Multiple integrals and iterated integration
Translate a region into correct bounds and identify when changing integration order is justified.
On this page
The question for today
What you’ll understand
- Construct iterated bounds from horizontal or vertical slices.
- Distinguish Tonelli's nonnegative case from Fubini's absolute-integrability case.
- Compute and compare two integration orders over a triangular region.
An integral over a region adds contributions across an area or volume. Iterated integration evaluates that sum by slicing, but the drawing and the hypotheses both matter. Incorrect slice bounds integrate over the wrong region even when every antiderivative is correct. Interchanging integrals also requires a theorem: nonnegative integrands and absolutely integrable signed integrands provide two especially important settings. These ideas turn geometric accumulation into a dependable computational method.
The region comes before the order
For a continuous function on a closed rectangle, the double integral can be evaluated by integrating one coordinate first and then the other. The inner integral treats the outer variable as fixed. On a nonrectangular region, each slice can have different endpoints. A vertically simple region is described by x between fixed endpoints and y between two functions of x. The integral then uses exactly those moving y bounds. Reversing order requires re-describing the same set using horizontal slices, rather than merely swapping the symbols dx and dy.
Drawing a rough boundary sketch and recording inequalities prevents many mistakes. For a triangle, a horizontal slice may begin at a sloping edge and end at a vertical one, whereas a vertical slice may begin at a horizontal edge. More complicated regions may need splitting because one formula cannot describe every slice. Boundaries of ordinary piecewise smooth planar regions have area zero, so including or excluding those boundaries does not affect an integrable area integral. This observation does not permit discarding a positive-area piece or ignoring an interior singularity.
Why two different interchange theorems are useful
Tonelli's theorem applies to a nonnegative measurable function on a product of sigma-finite measure spaces. Both iterated integrals equal the product integral, allowing the common value to be infinite. It is therefore an effective first tool for testing absolute integrability: apply it to the absolute value of a signed function. Fubini's theorem then guarantees that an absolutely integrable function has integrable slices almost everywhere and equal finite iterated integrals. Exceptional slices of measure zero do not invalidate that conclusion.
A conditional cancellation is not the same as absolute integrability. It can hide dependence on the order of limiting operations near a singularity or at infinity. If a signed integral is not absolutely integrable, computing two iterated expressions and seeing different answers signals that the standard Fubini guarantee was unavailable. One should specify the intended improper-limit convention or use another justified theorem, rather than declaring that integration order never matters. On compact ordinary regions, a continuous integrand avoids these issues because it is bounded and absolutely integrable.
Choose an order that exposes structure
Changing order is often the difference between a difficult antiderivative and an elementary one. For example, if an integrand depends only on y, integrating first in x can turn the problem into the integrand times the horizontal width of the region. This is not a trick: it is a geometric evaluation of equal slices. Conversely, a density concentrated near a curved boundary might be simpler after a coordinate change, which is a different operation requiring an area-scale factor and will be treated separately.
Multiple integrals retain units. A mass density in kilograms per square meter multiplied by area produces mass; integrating only over one coordinate gives a line density until the second integration is performed. Integrating the constant one gives region area, making it a useful check on bounds. Symmetry is another check: an integrable function odd in x on a region symmetric under x reversal has zero integral. Before relying on symmetry, verify that the transformation preserves the region and that the signed integral actually exists.
Work it through
Reverse the slices of a triangle
Compute the integral of x+y over D={0≤y≤x≤1} using both orders.
With vertical slices, x runs from 0 to 1 and y from 0 to x. The inner integral is x·x+x²/2=3x²/2, and integrating in x gives 1/2.
For horizontal slices, y runs from 0 to 1, while x begins at y and ends at 1. The inner integral is 1/2+y−3y²/2.
Integrating the second expression from 0 to 1 gives 1/2+1/2−1/2=1/2. The polynomial is continuous on the compact triangle, so absolute integrability justifies both evaluations.
Both orders give 1/2. Swapping the original differential order without changing the triangular slice bounds would describe a different calculation.
The living notebook
Explore the model
Interactive lab · function
A slice integral with an exact check
Reset, set a=1.5 and x=1. Predict the plotted area from 0 to x as a x³/3=0.5. Compare a fresh run with Reset, a=1.5, x=2, predicting area 4. This one-variable quadratic-area instrument checks a possible inner-integral calculation; it does not draw a two-dimensional region or swap its bounds.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forAn iterated integral evaluates one slice at a time, but its outer geometry must still be supplied correctly.
Pause & reason
Check your understanding
Your turn
Try it yourself
Evaluate ∫₀¹∫ₓ¹ e^(y²) dy dx by changing order. State why the change is justified and why attempting the original inner antiderivative is unnecessary.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The inequalities are 0≤x≤y≤1. For fixed y, find the length of the allowed x interval.
Compare with the solution
The continuous integrand is bounded on the compact triangular region, so it is absolutely integrable. With y from 0 to 1 and x from 0 to y, the integral becomes ∫₀¹ y e^(y²) dy. Substitution u=y² gives (e−1)/2. No elementary antiderivative for e^(y²) is needed: integrating first in x contributes the slice width y and produces an elementary remaining integral.
Keep the idea
Take it with you
- Integration order changes require new slice bounds for the same region.
- Tonelli handles nonnegative extended values; Fubini handles absolutely integrable signed functions.
- Units, area tests, and symmetry help check an iterated-integral setup.
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