Like

Report

The idea of the average value of a function, discussed earlier for functions of the form $y=f(x),$ can be extended to functions of more than one independent variable. For a function $z=f(x, y),$ the average value of $f$ over a region $R$ is defined as

$$\frac{1}{A} \iint_{R} f(x, y) d x d y$$

where $A$ is the area of the region $R .$ Find the average value for each function over the regions $R$ having the given boundaries.

$$

f(x, y)=e^{2 x+y} ; \quad 1 \leq x \leq 2,2 \leq y \leq 3

$$

$\approx 299.6947$

You must be signed in to discuss.

Campbell University

University of Nottingham

Idaho State University

Boston College

Okay, so we're giving the function after That's why which is equal to eat to the two X plus wife. And we want to find the average value of this function over the region are described by the set of all points x y that our corn appears or elements of r squared such that X is between one and two and wise between two and three. And this is just a little set notation It just describes all the points center within our region are. But let's just kind of right out, uh, for draw out Arabic Shin just to get a visual sense of what we're looking at. 123 Okay, so it's just gonna be this little square region right here. Uh, X equals one. X equals two. Now, in order to cafe for the average value, let's just recall the average value equation. What? That is just the double integral over our of f x y ta over the area of our our region are. And to generalize this just a little bit more, it's actually just gonna be the double integral over are after. That's why d a over, um, the double angle over our of one d A. And you may be thinking to yourself, What the heck, How does the double interval get you? The area of a region doesn't delectables always kindly for volume, right? That's what you may be asking yourself. And I'm just gonna draw a little analogy. You have the single integral A to B of one VX and not sure if you ever realize but whenever you did this, you're actually just solving for the length of the interval. It's just be safe. We just calculated for this explicitly, you have X from A to B, you just plug in your top lower limits using the second final terrible calculus. And you would get this justly like with the terrible. So you go from the area to the life of interval so two dimensional quantity to a one dimensional funny. And then similarly, for Dublin girls, you go from volume two a three dimensional wanted to a two dimensional plenty. When you're solving for a double integral for the function what? And for this problem, you don't necessarily have to use this doubling ableto can't wait for the area of our It would just be too time consuming because he can kind of just look at this and find the value. Are the area of this it's just lengthens with with which is one times 11 So either do this or spend your time doing a deal able to get one at your results. Uh, I find this to be a lot easier. No, I don't know about you, but this is just like the general formula. Just keeping back your head, Uh, anyways, so lets can't wait for the top in a rule. So eat two x plus y and they want to Let's just do t y t x doesn't really matter, just preference. So why ranges from 2 to 3? Sorry about that. And then X ranges from one to, uh, we're just use a little art over here. Just something a little bit. This is just eat a two x times piece. The why de y dx. And now you can regard this eat the two X as a constant because it's inside 80 y integral. So we're just gonna bring that out 23 of use of the why D y dx and no, you can actually just bring this entire D Y integral, um, out in front because this should actually just result in a numeric value. You can bring this whole integral to the front so you would get 2 to 3 of easily y d Y times enrolled from one to each of two x dx. And that's a whole lot simpler now because we're just completing for two single and giggles compared to calculating for a double role. Um, just make sure it's correct, so yeah, it looks good. So basically I turned a tube are double integral into two single intervals. And, um, usually you can only do this when you have a rectangular region are when you have numeric limits. Because if you had variable limits, then that would lead to variables in your answer. And that's a big no no. So when you have very building is you have to keep this it, keep it in an iterated fashion. But if you have a rectangular region with numeric limits, then you could do this. Okay, Um, all right, so let's solve for this easy peasy, lemon squeezy. Uh, so we have each of the why from 23 on each of the two x divide by two. I'm just using a little, uh, shortcut here. Um, I noticed that there's a linear function inside a exponential function. So we just integrate the exponential function as usual, and then we divide by the linear coefficient. So the linear coefficient is to in this case, So I have eaten two x over to you could verify, um, that this is, in fact, the anti derivative. If you just drive it back into this form, should just all work out nice and fine. Um, but let's just move on. So each of the third minus e squared times what we get when we plug in to serve you to the fourth over, two minus what you get when you plug in one just squared over two. Just make sure that's right. Okay. And then we do have to All right, so this is the value of the numerator, but we have to divide by the area of our region. Are, But in this case, it's just one. So this is in fact, the final answer. This is the average value after that's why over region are

Rutgers, The State University of New Jersey