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^{-5 y+3 x} ; \quad 0 \leq x \leq 2,0 \leq y \leq 2

$$

$\approx 6.7068$

You must be signed in to discuss.

Campbell University

Oregon State University

Harvey Mudd College

University of Nottingham

Okay, so we're giving the function affects why is equal to each the negative five y plus three x And we want to find the average value of this function over the region are just best five to be at the points. The set of all points. Um, set of all points. Let me see which points. Oh, okay. Says set of all points X y, which are elements of coordinate pairs or basically r squared such that above x and y in this case range from 0 to 2. Okay. And destroying this region out we get a as you can probably guess a square region Oh, my God. Frickin genius. Dude. Um, okay, now, just recalling how to actually, for the average value of a function just the double integral over our of our function after that's why d a over the area of our region are and just generalizing this a bit more. Um, actually, we can rewrite the denominator as the Dublin your goal over our of the function one d a. You may be asking what the heck How does dulling will have to have anything to do with kept waiting for the area right it always calculates for volume. How is this? How is this happening? Right? And, um, I'm just gonna create a little analogy here. Uh, if you recall from single variable Kallick, uh, may have, like, just No, you didn't take it second thought to it, but if you just do the single interval from A to B, But the function one dx it actually just caucus for the length of the interval, and we just can't pay for explicitly. You can see that this is the case results in the length of the air, Bowlby might say. And, um, I'm here to tell you that basically, for a single internal that usually accounts it calculates for area. If you do the if you do it that way. Obviously a counselor area. But, um, for the single integral for the function one, it actually scales down one dimension. It goes from area to the length of the line and summer literally for a double integral. That usually calculates for volume. If you do the double integral over for the function one, it goes from volume three dimensional. Um, quantity. Teoh, A two dimensional client being area. Okay, so in this case, we don't necessarily have to use the supplemental to capital for the area of our region are because it is a square. Region, in this case is pretty easy to happily for. It's just two times two times with just four, Um, But in other cases where the region looks like a blob, right, it's very hard to calculate for the area of this, and you're going to need to use a actually. In this case, you would have to use a few double integral Zagat pay for this, and there's reasons why you'd have to use a few. You can't just use one for this woman. It's not really relevant for this problem. Ah, anyways, so lets Cabaye for the numerator. So the doubling role of effects. Why? Which is e to negative five y plus three x de y dx Um, both of our variables ranged from zero to said. Those are limits and let's start capping for the six wisely. So the first step I'm going to do is I'm actually just gonna rewrite this exponents into a multiplication of two exponents, so use the mega five y times e to the three x de y dx And by doing this, I actually just take this each of three X out because office, inner, inner, enter that Inter Integral is, uh, 80 y interval. So you can just regard every other available as a constant. So we're just gonna take that each of three x out and then we have another ankle like, why do you want in TX? And, um, you may not see this, but basically this inner integral will actually evaluate too explicit numeric value. And then you can regard that as a constant within this DX and roll, so we could take this entire interval and put out as a constant. So now will be deserted. Teoh. Easy, negative. Five y d y times the angle from 0 to 2 of each, the three x ds. And now we've turned a double integral over a region are a particular region, are 22 single and rules over just intervals. Um, and you can usually do this if, um, the limits are numeric. If they're not in America than there would probably be a variable here. And if there's variables here on you tried to probably for that interval, you would be left with variables in your answer. So that would be pretty bad, because you're not getting explicit value, and you would only be able to get rid of those variables. It did. Iterated intervals. Okay, so this only works for, um, basically just rectangular regions where you have, um, numeric limits and just a generalized is the Basically, if you have a double, it'll where you have a function of X multiply fire function of Why do y dx or dxy one? In this case, it doesn't really matter, but that's gonna do it. DX dy y integral. You can take all your ex stuff and all your wife's stuff and separated out a to B f of x y f of x dx times in a row from CD of G of why the wife? Okay, so basically, DX goes with a B limits and after relaxes a X thingamabob, and then you have seen a d for your d Y integral, you just separated all out. So this would just be a more generalized form of this. Okay, so now going back to the problem, I just want to solve for this, which is pretty easy. In this case, we're just gonna use the, uh, use a little trick that you can see there's linear functions inside. A exponential function, so we can do is just great. This as usual and then just divided by the, um, linear coefficient. I just wanted value from 0 to 2. You can verify that. This is actually, if you drive this, you will get this. If you just want to verify it for yourself on, then we have three each of three X divided by our linear coefficient on then from 0 to 2 and that this should be equal to eats the negative 10 when we plug into five and then minus what we get when you plug in Surrounds. But it's just one. None. Re plug in to each the 6/3 minus 1/3. Yep. And then you could simplify a little bit. Um, we just have negative 1/5 e to the 10th. And then plus one thing, you both like this by eastern six minus 1/3. Um, but this point Ah. I mean, you could go ahead and just make commented nominators and then go from there, but, like, I'm just gonna leave it in this form. If you want to just plug into your calculator, you can You will get a numeric value and should be roughly, um, let me check. Should be just roughly 6.706 a m. So if you're getting that, that should be the correct answer for we. Wait. Sorry. So So I'm not done yet. Uh, we have to. Since we're solving for our value, we have to divide by the the area of our region. Sorry about that. We want to take this and just slop that division by four on there. And then that should be roughly equalled. 6.70 60. Um, yeah. And, uh, I believe that would be your final answer.

Rutgers, The State University of New Jersey