Double integrals and two dimensional accumulation

A business quantity can vary in two independent dimensions, such as production level and time or geographical position. Adding a rate over one dimension is not enough when each small rectangle contributes an amount. A double integral accumulates those rectangle contributions and makes their units explicit.

1 From rectangle sums to an integral

On $R=[a,b]\times[c,d]$ with $a<b$ and $c<d$, partition each coordinate interval. A small cell has area $\Delta x_i\Delta y_j$. Choose a sample point $\xi_{ij}$ in it and form

$$\sum_{i,j}f(\xi_{ij})\Delta x_i\Delta y_j.$$

If these sums approach one value as the largest cell diameter tends to zero, independently of the choices, that value is the double integral $\iint_Rf(x,y)dA$. Continuity on a closed rectangle is sufficient. To justify this, uniform continuity makes the maximum minus minimum of $f$ on every sufficiently small cell less than any prescribed positive number. The upper and lower sums then differ by at most this oscillation bound times the fixed rectangle area, forcing a common limit.

If $f$ has units currency per unit of $x$ per unit of $y$, multiplying by $dx\,dy$ produces currency. A table of total amounts already assigned to cells is not a density table; multiplying those amounts by area again would double count the cell sizes.

For nonnegative $f$, the integral is volume under its graph in the geometric interpretation. For signed $f$, it is signed accumulation, not necessarily a nonnegative volume.

2 Iterated integration and Fubini's theorem

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Compactness gives boundedness and uniform continuity

Let $K\subset\mathbb R^n$, $n\ge1$, be compact, and let $u:K\to\mathbb R$ be continuous relative to $K$. Compactness means the finite-subcover property. The empty set is harmless. At each $p\in K$, continuity supplies $r_p>0$ such that $|u(q)-u(p)|<1$ for $q\in K\cap B(p,r_p)$. A finite selection of these balls covers $K$; the largest of the finitely many numbers $|u(p)|+1$ bounds $|u|$.

For $\varepsilon>0$, instead choose $r_p>0$ such that $|u(q)-u(p)|<\varepsilon/2$ whenever $q\in K\cap B(p,r_p)$. Finitely many balls $B(p_i,r_{p_i}/2)$ cover $K$. Put $\delta=\min_i r_{p_i}/2>0$. If $x,y\in K$ and $\|x-y\|<\delta$, choose $i$ with $\|x-p_i\|<r_{p_i}/2$. Then both $x,y$ belong to $B(p_i,r_{p_i})$, so $|u(x)-u(y)|<\varepsilon$. This proves uniform continuity using the relative topology; no extension off $K$ was used. Closed bounded subsets of finite-dimensional Euclidean space are compact by the separately included Heine–Borel proof. This argument neither uses integration nor assumes a Fubini theorem.

Product-grid Darboux criterion and arbitrary rectangular tags

Let $R=[a,b]\times[c,d]$, with finite $a<b$, $c<d$, and let $f:R\to\mathbb R$ satisfy $|f|\le M<\infty$. A product grid consists of partitions $a=x_0<\cdots<x_r=b$, $c=y_0<\cdots<y_s=d$. Its cells $C_{ij}$ have areas $A_{ij}=\Delta x_i\Delta y_j$. Define $m_{ij}=\inf_{C_{ij}}f$, $M_{ij}=\sup_{C_{ij}}f$, and

$$L(f,P)=\sum_{i,j}m_{ij}A_{ij},\qquad U(f,P)=\sum_{i,j}M_{ij}A_{ij}.$$

Each tagged sum lies between these two numbers. Refinement increases $L$ and decreases $U$, because the subcell areas sum to the parent area. Two grids have a common refinement obtained by taking both sets of coordinate cuts. Consequently $L(f,P)\le U(f,Q)$ for all grids. Real completeness defines $I_-=\sup_P L(f,P)\le I_+=\inf_P U(f,P)$. These are finite. They coincide with a number $I$ if and only if, for every $\varepsilon>0$, some grid has $U-L<\varepsilon$. Indeed that gap bounds $I_+-I_-$; conversely choose an upper sum within $\varepsilon/2$ of $I_+$, a lower sum within $\varepsilon/2$ of $I_-$, and refine both.

Here is the bridge to arbitrary subrectangles in the original definition. A rectangular tiling is a finite cover of $R$ by closed axis-parallel positive-area rectangles with disjoint interiors; common edges may overlap. Its mesh is the largest cell diameter. Its lower and upper sums are defined in the same way. Taking every cell-edge coordinate gives a product-grid refinement, so its lower sum is at most $I$, and its upper sum is at least $I$.

Fix a product grid $P$ with gap less than $\varepsilon/2$. Let it have $v$ internal vertical lines and $h$ internal horizontal lines. In any tiling $Q$ of mesh less than $\delta$, call a cell bad if it meets one of these lines. A bad cell lies in a strip of width $2\delta$ about a line. Thus the sum of the bad-cell areas is at most

$$2\delta\bigl(v(d-c)+h(b-a)\bigr).$$

This is an elementary finite-area estimate: refining all rectangle and strip edges reduces it to addition of disjoint cell areas, without using an integral. Every other cell is contained in one cell of $P$. Its contribution to the oscillation sum is bounded by that parent-cell oscillation times its area; summing the good cells gives at most $U(f,P)-L(f,P)$. On bad cells the oscillation is at most $2M$. Hence

$$U(f,Q)-L(f,Q)\le U(f,P)-L(f,P) +4M\delta\bigl(v(d-c)+h(b-a)\bigr).$$

Choose $\delta$ so the last term is less than $\varepsilon/2$; when its coefficient is zero there is no restriction from that term. Every tag choice then gives a sum within $\varepsilon$ of $I$. This proves convergence for every sufficiently fine rectangular tiling, not only for grid refinements.

Conversely, if all tagged sums on sufficiently fine rectangular tilings approach $I$, this holds for product grids. On a fixed fine grid choose each tag within $\eta/|R|$ of its cell supremum in function value; the resulting sum differs from $U$ by at most $\eta$. Such tags exist by the definition of supremum; use infima for $L$. The assumed uniform control of every tag choice, followed by $\eta\downarrow0$, makes $U-L$ arbitrarily small. Thus the bounded-function Darboux criterion and the original arbitrary-diameter tagged definition agree.

The one-dimensional version is proved by the same construction with intervals and lengths. Explicitly, refine at the union of cuts; a fine interval that meets one of $q$ fixed interior cuts contributes to bad total length at most $2q\delta$. The same $2M$-oscillation estimate proves equivalence with all fine interval tags, and near-supremum/infimum tags give the converse. We use this proper one-dimensional Riemann integral below. Sums of tagged sums prove linearity; pointwise comparison proves monotonicity; refinement at a fixed cut proves interval additivity. In detail, if a function is integrable on the whole interval, refine a small-gap partition at the cut: each subinterval gap is at most the total gap, proving integrability of both restrictions. Conversely, join small-gap partitions of the two restrictions; lower and upper sums add, so the integral of the whole is the sum of the two limits. Constants have integral equal to value times length. These statements also follow directly by taking the just-proved tagged limits. A bounded change supported on finitely many points has integral zero: isolate those points in finitely many intervals of arbitrarily small total length, and bound the upper sum of its absolute value by its bound times that length. Therefore endpoint values do not change a proper integral.

If either side length of $R$ is zero, define every area sum and the double integral to be zero. An integral over a zero-length interval is also zero. The positive-area argument above is not applied after dividing by a zero area.

Fubini with an integrable function and all integrable sections

Assume $a<b$, $c<d$, that bounded $f:R\to\mathbb R$ is Riemann integrable, and that for every $x\in[a,b]$ the section $y\mapsto f(x,y)$ is properly Riemann integrable on $[c,d]$. Put $F(x)=\int_c^d f(x,y)\,dy$. Then $|F(x)|\le M(d-c)$. We prove that $F$ is properly Riemann integrable and

$$\iint_R f\,dA=\int_a^b F(x)\,dx.$$

For any product grid use its $m_{ij},M_{ij}$ above and set $A_i=\sum_jm_{ij}\Delta y_j$, $B_i=\sum_jM_{ij}\Delta y_j$. For every $x\in[x_{i-1},x_i]$, one-dimensional additivity and comparison give $A_i\le F(x)\le B_i$. Taking infima and suprema in that entire closed $x$-interval yields

$$L(f,P)\le L(F,X)\le U(F,X)\le U(f,P).$$

Given $\varepsilon>0$, integrability of $f$ supplies a grid with $U(f,P)-L(f,P)<\varepsilon$. The displayed inequality makes the outer Darboux gap less than $\varepsilon$, proving outer integrability rather than assuming it. Both $\iint_R f$ and $\int_a^b F$ lie in $[L(f,P),U(f,P)]$; their difference is at most $\varepsilon$, so they are equal. If instead all horizontal sections are properly integrable, the identical argument with the coordinate names exchanged gives the horizontal formula. Both orders are asserted only when both sets of sections exist. For a zero-width rectangle, the same equality holds whenever the displayed section integrals exist, since an outer or inner interval has zero length.

Continuous rectangle and interval integrability

Let $f$ be continuous relative to the closed rectangle $R$. If $|R|>0$, compactness and the finite-cover proof give boundedness and uniform continuity. For any $\varepsilon>0$, choose $\delta>0$ such that points less than $\delta$ apart have function values differing by less than $\varepsilon/(2|R|)$. A grid whose cell diameters are less than $\delta$ has

$$U(f,P)-L(f,P)\le\frac{\varepsilon}{2|R|}\sum_C |C|=\varepsilon/2.$$

The Darboux criterion proves integrability, and its tiling bridge proves the original arbitrary-rectangle tagged definition. On a compact interval the identical proof uses length instead of area, with zero-length intervals handled separately. Each fixed-coordinate section of continuous $f$ is therefore properly integrable. The proved section theorem gives both iterated formulas. Moreover $F(x)=\int_c^d f(x,y)\,dy$ is continuous: its difference at $x,x_0$ is bounded by

$$(d-c)\sup_{y\in[c,d]}|f(x,y)-f(x_0,y)|,$$

which tends to zero by uniform continuity. If a side of $R$ has zero length, continuous sections still have their one-dimensional proper integrals, and all relevant double/iterated integrals are zero. No nonsingular-change-of-variables or measure-theoretic theorem is involved.

For continuous real $p$ on $[a,b]$ and $q$ on $[c,d]$, the proved rectangular theorem gives

$$\iint_{[a,b]\times[c,d]}p(x)q(y)\,dA=\left(\int_a^bp(x)\,dx\right)\left(\int_c^dq(y)\,dy\right).$$

Indeed, the inner integral is $p(x)\int_c^dq(y)\,dy$ by one-dimensional linearity, and the remaining constant factors out of the outer integral. This states an integration identity, not probabilistic independence.

An inner integral treats the other variable as constant. For example,

$$\int_0^2\int_0^3(x+2y)dy\,dx=\int_0^2(3x+9)dx=24.$$

Reversing order gives $\int_0^3(2+4y)dy=24$. The equality comes from a theorem, not from assuming every order swap is valid. Singular improper integrands need additional hypotheses and are outside this continuous-rectangle result.

3 A production and time accumulation model

Suppose $w(q,t)=2+q+t$ is a model marginal cost density in currency per production unit per month, for $0\le q\le4$ and $0\le t\le3$. The accumulated quantity in that model is

$$C=\int_0^3\int_0^4(2+q+t)dq\,dt =\int_0^3(16+4t)dt=66.$$

Its average density over the $12$-unit parameter rectangle is $66/12=5.5$. This interpretation requires that the supplied object truly is a two-dimensional density. An ordinary monthly total cost function should be integrated only over time, not also over a production dimension without a stated density relation.

For a spatial storage-cost density $w(x,y)$ on a warehouse floor, $\iint_Dw\,dA$ gives total cost per unit time if $w$ has currency per area per time units. Multiplying that result by a duration gives accumulated cost only when the time dependence is appropriately accounted for.

4 Nonrectangular feasible regions

Let $D=\{(x,y):0\le x\le2,\ 0\le y\le2-x\}$. Its upper boundary couples the two quantities. For a continuous density on this triangle, extend it by zero to a containing rectangle. The extension may jump on the three boundary segments, but those can be covered by rectangles of arbitrarily small total area. Boundedness controls those cells, and uniform continuity controls the others, giving integrability. Slicing therefore yields

$$\iint_Df\,dA=\int_0^2\int_0^{2-x}f(x,y)dy\,dx.$$

In the opposite order the bounds are $0\le y\le2$ and $0\le x\le2-y$. Bounds must be reconstructed from the region, not swapped mechanically.

For $f=x+y$, compute

$$\int_0^2\left[x(2-x)+\frac{(2-x)^2}{2}\right]dx =\int_0^2\left(2-\frac{x^2}{2}\right)dx=\frac83.$$

The triangle's area is 2, so its average value is $4/3$. An average over a region is $\iint_Df/\operatorname{area}(D)$ when area is positive, not simply an average of corner values.

5 Weighted centers

If $w\ge0$ describes a positive allocation density and $M=\iint_Dw>0$, define its weighted center by

$$\bar x=\frac{\iint_Dxw\,dA}{M},\qquad\bar y=\frac{\iint_Dyw\,dA}{M}.$$

For $w=1+x$ on the unit square, $M=3/2$, $\bar x=5/9$, and $\bar y=1/2$. More weight lies on the right, shifting only the horizontal coordinate. A center computed from signed weights need not have this nonnegative-weight interpretation.

6 Graded practice

E1 Foundation. Evaluate $\iint_{[0,1]\times[0,2]}(3x+y)dA$ in both orders and find the average value.

E2 Region bounds. Reverse the order of $\int_0^1\int_x^1xy\,dy\,dx$ and evaluate it.

E3 Application. A floor has $0\le x\le2$, $0\le y\le1$ and density $w=4+x$ currency per square meter per day. Find total daily cost and its weighted horizontal center.

E4 Signed accumulation. For $f=x-y$ on the unit square, find its integral. Does a zero answer imply the function is zero everywhere?

7 Full solutions

S1. Integrating $y$ first gives $\int_0^1(6x+2)dx=5$. Integrating $x$ first gives $\int_0^2(3/2+y)dy=5$. Divide by area 2 to obtain average $5/2$.

S2. The region is $0\le x\le y\le1$. Reverse to $\int_0^1\int_0^yxy\,dx\,dy=\int_0^1y^3/2\,dy=1/8$. Direct original-order integration gives the same value.

S3. Daily cost is $\int_0^2(4+x)dx=10$ currency per day. Its horizontal first moment is $\int_0^2x(4+x)dx=8+8/3=32/3$, so $\bar x=16/15$ meters. The center lies right of the geometric midpoint 1 because density increases with $x$.

S4. By symmetry or direct integration, $\iint(x-y)dA=1/2-1/2=0$. The function is positive above one coordinate ordering and negative below it; cancellation, not pointwise vanishing, causes the zero net value.

Summary

Double integrals accumulate density over two independent dimensions. For continuous rectangle data, Fubini justifies successive one-variable integrations. Feasible-region geometry determines variable bounds, while units, sign and weight assumptions determine what the resulting total or average means.