\( \newcommand\der{\operatorname{der}} \newcommand\Der{\mathrm{D}} \newcommand\dd{\operatorname{d}} \newcommand\ang[1]{#1^\circ} \newcommand\parenthesize[1]{\left(#1\right)} \newcommand\dif{\mathrm{d}} \newcommand\Dif{\Delta} \newcommand\LinearInterpolation{\text{LInterp}} \)

1Prerequisites

1.1? Linear interpolation

Linear interpolation:

\[\begin{align*} \LinearInterpolation(a,b,0) &= a \\ \LinearInterpolation(a,b,1) &= b \\ \LinearInterpolation(a,b,p) &= (1-p)a + pb \end{align*} \]

1.2Measure

A measure space \((S,\mu)\) is a space \(S\) and a measure \(\mu : S \to \Real\).

"Measure" generalizes the concept of length, area, volume, and hypervolume.

Usually, at least in physics:

\[\begin{align*} \mu([a,b]) &= b-a \\ \mu([a,b)) &= b-a \\ \mu((a,b]) &= b-a \\ \mu((a,b)) &= b-a \\ \mu(A \times B) &= \mu(A) \times \mu(B) \\ \mu(A \cup B) &= \mu(A) + \mu(B) - \mu(A \cap B) \\ \mu\left(\prod_k A_k\right) &= \prod_k \mu(A_k) \end{align*} \]

If we allow infinities, \( \mu(\Real) = \infty \); otherwise \( \mu(\Real) \) is undefined.

2Riemann integrals

2.1In one dimension

Let \( (\Real,\mu) \) be a measure space.

Let \( f : \Real \to \Real \) be the integrand.

Let \( X = [a,b] \) be the domain of integration.

Let \(n\) be the number of partitions.

Partition \(X\) evenly into \(n\) partitions \(p_0, \ldots, p_{n-1}\) such that, for each \(k\): \[ \mu(p_k) = \frac{\mu(X)}{n} = \frac{b-a}{n} \]

We write \( p_k^* \) to mean "any point in \(p_k\)".

A Riemann summand is the area of a rectangle: \[ s_k = f(p_k^*) \cdot \mu(p_k) \]

The Riemann sum is the sequence \( S_1, S_2, \ldots \) where: \[ S_n = \sum_{k=0}^{n-1} s_k \]

The Riemann integral is the limit of the Riemann sum: \[ I_X f = \int_X f(x)~\dif x = \int_a^b f(x) ~ \dif x = \lim_{n\to\infty} S_n \]

As \(n\) grows unbounded, the sum involves more numerous but smaller partitions, and the sum approaches the area under the curve.

The integral sign \(\int\) is a stylized letter "S" that stands for "sum".

The notation \(\int_a^b f(x)~\dif x\) means "the area enclosed by the line \(x=a\), the line \(x=b\), the line \(y=0\), and the curve \(y=f(x)\)". Negative area means that the curve \(y=f(x)\) goes below the line \(y=0\). See picture. See Wikipedia.

2.2In many dimensions

Let \( f : \Real^r \to \Real \) replace the integrand.

Let \( X = X_1 \times \ldots \times X_r \) replace the domain of integration.

Let \( X_i = [a_i,b_i] \) for each \(i\).

Then the Riemann integral is replaced by:

\[\begin{align*} I_X f &= \int_X f(x)~\dif x \\ &= \int_{X_r} \ldots \int_{X_1} f(x_1,\ldots,x_r) ~ \dif x_1 ~ \ldots ~ \dif x_r \\ &= \int_{a_r}^{b_r} \ldots \int_{a_1}^{b_1} f(x_1,\ldots,x_r) ~ \dif x_1 ~ \ldots ~ \dif x_r \end{align*} \]

3Arc length

See Wikipedia.

Let \(f : [a,b] \to \Real\).

Let \(f'\) be the derivative of \(f\).

The arc length of the curve described by the graph of \(f\) is \[ \int_a^b \sqrt{1 + (f'(x))^2}~\dif x \]

4Line integral

4.1Of scalar fields

See derivation.

See the picture in Wikipedia.

Begin with the basic case.

Let \(X = [a,b]\).

Let \(c : X \to Y\) be a curve.

Let \(f : Y \to \Real\) be the integrand (a scalar field).

Partition \(X\) evenly to \(n\) subintervals \( p_0, \ldots, p_{n-1} \).

We write \(c[A]\) to mean \(\SetBuilder{c(x)}{x \in A}\).

If we compare a line integral summand and a Riemann summand as follows, we see that a Riemann integral is a line integral along an identity curve described by \(c(x) = x\) for all \(x \in X\):

\[\begin{align*} s_k &= f(p_k^*) \cdot \mu(c[p_k]) & \text{(line integral summand)} \\ s_k &= f(p_k^*) \cdot \mu(p_k) & \text{(Riemann summand)} \end{align*} \]

By \(\mu(c[p_k])\), we mean "the arc length of the subcurve \(c[p_k]\)"?

The magical step is the application of the mean value theorem?

We write \( I_c f \) to mean "the line integral of \(f\) along \(c\)".

Example: work done by force on a particle?

If \(X=[a,b]\), then "Riemann integral of \(f\) in \(X\)" is the same as "line integral of \(f\) along \(\text{id}_X\)", where \( \text{id}_X : X \to X \) and \( \text{id}_X(x) = x \).

This table compares a Riemann integral and a line integral:

term Riemann integral line integral
of \(f\) in \(X\) of \(f\) along \(c\)
notation \( I_{[a,b]} f = \int_a^b f(x) ~ \dif x \) \(I_c f\)
type of integrand \(f\) \(X \to \Real\) \(Y \to \Real\)
type of curve \(c\) \(X \to X\) \(X \to Y\)
curve \(c\) \(c(x)=x\) anything
domain of integration \(X\) \([a,b]\) \([a,b]\)
partition \(p_k\) \(X_k\) \(c[X_k]\)
partition measure \(\mu(p_k)\) \(\frac{b-a}{n}\) arc length of \(p_k\)
summand \(s_k\) \(f(p_k^*) \cdot \mu(p_k)\) \(f(p_k^*) \cdot \mu(c[p_k])\)

4.2? Of vector fields

5? Hypervolume integral

Surface integral, volume integral, etc.

6Notation for multiple integrals

I propose another notation for multiple integrals at the end of this section.

Suppose that \(f : \Real^3 \to \Real\).

We write \(f(x,y,z)\) to mean \(f(xi+yj+zk)\) where \(i,j,k\) are the standard Cartesian basis vectors.

Let \( V = X \times Y \times Z \).

The common notation for multiple integrals exposes too much detail: \[ \iiint_V f(x,y,z) ~ \dif x ~ \dif y ~ \dif z = \int_Z \int_Y \int_X f(x,y,z) ~ \dif x ~ \dif y ~ \dif z \]

The integral sign should work with all measure spaces, not only with \( \Real \), because the Riemann integral (the following equation) accepts all measure spaces: \[ I_V f = \int_V f(x) ~ \dif x = \lim_{n \to \infty} \sum_{p \in P(V,n)} f(p^*) \cdot \mu(p) \] where \(P(V,n)\) is a set of \(n\) partitions of \(V\), and \( p^* \) is a point in partition \(p\), and \( \mu \) is the measure in context.

(See generalized Riemann sum. A Riemann integral is the limit of a Riemann sum as the partitions vanish.)

I propose this notation, which treats all measure spaces such as \(V\) uniformly: \[ \int_V f(v^*) ~ \dif v \] where \(v^*\) mean a point in an infinitesimal subspace \(\dif v\) of \(V\).

7? Comparison of some integrals

Let \((\Real,\mu)\) be a measure space.

Let \(f : \Real \to \Real\) be a function (the "integrand").

Let \(X \subseteq \Real\) be the domain of integration.

Let \(Y = \{ f(x) ~\vert~ x \in X\}\).

Let \(X_1,\ldots,X_{n+1}\) be a partitioning of \(X\).

Let \(Y_1,\ldots,Y_{n+1}\) be a partitioning of \(Y\).

For each \(k\), let \(x_k \in X_k\) and \(y_k \in Y_k\). (Pick a point from each partition.)

An integral of function \( f \) in space \( X \) is \(I_X f = \lim_{n\to\infty} \sum_{k=1}^{n} a_k\) where each \(a_k = \mu(X_k) \cdot \mu(Y_k)\) is a rectangular part of the total area. There are several kinds of integrals depending on how the area is divided. See the table.

Name \(X_k\) \(Y_k\)
Riemann integral \([x_k,x_{k+1}]\) \([0,f(x_k)]\)
Stieltjes integral with respect to \(g\)? \([g(x_k),g(x_{k+1})]\) \([0,f(x_k)]\)
Lebesgue integral \( \{x ~\vert~ f(x) \in Y_k\} \) \([y_k,y_{k+1}]\)

In Riemann integral, you divide \(X\) freely.

In Lebesgue integral, you divide \(Y\) freely.

8Digression

(Archimedes? Riemann?)

A simple closed path divides a space into two: the space inside the shape and the space outside the shape.

We want to compute the amount of space inside the shape (the area enclosed by the shape).

To approximate the area inside a shape, there are two approaches:

  • place increasingly-smaller rectangles into the shape, or
  • carve out increasingly-smaller rectangles from the outside the shape.

One does not have to use rectangles; any simple shape of a known area will do.

What is the history of the word "integral" in mathematics? What was the reason for choosing that word? Wikipedia has a history of the concept but not a history of the word.1

The syntax and types of the concepts involved are:

  • If \( f : A \to B \) (the integrand) is a function
  • and \( A \) (the domain of the integrand) is a metric space with metric \( M \)
  • and \( V \) (the domain of integration) is a subset of \( A \)
  • and addition is defined for \( B \) (the codomain of the integrand)
  • then \( I_V f : B \) is called "the integral of \(f\) in \(V\)".

There is an alternative syntax for \( I_V f \): we can write \( [I f](V) \) when we want to think of integration as an operator.

Here we define \( I_V f \).

First, we pick a function \(P\) that divides a given set into a given number of partitions. That is, \( P(S,n) \) is a set of \(n\) partitions of the set \(S\).

\[\begin{align*} I_V f &= \lim_{n \to \infty} \sum_{p \in P(V,n)} f(R(p)) \cdot M(p) \end{align*} \]

\( R(p) \) is any point in partition \( p \).

\( M(p) \) is the measure of partition according to the domain of the integrand. (Recall that the domain of the integrand is a metric space.)

Do not confuse between "the measure of a set according to a metric" and "the cardinality of a set".

We write it \( I_V f \) to mean the integral of \( f \) in \( V \).

Example: Evaluate \( I_{[1,2]} (x \mapsto x^2) \).

Observe that a line segment \( [a,b] \) can be divided into \( n \) segments \( [p_0,p_1], ~ [p_1,p_2], ~ \ldots, ~ [p_n,p_{n+1}] \) by defining these \( n + 1 \) points (including the endpoints): \[ p_k = a + k \cdot \frac{b-a}{n+1} \] for \( k = 0, 1, 2, \ldots, {n+1} \).

Thus we define \( P([a,b],n) = \Set{[p_1,p_2], [p_2,p_3], \ldots, [p_n,p_{n+1}]} \) where \( p_k = a + k \cdot \frac{b-a}{n+1} \).

We define \( R([a,b]) = a \).

We define \( M([a,b]) = b-a \).

Here we are going to appreciate the fundamental theorem of calculus.

Imagine the hardship:

\[\begin{align*} I_{[1,2]} (x \mapsto x^2) &= \lim_{n \to \infty} \sum_{p \in P([1,2],n)} (x \mapsto x^2)(R(p)) \cdot M(p) \\ &= \ldots \end{align*} \]

Imagine even more hardship in evaluating integrals of higher-power monomial functions such as \( I_{[2,3]} (x \to x^9) \).

The fundamental theorem of calculus relates integral and derivative: \( I_{[a,b]}(Df) = f(b) - f(a) \).

The fundamental theorem of calculus makes integration so much easier that integrating monomial functions become embarrassingly trivial because the theorem enables us to derive the power rule for integrals from the power rule for derivatives. The theorem enables technologies that reduce many problems from complicated algebraic juggling to simple pattern matching.

A mathematical theorem is a technology too, because it increases productivity (it makes life easier, it reduces the effort required to accomplish something).

We take this for granted, but we should not forget that for hundreds of years people did not know that and they painstakingly did things so that we do not have to, and for that we should be grateful.

  1. <2019-11-07> not about the history of the word https://en.wikipedia.org/wiki/Integral#History↩