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Use the region $R$ with the indicated boundaries to evaluate each double integral.

$$

\iint_{R} e^{2 y / x} d y d x ; \quad R \text { bounded by } y=x^{2}, y=0, x=2

$$

$\frac{3 e^{4}-7}{8}$

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Missouri State University

Campbell University

Harvey Mudd College

Boston College

Okay, So working in the double integral, um, a region are the function e to the too wide divide by X Do I d. X. Given that the region is described by order bounded by these three functions y equals X squared y equals zero and y equals r Sorry, X equals two. Great. So the start off by drawing out what our lives like because then that will give us, like, an idea about how to write the limits of our double integral. All right, so people's too to slide this vertical line here. Michael zeros here. All right, for so it should be passing through here a perfect. And this is y equals X squared. Obviously, there's a lefty inside, but, um, that part doesn't really matter, because that's not what's being bounded. Um, but this little Kirby triangle right here is what I read in our looks. Think so. We can get a sense of what kind of limits we want to write for our double integral. So the double integral I right over here is a d Y DX intervals. So we want first, right? How Why is ranging. And then we want to write about how excess ranging. So for this Inter integral, uh, you see how Why ranges? And what I like to do is kind of draw imaginary line here and see, like passes through white was reversed. Always So are lower limit is zero for why, and then passes all the way to y equals X squared. So export is their upper limit. And then we have this these vertical lines passing through both of these. Why limits for all X equals zero to X equals two. So that would be our ex elements. Okay. And then this is basically the properly set up a double integral and start calculating and explicitly solving for him. So selling for the inner an inner inner girl. Um, just gonna rewrite this a little bit. So it makes it easier to see how we're gonna, um, integrate this because of those. A little tricky, But trust me, it's not. It's not too bad. So writing it this way, it helps show, like what the constants of our exponents are like the constant values of our exponents. Meaning are like basically in this case, would be to over x on. And why is r variable? Because we're integrating. Specialize. So every other variable that's not wise, seen as a constant in this case. So what we can do is we can use a little trick where, since we have basically a linear function inside of a exponential function Um, the trick I use is nor to integrate this I integrated as if you were regular like each the three x where it's of the actually, lemon, let me just show you what I do. So I integrated as usual, as if you were like you for the X, right? So I say each the two divide by X times y and then what you need to do is need account for this part. So because this is a linear function, all we need to do is just divide by that linear coefficient. So in this case, would be just to divide by X on. This would be, uh, settlements are users who x squared. And just to verify that this actually works, we can do is that basically we different or we integrate respect. Why so in order to see if, like we can get back to this, But we just needed to is partially drive it respect to why? Just make sure that it's correct. So by to about eggs, all right. And if we try to do this, it's basically you take this, um, e to the to divide by x times by everything stays the same because it's basically how you differentiate you, right? This is just a constant. So this comes along for the ride, and then you have to use the chain rule and we have to drive this part, and it should just be two over X. And then basically, this two over X canceled out with this two over X, leaving you with the original function I have. So just raise everything. So that's just verification that this little trick. So basically, if you ever have ah, linear function inside of another or complex function that you do know how to in a great you can use this like little division trip waits divide by the, um, Lear linear coefficient up done. And it should all work out. Right. So now just gonna be right this a little bit, um x over to each the two divide by X zero x squared. Um, and then we needed a plug in our limits since we integrated respect to why we're gonna be playing these limits into why not X? So just be careful of that X e to the to divide the X plug in X squared bye bye to minus what we get when we plug in zero So x each to divide by X times zero bye bye to this whole thing should become one because zero times any number is still zero and then e to any number hurt e r any number races or of power is always one. So one times X over two is just sort of So we have X e This should just some fight to this. Bye bye to on that minus X divide right to So this is what we get from the inner limit or the inner integral. And now we need a please start on the curb place in the outside integral. So, sir, to two of this quantity X b to the two X survive by two minus X over to d X. Um Okay, so what I'm gonna do first is gonna just split the interval as so minus two. Two d x and this just allows me to, like, separate out. Um, I looked at this part, and it looks like we need to use integration by parts. So just to take this out of the way, I'm just splitting up the interval. So then I can work on this piece separately and this piece of relief and not having them combined together. Um Okay, so let's tackle the harder part. So the integration by parts, Um, I'm sure you guys know about the acronym lying where you set you to be, whatever it comes for. So l for logs, I for inverse trig A for algebraic or basically a polynomial, uh, t for trick and e for exponential. So basically, you set you in this priority. So you had sent you to either a log first. If you see a log here, uh, you to be a inverse trick if there is inverse trick and then so on. So forth. Um, this case, I see that there's an aggravated component x over to first. So let's just set you to be we're after Is this that you to be x and then the other company would be devi devious, set to everything else. That's not included. So Devi, which is B E to two X, abide by two DX. Let me just make sure you too. Yep. Okay. And then reason you find to you, do you, um we were just different shades. Respect to X On the right hand side, we just get DX and then we integrate for the Devi part, So V is equal to we integrate this part, and it would just be e to two x divided by two. So I'm just in agreeing as if it were a regular e to the X and then we just need to divide by that, um, linear coefficient inside some using the truth I used before And that would be V. And this would just simplified seats that two x two bye bye for Okay, So once we have all this, we know that the integral of yeah you d v is equal. Teoh u times v my integral from are integral of e d u. Okay, so applying it to this integral we get is the integral of 0 to 2 of x e 22 x divide by two DX since you is just X and V is part devi is this whole thing. It means that, um, it's equal to you. Tens be so x Times eats the two x divided by four from zero to minus. Then grow from zero to a V D u So v do you ve is this to eat the two X divide by four and d you is just steps, okay. And then we just need to sulphur this entire thing here. So this is gonna be equal. Teoh to e to the fourth over. Four minus zero. I just clicked into top limit first and then the bottom limit second. And then when you plug in the bottom limit, you get zero because yes, would become zero. And any number of times zero is zero minus eat the two x divided by eight again. I'm just using that Lanier trick. I just divided or I brought e 22 x over four first and then I divide by two. And then I did the all in one step, and then this is evaluated from zero to. This will be equal to to eat 1/4 over four minus. We get when we plug in to so each the fourth over eight minus what we get when we plugged in zero, which is just 1/8 this is equal to you. Said for over to each of the fourth over eight. Plus one teeth. Right. You think? Let me just make sure. Um yeah, okay. And then this would just simplify to four years the fourth over eight, East, fourth over eight and then lost 20. So this would result in three E to the fourth plus one over a. Okay, so we evaluated this integral. And now we just need todo with this simple and a girl pretty easy. Um, we do that here. So negative of the quantity X squared over four from our reverse power rule here, Um, from 0 to 2, there should be able to and plug into 4/4 minus zero. So you get a negative one here, So basically, we just need a subtract one from this result. So now we have, and that should just result in Let me just simplify a little bit more, so there should just be three each of the fourth, minus 7/8. Andi, I think that is our final answer. Yeah, that is our final answer.

Rutgers, The State University of New Jersey