Additivity Across Rectangular Subdivisions

Why Subdivide a Rectangle?

A single double integral can represent a total accumulated quantity over an entire rectangular region. Sometimes it is easier to understand or calculate that total by considering smaller rectangular pieces: a material may have different densities in different zones, or separate parts of a surface may contribute different signed amounts.

Additivity gives a precise rule for this procedure. When the pieces form a genuine subdivision of the original rectangle, the integral over the whole is the sum of the integrals over the pieces. The subdivision must cover the original rectangle without gaps, and the pieces must not overlap in their interiors.

Key idea: A rectangular subdivision lets us calculate a total on a large rectangle by adding the totals on its smaller rectangular pieces.

Rectangles and Rectangular Subdivisions

Write $R=[a,b]\times[c,d]$, where $a<b$ and $c<d$, for the rectangle with horizontal coordinates from $a$ to $b$ and vertical coordinates from $c$ to $d$. Its side lengths are $b-a$ and $d-c$. We use $\iint_R f(x,y)\,dA$ to denote the double integral of $f$ over $R$.

A grid subdivision is made by choosing cut coordinates in each direction. For example, choose $a=x_0<x_1<\cdots<x_m=b$ and $c=y_0<y_1<\cdots<y_n=d$. The resulting smaller rectangles are

R_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j], \qquad 1\leq i\leq m,\quad 1\leq j\leq n.

The smaller rectangles cover $R$. Neighboring rectangles share edges, but their interiors do not overlap. More general finite subdivisions can be handled by making additional cuts so that the pieces form a grid.

Definition: A finite rectangular subdivision of $R$ is a finite collection of rectangles whose union is $R$ and whose interiors are pairwise disjoint. Shared edges or corners are allowed.

The Additivity Rule

For a continuous function on a rectangle, the double integral exists. The same additivity rule applies whenever the function is Riemann integrable on the whole rectangle and on the pieces. Continuity is a standard sufficient condition for these integrability requirements.

Theorem: If $f$ is integrable on $R$, and $R$ is divided into finitely many rectangles $R_1,\ldots,R_k$ with disjoint interiors, then $$ \iint_R f(x,y)\,dA = \sum_{r=1}^{k}\iint_{R_r} f(x,y)\,dA. $$ The rectangles must cover $R$; shared edges do not count as overlaps of positive area.

This rule applies to a single cut as well as to a grid. For example, if a vertical cut at $x=s$ divides $R=[a,b]\times[c,d]$ into a left rectangle and a right rectangle, then

\iint_R f(x,y)\,dA = \iint_{[a,s]\times[c,d]} f(x,y)\,dA + \iint_{[s,b]\times[c,d]} f(x,y)\,dA.

A horizontal cut gives the corresponding sum of the integrals on the lower and upper rectangles. Repeating cuts produces the rule for a grid.

Why Additivity Holds

The reason can be seen directly from rectangular Riemann sums. A Riemann sum divides the domain into small rectangles, selects a sample point in each one, and adds the function value at that point multiplied by the small rectangle’s area. Its limit, as the rectangles become arbitrarily small, is the double integral.

For a subdivision, choose finer rectangular partitions inside every piece. Taken together, those fine rectangles partition the original rectangle. The sum for the whole rectangle is therefore exactly the sum of the Riemann sums for the individual pieces: every small rectangle contributes its function value times its area once, on either side of the equality.

Partition each piece.
Choose fine rectangular partitions within the smaller rectangles, with the maximum cell width and height tending to zero.

Combine the sums.
The fine cells from all pieces together partition the original rectangle. The total Riemann sum is the sum of the piecewise Riemann sums.

Take limits.
Integrability ensures that each piecewise sum approaches its integral and the combined sum approaches the integral over the whole rectangle. Since there are only finitely many pieces, the limit of the sum is the sum of the limits.

Thus the equality of integrals follows from the equality of the corresponding sums. In more detail, let $S_r$ be the Riemann sum on piece $R_r$, using the fine cells and chosen sample points there. The sum on the whole rectangle is $S=S_1+\cdots+S_k$. As the partitions are refined, $S_r$ tends to $\iint_{R_r} f(x,y)\,dA$, and $S$ tends to $\iint_R f(x,y)\,dA$. Therefore

\iint_R f(x,y)\,dA = \lim S = \lim(S_1+\cdots+S_k) = \sum_{r=1}^{k}\iint_{R_r} f(x,y)\,dA.

Adjacent closed rectangles share boundary points. Those shared edges have area zero, so they do not contribute an additional area-weighted part to a double integral. The theorem concerns coverage and overlap in the interiors, not whether boundary points are written into both rectangle descriptions.

Worked Examples

Worked Example: A Constant Function on Four Pieces

Let $f(x,y)=3$ on $R=[0,4]\times[1,3]$. Cut at $x=1$ and $y=2$. Find the integral by adding the four smaller-rectangle integrals.

The cut coordinates make widths $1$ and $3$, and heights $1$ and $1$. Since the integral of a constant over a rectangle is the constant multiplied by the rectangle’s area, the four contributions are

\begin{aligned} \text{lower left: }&3(1)(1)=3,\\ \text{lower right: }&3(3)(1)=9,\\ \text{upper left: }&3(1)(1)=3,\\ \text{upper right: }&3(3)(1)=9. \end{aligned}

Adding the contributions gives $3+9+3+9=24$. As a check, the original rectangle has width $4$, height $2$, and area $8$. Thus its integral is $3(8)=24$, agreeing with the sum over the pieces.

Worked Example: Combining Signed Subtotals

Suppose a rectangle $R$ is divided into three smaller rectangles $R_1,R_2,R_3$, and the integrals on them have already been evaluated as $7$, $-5$, and $2$, respectively. Find the integral on $R$.

The pieces form a subdivision, so additivity gives

\iint_R f(x,y)\,dA = \iint_{R_1} f(x,y)\,dA + \iint_{R_2} f(x,y)\,dA + \iint_{R_3} f(x,y)\,dA.

Substitute the given values and keep the negative sign:

\iint_R f(x,y)\,dA = 7+(-5)+2 = 2+2 = 4.

Check the arithmetic by grouping the positive contributions and the negative contribution: $7+2=9$, and $9-5=4$. The total is $4$. A negative integral on one piece is a signed contribution; it must be added with its sign, not replaced by its absolute value.

Worked Example: Adding Four Constant Contributions

Consider the grid formed by $x=0,1,3$ and $y=-1,1,2$. On its four smaller rectangles, suppose $f$ is constant as follows: $4$ on $[0,1]\times[-1,1]$, $-1$ on $[1,3]\times[-1,1]$, $3$ on $[0,1]\times[1,2]$, and $2$ on $[1,3]\times[1,2]$. Find the integral over the whole rectangle.

The widths are $1$ and $2$; the heights are $2$ and $1$. For each piece, multiply its constant value by its area:

\begin{aligned} [0,1]\times[-1,1]:&\quad 4(1)(2)=8,\\ [1,3]\times[-1,1]:&\quad (-1)(2)(2)=-4,\\ [0,1]\times[1,2]:&\quad 3(1)(1)=3,\\ [1,3]\times[1,2]:&\quad 2(2)(1)=4. \end{aligned}

Add the four values: $8+(-4)+3+4=4+3+4=11$. As a check, the lower row contributes $8-4=4$, and the upper row contributes $3+4=7$; their sum is also $4+7=11$. The subdivision accounts for the whole rectangle, and each piece’s area has been included once.

A Common Mistake: Adding Overlapping Regions

Additivity does not say that the integrals over any collection of rectangles can be added to obtain the integral over their union. The pieces must form a subdivision: they must cover the target region without overlapping interiors. Otherwise, part of the union may be counted more than once.

For a concrete example, let $f(x,y)=1$, and consider $R=[0,3]\times[0,1]$. Its integral is its area, $3$. Now consider the two rectangles $[0,2]\times[0,1]$ and $[1,3]\times[0,1]$. Each has area $2$, so each integral is $2$. Adding these gives $2+2=4$, which is not the integral over $R$. The overlap $[1,2]\times[0,1]$, of area $1$, has been counted twice.

Correction: Before applying additivity, check that the rectangles cover the intended region and have disjoint interiors. If rectangles overlap, first form a non-overlapping subdivision, or account for the overlap separately.

When This Property Is Useful

Subdivision is useful when different parts of a rectangle have different known totals, or when a function is easier to handle on separate pieces. It also makes sign changes manageable: contributions where the function is negative remain negative, and the final integral is their signed sum with the other contributions.

The rule applies to every finite rectangular grid, regardless of the number of cuts. It does not require equal widths or equal heights. The cuts can be placed wherever convenient, provided the resulting rectangles really do cover the original domain without interior overlap. Additivity says how the totals combine; it does not by itself calculate the integral on each smaller rectangle.

Check Your Understanding

Use the definitions and additivity rule above to answer these questions.

What two geometric conditions must a collection of rectangles satisfy to be a subdivision?
A rectangle is divided into four pieces with integrals $3$, $-2$, $5$, and $-1$. What is the integral over the whole rectangle?
Why do shared edges between neighboring rectangles not invalidate additivity?
Two rectangles overlap in a region of positive area. Why can their integrals not automatically be added to obtain the integral over their union?
In the proof from Riemann sums, why can the sum over the whole rectangle be written as the sum of the sums over its pieces?