Back to the on-screen lesson ·
The double integral as a limit of sums and as a signed volume, the iterated integral that computes it, Fubini's theorem and what it needs, separable integrands, and the average value of a function over a rectangle.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to read a double integral over a rectangle as a signed volume, write it as an iterated integral with the four limits in the right places, evaluate it in either order, use the factoring shortcut when the integrand is separable, and find the average value of a function over a rectangle. You will also be able to say what Fubini's theorem assumes.
The definite integral of one variable as a limit of Riemann sums, and partial differentiation, where one variable is held constant while the other moves. This lesson uses both: the sum is the definition, and holding a variable constant is how every double integral is actually evaluated.
A double integral $\int\int_R f\,dA$ adds up $f$ times a small piece of area over a region $R$. An iterated integral is two ordinary integrals nested one inside the other, and it is how a double integral is computed. The inner integral is the one written next to the function; the outer is the one written outside it. Fubini's theorem is the statement that the two are equal and that the order may be swapped. The area element $dA$ is the small piece of area, which over a rectangle is $dx\,dy$.
Cut the rectangle $R = [a,b] \times [c,d]$ into $mn$ small rectangles of area $\Delta A$, pick a point in each, and add:
$$\int\!\!\int_R f(x,y)\,dA = \lim_{m,n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^, y_{ij}^)\,\Delta A.$$
When $f \ge 0$ this is the volume under the graph and above $R$. When $f$ takes both signs it is a signed volume, with the part below the plane counting negatively — exactly as a one-variable integral is a signed area.
How it is actually computed. Nobody evaluates that limit. Instead:
$$\int\!\!\int_R f\,dA = \int_a^b\!\!\left(\int_c^d f(x,y)\,dy\right)dx = \int_c^d\!\!\left(\int_a^b f(x,y)\,dx\right)dy.$$
The inner integral treats the outer variable as a constant, so it is an ordinary one-variable integral; substituting its limits leaves a function of the outer variable, and integrating that finishes the job. The differentials read from the inside out: $dy\,dx$ means $y$ first.
Fubini's theorem says the two orders agree, and it needs a hypothesis: $f$ continuous on $R$ is enough (bounded with only mild discontinuity is enough too). Over a rectangle this is almost never in doubt, which is why the theorem feels like bookkeeping here; on an unbounded region, or with an unbounded integrand, the two orders genuinely can disagree.
Separable integrands. If $f(x,y) = g(x)h(y)$ and the region is a rectangle, then
$$\int\!\!\int_R g(x)h(y)\,dA = \left(\int_a^b g\,dx\right)\left(\int_c^d h\,dy\right).$$
Both conditions are needed. The moment the region stops being a rectangle the factorisation fails, because the inner limits start depending on the outer variable.
Average value. The average of $f$ over $R$ is the total divided by the size of the region:
$$f_{\text{avg}} = \frac{1}{\text{area}(R)}\int\!\!\int_R f\,dA.$$
Another way: picture
Stand the graph of $f$ over the rectangle like a lid on a box. The inner integral slices that solid with a knife held at one value of $x$ and reports the area of the slice; the outer integral slides the knife from $x = a$ to $x = b$ and adds the slices up. Slicing the other way — knife held at one $y$ — gives different slices and the same total, which is all Fubini's theorem says.
Another way: steps
To evaluate a double integral over a rectangle:
The notation $\int_a^b\!\int_c^d f\,dy\,dx$ packs four numbers and two variables into one line, and reading it wrongly is the commonest early error. The rule is that the differentials and the limits are both read inside out: $dy$ is the inner differential, so $c$ and $d$ are the $y$ limits, and $dx$ is the outer one, so $a$ and $b$ are the $x$ limits.
A reliable way to keep it straight is to write the brackets in:
$$\int_a^b\left(\int_c^d f(x,y)\,dy\right)dx.$$
Now the inner integral is visibly a complete object that produces a function of $x$, and the outer integral is visibly an ordinary integral of that function. Once the pattern is second nature the brackets come off again, but there is no prize for dropping them early.
Over a rectangle the four limits are all constants, so the order of the two differentials changes nothing except which one-variable integral you meet first — and that can still matter enormously in practice, because one order may be elementary and the other may not be.
Matching the limits to the wrong differential. $\int_0^2\!\int_0^3 f\,dy\,dx$ has $y$ running to $3$, not $x$. Writing the brackets in for a few weeks fixes this permanently.
Letting the outer variable escape. After the inner integral has been evaluated and its limits substituted, no $y$ may remain. A leftover $y$ means a limit was not substituted, and the final answer will be a function when it should be a number.
Factoring a non-separable integrand. $\int\int (x + y)\,dA$ does not split into a product; only a genuine product $g(x)h(y)$ does.
Dividing by the wrong thing to get an average. The denominator is the area of the region, not the length of a side.
Reading a zero integral as a zero function. An orientation, an order of integration and an order of factors are part of the answer, not part of the handwriting. Reversing a curve flips the sign of the work along it, swapping the factors of a cross product flips the vector, and exchanging the two limits of an inner integral changes what region was integrated over. Say which one you chose, every time.
It is one limit of one sum, and the iterated integral is a theorem about how to compute it rather than its definition. That distinction is invisible over a rectangle with a continuous integrand, which is exactly why it is worth naming here: it is the last moment at which everything works and nothing needs checking.
The moment the region stops being a rectangle, the inner limits become functions of the outer variable and the two orders stop looking alike. The moment the integrand is unbounded, or the region is, Fubini's theorem acquires teeth and the two orders can give genuinely different numbers. Treating the iterated form as the definition leaves you with no way to even ask whether that has happened.
Evaluate $\int_0^2\!\int_0^3 (x + 2y)\,dy\,dx$. Inner first, with $x$ frozen: $\int_0^3 (x + 2y)\,dy = \left[xy + y^2\right]_0^3 = 3x + 9$.
The inner integral leaves a function of $x$ alone.
Now the outer: $\int_0^2 (3x + 9)\,dx = \left[\tfrac{3x^2}{2} + 9x\right]_0^2 = 6 + 18 = 24$.
An ordinary one-variable integral finishes it.
The other order: $\int_0^2 (x + 2y)\,dx = \left[\tfrac{x^2}{2} + 2xy\right]_0^2 = 2 + 4y$, and $\int_0^3 (2 + 4y)\,dy = 6 + 18 = 24$. Same answer, different intermediate function.
Fubini's theorem, checked rather than quoted.
Evaluate $\int_0^1\!\int_0^{\pi} x\sin y\,dy\,dx$. The integrand is a function of $x$ times a function of $y$, and the region is a rectangle.
Both conditions for factoring hold.
So the answer is $\left(\int_0^1 x\,dx\right)\left(\int_0^{\pi} \sin y\,dy\right) = \tfrac12 \times 2 = 1$.
Two single integrals, multiplied.
Doing it the long way gives $\int_0^1 x\left[-\cos y\right]_0^{\pi} dx = \int_0^1 2x\,dx = 1$ — the same, and the factoring is simply that calculation with the constant pulled out early.
The shortcut is not a different method, only a tidier one.
The integral is $\left(\int_0^3 x^2\,dx\right)\left(\int_0^2 y\,dy\right) = 9 \times 2 = 18$, since the integrand is separable and the region is a rectangle.
Factor first; it is much less writing.
The area of the rectangle is $3 \times 2 = 6$.
So the average is $18/6 = 3$. Sanity check: $f$ runs from $0$ up to $9 \times 2 = 18$ on this rectangle, and $3$ sits low in that range — which is right, because $x^2y$ is small over most of the rectangle and large only in one corner.
Put the stages of evaluating $\int_{0}^{2}\!\int_{0}^{3} f(x,y)\,dy\,dx$ in order.
Number the steps in order (write the number in the box):
The rectangle $R$ has $-3 \le x \le 9$ and $-5 \le y \le 3$. Give the four limits of $\int\int_R f\,dy\,dx$.
| Value | |
|---|---|
| The inner lower limit | |
| The inner upper limit | |
| The outer lower limit | |
| The outer upper limit |
Evaluate $\displaystyle\int_{0}^{5}\!\int_{0}^{6} xy\,dy\,dx$.
Answer:
What is the average value of $f(x,y) = xy$ over the rectangle $0 \le x \le 4$, $0 \le y \le 4$?
Answer:
Put the stages of evaluating $\int_{0}^{6}\!\int_{0}^{5} f(x,y)\,dy\,dx$ in order.
Number the steps in order (write the number in the box):
On the rectangle $-3 \le x \le 3$, $0 \le y \le 1$, a continuous $f$ is never zero, yet $\int\int_R f\,dA = 0$. Is that possible?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The rectangle $R$ has $0 \le x \le 4$ and $0 \le y \le 8$. Give the four limits of $\int\int_R f\,dy\,dx$.
| Value | |
|---|---|
| The inner lower limit | |
| The inner upper limit | |
| The outer lower limit | |
| The outer upper limit |
You can turn a double integral over a rectangle into an iterated integral, evaluate it inside out, and say what its sign means. Next: regions that are not rectangles, where the inner limits stop being constants.
10. Your turn: the average value of $f(x,y) = x^2 y$ over $[0,3] \times [0,2]$, step 3