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Section 1.2 Iterated integrals and Fubini's theorem

As we saw in Example 1.1.8, computing double integrals directly using the limit definition (Definition 1.1.3) is rather onerous. Fubini's theorem provides us a more convenient means of integrating using iterated integrals.

Definition 1.2.1. Doubly iterated integral over rectangle.

Let \(f\) be defined on the rectangle \(R=[a,b]\times [c,d]\) and suppose the single-variable functions

\begin{align*} g(x) \amp =\int_c^d f(x,y)\, dy \amp h(y) \amp =\int_a^b f(x,y)\, dx \end{align*}

are integrable on \([a,b]\) and \([c,d]\text{,}\) respectively. We define the two iterated integrals \(\int_a^b\int_c^d f(x,y)\, dx\, dy\) and \(\int_c^d\int_a^bf(x,y)\, dx\, dy\) as follows:

\begin{align} \int_a^b\int_c^d f(x,y)\, dy\, dx\amp =\int_a^b g(x)\, dx=\int_a^b\left(\int_c^df(x,y)\, dy\right)dx \tag{1.2.1}\\ \int_c^d\int_a^b f(x,y)\, dx\, dy\amp =\int_c^d h(y)\, dy=\int_c^d\left(\int_a^bf(x,y)\, dx\right)dy \text{.}\tag{1.2.2} \end{align}

Remark 1.2.3.

Observe that Fubini's theorem tells us two interesting facts: (1) the two iterated integrals are equal in value, and (2) in fact both are equal to the double integral of \(f\) over \(R\text{.}\)

Remark 1.2.4.

Along similar lines, Fubini's theorem gives us two potential ways of computing an integral: integrate first with respect to \(x\text{,}\) and then with respect to \(y\text{;}\) or integrate first with respect to \(y\text{,}\) and then with respect to \(x\text{.}\) Oftentimes one of these choices will lead to easier integration techniques than the other. Choose wisely!

Example 1.2.5. Volume below graph revisited.

Let \(f(x,y)=x+y^2\text{,}\) and let \(\mathcal{S}\) be the region lying above the rectangle \(R=[0,1]\times[0,1]\) and below the graph of \(f\text{.}\) Compute \(\operatorname{vol} \mathcal{S}\) using Fubini's theorem.

Solution.

We have

\begin{align*} \operatorname{vol} S \amp =\iint\limits_Rf(x,y)\, dA \amp (\text{def. of vol})\\ \amp=\int_0^1\int_0^1f(x,y)\, dx\, dy \amp (\text{Fubini's thm.}) \\ \amp =\int_0^1\left(\int_0^1 x+y^2\, dx\right) dy \\ \amp=\int_0^1\left(\left(\frac{1}{2}x^2+xy^2\right)\Bigr\vert_{x=0}^{x=1}\right)dy \\ \amp = \int_0^1\frac{1}{2}+y^2\, dy\\ \amp =\left(\frac{1}{2}y+\frac{1}{3}y^3\right)\Bigr\vert_0^1\\ \amp =\frac{5}{6}\text{.} \end{align*}

Phew, this is indeed the same answer we got in Example 1.1.8! For kicks try the other choice of iterated integral:

\begin{equation*} \int_0^1\int_0^1f(x,y)\, dy\, dx\text{.} \end{equation*}

Example 1.2.6. Fubini examples.

Compute the following integrals using Fubini's theorem.

  1. \(\iint\limits_R x\cos(xy)\, dA\text{,}\) \(R=[1,2]\times [0,\pi]\)

    Solution.

    By Fubini's theorem we have

    \begin{align*} \iint\limits_R x\cos(xy)\, dA \amp=\int_1^2\int_0^\pi x\cos(xy)\, dy\, dx \\ \amp=\int_1^2(\sin(xy))\Bigr\vert_{y=0}^{y=\pi}\, dx\\ \amp = \int_1^2 \sin\pi x \\ \amp = -\frac{1}{\pi}\cos\pi x\Bigr\vert_1^2\\ \amp = -\frac{1}{\pi}(\cos 2\pi-\cos\pi)\\ \amp =-\frac{2}{\pi}\text{.} \end{align*}

    Note that if we integrated with respect to \(x\) first we would have needed to use integration by parts.

  2. \(\iint\limits_R \frac{x}{1+xy}\, dA\text{,}\) \(R=[0,1]\times [1,2]\)

    Solution.

    By Fubini's theorem we have

    \begin{align*} \iint\limits_R \frac{x}{1+xy}\, dA \amp=\int_0^1\int_1^2 \frac{x}{1+xy}\, dy\, dx \\ \amp=\int_0^1\ln(1+xy)\Bigr\vert_1^2 dx\\ \amp = \int_0^1 \ln(1+2x)-\ln(1+x)\, dx \\ \amp = \left(\frac{1}{2}((1+2x)\ln(1+2x)-(1+2x))-((1+x)\ln(1+x)-(1+x))\right)\Bigr\vert_0^1 \\ \amp = \frac{3}{2}\ln 3-2\ln 2\text{.} \end{align*}

    The penultimate equality uses the indefinite integral formula \(\int \ln u\, du=u\ln u-u\text{.}\) See what happens when you first integrate with respect to \(x\text{.}\)