Questions: Iterated Integrals and Fubini's Theorem

Question 1

Consider ∫₀¹ ∫ₓ¹ e^(y²) dy dx. Why is switching the order of integration beneficial here?

• A. The region is more naturally expressed in polar coordinates after switching
• B. e^(y²) has no elementary antiderivative, so integrating y first is impossible; integrating x first gives a tractable inner integral
• C. The outer integral must always correspond to the variable with the simpler bounds
• D. Switching order changes the value of the integral, producing a simpler number

Answer: (B)

This is the canonical example of why switching order matters: ∫ₓ¹ e^(y²) dy has no closed form because e^(y²) has no elementary antiderivative. Reversing the order — for fixed y, x runs from 0 to y, then y from 0 to 1 — gives inner integral ∫₀^y e^(y²) dx = y·e^(y²), which is trivially evaluated, and the outer integral ∫₀¹ y·e^(y²) dy = (e−1)/2 by substitution. Fubini's theorem guarantees the value is unchanged; only the computability changes.

Question 2

Fubini's theorem guarantees that the two orders of integration give the same value as the double integral. For which of the following is this guarantee strongest?

• A. Any integrable function on any bounded region
• B. Any continuous function on a closed bounded rectangular region
• C. Any function where both iterated integrals exist and are finite
• D. Any function where the outer integral converges absolutely

Answer: (B)

Fubini's theorem in its basic form applies to continuous functions on closed bounded rectangles — this gives the cleanest guarantee with no additional conditions. The theorem extends to larger classes (bounded measurable functions, absolutely integrable functions) under stronger hypotheses, but those require Lebesgue integration theory. The key failure mode: a bounded discontinuous function can have iterated integrals that both exist but give different values, which is why 'any bounded function' is too broad without further conditions.

Question 3

When switching the order of integration in a double integral over a non-rectangular region, the limits of integration must be recomputed by re-describing the same geometric region with the variables in the reversed order.

• T. True
• F. False

Answer: True

Switching order is not as simple as swapping limit symbols — you must re-read the region's boundary curves with the roles of x and y exchanged. For a triangular region described as 'x from 0 to 1, y from x to 1,' the reversed description is 'y from 0 to 1, x from 0 to y.' Failing to redraw and re-derive the limits is the most common error in switching integration order, and it produces incorrect integrals even when the theorem applies.

Question 4

For any bounded function defined on a closed rectangular region, the two orders of integration generally produce the same value.

• T. True
• F. False

Answer: False

This is false. Fubini's theorem requires more than boundedness. A classic counterexample involves functions that are discontinuous in a way that affects the iterated integrals differently: ∫₀¹ (∫₀¹ f(x,y) dx) dy ≠ ∫₀¹ (∫₀¹ f(x,y) dy) dx for certain pathological bounded functions. Continuity (or absolute integrability in the Lebesgue sense) is the operative condition. Assuming the theorem applies without checking conditions is a common error in advanced applications.

Question 5

Explain why switching the order of integration is a practically important skill, and give an example of when it converts an impossible computation into a tractable one.

Model answer: Switching order matters because one order may yield an inner integral with no closed-form antiderivative while the other is easily computable. Example: ∫₀¹ ∫ₓ¹ e^(y²) dy dx cannot be evaluated with y as the inner variable since e^(y²) has no elementary antiderivative. Rewriting the triangular region as 'x from 0 to y, y from 0 to 1' gives inner integral y·e^(y²), and the outer integral equals (e−1)/2 by substitution.

The geometric key is that both orders describe the same region — one with horizontal slices, one with vertical slices. The analytic key is that the integrand may factor or simplify in one orientation. Recognizing 'switch order' as a solution strategy comes from understanding that the order of integration is flexible (when Fubini applies), not fixed by convention.