Partitions and Sample Points for Rectangular Riemann Sums

From One Rectangle to a Grid

A rectangular region can be divided into smaller rectangles by placing vertical and horizontal cuts through it. This simple construction is the geometric groundwork for approximating the accumulated values of a function over a region. Each small rectangle has an area, and a point chosen inside it can represent the function’s value on that cell.

To describe the construction without ambiguity, we keep track of the cut locations in each coordinate direction. The cuts need not be equally spaced: a grid with unequal widths and heights is still a valid partition.

Definition: Let $R=[a,b]\times[c,d]$, where $a<b$ and $c<d$. Choose increasing lists $a=x_0<x_1<\cdots<x_m=b$ and $c=y_0<y_1<\cdots<y_n=d$. These lists divide $R$ into $mn$ smaller rectangular cells. The collection of cells is a rectangular partition of $R$.

The $x$-coordinates determine the vertical cuts, and the $y$-coordinates determine the horizontal cuts. The indices $i$ and $j$ identify the intervals in the two directions. For $1\leq i\leq m$ and $1\leq j\leq n$, write the cell as

R_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j].

Adjacent cells share edges, so their closed rectangles are not disjoint as sets. That causes no problem: the grid still covers the original rectangle, with no gaps, and the shared edges have zero area. When describing the cells’ areas, each shared edge contributes nothing extra.

Widths, Heights, and Cell Areas

The width of cell $R_{ij}$ is the change in its $x$-coordinate, and its height is the change in its $y$-coordinate. Define

\Delta x_i=x_i-x_{i-1}, \qquad \Delta y_j=y_j-y_{j-1}.

Because the coordinate lists are strictly increasing, every $\Delta x_i$ and every $\Delta y_j$ is positive. A rectangle’s area is its width multiplied by its height, so the area of $R_{ij}$ is

\Delta A_{ij}=\Delta x_i\,\Delta y_j.

The area can depend on both indices: when widths or heights vary from one interval to another, different cells need not have the same area. In a uniform grid, by contrast, all vertical intervals have one common width and all horizontal intervals have one common height, so all cells have the same area.

Useful check: The widths must add to the full horizontal side length, and the heights must add to the full vertical side length: $$ \sum_{i=1}^{m}\Delta x_i=b-a, \qquad \sum_{j=1}^{n}\Delta y_j=d-c. $$

Why the Cell Areas Add to the Whole Area

A partition should account for exactly the area of the original rectangle. The coordinate differences make this check transparent. Adding the areas of all the cells gives

\sum_{i=1}^{m}\sum_{j=1}^{n}\Delta A_{ij} = \sum_{i=1}^{m}\sum_{j=1}^{n}\Delta x_i\,\Delta y_j = \left(\sum_{i=1}^{m}\Delta x_i\right) \left(\sum_{j=1}^{n}\Delta y_j\right).

The factorization is valid because $\Delta x_i$ depends only on $i$, while $\Delta y_j$ depends only on $j$. The coordinate differences telescope: each interior cut appears once with a positive sign and once with a negative sign. Thus the two sums equal $b-a$ and $d-c$, respectively.

Justification: Since $\sum_{i=1}^{m}(x_i-x_{i-1})=x_m-x_0=b-a$ and $\sum_{j=1}^{n}(y_j-y_{j-1})=y_n-y_0=d-c$, the total cell area is $$ \sum_{i=1}^{m}\sum_{j=1}^{n}\Delta A_{ij} =(b-a)(d-c), $$ the area of $R$. Therefore the grid neither omits area nor adds area.

Choosing a Sample Point in Each Cell

After partitioning the rectangle, choose one point from each cell. For $R_{ij}$, denote that point by $(x_{ij}^{*},y_{ij}^{*})$. It must satisfy

x_{i-1}\leq x_{ij}^{*}\leq x_i, \qquad y_{j-1}\leq y_{ij}^{*}\leq y_j.

Any point satisfying these inequalities is a valid sample point. It may be in the interior of the cell or on one of its edges or corners. A common systematic choice is the midpoint of each cell:

x_{ij}^{*}=\frac{x_{i-1}+x_i}{2}, \qquad y_{ij}^{*}=\frac{y_{j-1}+y_j}{2}.

For a function $f$ defined on $R$, the value $f(x_{ij}^{*},y_{ij}^{*})$ is paired with the area $\Delta A_{ij}$. This produces the weighted sum

\sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^{*},y_{ij}^{*})\,\Delta A_{ij}.

This expression is a rectangular Riemann sum. At this stage, the important task is to specify the cells, their areas, and valid sample points correctly. A Riemann sum is a finite sum, and a different valid selection of sample points can produce a different finite value.

Worked Examples

Worked Example: A Nonuniform Partition

Partition $R=[0,4]\times[1,5]$ using $x$-cuts $0,1,4$ and $y$-cuts $1,3,5$. Find the widths, heights, and areas of all cells, and choose a sample point in each.

There are two intervals in each direction, so $m=2$ and $n=2$, giving $mn=4$ cells. In the $x$-direction,

\Delta x_1=1-0=1, \qquad \Delta x_2=4-1=3.

In the $y$-direction,

\Delta y_1=3-1=2, \qquad \Delta y_2=5-3=2.

Using $\Delta A_{ij}=\Delta x_i\Delta y_j$, the four cell areas are

\Delta A_{11}=1\cdot2=2,\quad \Delta A_{12}=1\cdot2=2,\quad \Delta A_{21}=3\cdot2=6,\quad \Delta A_{22}=3\cdot2=6.

One valid choice of sample points is the midpoint of each cell. The first coordinate is the midpoint of its $x$-interval, and the second is the midpoint of its $y$-interval:

(x_{11}^{*},y_{11}^{*})=(0.5,2),\quad (x_{12}^{*},y_{12}^{*})=(0.5,4),\quad (x_{21}^{*},y_{21}^{*})=(2.5,2),\quad (x_{22}^{*},y_{22}^{*})=(2.5,4).

For instance, $0.5\in[0,1]$ and $2\in[1,3]$, so $(0.5,2)\in R_{11}$. The other three points satisfy the corresponding cell bounds. As a check, the cell areas add to $2+2+6+6=16$, which matches the area $(4-0)(5-1)=16$ of $R$.

Worked Example: A Uniform Grid and Its Midpoints

Divide $R=[-1,2]\times[0,3]$ into three equal intervals in the $x$-direction and two equal intervals in the $y$-direction. Find the cell dimensions and midpoint sample points.

The horizontal side length is $2-(-1)=3$. Dividing it into three equal parts gives width $3/3=1$. Thus the $x$-cuts are $-1,0,1,2$, and each $\Delta x_i=1$. The vertical side length is $3-0=3$. Dividing it into two equal parts gives height $3/2$. Thus the $y$-cuts are $0,1.5,3$, and each $\Delta y_j=1.5$.

There are $3\cdot2=6$ cells, each with area

\Delta A_{ij}=1\cdot1.5=1.5.

The $x$-midpoints are $-0.5,0.5,1.5$, and the $y$-midpoints are $0.75,2.25$. Pairing each $x$-interval with each $y$-interval gives the six sample points:

(-0.5,0.75),\ (0.5,0.75),\ (1.5,0.75),\ (-0.5,2.25),\ (0.5,2.25),\ (1.5,2.25).

Each point lies within the cell determined by its two midpoint coordinates. The total area check is $6\cdot1.5=9$. The original rectangle also has area $(2-(-1))(3-0)=3\cdot3=9$, as required.

Worked Example: Building a Sum from Chosen Points

Let $R=[0,2]\times[0,2]$, with cuts $0,1,2$ in both directions. For $f(x,y)=x+2y$, choose the lower-left corner of each cell as its sample point and write the resulting Riemann sum.

Each interval has length $1$, so all four cells have area $\Delta A_{ij}=1\cdot1=1$. The chosen sample points are $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. They are valid because each is a corner of its corresponding cell. Evaluate the function at these points:

f(0,0)=0+2(0)=0,\quad f(0,1)=0+2(1)=2, $$ $$ f(1,0)=1+2(0)=1,\quad f(1,1)=1+2(1)=3.

Multiply each function value by the area of its cell and add the four terms:

0\cdot1+2\cdot1+1\cdot1+3\cdot1=0+2+1+3=6.

Thus this particular partition and sample-point choice produce the finite sum $6$. The geometric checks are that there are four cells, each of area $1$, and their total area $4$ agrees with $(2-0)(2-0)=4$.

Sample-Point Choices and a Common Pitfall

The sample point is selected separately for each cell; it need not be the same relative point in every cell. Midpoints are convenient, but they are not required. In the preceding example, choosing midpoints instead would give the points $(0.5,0.5)$, $(0.5,1.5)$, $(1.5,0.5)$, and $(1.5,1.5)$. The function values are $1.5$, $3.5$, $2.5$, and $4.5$, so the associated sum is

$$ 1.5(1)+3.5(1)+2.5(1)+4.5(1)=12. $$

This differs from the sum $6$ because the sample points differ. Both sums use the same valid partition and cell areas. They are simply two different Riemann sums.

Common pitfall: A sample point must lie in the cell whose function value it represents. For example, $(1.5,0.5)$ is not in the lower-left cell $[0,1]\times[0,1]$, because $1.5\notin[0,1]$. Pairing that point with the lower-left cell’s area would not be a valid choice for that cell. Choose a point satisfying both coordinate bounds for each cell, then multiply its function value by that cell’s area.

Another frequent error is to give every cell the same area in a nonuniform partition. In the first example, the cells in the left column have area $2$, while those in the right column have area $6$. Their widths differ, so multiplying every function value by one common area would misrepresent the geometry.

A Reliable Construction Process

When setting up a rectangular Riemann sum, keep the geometric decisions in order. First specify the original rectangle and the coordinate cuts. Then compute the interval lengths and cell areas. Only after the cells are fixed should you select sample points and evaluate the function there.

List the cuts.
Write the increasing $x$- and $y$-coordinate lists, including the endpoints of the rectangle.

Identify each cell.
Use $R_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j]$ for the cell in column $i$ and row $j$.

Calculate its area.
Find $\Delta x_i=x_i-x_{i-1}$ and $\Delta y_j=y_j-y_{j-1}$, then multiply them.

Choose a point in each cell.
Check that both coordinates lie within the appropriate interval; midpoints are one common option.

Pair function values with cell areas.
For each cell, use its own value $f(x_{ij}^{*},y_{ij}^{*})$ and its own area $\Delta A_{ij}$.

Why This Setup Matters

A partition gives a finite geometric model of a rectangle. The cell areas record how much of the region each cell occupies, while the sample points provide locations at which to inspect the function. Together, these choices determine the terms in a rectangular Riemann sum.

Accuracy in this setup matters. A misplaced cut changes the cells; a mistaken width or height changes an area; and a point outside its assigned cell is not a valid sample point for that cell. Useful checks are to verify that the coordinate intervals cover the full side lengths, that the cell areas sum to the full area, and that every chosen point satisfies the bounds of its cell.

The next step in studying double integrals is to examine Riemann sums for functions of two variables in greater detail. For now, the essential construction is precise: divide the rectangle with coordinate cuts, compute each cell’s area, choose a point in each cell, and pair its function value with that area.

Check Your Understanding

Use the definitions and checks in this tutorial to answer the following questions.

For $R=[-2,4]\times[1,5]$, how many cells result from four $x$-intervals and three $y$-intervals?
If one cell has $x$-interval $[2,4]$ and $y$-interval $[1,1.5]$, what are its width, height, and area?
What inequalities must the coordinates of a sample point in $R_{ij}$ satisfy?
Why can two Riemann sums using the same partition have different values?
What calculation checks that the areas of all cells add to the area of the original rectangle?