Submitted to HEJ Manuscript no.: ANM-010201-A

Basic properties

The proofs can be found in any textbook on calculus.

(a)

$f$ is integrable over $[a,b]$ if and only if for every $\varepsilon>0$ there exists a subdivision $S$ of $[a,b]$ such that upper$(f,S) -$lower $(f,S)<\varepsilon$. (The existence of a real number $I$ with the property lower $(f,S) \leq I\leq \mbox{\bf upper}(f,S)$ is a consequence.)

(b)

In the definition of ``integrable'' the notions subdivision and regular subdivision are equivalent. To be more precise the subintervals can have different length, provided only that as $n$ increases the length of the longest subinterval tends to $0$.

(c)

If $f$ is continuous then $f$ is integrable. (It is a consequence of uniform continuity of $f$ on $[a,b]$.)

(d)

Every monotonic function is integrable. (upper$(f,S)-$lower$(f,S)$ can easily be estimated.)

(e)

If $f$ and $g$ are integrable over $[a,b]$ then $f+g$ is integrable and

$$\int_{a}^{b}(f(x) + g(x))\,dx = \int_{a}^{b} f(x)\,dx + \int_{a}^{b} g(x)\,dx.$$

(f)

If $f$ is integrable over $[a,b]$ and $c$ is a real number then $c\cdot f$ is integrable and

$$\int_{a}^{b}c\,f(x)\,dx= c \int_{a}^{b}f(x)\,dx.$$

(g)

Let $f$ be integrable over $[a,b]$ and $\varepsilon>0$. Then there exists $\delta>0$ with the following property: for any subdivision $S=[a_{0},\ldots,a_{n}]$ into subintervals of length smaller than $\delta$ and any $b_{1},\ldots,b_{n}$ satisfying $a_{i - 1}\leq b_{i}\leq a_{i}$ the difference of the ``Riemann sum'' from the integral of $f$ is less than epsilon, i.e.

$$\bigg\vert\sum_{i=1}^{n}\,({a_{i}} - {a_{i - 1}})\,f({b_{i}}) - \int_{a}^{b} f(x)\,dx\ \bigg\vert < \varepsilon.$$

(h)

(Immediate consequence of g.) Let $f$ be integrable over $[a,b]$ and $\varepsilon>0$. Then there exists a positive integer $n$ such that for all $b_{1},\ldots,b_n$ satisfying $b_{i} \in [a_{i - 1},a_{i}]$

$$\bigg\vert \big(\sum_{i=1}^{n}\,f({b_{i}})\big) \big/ n - \int_{a}^{b}f(x)\,dx\bigg\vert<\varepsilon.$$

(i)

(another important consequence of g.) Let $f$ be continuous on $[a,b]$. Then there exists a $c\in [a,b]$ such that

$$\int_{a}^{b}f(x)\,dx = (b-a) f(c).$$

It is often called the ``mean value theorem'' of the integral calculus.

EXAMPLE 2.2.

(c) shows that in EXAMPLE 1.4 the function

$$f\ :\ x\rightarrow x\ {\rm sin}(x)^{2}$$

is integrable, whereas (h) implies that int_step($S,V$) goes to $\int _{1}^{3}x\,{\rm sin}(x)^{2}\,dx$ for $n \rightarrow \infty$. Hence

$$\lim_{n\rightarrow \infty}\,\sum_{i=1}^{n}\,\frac{(1 + \frac ... ...\frac {2\,i}{n})^{2}}{n} = \int_{1}^{3}x\,{\rm sin}(x)^{2}\,dx.$$

Exercise 2.1.

Prove that

$$\lim_{n\rightarrow \infty }\,\sum _{i=1}^{n}\,\frac{\sqrt{n^{2} - i^{2}}}{n^{2}} = \pi/4.$$

Hint: Look at $\int _{0}^{1}\sqrt{1 - x^{2}}\,dx$.

Submitted to HEJ Manuscript no.: ANM-010201-A