Basic Properties

Introduction 5.1.1  
The basic idea behind integration is a very simple one, namely to measure the area between the $x$-axis and the graph of a function by approximating it with a series of rectangles. When approximating the function from below, the total surface of the rectangles increases when refining the mesh (this is called the lower Riemann sum). When approximating the function from above, the total surface of the rectangles decreases when refining the mesh (this is called the upper Riemann sum). As often seems to be the case in calculus, two pictures can say a thousand words:

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lower Riemann sums		upper Riemann sums

The natural question to ask now is if these sums converge at all and if so if they converge to the same number which would then be the area bounded by the $x$-axis and $f(x)$. The answer to this question is worded in the integrability theorem:

• Integrability Theorem: Let $f$ be a real-valued function over $[a,b]$ such that $f$ is bounded and that $f$ is continuous except at a finite number of points, then the upper Riemann sums and the lower Riemann sums will converge to the same number. Such a function is then called integrable.

Note that integrability is considered over the closed interval $[a,b]$ and not over the open interval $]a,b[$ to avoid problems with arbitrarily large values, like e.g. the values of $\frac{1}{x}$ close to 0. It is possible to prove integrability for much ``wilder" functions, but this would lead to a lot of unnecessary complications. Yet, there are quite a few functions that fit this ``restrictive" conditions:

• all polynomials
• $\sin$ and $\cos$ functions
• rational functions, provided $[a,b]$ does not contain a point where the denominator is 0. (Note that the tangent function has essentially the same problem with $x=\frac{\pi}{2}+k\pi$.)
• exponential functions
• logarithmic functions when the interval is fully part of the positive half of the $x$-axis

It is not difficult to come up with functions that are not integrable. Again, as was the case for sequences, functions and derivatives, there are basically two reasons why a function could not be integrable: because the area below the function is not finite, or because the upper and lower Riemann sum do not converge to the same number.

• The Area is Infinite: take e.g. the interval $[-2,2]$ and the function

  $$f: \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto \frac{1}{x^4}$$ (5.1.1)

  
  
  Just looking at the picture already drops a strong hint that the area below the graph of the function is infinite. Proving this will be an easy calculation once the integral shows its true colours, namely the ``anti-derivative".
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• The Riemann sum does not converge properly for the function

  $$f: \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: \b... ... x \mapsto 0 & x \not\in \hbox {\hskip 4.1pt l \hskip -8.2pt {Q}}\\ \end{cases}$$ (5.1.2)

  
  
  over the interval $[0,1]$ (or any other interval for that matter). The fundamental reason is that for every rational number there is an arbitrarily close irrational number and vice versa. So for the Riemann upper sum, we can always choose a rational number, so $f(x_i)$ will always be $1$, while for the lower Riemann sum, we can always choose an irrational number, so $f(x_i)$ will always be 0. Therefore the upper Riemann sum of $f$ over $[0,1]$ will be $1$ and the lower Riemann sum will be 0. So there is no integral. Note that this function is bounded, but discontinuous in an infinite number of points.

Definition 5.1.2  
Let $f$ be an integrable function over the interval $[a,b]$ then the integral of $f$ over $[a,b]$ is defined as follows:

$$\int_a^b f(x)dx = \lim_{n \to \infty}\sum_{i=1}^n f(x_i) \delta_n$$ (5.1.3)

In this formula, $\delta_n$ is the width of the rectangle in a Riemann sum, namely $\frac{b-a}{n}$. The idea is that if $n$ gets bigger, $\delta_n$ gets smaller and the approximation becomes more precise.
$x_i$ is any choice on the width of the rectangle of a Riemann sum, more precisely, $x_i$ is any choice in the interval $[(i-1)\frac{b-a}{n},i\frac{b-a}{n}]$. If you choose $x_i$ so that $f(x_i)$ is the highest value for that little interval then you get the upper Riemann sum, if $f(x_i)$ is the lowest value for that little interval, then you get the lower Riemann sum. Any other choice is somewhere in between, but this doesn't matter at all, because for integrable functions, any choice converges to the same number.

Properties 5.1.3  
Let $f$ and $g$ be integrable functions and take $a<b<c$ on the real line. Let $\alpha$ and $\beta$ be real numbers. Then the following equalities hold:

1. Concatenation of Intervals

   $$\int_a^b f(x) dx + \int_b^c f(x)dx = \int_a^c f(x)dx$$ (5.1.4)

   
   

2. Linearity of the Integral

   $$\int_a^b [\alpha \cdot f(x) + \beta g(x)] dx = \alpha \cdot \int_a^b f(x)dx + \beta \cdot \int_a^b g(x)dx$$ (5.1.5)

   
   

3. Order Preserving Property of the Integral

   $$\hbox{If $f\leq g$, then } \int_a^b f(x)dx \leq \int_a^b g(x)dx$$ (5.1.6)

   
   

The proofs of these properties are quite trivial. All three boil down to the fact that the same relations hold for limits and sums. It is also clear that these properties hold when looking at the integral as a measure of area.

Theorem 5.1.4 (Anti-Derivative)  
Let $f$ be an integrable function over $[a,b]$. Then the following relation holds for any $x$ in $[a,b]$:

$$\frac{d{}}{dx} \int_a^x f(t)dt = f(x)$$ (5.1.7)

Proof.
By the definition of derivative and by property 5.1.4, we have

$$\begin{aligned}\frac{d{}}{dx} \int_a^x f(t)dt &= \lim_{y \to x}... ... &= \lim_{y \to x} \frac{1}{y-x} \int_x^y f(t)dt \\ \end{aligned}$$

Furthermore, note that by the definition of upper Riemann and lower Riemann sums,

$$(y-x) \cdot \min_{t \in [x,y]} f(t) \leq \int_x^y f(t)dt \leq (y-x) \cdot \max_{t \in [x,y]} f(t)$$ (5.1.9)

If you need convincing, have a look at the following graphs:

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Therefore, we have:

$$\min_{t \in [x,y]} f(t) \leq \frac{1}{y-x} \int_x^y f(t)dt \leq \max_{t \in [x,y]} f(t)$$ (5.1.10)

If $y$ approaches $x$ then naturally the minimum and maximum over $[x,y]$ approach each other and thus:

$$\lim_{n \to \infty}\frac{1}{y-x} \int_x^y f(t)dt = f(x)$$ (5.1.11)

$\qedsymbol$

Remark 5.1.5  
Theorem (5.1.4) is sometimes called the fundamental theorem of calculus because it establishes a very intricate link between integration and differentiation. Although properties (5.1.4), (5.1.5) and (5.1.6) are quite useful, without the fundamental theorem of calculus, you would still have to calculate any given integral by writing down the Riemann sums and taking the limit. A very cumbersome procedure to go through, even for the simplest of integrals, while now, we can simply state:

$$\int_a^x \cos(t) dt = \sin(x) + C$$ (5.1.12)

for any constant $C$ because

$$\frac{d{}}{dx} \int_a^x \cos(t) dt = \cos(x)$$ (5.1.13)

This can easily be extended as follows:

$$\begin{aligned}\int_x^y \cos(t) dt &= \int_x^a \cos(t)dt + \int... ...\sin(y) + C) - (sin(x) + C) \\ &= \sin(y) - \sin(x) \end{aligned}$$

for any constant $C$. This equality holds for all $x$ and $y$. As it is true for any chosen boundaries, we might as well write it as

$$\int \cos(t) dt = \sin(t)$$ (5.1.15)

while we actually mean (5.1.14) for any $x$ and $y$. Integrals with boundaries like (5.1.12) are called definite integrals while integrals without, like (5.1.15) are called indefinite integrals.
Also note that the more pedantic among us will insist on writing

$$\int \cos(t) dt = \sin(t) + C ~~~~\forall C \in \hbox {I \hskip -5.2pt {R}}$$ (5.1.16)

And, quite frankly, the pedantic among us are right: this relation holds for any constant $C$ because $\frac{dC}{dx}=0$ and not just for $C=0$. When calculating integrals as such, this is not that important, as long as you are very well aware of the fact that a constant should be added. When getting into the vast realm of differential equations, this will be of crucial importance, because the solution of a differential equation is invariably of the form

$$C_1 \cos(x) + C_2 \sin(x)$$ (5.1.17)

so assuming your constants to be 0 in this case makes you throw away virtually all your solutions.

Remark 5.1.6  
Note that there are functions that are integrable, but for which the integral is not defined. This is not a contradiction: take e.g.

$$\hbox{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt$$ (5.1.18)

There is no closed form for this function, meaning that it is not possible to write it without integrals or infinite sums. Yet, $e^{-t^2}$ is clearly an integrable function (bounded by $1$ and continuous). You should not think that this is a rare situation. In fact, very standard functions as $\sin(x)$ and $\cos(x)$ cannot be written without infinite sums or in an implicit way as solutions of a differential equation.
It is also not the case that above function is a very exotic and useless one. In fact, this function also known as ``normal distribution" or ``Gauss" function is one of the standard cases in probability and statistics.