From Rectangular Sums to Iterated Integrals

From Cells to Coordinate Sums

A double integral over a rectangle begins with contributions from small rectangular cells. Each contribution is a sampled function value multiplied by the area of its cell. When the cells form a grid, the same finite sum can be organized in two stages: add the contributions along one coordinate, then add those results across the other coordinate.

This regrouping is the basic idea behind a repeated integral. It does not change the finite sum. Rather, it reveals how the sum is connected to two successive one-dimensional integrations, each with constant bounds determined by a side of the rectangle.

Rectangular Sums on a Grid

Let \(R=[a,b]\times[c,d]\), where \(a<b\) and \(c<d\). Partition the \(x\)-interval at points \(a=x_0<x_1<\cdots<x_m=b\), and the \(y\)-interval at points \(c=y_0<y_1<\cdots<y_n=d\). Write \(\Delta x_i=x_i-x_{i-1}\) and \(\Delta y_j=y_j-y_{j-1}\). The cell in column \(i\) and row \(j\) has area \(\Delta x_i\Delta y_j\).

Choose one sample coordinate \(\xi_i\) from each \(x\)-interval and one sample coordinate \(\eta_j\) from each \(y\)-interval. The resulting sample point for cell \((i,j)\) is \((\xi_i,\eta_j)\). For a function \(f\), the rectangular Riemann sum is

The area factor matters: \(\Delta x_i\) is a width, \(\Delta y_j\) is a height, and their product is the area of the cell. The sum adds the sampled height \(f(\xi_i,\eta_j)\), weighted by that area, over every cell.

Why the Regrouping Leads to Repeated Integrals

Fix a column \(i\). The value \(\xi_i\) is now fixed, so the function values in the inner sum are samples of the one-variable function \(y\mapsto f(\xi_i,y)\). Thus

is a one-dimensional Riemann sum along the vertical segment from \(y=c\) to \(y=d\). As the \(y\)-partition becomes finer, this sum approaches the integral of that slice, when the slice is integrable. The outer sum then weights each slice total by the width \(\Delta x_i\) and adds across \(a\le x\le b\).

For continuous \(f\) on the closed rectangle, the slices are integrable. Moreover, continuity on a closed and bounded rectangle gives uniform continuity: small changes in the coordinates make uniformly small changes in \(f\). This lets the vertical sums approximate the slice integrals as the \(y\)-widths shrink. The resulting slice integral varies continuously with \(x\), so its outer Riemann sums have a well-defined limit as the \(x\)-widths shrink.

This explains the repeated-integral notation suggested by the two stages:

The inner bounds \(c\) and \(d\) describe the full \(y\)-side of the rectangle; the outer bounds \(a\) and \(b\) describe its \(x\)-side. In this order, \(y\) is integrated first while \(x\) is held fixed, and then the result is integrated in \(x\). The finite regrouping explains the form of the notation. The theorem that formally equates the double integral and this repeated integral for continuous functions is a separate result.

A Derivation from Finite Sums

The key algebraic step is exact, before any limit is taken. For each fixed \(i\), multiply every term in the \(j\)-sum by the same width \(\Delta x_i\), and then add the column sums. This gives

There is no approximation in this equality: it is just a rearrangement of a finite sum. The approximation enters when the inner sums and then the outer sum are interpreted as Riemann sums. For example, if the function is continuous, the error in an inner sum is controlled by how much \(f(\xi_i,y)\) can vary within each \(y\)-interval. Uniform continuity makes that control work uniformly over the \(x\)-coordinates. This is why taking finer rectangular grids leads naturally to a one-dimensional integral of the slice totals.

Worked Example: A Constant Height

Worked Example: Summing \(x+y\) Along the Grid

Worked Example: Unequal Cell Widths

Bounds, Order, and a Common Mistake

On a rectangle, each vertical slice runs over the same \(y\)-interval, so its bounds are always \(c\) and \(d\). The outer coordinate runs over the same \(x\)-interval, so its bounds are \(a\) and \(b\). If instead the sum is grouped by rows, the \(x\)-sum comes first and the repeated-integral form is \(\int_c^d\left(\int_a^b f(x,y)\,dx\right)dy\). Both descriptions use the rectangle's constant side bounds; this tutorial's grouping focuses on summing along \(y\) first.

A common mistake is to drop one width when regrouping. For a cell with widths \(\Delta x_i=1\) and \(\Delta y_j=1/2\), and sampled value \(f(\xi_i,\eta_j)=6\), the cell contributes \(6(1)(1/2)=3\), not \(6(1)=6\). Omitting \(\Delta y_j\) doubles this cell's contribution. The correction is to keep the full area factor in every term, then factor out only the width that is constant across the inner sum.

Another source of confusion is changing sample coordinates within a column. To interpret the inner sum as a Riemann sum of one fixed slice, the \(x\)-coordinate must stay fixed while \(j\) varies. Choosing \(\xi_i\) for a whole column and \(\eta_j\) for each row gives precisely that arrangement. The finite sum is then organized both as a sum over cells and as a sum of vertical slice approximations.

Why This Viewpoint Matters

The rectangular sum records contributions cell by cell, while the repeated-integral form records them slice by slice. Moving between these descriptions is useful because the rectangular sum explains what a double integral accumulates, and the repeated form gives an organized route to computing it. The constant bounds also have a geometric explanation: every slice across a rectangle has the same endpoints.

The main lesson is not that a double integral is automatically two calculations. It is that a grid sum already contains the two-stage structure: each column has a one-dimensional sum, and those column totals form an outer sum. Refining the grid is what turns that structure into repeated integration.