CalculusWithJuliaNotes.jl/CwJ/integrals/mean_value_theorem.jmd

# Mean value theorem for integrals

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:

```math
\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:

```math
\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

```math
\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 $[0, \pi]$?

```math
\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 $\log(2)$?

```math
\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)}.
```

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

```math
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:

```math
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:

```math
\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 $[x, x+h]$:

```math
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``"
]
ans = 1
radioq(choices, ans)
```


###### 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 $g(x) = \lvert x \rvert$."]
ans = 2
radioq(choices, ans)
```



###### 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)
ans = 1 + (n1 < n2)
radioq(choices, ans)
```

###### 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:

```math
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"
]
ans = 1
radioq(choices, ans)
```

* 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"
]
ans = 3
radioq(choices, ans)
```

###### 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 $\log(f)$"
]
ans = val1 > val2 ? 1 : 2
radioq(choices, ans)
```