Double Integrals and Fubini's Theorem

Four cases, chosen because each one settles a different question about what the order and the region cost.

---

1. A separable integrand, where the double integral is a product. Over $R=[0,2]\times [0,3]$ with $f(x,y)=x{y}^2$, the inner integration carries the $x$ through as a constant:

But because the integrand factors as $g(x)h(y)$ and the limits are constants, the whole thing factors too:

The factorisation needs both conditions. ${\int }_0^2{\int }_0^3(x+y^2)dydx=24$, and the corresponding product of separate integrals is $(2)(9)=18$, which is not it: a sum does not factor. And over a non-rectangular region the limits carry the outer variable, so even a product integrand does not separate.

---

2. A region where the order decides how much work it is. Consider ${\displaystyle {\int }_0^1{\int }_y^1{e}^{x^2}dxdy}$.

As written, the inner integral $\int e^{x^2}dx$ has no elementary antiderivative, and the calculation stops. Reversing the order rescues it. The region is $\{(x,y):0\le y\le 1,y\le x\le 1\}$, which is the triangle below the line $y=x$; described the other way, $x$ runs over $[0,1]$ and for each $x$, $y$ runs from $0$ to $x$:

The inner integration produced the factor $x$ that the substitution needed. Both orders are valid and give the same number; only one is computable in closed form. Check: a $3000\times 3000$ midpoint sum over the triangle returns $0.85914\dots$.

---

3. A region needing two pieces in one order and one in the other. Let $S$ be bounded by $y=x$, $y=2-x$ and $y=0$: a triangle with vertices $(0,0)$, $(1,1)$, $(2,0)$.

Sweeping in vertical strips, the upper boundary changes at $x=1$, so $dydx$ needs two integrals:

Sweeping in horizontal strips, each strip at height $y$ runs from $x=y$ to $x=2-y$ throughout, so $dxdy$ needs one:

Both give the triangle's area of 1. The choice of order is not about correctness; it is about how many pieces the region breaks into.

---

4. Where Fubini has nothing to say. Over $[0,1]\times [0,1]$ take

The two iterated integrals both exist and come to $\pi /4$ and $-\pi /4$. They disagree, and the double integral does not exist at all, because $\iint |f|dA$ diverges near the origin where $f$ is unbounded.

Nothing in the arithmetic of either iterated integral signals this. Each is a perfectly ordinary calculation returning a finite number. Only checking the hypothesis does, and the hypothesis fails because $f$ is not continuous on the closed rectangle and not absolutely integrable over it. This is the case that makes Fubini a theorem rather than a convention.

---

What the four have in common. The region and the integrand together decide what is possible: whether the integral factors, whether either order is computable, how many pieces are needed, and whether the orders agree at all. None of that is visible from the notation ${\iint }_{R}fdA$, which is why the first step is always to describe the region and check the integrand rather than to start integrating.

Swapping the limits of a non-rectangular double integral. Over a rectangle the limits are constants and either order works. Over a region where the inner limits depend on the outer variable, exchanging the order requires describing the region the other way round, reversing the written limits produces a different region and a different number.

Treating Fubini as unconditional. Over a rectangle with a continuous integrand the two orders agree, and disagreement means an arithmetic slip. That is a consequence of the hypothesis, not a general fact about integration: without continuity or absolute integrability the two iterated integrals can genuinely differ, and the two-dimensional integral may not exist at all. Quoting "either order works" without the rectangle and the continuity is quoting a conclusion without its premise.

Reading the inner limits as fixed when they are not. Over a region bounded by $y=x^2$ and $y=x$, the inner limits in $\int \int dydx$ run from $x^2$ to $x$, both functions of the outer variable. Substituting them leaves an expression in $x$ alone, which is the point. Writing constant inner limits there integrates over a rectangle the region does not fill.