Change order of integration examples

Given a double integral

$\displaystyle \iint_{D}^{}$f (x, y) dA    

of a function f (x, y) over a region A, you may be able to write it as two different iterated integrals. You can integrate with respect to x first, or you can iterate with respect to y first. If you integrate with respect to x first, you will obtain an integral that looks something like

$\displaystyle \iint_{D}^{}$f (x, y) dA = $\displaystyle \int_{{\boxempty}}^{{\boxempty}}$$\displaystyle \left(\vphantom{\int_{\boxempty}^{\boxempty} f(x,y) dx }\right.$$\displaystyle \int_{{\boxempty}}^{{\boxempty}}$f (x, y) dx$\displaystyle \left.\vphantom{\int_{\boxempty}^{\boxempty} f(x,y) dx }\right)$dy, (1)

and if you integrate with respect to y first, you will obtain an integral that looks something like

$\displaystyle \iint_{D}^{}$f (x, y) dA = $\displaystyle \int_{{\boxempty}}^{{\boxempty}}$$\displaystyle \left(\vphantom{\int_{\boxempty}^{\boxempty} f(x,y) dy }\right.$$\displaystyle \int_{{\boxempty}}^{{\boxempty}}$f (x, y) dy$\displaystyle \left.\vphantom{\int_{\boxempty}^{\boxempty} f(x,y) dy }\right)$dx. (2)

The difficult part often is how to determine what the limit of integration are, i.e., to determine what to put where the boxes $ \boxempty$ are. In particular, sometimes we start with an iterated integral of form (1) and want to change it to an iterated integral of form (2) (or vice versa). This process is called changing the order of integration.

Changing the order of integration is slightly tricky because its hard to write down a specific algorithm for the procedure. The easiest way to accomplish the task is through drawing a picture of the region D. From the picture, you can determine the corners and edges of the region D, which is what you need to write down the limits of integration.

Example 1

This process is shown best by an example. Let's say you are given the integral

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{1}^{{e^y}}$f (x, y)dx dy.    

Since for this example, we are just dealing with the limits of integration, it doesn't matter what the function f (x, y) is. From the above integral we can determine that the region D of integration is defined by

0$\displaystyle \le$y$\displaystyle \le$1    
1$\displaystyle \le$x$\displaystyle \le$ey.    

If we draw this region, it looks like

\includegraphics[width=2.5in]{intchange.eps}

We've simply drawn that y ranges from 0 to 1, and for a given value of y, the variable x ranges from 1 to ey. The curvey edge of the region is given by x = ey. The left edge is x = 1 and the top edge is y = 1. The upper-right corner is the point where both y = 1 and x = ey. This implies that x = e1 = e, so the point is (e, 1).

If we move x to be the outer integral, its limits must be constant. The limits will be the largest range of x. From picture, we see that 1$ \le$x$ \le$e.

For a given value of x, how far does y range? It must be less than the top edge y = 1 and greater than the curvey edge x = ey. We can write the curvey edge as y = log x. Hence, the range of y is log x$ \le$y$ \le$1.

Integral with integration order reversed is

$\displaystyle \int_{1}^{e}$$\displaystyle \int_{{\log x}}^{1}$f (x, y)dy dx.    

Example 2

Sometimes you need to change the order of integration to get a tractable integral. For example, if you tried to evaluate

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{x}^{1}$ey2dy dx    

directly, you would run into trouble. There is no antiderivative of ey2, so you get stuck trying to compute the integral with respect to x. But, if we change the order of integration, then we can integrate with respect to x first, which it doable. And, it turns out that the integral with respect to y also becomes possible after we finish integrating with respect to x.

In this case, the region of integration is

0$\displaystyle \le$x$\displaystyle \le$1    
x$\displaystyle \le$y$\displaystyle \le$1,    

as shown by the follow picture.
\includegraphics[width=3in]{dintex.eps}
We can also describe the region by

0$\displaystyle \le$y$\displaystyle \le$1    
0$\displaystyle \le$x$\displaystyle \le$y.    

Hence, the integral with the order changed is

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{x}^{1}$ey2dy dx = $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{y}$ey2dx dy    

We can compute the second integral, first with respect to x

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{y}$ey2dx dy = $\displaystyle \int_{0}^{1}$xey2$\displaystyle \big\vert _{{x=0}}^{{x=y}}$dy = $\displaystyle \int_{0}^{1}$yey2dy    

Now, we can compute the integral with respect to y by using a u-substitution, u = y2, so du = 2y dy, and we integrate u from 0 to 1.

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{y}$ey2dx dy = $\displaystyle \int_{0}^{1}$yey2dy = $\displaystyle \int_{0}^{1}$$\displaystyle {\frac{{1}}{{2}}}$eudu = $\displaystyle {\frac{{1}}{{2}}}$eu$\displaystyle \big\vert _{0}^{1}$ = $\displaystyle {\frac{{1}}{{2}}}$(e - 1)    



Duane Nykamp
nykamp@math.umn.edu
2005-09-28