sketch the region of integration and change the order of integration.

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November 1st 2010, 04:40 PM
Evan.Kimia

sketch the region of integration and change the order of integration.

Heres the problem: http://i52.tinypic.com/2jd2paa.png

HAving a hard time understanding what this question is asking. When they say change the order of integration, it seems like i am supposed to integrate dx with the 4x to 4 bound and integrate dy with the 0 to 1 bound...but im pretty sure im wrong. Any help is much appreciated.
November 1st 2010, 05:01 PM
Ackbeet

What does the region look like? When changing the order of integration, the limits will likely change as well. That is, in general you do NOT have this:

$\displaystyle\int_{a}^{b}\int_{f(x)}^{g(x)}h(x,y)\ ,dy\,dx=\int_{f(x)}^{g(x)}\int_{a}^{b}h(x,y)\,dx\, dy.$

You have to sketch out the region. In the first integral, which is like yours, you view the inner integral as happening first. In addition, you think of the region first as being bounded below and above by two different functions of x. Then, the outer integral gives you a number by integrating from one number to another.

After you've switched the order of integration, you have something that looks more like this:

$\displaystyle\int_{c}^{d}\int_{j(y)}^{k(y)}h(x,y)\ ,dx\,dy,$

and now you must think of the inner integral's limits as enclosing the left function up to the right function. Then you get a y-interval in the outer integral to close it off.

I don't feel like I'm explaining this very clearly. Does this make sense?
November 2nd 2010, 04:57 AM
HallsofIvy

As Ackbeet suggests, as does the title of your post, first sketch the region. $0\le x\le 1$ so draw two vertical lines at x= 0 and x= 1. The upper bound in the y-integral is 4 so draw the line y= 4. The lower bound is 4x so draw the line y= 4x.

You should see a triangle with vertices at (0, 0), (0, 4) and (1, 4). The region over which you want to integrate is the inside of that triangle. Reversing the order of integration, you will be integrating with respect to y last so its limits of integration must be constants. You can see from your sketch that the lowest value of y is 0 and the highest is 4. Those must be the limits of integration for y. Now, I recommend drawing a horizontal line, representing a given value of y, inside the triangle. You can see that the left end is on the vertical line x= 0 while the right end is on y= 4x or x= y/4. The limits of integration on the inside, x, integral are x= 0 to x= y/4.