Practice: Double Integrals and Fubini's Theorem

(a) The rectangle, both orders. Inner over $y$, holding $x$ fixed:

Then over $x$:

Reversed, inner over $x$, holding $y$ fixed:

Both give 21 ✓, and a $1500\times 1500$ midpoint sum returns $21.00000000$ ✓.

What guarantees agreement. Fubini's theorem: for a function continuous on a rectangle, the double integral equals either iterated integral. The integrand $2x+y$ is a polynomial, hence continuous on $[0,2]\times [0,3]$, so the hypothesis holds. This is a theorem rather than a definition: the double integral is a two-dimensional limit, the iterated integrals are two one-dimensional limits taken in sequence, and that they coincide needs proving. Without the hypothesis it can fail outright, and the two iterated integrals can exist with different values while the double integral does not exist at all.

(b) The region $T$, in the order $dydx$. For $0\le x\le 1$ the two boundaries are $y=x^2$ and $y=x$, and on that interval $x^2\le x$, so for a fixed $x$ the vertical strip runs from $y=x^2$ up to $y=x$:

Inner integration, holding $x$ fixed:

Outer integration:

So the integral is $\boxed{{\displaystyle 3/20}}=0.15$. Check: a $2000\times 2000$ midpoint sum restricted to $T$ gives $0.14999\dots$ ✓.

(c) The same region, in the order $dxdy$. The limits must be rederived from the region, not exchanged. Both boundary curves are rewritten as functions of $y$: $y=x^2$ gives $x=\sqrt{y}$, and $y=x$ gives $x=y$. As $y$ ranges over $[0,1]$, the horizontal strip at height $y$ runs from the line to the parabola, and on $[0,1]$ we have $y\le \sqrt{y}$, so

Inner integration, holding $y$ fixed:

Outer integration:

The same $3/20$, as it must be.

Why the limits had to be rederived. Over a rectangle the limits are constants, the region is a product of two intervals, and describing it by $x$ then $y$ or by $y$ then $x$ names the same set, so the written limits can simply be swapped. $T$ is not a product. Its inner limits are functions of the outer variable, and the two orders are genuinely different decompositions of the region: part (b) sweeps it in vertical strips between a parabola and a line, part (c) in horizontal strips between a line and a parabola. Exchanging the written limits would have produced ${\int }_0^1{\int }_{x^2}^x(x+y)dxdy$, whose inner limits still mention $x$ while integrating with respect to $x$, which is not a well-formed expression, and reading them as constants describes a rectangle $T$ does not fill.

Note what did not change: the value. Fubini guarantees the orders agree wherever the integral exists, and over $T$ with a continuous integrand it does. What the non-rectangular region costs is not the agreement but the convenience: the limits are work in both directions, and for some regions one order needs the region split into pieces while the other does not.