Double integrals over rectangles — Fubini's theorem: Question bank

True or false — justify

Each answer restates the reasoning, never just the label.

On a rectangle

R = [a, b] \times [c, d]

, you may swap

d x d y

d y d x

and keep the same numbers in the limit slots.

True — on a rectangle every limit is a pure constant, so the numbers

a, b

always pair with

x

and

c, d

always pair with

y

no matter the order. This is only safe because the region is a box.

The same limit-swapping habit is safe on a non-rectangular region.

False — there the inner limits depend on the outer variable, so blindly moving them to the other slot mixes a variable into its own bound; you must re-read the region and re-derive the limits.

\iint_{R} f d A

always equals a positive number.

False — it is a signed volume. Where

f < 0

the surface is below the

x y

-plane and contributes negatively, so the total can be zero or negative.

f (x, y) = 0

at every sample point of some partition, the double integral must be

0

False in general — a single partition proves nothing;

f

could be nonzero between the sample points. The integral is a limit over all shrinking partitions, not one lucky choice.

Fubini's theorem is essentially the single-variable volume-by-slicing formula applied twice.

True — slicing at fixed

x

gives cross-sectional area

A (x) = \int_{c}^{d} f d y

, and

\int_{a}^{b} A (x) d x

is the ordinary "integrate the cross-section" formula; slicing the other way just repeats the idea with

x

and

y

swapped.

For a separable integrand

g (x) h (y)

, the double integral equals the product of two single integrals for any region

R

False — the split only works on a rectangle, where the inner limits are constants so

g (x)

(and the numeric inner result) can be pulled cleanly outside; on a general region the inner limits depend on the outer variable and the product form breaks.

If both iterated integrals of

f

over a rectangle exist and are finite, they are automatically equal.

Not guaranteed by that fact alone — for a genuinely non-integrable

f

the two orders can differ even when each exists; equality is guaranteed once

f

is integrable (e.g. continuous, or

\iint ∣ f ∣ < \infty

by Fubini–Tonelli theorem).

Changing which variable you integrate first can change the answer on a continuous rectangle problem.

False — for continuous

f

on a rectangle the answer is fixed; only the effort changes. Choosing the easier order is the whole practical payoff of Fubini.

Spot the error

Each reveal names the flawed step and repairs it.

"To do the inner integral

\int_{c}^{d} f (x, y) d y

, I integrate both

x

and

y

because both appear."

Wrong — the inner integral integrates only its own variable

y

; here

x

is a frozen constant. Treating

x

as variable produces the wrong antiderivative (e.g. dropping the

x y

term).

\int_{0}^{2} \int_{1}^{3} (x + 2 y) d y d x

: the inner gives

\int_{1}^{3} 2 y d y = [y^{2}]

and I forgot the

x

term, so inner

= 8

Error —

x

is a constant, not zero, so its antiderivative in

y

x y

, not nothing. The inner is

[x y + y^{2}]_{1}^{3} = 2 x + 8

, and the

x

term matters in the outer step.

"Because Fubini says order is free, I can move the outer limits inside without changing anything, even here on a triangle."

Error — Fubini's freedom to swap the order does not preserve limit slots on a non-rectangle. The order-swap and the limit values are two separate things; on a triangle you must recompute both bounds.

"The integrand blows up somewhere on

R

, but Fubini is a theorem so I'll trust equal orders anyway."

Error — Fubini requires

f

integrable (continuous, or absolutely integrable). If

f

is unbounded/non-integrable, the hypothesis fails and the two orders can legitimately disagree.

f = x e^{y}

is separable, so on

R = {y \leq x} \cap [0, 1]^{2}

I'll write it as

(\int x d x) (\int e^{y} d y)

Error — separability of the function is not enough; the region must also be a rectangle for the product split. Here

y \leq x

couples the limits, so the shortcut is invalid.

"I sampled the corner points of each sub-rectangle and got one value, so that's the double integral."

Error — the double integral is the limit as all sub-rectangles shrink to zero, independent of which sample point you pick. One partition with one sample choice is just a Riemann sum approximation, not the exact value.

Why questions

Each reveal is the mechanism, not a slogan.

Why does freezing

x

turn the inner integral into an ordinary single-variable integral?

With

x

fixed at

x_{0}

, the map

y \mapsto f (x_{0}, y)

is a function of one variable, so its integral over

[c, d]

is a plain 1-D area — the cross-section of the solid at that

x_{0}

Why does

A (x) = \int_{c}^{d} f (x, y) d y

deserve the name "cross-sectional area"?

The vertical plane at

x = x_{0}

slices the solid; the sliced face is bounded below by the plane and above by the curve

z = f (x_{0}, y)

, and

A (x_{0})

is exactly the area of that face.

Why can

g (x)

be pulled outside the inner

y

-integral in the separable case?

Relative to the inner integration variable

y

g (x)

is a constant, and constants factor out of any integral — leaving

g (x)

times a fixed number.

Why does the "same solid, two slicings" argument force the two orders to be equal?

Both iterated integrals reconstruct the volume of one and the same solid; a solid has a single volume, so the two computations must land on the same number.

Why is choosing the integration order a real practical tool and not just aesthetics?

One order may hand you an antiderivative for free (as when a factor is exactly

\frac{d}{d x}

of something), while the other forces integration by parts — same answer, far more labour, via Change of order of integration.

Why do we need the limit-of-Riemann-sums definition at all, instead of just declaring the iterated integral to be the answer?

The double integral is defined as the volume-approximating limit; Fubini is a theorem proving the iterated integral equals that limit. Without the definition there is nothing for Fubini to be true about.

Edge cases

Boundary and degenerate scenarios the theorem quietly covers.

What is

\iint_{R} f d A

when

R

is degenerate, e.g.

[a, a] \times [c, d]

(zero width)?

It is

0

— the region has zero area, every sub-rectangle base

Δ A = 0

, so every Riemann-sum term vanishes; there is no solid to have volume.

f (x, y) = k

is constant on

R = [a, b] \times [c, d]

, what does the double integral give?

k (b - a) (d - c)

— a flat slab of height

k

over the box, i.e. height times base area, exactly what both iterated orders return.

Can Fubini handle an

f

with a jump discontinuity along a line inside

R

Yes — bounded

f

with finitely many discontinuities (a measure-zero set) is still integrable, so both orders agree; continuity is sufficient but not strictly necessary.

f

is odd about the centre of a symmetric rectangle (e.g.

f (x, y) = x

[- 1, 1] \times [c, d]

), what happens?

The integral is

0

— positive contributions on one half cancel equal negative contributions on the mirror half, which the signed-volume interpretation makes visible instantly.

Does Fubini's rectangular statement extend to three variables and boxes?

Yes — for continuous

f

on a box you may iterate the three single integrals in any of the orders, which is the Triple integrals generalization of the same slicing logic.

What if

f

is continuous but the rectangle is unbounded, e.g.

[0, \infty) \times [0, 1]

Then it is an improper double integral; Fubini's plain form does not directly apply. You need the Fubini–Tonelli theorem with

\iint ∣ f ∣ < \infty

to justify swapping orders.

Connections

Riemann sums — every trap about "one partition isn't the answer" traces back here.
Volume by slicing (single-variable) — the engine behind the "why" answers.
Double integrals over general regions — where the limit-swap habit becomes dangerous.
Change of order of integration — the "choose the easy order" payoff.
Fubini–Tonelli theorem — the safety net for unbounded/improper cases.
Triple integrals — the edge case extending to boxes.