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The Order of Integration and Fubini’s Theorem

April 12th, 2009 |

Author: Merrin

Any double integral can be represented as an integral over an area

$\displaystyle\iint \nolimits _{R}f(x,y)dxdy$

There is no rule specifying the order of integration with respect to x or y. This integral is just as good.

$\displaystyle \iint \nolimits _{R}f(x,y)dydx$

It should come as no surprise that, one method of manipulating multiple integrals is to change the order of integration.

There are a variety of reasons why interchanging the order of integration would be useful. Sometimes the integral can only be computed with elementary functions in one particular order. Along the same lines, it may be possible to identify a good substitution with a particular order of integration. If integration in one order works but the other doesn’t this means a new integral identity has been derived. These results can be tabulated. Interchanging the order of integration over rectangular regions is always possible and is known as Fubini’s theorem.

Theorem 1. Fubini’s Theorem. When integrating over the rectangular region $latex R:\{a\le x\le b,c\le y\le d\} $ the order of integration can be reversed and the limits of integration unmodified.

$\displaystyle \int _{a}^{b}\left[\displaystyle \int _{c}^{d}f(x,y)dy \right] dx=\displaystyle \int _{c}^{d}\left[\displaystyle \int _{a}^{b}f(x,y)dx \right] dy$

Example 1. Verify Fubini’s theorem for a simple multiple integral.

$I=\displaystyle \int _{0}^{1}\displaystyle \int _{0}^{2}(x+2y^{2} )dydx$

Solution 1. We will try the integral both ways, first integrating y then x, or second integrating x then y.

${\displaystyle \int _{0}^{1}\displaystyle \int _{0}^{2}(x+2y^{2} )dydx }={\displaystyle \int _{0}^{1}(xy+\cfrac{2}{3} y^{3} \left. )\right|_{y=0}^{y=2}dx =\displaystyle \int _{0}^{1}(2x+\cfrac{16}{3} ) dx} \\={(x^{2} + \cfrac{16}{3} x\left. )\right|_{x=0}^{x=1} =\cfrac{19}{3} } \\ {\displaystyle \int _{0}^{2}\displaystyle \int _{0}^{1}(x+2y^{2} )dxdy } = {\displaystyle \int _{0}^{2}(\cfrac{x^{2} }{2} +2y^{2} x\left. )\right|_{x=0}^{x=1} dy =\displaystyle \int _{0}^{2}(\cfrac{1}{2} +2y^{2} )dy } \\=( \cfrac{1}{2}y+\cfrac{2}{3}y^3) \left|_{y=0}^{y=2}=1+16/3=19/3 \right.$

Fubini’s theorem is verified for this example. This particular integral is a sum of separable integrals and not very impressive. If I wanted to calculate the integral as fast as possible then I would write.

$I=\displaystyle \int _{0}^{1}xdx \displaystyle \int _{0}^{2}dy +2\displaystyle \int _{0}^{1}dx \displaystyle \int _{0}^{2}y^{2} dy =\cfrac{1}{2} 2+2(1)(\cfrac{8}{3} )=\cfrac{19}{3}$

Example 2. Attempt to verify Fubini’s theorem and which order is it easier to evaluate the integral.

$I=\displaystyle \int _{0}^{1}\displaystyle \int _{0}^{1}2xye^{xy^{2} } dydx$

Solution 2. We see that we can make a u substitution when the integral is first done in terms of y.

${I}={\displaystyle \int _{0}^{1}\displaystyle \int _{0}^{1}2xye^{xy^{2} } dydx =\displaystyle \int _{0}^{1}\displaystyle \int _{0}^{x}e^{u} du dx=\displaystyle \int _{0}^{1}(e^{x} -1)dx } \\=e^1-e^0-(1-0)=e-2$

If we had integrated in terms of x first then we would have gotten

$I=\displaystyle \int _{0}^{1}\displaystyle \int _{0}^{1}2xye^{xy^{2} } dxdy =\displaystyle \int _{0}^{1}\cfrac{2}{y^{3} } (1-e^{y^{2} } +y^{2} e^{y^{2} } )dy$

Now the last integral we end up with after integrating by parts is by no means trivial.

$\int _{0}^{1}\cfrac{2}{y^{3} } (1-e^{y^{2} } +y^{2} e^{y^{2} } )dy \quad \, u=y^{2} du=2ydy y=\sqrt{u} } \\ {\displaystyle \int _{0}^{1}\cfrac{2\sqrt{u} }{u^{2} } (1-e^{u} +ue^{u} )\cfrac{du}{2\sqrt{u} } =\displaystyle \int _{0}^{1}\cfrac{1}{u^{2} } (1-e^{u} +ue^{u} )du$

I’m not quite sure how to proceed in evaluating this improper integral, but Maple evaluates it as \textit{e}-2 which is what we got for the other integral. Solving this integral requires a new method. The main point is that Fubini’s theorem appears to hold, even for difficult integrals

So we couldn’t solve it in this order, but on the flip side, interchanging the order of integration can give us a result for a really difficult integral because Fubini’s theorem applies. Such results can be kept in a table of

integration. One thing is for sure, the order integration matters in practice.

Example 3. Calculate the following integral which is over a rectangular region.

$\displaystyle \int _{0}^{1}\displaystyle \int _{a}^{b}x^{y} dydx =\displaystyle \int _{a}^{b}\displaystyle \int _{0}^{1}x^{y} dxdy$

Solution 3. We can use Fubini’s theorem in evaluating the integral. We can choose either order of the integration. In the order dxdy, we find an answer to the integral. In the order dydx, we are left with an integral that we do not know how to evaluate by our repertoire of methods. Equating the two integrals by Fubini’s theorem we derive a result for a curious looking integral.

$\int _{a}^{b}\displaystyle \int _{0}^{1}x^{y} dxdy } = {\displaystyle \int _{a}^{b}\left(\cfrac{x^{y} }{1+y} \right) _{x=0}^{x=1} =\displaystyle \int _{a}^{b}\cfrac{dy}{1+y} =\ln \left(\cfrac{1+b}{1+a} \right)} \\ {\displaystyle \int _{0}^{1}\displaystyle \int _{a}^{b}x^{y} dydx } = {\displaystyle \int _{0}^{1}\cfrac{x^{b} -x^{a} }{\ln x} dx=?} \\ {\displaystyle \int _{0}^{1}\cfrac{x^{b} -x^{a} }{\ln x} dx} = {\ln \left(\cfrac{1+b}{1+a} \right)$

The integrals we examined up to now have dealt with constant limits of integration. Now we will introduce variable limits of integration. For a double integral integration over a simple region can be represented as

$\displaystyle \int _{x=a}^{x=b}\displaystyle \int _{y=g(x)}^{y=h(x)}f(x,y)dydx$

Or written for short,

$\displaystyle \int _{a}^{b}\displaystyle \int _{g(x)}^{h(x)}f(x,y)dydx$

Here g(x) and h(x) represent the boundary of a region. Sometimes it is possible to interchange the order of integration and the limits enclose the entire region. Other times the integral over the total area has to be broken into pieces.

$\displaystyle \int _{c}^{d}\displaystyle \int _{r(x)}^{s(x)}f(x,y)dxdy$

Draw a picture of the region to calculate the new variable limits. In general, this is a bit messy, but there is one special region which is easy to reverse, a rectangular region.

The most general triple integral over a volume is

$\displaystyle \int _{a}^{b}\displaystyle \int _{m(x)}^{n(x)}\displaystyle \int _{g(x,y)}^{h(x,y)}f(x,y,z)dzdydx$

After successive integrations, dummy variables disappear. In principle, b and a can be functions themselves but they shouldn’t have any of the dummy variables x, y, or z in them.

Example 4. Write out an integral for $latex f(x,y)$ over a circle of radius R centered at the origin. Reverse the order of integration and write the new integral.

Solution 4. A circle has an upper and lower branch enclosing its region.

${x^{2} +y^{2} =R^{2} } \\ {y=\pm \sqrt{R^{2} -x^{2} } }$

The integral of $latex f(x,y)$ over the circle can be written as.

$\displaystyle \int _{-R}^{R}\displaystyle \int _{-\sqrt{R^{2} -x^{2} } }^{\sqrt{R^{2} -x^{2} } }f(x,y)dydx$

Since a circle is a simple region that can be reversed without splitting it into pieces we can write the interchanged integral as

$\displaystyle \int _{-R}^{R}\displaystyle \int _{-\sqrt{R^{2} -y^{2} } }^{\sqrt{R^{2} -y^{2} } }f(x,y)dxdy$

Example 5. Show how to define and interchange the limits of integration when integrating over a triangle with vertices situated at (0, 0) to (0, 1) to (1, 1).

Solution 5. When we integrate with respect to y and then with respect to x, the top of the triangle is parameterized by a sloped line and the bottom of the triangle is the x axis.

$y_{2} =x\quad \quad \quad y_{1} =0$

This is the first parameterization of the integral. In the second parameterization of the integral, the left and right curves are

$x_{2} =1\quad \quad \, x_{1} =y$

We can write an integral over this region two ways

$\displaystyle \int _{x=0}^{x=1}\displaystyle \int _{y=0}^{y=x}f(x,y)dydx=\displaystyle \int _{y=0}^{y=1}\displaystyle \int _{x=y}^{x=1}f(x,y)dxdy$

Example 6. Calculate the area of the triangle in the previous two parameterizations and show that they are equal.

Solution 6. For finding the area $latex f(x,y)=1$. We have

${I} = {\displaystyle \int _{x=0}^{x=1}\displaystyle \int _{y=0}^{y=x}dydx =\displaystyle \int _{x=0}^{x=1}(x-0)dx =\cfrac{1}{2} } \\ {I} = {\displaystyle \int _{y=0}^{y=1}\displaystyle \int _{x=y}^{x=1}dxdy =\displaystyle \int _{y=0}^{y=1}(1-y)dy =\left. (y-y^{2} /2)\right|_{0}^{1} =1-\cfrac{1}{2} =\cfrac{1}{2} }$

The area of the triangle is 1/2 in both cases.

Example 7. Integrate the area of the region bounded by the two functions given and then reverse the order of integration and get the same result.

$y=\sqrt{x} \qquad y=x^{2} \qquad x\in [0,1]$

Solution 7.

$\displaystyle \int _{0}^{1}\displaystyle \int _{x^{2} }^{\sqrt{x} }dydx=\displaystyle \int _{0}^{1}(\sqrt{x} -x^{2} )dx=(\cfrac{2}{3} x^{3/2} -\cfrac{1}{3} x^{3} \left. )\right|_{0}^{1} =\cfrac{2}{3} -\cfrac{1}{3} =\cfrac{1}{3}$

Reversing the equations for x and y allows the two new bounds to be found. This region is symmetric so the two bounds are.

$x=y^{2} \qquad x=\sqrt{y} , y\in [0,1]$

The integral for the area is the same with the symbols reversed.

$\displaystyle \int _{0}^{1}\displaystyle \int _{y^{2} }^{\sqrt{y} }dxdy=\displaystyle \int _{0}^{1}(\sqrt{y} -y^{2} )dy=(\cfrac{2}{3} y^{3/2} -\cfrac{1}{3} y^{3} \left. )\right|_{0}^{1} =\cfrac{2}{3} -\cfrac{1}{3} =\cfrac{1}{3}$

Example 8. The way in which a region is specified may dictate which order to setup the integration for. Suppose our region is bounded by the hyperbola

$x^{2} -y^{2} =1\quad x\in [-\sqrt{2} ,\sqrt{2} ]\quad y\in [-1,1]$

Evaluate the area of the region.

Solution 8. Notice this area is hourglass shaped. We can’t first integrate with respect to x because the boundary of the region does not pass a vertical line test. Say out near x = 1.1 the line x = 1.1 crosses the hyperbola twice and crosses the lines $latex y=\pm \sqrt{3} $ also. Since it crosses four times you can’t write this as an integral. We can integrate in one piece with respect to y first.

$\displaystyle\iint \nolimits _{A}dydx$

The integral looks like this

$\int _{-1}^{1}\displaystyle \int _{-\sqrt{1+y^{2} } }^{\sqrt{1+y^{2} } }dxdy =\displaystyle \int _{-1}^{1}2\sqrt{1+x^{2} } dx =\left. \left(x\sqrt{x^{2} +1} +\text{arcsinh }\,x\right)\right|_{-1}^{1} } \\ {=2\sqrt{2} +\text{arcsinh}\,(1)-\text{arcsinh}\,(-1)} \\ {=2\sqrt{2} +2\ln (1+\sqrt{2} )$

I leave integration details as an exercise. This example illustrates when with variable limits, integration in one step may only be possible with a certain order of integration.

We have studied some practical matters of evaluating multiple integrals. If the limits of integration are constant then the order of integration can be reversed. This is known as Fubini’s theorem. Variable limits of integration correspond to more general regions. Here also the order of integration can be interchanged, but for certain regions the double integral must be broken into pieces. When an integral can be evaluated in one order, but not the other this can be regarded as a new integration technique called integration under the integral sign.

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One Response to “The Order of Integration and Fubini’s Theorem”

someone_who_saw_an_error says:

June 4, 2010 at 10:30 am

“When integrating over the rectangular region $latex R:\{a\le x\le b,c\le y\le d\} $ the order of integration can be reversed and the limits of integration unmodified.”

This is wrong, see http://en.wikipedia.org/wiki/Fubini%27s_theorem#Evaluation.

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The Order of Integration and Fubini’s Theorem

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