Double integrals over rectangles — Fubini's theorem: Exercises

Recall the three-step ritual from the parent, our compass for every problem:

Level 1 — Recognition

These test one thing: can you read an iterated integral and see which variable is frozen?

Exercise 1.1

In the integral $\int_{0}^{2} \int_{1}^{4} (x^{2} y) d y d x$ , which variable is integrated first, and which is treated as a constant during that first step?

Recall Solution 1.1

The inner differential is $d y$ , so we integrate over $y$ first. During that inner step, $x$ is frozen — treated exactly like a fixed number (say $7$ ). Concretely the inner step is $\int_{1}^{4} x^{2} y d y = x^{2} [\frac{y ^{2}}{2}]_{1}^{4} = x^{2} \cdot \frac{16 - 1}{2} = \frac{15}{2} x^{2},$ where $x^{2}$ slid out because it is constant with respect to $y$ . Answer: $y$ first; $x$ constant.

Exercise 1.2

Is $f (x, y) = x^{3} e^{2 y}$ separable (of the form $g (x) h (y)$ )? If so, name $g$ and $h$ .

Recall Solution 1.2

Yes. It is a product of a function of $x$ alone and a function of $y$ alone: $g (x) = x^{3}, h (y) = e^{2 y} .$ Because it separates, over a rectangle we may split the double integral into a product of two 1-D integrals (parent shortcut). Contrast with $x^{3} e^{2 x y}$ : the exponent mixes $x$ and $y$ , so it is not separable.

Level 2 — Application

Turn the crank with the ritual. Check your arithmetic.

Exercise 2.1

Compute $\iint_{R} (3 x^{2} + 2 y) d A$ , where $R = [0, 1] \times [0, 2]$ .

Recall Solution 2.1

FREEZE $x$ , FILL over $y$ (inner): $\int_{0}^{2} (3 x^{2} + 2 y) d y = [3 x^{2} y + y^{2}]_{0}^{2} = 6 x^{2} + 4.$ Here $3 x^{2}$ is constant in $y$ , giving $3 x^{2} y$ ; the antiderivative of $2 y$ is $y^{2}$ . FILE into the outer $x$ -integral: $\int_{0}^{1} (6 x^{2} + 4) d x = [2 x^{3} + 4 x]_{0}^{1} = 2 + 4 = 6.$ Answer: $6$ .

Exercise 2.2

Compute $\iint_{R} x e^{y} d A$ , $R = [1, 3] \times [0, 1]$ , using the separable shortcut, then confirm it equals the iterated integral.

Recall Solution 2.2

$f = x \cdot e^{y}$ is separable, so $(\int_{1}^{3} x d x) (\int_{0}^{1} e^{y} d y) = [\frac{x ^{2}}{2}]_{1}^{3} \cdot [e^{y}]_{0}^{1} = \frac{9 - 1}{2} \cdot (e - 1) = 4 (e - 1) .$ Confirm by iterating (inner over $y$ , $x$ frozen): $\int_{1}^{3} x (\int_{0}^{1} e^{y} d y) d x = \int_{1}^{3} x (e - 1) d x = (e - 1) \cdot \frac{9 - 1}{2} = 4 (e - 1) . ✓$ Answer: $4 (e - 1) \approx 6.873$ .

Exercise 2.3

Compute $\int_{0}^{π /2} \int_{0}^{1} y sin x d y d x$ .

Recall Solution 2.3

Separable with $g (x) = sin x$ , $h (y) = y$ : $(\int_{0}^{π /2} sin x d x) (\int_{0}^{1} y d y) = [- cos x]_{0}^{π /2} \cdot [\frac{y ^{2}}{2}]_{0}^{1} = (0 - (- 1)) \cdot \frac{1}{2} = \frac{1}{2} .$ Answer: $\frac{1}{2}$ .

Level 3 — Analysis

Now the order of integration is a choice. Pick the one that avoids extra work.

Exercise 3.1

Compute $\iint_{R} y e^{x y} d A$ , $R = [0, 1] \times [0, 1]$ . Choose the order that avoids integration by parts.

Recall Solution 3.1

Notice $\frac{\partial}{\partial x} e^{x y} = y e^{x y}$ — exactly the integrand. So integrating over $x$ first (with $y$ frozen) is trivial: $\int_{0}^{1} y e^{x y} d x = [e^{x y}]_{x = 0}^{x = 1} = e^{y} - 1.$ FILE into the $y$ -integral: $\int_{0}^{1} (e^{y} - 1) d y = [e^{y} - y]_{0}^{1} = (e - 1) - (1 - 0) = e - 2.$ Had we integrated over $y$ first, $\int y e^{x y} d y$ needs integration by parts — same answer, more labour. Answer: $e - 2 \approx 0.718$ .

Exercise 3.2

Compute $\iint_{R} x cos (x y) d A$ , $R = [0, π /2] \times [0, 1]$ (here $x \in [0, π /2]$ , $y \in [0, 1]$ ). Pick the clean order.

Recall Solution 3.2

Because $\frac{\partial}{\partial y} sin (x y) = x cos (x y)$ , integrating over $y$ first (freeze $x$ ) undoes the derivative directly: $\int_{0}^{1} x cos (x y) d y = [sin (x y)]_{y = 0}^{y = 1} = sin x - 0 = sin x .$ FILE: $\int_{0}^{π /2} sin x d x = [- cos x]_{0}^{π /2} = 0 - (- 1) = 1.$ Answer: $1$ . (Integrating over $x$ first would force parts.)

Level 4 — Synthesis

Combine slicing, separability, and geometry.

Exercise 4.1

Find the volume of the solid under $z = 4 - x^{2} - y^{2}$ over $R = [- 1, 1] \times [- 1, 1]$ . (See the figure.)

Figure — Double integrals over rectangles — Fubini's theorem

Recall Solution 4.1

Volume $= \iint_{R} (4 - x^{2} - y^{2}) d A$ . Split the sum: $\iint_{R} 4 d A - \iint_{R} x^{2} d A - \iint_{R} y^{2} d A .$ First piece: $4 \cdot area (R) = 4 \cdot (2 \cdot 2) = 16.$ Second piece (separable, $x^{2} \cdot 1$ ): $(\int_{- 1}^{1} x^{2} d x) (\int_{- 1}^{1} 1 d y) = \frac{2}{3} \cdot 2 = \frac{4}{3} .$ Third piece is identical by symmetry: $\frac{4}{3} .$ $V = 16 - \frac{4}{3} - \frac{4}{3} = 16 - \frac{8}{3} = \frac{40}{3} .$ The surface stays above the $x y$ -plane on $R$ (minimum value at corners: $4 - 1 - 1 = 2 > 0$ ), so this is a true positive volume — no cancellation. Answer: $\frac{40}{3} \approx 13.33$ .

Exercise 4.2

Compute the average value of $f (x, y) = x y$ over $R = [0, 2] \times [0, 3]$ . (Average value $= \frac{1}{area ( R )} \iint_{R} f d A$ .)

Recall Solution 4.2

$area (R) = 2 \cdot 3 = 6.$ The integral (separable): $\iint_{R} x y d A = (\int_{0}^{2} x d x) (\int_{0}^{3} y d y) = \frac{4}{2} \cdot \frac{9}{2} = 2 \cdot \frac{9}{2} = 9.$ Average value: $\frac{1}{6} \cdot 9 = \frac{3}{2} .$ Answer: $\frac{3}{2}$ .

Level 5 — Mastery

A problem designed to catch you. Slow down.

Exercise 5.1

Compute $\int_{0}^{1} \int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d y d x$ and $\int_{0}^{1} \int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d x d y$ . Do the two orders agree? Explain what this shows about Fubini.

Recall Solution 5.1

Key antiderivative fact: $\frac{\partial}{\partial y} (\frac{- y}{x ^{2} + y ^{2}}) = \frac{- ( x ^{2} + y ^{2} ) + y \cdot 2 y}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = - \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$ So $\frac{\partial}{\partial y} (\frac{y}{x ^{2} + y ^{2}}) = \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ , exactly our integrand.

Order $d y$ first (freeze $x > 0$ ): $\int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d y = [\frac{y}{x ^{2} + y ^{2}}]_{0}^{1} = \frac{1}{x ^{2} + 1} .$ Then $\int_{0}^{1} \frac{1}{x ^{2} + 1} d x = [arctan x]_{0}^{1} = \frac{π}{4} .$

Order $d x$ first: the integrand is antisymmetric under swapping $x \leftrightarrow y$ (it flips sign). By the same antiderivative in $x$ , $\int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d x = [\frac{- x}{x ^{2} + y ^{2}}]_{0}^{1} = \frac{- 1}{1 + y ^{2}},$ then $\int_{0}^{1} \frac{- 1}{1 + y ^{2}} d y = - \frac{π}{4} .$

The two orders give $+ \frac{π}{4}$ and $- \frac{π}{4}$ — they DISAGREE.

Why Fubini does not save us here: the integrand blows up near the corner $(0, 0)$ ; in fact $\iint_{R} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d A = \infty$ . Fubini requires $f$ integrable (continuous on the closed rectangle, or $\iint ∣ f ∣ < \infty$ ). This $f$ is unbounded and not absolutely integrable, so equality of orders is not guaranteed — and indeed fails. This is the textbook example that the parent's third mistake callout warned about. Answer: $+ \frac{π}{4}$ (order $d y d x$ ), $- \frac{π}{4}$ (order $d x d y$ ); Fubini fails because $f$ is not integrable near $(0, 0)$ .

Connections

Riemann sums — the limit these integrals secretly are.
Volume by slicing (single-variable) — the engine behind Exercise 4.1.
Change of order of integration — Level 3 is its warm-up on rectangles.
Double integrals over general regions — where swapping limits gets subtle.
Fubini–Tonelli theorem — the exact hypothesis Exercise 5.1 violates.
Triple integrals — same slicing idea, one dimension deeper.

Flashcards

\int_{a}^{b} \int_{c}^{d} f d y d x

, which integral is done first?

The inner one, over

y

(the differential nearest the integrand).

Average value of

f

over

R

\frac{1}{area ( R )} \iint_{R} f d A

\iint_{[0, 1] \times [0, 1]} y e^{x y} d A = ?

e - 2

Why can the two orders of

\int\int \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}

differ?

The integrand is unbounded / not absolutely integrable near

(0, 0)

, so Fubini's hypothesis fails.