (v(0) + v(u))/2 du = v(0)/2\\cdot t + x(u)/2\\ \\big|_0^t``",
"``(v(0) + v(t))/2 \\cdot \\int_0^t du = (v(0) + v(t))/2 \\cdot t``"
]
answ = 1
radioq(choices, answ)
```

###### Question


Find the average value of $\cos(x)$ over the interval $[-\pi/2, \pi/2]$.


```{julia}
#| hold: true
#| echo: false
f(x) = cos(x)
a,b = -pi/2,pi/2
val, _ = quadgk(f, a, b)
val = val/(b-a)
numericq(val)
```

###### Question


Find the average value of $\cos(x)$ over the interval $[0, \pi]$.


```{julia}
#| hold: true
#| echo: false
f(x) = cos(x)
a,b = 0, pi
val, _ = quadgk(f, a, b)
val = val/(b-a)
numericq(val)
```

###### Question


Find the average value of $f(x) = e^{-2x}$ between $0$ and $2$.


```{julia}
#| hold: true
#| echo: false
f(x) = exp(-2x)
a, b = 0, 2
val, _ = quadgk(f, a, b)
val = val/(b-a)
numericq(val)
```

###### Question


Find the average value of $f(x) = \sin(x)^2$ over the $0$, $\pi$.


```{julia}
#| hold: true
#| echo: false
f(x) = sin(x)^2
a, b = 0, pi
val, _ = quadgk(f, a, b)
val = val/(b-a)
numericq(val)
```

###### Question


Which is bigger? The average value of $f(x) = x^{10}$ or the average value of $g(x) = \lvert x \rvert$ over the interval $[0,1]$?


```{julia}
#| hold: true
#| echo: false
choices = [
L"That of $f(x) = x^{10}$.",
L"That of $g(x) = \lvert x \rvert$."]
answ = 2
radioq(choices, answ)
```

###### Question


Define a family of functions over the interval $[0,1]$ by $f(x; a,b) = x^a \cdot (1-x)^b$. Which has a greater average, $f(x; 2,3)$ or $f(x; 3,4)$?


```{julia}
#| hold: true
#| echo: false
choices = [
"``f(x; 2,3)``",
"``f(x; 3,4)``"
]
n1, _ = quadgk(x -> x^2 *(1-x)^3, 0, 1)
n2, _ = quadgk(x -> x^3 *(1-x)^4, 0, 1)
answ = 1 + (n1 < n2)
radioq(choices, answ)
```

###### Question


Suppose the average value of $f(x)$ over $[a,b]$ is $100$. What is the average value of $100 f(x)$ over $[a,b]$?


```{julia}
#| hold: true
#| echo: false
numericq(100 * 100)
```

###### Question


Suppose $f(x)$ is continuous and positive on $[a,b]$.


  * Explain why for any $x > a$ it must be that:


$$
F(x) = \int_a^x f(x) dx > 0
$$

```{julia}
#| hold: true
#| echo: false
choices = [
L"Because the mean value theorem says this is $f(c) (x-a)$ for some $c$ and both terms are positive by the assumptions",
"Because the definite integral is only defined for positive area, so it is always positive"
]
answ = 1
radioq(choices, answ)
```

  * Explain why $F(x)$ is increasing.


```{julia}
#| hold: true
#| echo: false
choices = [
L"By the extreme value theorem, $F(x)$ must reach its maximum, hence it must increase.",
L"By the intermediate value theorem, as $F(x) > 0$, it must be true that $F(x)$ is increasing",
L"By the fundamental theorem of calculus, part I, $F'(x) = f(x) > 0$, hence $F(x)$ is increasing"
]
answ = 3
radioq(choices, answ)
```

###### Question


For $f(x) = x^2$, which is bigger: the average of the function $f(x)$ over $[0,1]$ or the geometric mean which is the exponential of the average of the logarithm of $f$ over the same interval?


```{julia}
#| hold: true
#| echo: false
f(x) = x^2
a,b = 0, 1
val1 = quadgk(f, a, b)[1] / (b-a)
val2 = exp(quadgk(x -> log(f(x)), a, b)[1] / (b - a))

choices = [
L"The average of $f$",
L"The exponential of the average of $\log(f)$"
]
answ = val1 > val2 ? 1 : 2
radioq(choices, answ)
```