# Question about the definite integral

Consider the area of the region bound by $y=f(x)$, the x axis, and joining vertical segments $x=a$ and $x=b$. Subdivide the interval $a \leq x \leq b$ into n subintervals by means of the points $x_1, x_2,....x_{n-1}$, chosen arbitrarily. In each of the new intervals, choose points $\zeta _1, \zeta _2 ... \zeta_n$ arbitrarily. With $x_0 = a$, $x_n =b$ and $(x_k - x_{k-1}) = \Delta x_k$ this can be written as $\displaystyle\sum\limits_{k = 1}f (\zeta_k)\Delta x_k$, which represents the total area of all rectangles.

My question is, why?Especially the need for the arbitrary points $\zeta$ are unclear to me

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It looks like this is just a general definition of a Riemann sum, which is why everything is arbitrary. For example when you choose arbitrary $\zeta_1, \zeta_2, \ldots, \zeta_n$ this is to account for all the different possible ways of measuring the rectangles. For example if you decided to use the left endpoints (often known as a "left Riemann sum") you would have $\zeta_1 = x_0, \ldots, \zeta_n = x_{n-1}$. If you decided to use the right endpoints ("right Riemann sum") you would have $\zeta_1 = x_1, \ldots, \zeta_n = x_n$.
To see why that choice matters for sums like these, consider $f(x) = x$ on $[0,1]$. We can subdivide $[0,1]$ into subintervals $[0,1/2]$ and $[1/2,1]$. But the sums are different for $\zeta_1 = 0, \zeta_2 = 1/2$ (left) and $\zeta_1 = 1/2, \zeta_2 = 1$ (right).
Ahh, I'm a bit stupid, I didn't realize that all the $\zeta$ where $f(\zeta)$, thank you for your swift reply –  Ylyk Coitus Feb 23 '13 at 16:00
Each rectangle as area width times height. The width is $x_k-x_{k-1}$ and the hight is $f(\zeta_k)$, which seems to be somewhat arbitrary at the moment. However, some choices of $\zeta_k$ suggest themselves: If $\zeta_k$ is always chosen at a minumium of $f$ within $[x_{k-1},x_k]$ (provided such a point exists, for example because $f$ is continuous), then clearly the area of the rectangle is at most as big as the area under the graph between $x=x_{k-1}$ and $x=x_k$ because the rectangle is fully contained in it. Similarly, if we choose $\zeta_k$ at a maximum of $f$ within $[x_{k-1},x_k]$, the area of the rectangle is at least as big as the graph area because the rectangle fully contains the graph area. Other choices for $\zeta_k$, e.g. $\zeta_k=x_{k-1}$ or $\zeta_k=x_k$ may suggests themselves, but they may not always be the "best" choice - and if we chould allow the choce between the end points of $[x_{k-1},x_k]$, why not allow arbitrary choice in the interval if it doesn't make the treatment more complicated (and on the contrary, it gets simplified)?
In introducing the Riemann integral, you will contemplate what happens as $\max \Delta x_k\to 0$ and whethet this depends on the choice of $\zeta_k$ and how one can get good estimates for the integral by suitable choices of the $\zeta_k$ (e.g. for monotonic functions).