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# Area integral

Riemann integral defined over intervals can be generalized to functions of two variables integrated over rectangles and regions made of rectangles.

We consider

$$f: {\bf R}^2 \rightarrow {\bf R}.$$

The are of the rectangle

$$R = [ a, b ] \times [ c, d ]$$

is

$$A = (b - a)(d - c),$$

i.e. the area of the product of $2$ intervals is the product of their lengths.

In general, the area of each little rectangle is

$\Delta A = \frac{(b - a)(d - c)}{k^2} \rightarrow 0$ as $k \rightarrow \infty$.

The Riemann sum of $f: {\bf R} \rightarrow {\bf R}$ for this partition:

Each side is divided into $k$ intervals, and each interval has a point in it, call it $c_{ii}$.

In dim = $2$:

$\displaystyle\sum_{i,j} f( c_{ij} ) \Delta A$, where the $3$-volume of each box is $f( c_{ij} ) \Delta A$, and the height is $f( c_{ij} )$, and the base is $\Delta A$.

Recall dim $= 1$. Consider

$$\displaystyle\lim_{k \rightarrow \infty} \displaystyle\sum_{i=1}^k f( c_i ) \Delta x$$

(or $\displaystyle\lim_{\Delta x \rightarrow 0}$ ...),

where $c_i$ denotes the midpoint of the respective intervals. If this limit exists, it is called the Riemann integral of $f$ from $a$ to $b$. We write

$$\displaystyle\int_a^b f(x) dx,$$

and call $f$ an integrable function on $[ a, b ]$.

For dim $= 2$ it's all very similar

$$\displaystyle\lim_{k \rightarrow \infty} \displaystyle\sum_{i=1}^k f( c_{ij} ) \Delta A = \displaystyle\int\displaystyle\int_R f(x) dA,$$

where

$$R = [ a, b ] \times [ c, d ].$$

Example. Let $f(x) = x, a = 0, b = 1$, and

$x_i = \frac{1}{k} i$ the endpoints for $i = 1, ..., k$ ($k$ intervals).
$c_i = \frac{1}{k} i$ chosen points for $i = 0, ..., k-1$.

Then

$$\begin{array}{} \displaystyle\lim_{k \rightarrow \infty} \displaystyle\sum_{i=1}^k f( c_i ) \Delta x \\ &= \displaystyle\lim_{k \rightarrow \infty} \displaystyle\sum_{i=1}^k c_i \Delta x \\ &= \displaystyle\lim_{k \rightarrow \infty} ( c_1 + ... + c_k ) \frac{1}{k} \\ &= \displaystyle\lim_{k \rightarrow \infty} ( 0 + (\frac{1}{k}) + (\frac{2}{k} ) + ... + (\frac{k-1}{k} ) k \\ &= \displaystyle\lim_{k \rightarrow \infty} ( (\frac{1}{k}^2) + (\frac{2}{k}^2) + ... + (\frac{k-1}{k}^2) ) {\rm \hspace{3pt} arithmetic \hspace{3pt} progression} \\ &= \frac{k \frac{k-1}{2}}{k^2} \\ &= \frac{1}{2} \end{array}$$

So,

$$\begin{array}{} \displaystyle\int_{[0,1]} x dx &= \displaystyle\int_0^1 x dx \\ &= \frac{x^2}{2}|_0^1 \\ &= \frac{1}{2}. \end{array}$$

Related: areas of regions in digital images.