Reading Iterated Integral Bounds
An iterated integral records two integrations in a particular order. Its notation tells you which variable is integrated first, which variable is integrated second, and the limits for each variable. Reading those pieces accurately is essential: changing the order of integration changes where the bounds appear, even when the rectangular domain stays the same.
This tutorial focuses on rectangles whose sides are parallel to the coordinate axes. A rectangle of this kind is determined by an interval of \(x\)-values and an interval of \(y\)-values. Each variable ranges independently over its interval, so all four endpoints in either order are constants.
Inner and Outer Variables
In an iterated integral, the differential closest to the integrand identifies the inner variable: that variable is integrated first. The other differential identifies the outer variable: it is integrated after the inner integral has been formed.
For example, in \(\int_c^d\int_a^b f(x,y)\,dx\,dy\), the \(dx\) immediately follows the inner integral, so \(x\) is the inner variable. Its limits are \(a\) and \(b\). The outer differential is \(dy\), so \(y\) is the outer variable, with limits \(c\) and \(d\).
Read the nested notation from the inside outward: first integrate with respect to \(x\), holding \(y\) fixed; then integrate the resulting expression with respect to \(y\). For this rectangle, the \(x\)-limits do not depend on \(y\), and the \(y\)-limits do not depend on \(x\).
In the other order, \(y\) is integrated first and \(x\) second. The bounds must follow their variables: the inner \(y\)-integral uses \(c\) and \(d\), and the outer \(x\)-integral uses \(a\) and \(b\).
Both expressions describe integration over the same rectangle. The integrand remains \(f(x,y)\); what changes is the order of the differentials and the placement of the corresponding limits.
Writing Both Orders for a Rectangle
To write both iterated integrals for a given rectangle, first identify its \(x\)-interval and \(y\)-interval. Then assign the \(x\)-interval to \(x\) and the \(y\)-interval to \(y\), regardless of which variable is integrated first. Finally, place the inner integral inside the outer integral and put each differential beside its own limits.
Record the interval of \(x\)-values and the interval of \(y\)-values separately.
Its two constant endpoints become the limits of the inner integral.
Its endpoints become the limits of the outer integral.
Confirm that each differential is paired with the correct limits and that both variables cover the stated rectangle.
Worked Example: Read an Integral and Identify Its Rectangle
Consider \(\int_1^4\int_{-2}^3 f(x,y)\,dx\,dy\). Identify the inner and outer variables, then state the rectangle.
The inner differential is \(dx\), so \(x\) is integrated first. Its limits are \(-2\) and \(3\), giving \(-2\le x\le3\). The outer differential is \(dy\), so \(y\) is integrated second, with \(1\le y\le4\).
As a check, the two sets of limits describe independent intervals, so their combination is exactly the stated rectangle. Its side lengths are \(3-(-2)=5\) and \(4-1=3\); thus its area is \(5\cdot3=15\). Integrating the constant function \(1\) using these bounds gives \(\int_1^4\int_{-2}^3 1\,dx\,dy=\int_1^4 5\,dy=15\), confirming the area.
Writing the Same Rectangle in Both Orders
For a rectangle \(R=[a,b]\times[c,d]\) and a continuous function \(f\) on \(R\), both orders use the same \(x\)-interval \([a,b]\) and \(y\)-interval \([c,d]\). Fubini’s theorem for continuous functions on rectangles, established in the earlier material, says that the double integral can be represented by either iterated integral. The important task here is to place the constant endpoints with their correct variables.
Here \(dA\) denotes area in the plane. The two iterated forms correspond to the same set of points because the conditions \(a\le x\le b\) and \(c\le y\le d\) hold together in either description. For each fixed \(y\) between \(c\) and \(d\), the horizontal slice runs through all \(x\)-values from \(a\) to \(b\). For each fixed \(x\) between \(a\) and \(b\), the vertical slice runs through all \(y\)-values from \(c\) to \(d\). Neither slice has an endpoint that depends on the other variable.
This slice interpretation explains why the bounds are constant for rectangles. It also explains why the bounds cannot be assigned from position alone: the differential tells you which coordinate’s interval the limits describe.
Worked Example: Write Both Orders from a Domain
Let \(R\) be the rectangle with \(0\le x\le2\) and \(-1\le y\le5\). Write iterated integrals of \(f(x,y)\) in both orders.
The \(x\)-interval is \([0,2]\), and the \(y\)-interval is \([-1,5]\). If \(x\) is inner, put its limits beside \(dx\), then use the \(y\)-limits outside with \(dy\):
If \(y\) is inner, put its limits beside \(dy\), then use the \(x\)-limits outside with \(dx\):
To check, both expressions impose \(0\le x\le2\) and \(-1\le y\le5\). The rectangle’s side lengths are \(2-0=2\) and \(5-(-1)=6\), so its area is \(2\cdot6=12\). The first-order area check is \(\int_{-1}^{5}\int_0^2 1\,dx\,dy=\int_{-1}^{5}2\,dy=12\). The reversed-order check is \(\int_0^2\int_{-1}^{5}1\,dy\,dx=\int_0^2 6\,dx=12\). Both agree with the rectangle’s area.
Why the Two Orders Describe the Same Domain
The bounds in an iterated integral are a compact way to state which coordinate pairs belong to the region. The rectangle can be described by either of these equivalent conditions:
Thinking of \(y\) as the outer variable means choosing a value of \(y\) from \([c,d]\) and then allowing \(x\) to range over all of \([a,b]\). Thinking of \(x\) as the outer variable means choosing \(x\) from \([a,b]\) and then allowing \(y\) to range over all of \([c,d]\). Both descriptions include every point satisfying the two inequalities, and neither includes a point violating either inequality.
This is the domain-level reason the same rectangle admits both orders. For a continuous integrand, the earlier result on Fubini’s theorem also ensures the two iterated integrals have the same value. That conclusion relies on the function being continuous on the rectangle (or on other hypotheses of an applicable version of Fubini’s theorem); it is not a license to ignore the bounds or differentials.
Worked Example: Reverse the Order Without Changing the Rectangle
Start with \(\int_{-2}^{2}\int_{-3}^{-1} f(x,y)\,dy\,dx\). Write the same domain with the order reversed, and check the result.
The inner differential is \(dy\), so the inner variable is \(y\), with \(-3\le y\le-1\). The outer differential is \(dx\), so the outer variable is \(x\), with \(-2\le x\le2\). Thus the domain is \(-2\le x\le2\) and \(-3\le y\le-1\). Reversing the order makes \(x\) inner and \(y\) outer. The \(x\)-interval belongs beside \(dx\), and the \(y\)-interval belongs beside \(dy\):
Check by translating each integral back into inequalities. The original has \(-2\le x\le2\) and \(-3\le y\le-1\); the rewritten integral has those very same inequalities. The side lengths are \(2-(-2)=4\) and \(-1-(-3)=2\), so the area is \(4\cdot2=8\). In the original order, integrating \(1\) gives \(\int_{-2}^{2}\int_{-3}^{-1}1\,dy\,dx=\int_{-2}^{2}2\,dx=8\). In the reversed order, it gives \(\int_{-3}^{-1}\int_{-2}^{2}1\,dx\,dy=\int_{-3}^{-1}4\,dy=8\). Both checks match.
A Common Bounds Mistake
A frequent error is to reverse the differentials but leave the limits in their old positions. For the rectangle \(0\le x\le2\), \(-1\le y\le5\), the expression \(\int_0^2\int_{-1}^{5}f(x,y)\,dx\,dy\) is not a correctly written iterated integral for this rectangle. The inner differential is \(dx\), but its limits \(-1\) and \(5\) are the \(y\)-endpoints, not the \(x\)-endpoints. The outer differential is \(dy\), but its limits \(0\) and \(2\) are the \(x\)-endpoints, not the \(y\)-endpoints.
The correction is to match limits to variables, not to the visual position where you expect them to appear:
In the first expression, \(dx\) uses the \(x\)-interval and \(dy\) uses the \(y\)-interval. In the second, the same pairings hold, although the inner and outer roles have switched. A reliable check is to write the inequalities for \(x\) and \(y\) explicitly before writing either integral.
Why Careful Bound Reading Matters
Bounds are more than symbols surrounding an integrand. They specify the region being accumulated over. A correctly computed integral with incorrectly assigned bounds answers a different question. For rectangles, the independent constant intervals make the structure especially clear: each variable has its own endpoints, and changing the order only changes which interval is inner and which is outer.
Before evaluating an iterated integral, identify the inner variable, the outer variable, and the interval attached to each one. If the domain is given as inequalities, those inequalities provide the direct check. If the domain is given as a rectangle, its \(x\)- and \(y\)-side intervals determine both orders without changing the region.
Check Your Understanding
Use the variables and bounds to describe the rectangle, then check your reasoning against the stated intervals.
- In \(\int_2^6\int_{-1}^3 f(x,y)\,dx\,dy\), which variable is inner, and what inequalities describe the domain?
- Write both orders for the rectangle \(-2\le x\le1\), \(3\le y\le4\).
- For \(\int_0^5\int_1^2 f(x,y)\,dy\,dx\), identify the inner and outer variables and their intervals.
- A student writes \(\int_0^2\int_{-1}^5 f(x,y)\,dx\,dy\) for \(0\le x\le2\), \(-1\le y\le5\). Explain the mismatch and write one correct order.
- Why do both orders of integration describe the same rectangular domain when the bounds are assigned to their matching variables?