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We know that the derivative of a function called $f(x)$ can be written as a limit, just like here:$$\frac{d}{dx}f(x)=\lim\limits_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ but

Is there any definition of integrals in the form of limits?

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There are many different definitions. Pick your favorite three maybe? :-) –  anon May 8 '12 at 18:59
Maybe look in your textbook? –  copper.hat May 8 '12 at 19:17
There is whole lot of Integration theory . Enjoy :) –  Theorem May 8 '12 at 19:18
Not only is there a definition of integrals in the form of limits, the integral is defined in terms of certain limits existing and being equal. –  Arturo Magidin May 8 '12 at 19:20

2 Answers 2

up vote 14 down vote accepted

Recall that a partition of an interval $[a,b]$ is simply a set of points of the interval, one which is $a$ and the other which is $b$.

Darboux's Integral The construction is as follows. We define the upper and lower sums, $U$ and $L$, of $f$ for the partition $P$ as follows: $U(f,P) =\sum_{i=1}^n (a_{i}-a_{i-1})M_i $. $L(f,P) =\sum_{i=1}^n (a_{i}-a_{i-1})m_i $ where $M_i = \sup\limits_{[a_{i-1},a_i]} f(x)$ and $m_i = \inf\limits_{[a_{i-1},a_i]} f(x)$. If $f$ happens to be continuous we can just define them as the maximum and minimum of $f$ in the interval $[a_{i-1},a_i]$. Many properties are derived, but an important result is the following (Spivak's Calculus, Ch. 13, p.355):

If it is the case that $$\inf \left\{ U\left( {f,P} \right);\text{ P is a partition of I} \right\} = \sup \left\{ L\left( {f,P} \right); \text{ P is a partition of I} \right\}$$ then we say $f$ is integrable on $I$ and denote this common value by $$\int_a^b f(x) dx$$

The second important theorem is (Spivak's Calculus, Ch. 13, p.356)

If $f$ is bounded on an interval $I$, then it is integrable on $I$ if for any $\epsilon >0$ there exists a partition $P$ of $I$ such that $$U(f,P)-L(f,P)<\epsilon$$

An example would help. Take $f:[0,1]\to\Bbb R$ with $x\mapsto x^2$. We partition $[0,1]$ into $n$ intervals. $[0,1/n],[1/n,2/n],[2/n,3/n],\dots,[(n-1)/n,n/n]$. Note we have that $\Delta P = 1/n$. Since our function is monotone increasing, we can choose the $M$ and $m$ as the maximum and minimums. That is $$M_i = f(a_i)= \frac{i^2}{n^2}$$

$$m_i = f(a_i)= \frac{(i-1)^2}{n^2}$$

The sums are then

$$\eqalign{ & U\left( {f,P} \right) = \sum\limits_{i = 1}^n {\frac{1}{n}\frac{{{i^2}}}{{{n^2}}}} = \frac{1}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} \cr & L\left( {f,P} \right) = \sum\limits_{i = 1}^n {\frac{1}{n}\frac{{{{\left( {i - 1} \right)}^2}}}{{{n^2}}}} = \frac{1}{{{n^3}}}\sum\limits_{i = 1}^{n - 1} {{i^2}} \cr} $$

Those sums can be evaluated to yield

$$\eqalign{ & U\left( {f,P} \right) = \frac{1}{{{n^3}}}\frac{{n\left( {2n + 1} \right)\left( {n + 1} \right)}}{6} \cr & L\left( {f,P} \right) = \frac{1}{{{n^3}}}\frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{6} \cr} $$

We have then that

$$\frac{1}{{{n^3}}}\frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{6} < \int\limits_0^1 {{x^2}dx} < \frac{1}{{{n^3}}}\frac{{n\left( {2n + 1} \right)\left( {n + 1} \right)}}{6}$$

Note that we can make the difference of the sums arbitrarily close to each other taking finer and finer partitions, that is $U\left( {f,P} \right) - L\left( {f,P} \right) < \frac{1}{n}$ and since:

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3}}}\frac{{n\left( {2n + 1} \right)\left( {n + 1} \right)}}{6} = \frac{1}{3} \cr & \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3}}}\frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{6} = \frac{1}{3} \cr} $$

The above allows us to conclude that

$$\int\limits_0^1 {{x^2}dx} = \frac{1}{3}$$

A final trickery can be used now that we have this result to obtain a new one:

$$\int\limits_0^a {{x^2}dx} = {a^3}\int\limits_0^a {{{\left( {\frac{x}{a}} \right)}^2}d\left( {\frac{x}{a}} \right)} = {a^3}\int\limits_0^1 {{u^2}du} = \frac{{{a^3}}}{3}$$

And then ($b>a$)

$$\int\limits_a^b {{x^2}dx} = \int\limits_0^b {{x^2}dx} - \int\limits_0^a {{x^2}dx} = \frac{{{b^3}}}{3} - \frac{{{a^3}}}{3}$$

Riemann's Integral. Remember given a partition $P=\{x_i\}_{1\leqslant i\leqslant n}$ we call collection of numbers for which $x_{i-1}\leqslant t_i\leqslant x_i$ a set of tags for the partition $P$. In this case we have no upper or lower sums, but a sum:

$$R(f,P)=\sum_{i=1}^n f(t_i) (x_i-x_{i-1})$$

We say that $f$ is Riemann integrable in $I$ and denote $$\int_a^b f(x) dx = \mathcal I$$ if, for any $\epsilon >0$, there exists a $\delta >0$ such that, if $\Delta P < \delta$, then

$$\left| R(f,P)- \mathcal I\right|<\epsilon$$

Note that for any Riemann sum, we have that

$$L(f,P) \leqslant R(f,P) \leqslant U(f,P)$$

This can be used to prove the assertion that

$f$ is Darboux integrable $\iff$ $f$ is Riemann integrable.

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How about Lebesgue's approach in the Bourbaki way? –  Jonas Teuwen May 10 '12 at 21:55
@JonasTeuwen BTW; no upvote? =) –  Pedro Tamaroff May 10 '12 at 21:58
Yes I did. No problem! –  Jonas Teuwen May 11 '12 at 12:30

Yes, a definite integral is also defined as a limit. (There are several ways of doing this)

$$ \int_a^b f(x) \; dx = \lim_{n\to \infty} \sum_{i=1}^{n} f(x_{i})\Delta x $$ where $$\Delta x = \frac{b-a}{n}$$ and $$x_i = a + i\Delta x.$$

Now to get an understanding of why this makes sense, you can for example take a look at this Wikipedia article:

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Would this yield all Riemann integrable functions? –  Jonas Teuwen May 8 '12 at 19:23
@JonasTeuwen I am not sure that I understand what you mean by "yield"? –  Thomas May 8 '12 at 19:29
Does this limit exist for all Riemann integrable functions? –  Jonas Teuwen May 8 '12 at 20:10
This limit exists for some non-integrable functions as well. –  Santiago Canez May 9 '12 at 22:39

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