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

$$

\iint_{R} \frac{1}{y} d y d x ; \quad R \text { bounded by } y=x, y=\frac{1}{x}, x=2

$$

$\approx 0.7726$

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Missouri State University

Campbell University

Harvey Mudd College

Boston College

All right, So we're given the double integral over region are of the function one over y t y dx, and we have the region are to be basically inside the functions. Why equals X Y equals one over X and X equals two. So let's first see what this region are. Looks like we have y equals X. So this diagonal line here next. Why, Um, we have X equals two. So 12 and then y equals one over X, and the first quadrant should just look like this. Um, so basically, it's this enclosed region right here. Um, yeah. So the first order of business is to define our region, are, and in order to do so, we coming to find this intersection point right here. Um, this intersection is between y equals one over x end, so we just set them equal to each other to find the, uh, intersection. So we have X is equal to a plus or minus one. And the reason is plus or minus one, because is because we have a second part of y equals one of Rex and the third quadrant, which is that X equals negative one. But that one doesn't really matter to us. We're only focusing on the first water. So we're just going to say that this intersection occurs at X equals one, and then when X is equal to one, um, why is equal to one as well? Because why is equal to X? So we have a intersection of 11 Okay. And now we can write the limits in order to evaluate stubble ankle. So since we're doing a D Y V X liberal, we want think about how y ranges and then how exchanges going from the inside out. So how does why range? So we think about a vertical line, and then it passes through that line and then comes out over there. So it basically ranges from our lower function one over X to our upper function X. And then our exes range from 1 to 2. Because it starts at this intersection and results are ends at X equals two. Okay, so this is our double integral here, and we want it now evaluated. Um, and this part was a little tricky to do myself, but I'll go through it step by step and explain how I did it. Uh, so it's to the inside. First, we have the interval from 11 over. Access acts of one over. Why do Why, Um so this should just result in Ln of asset value. Why? From whatever x two X playing the century of Ellen of X minus Ellen up one over X and using the rules of logs, he have X divide by one over X. This should just be Ellen of X squared. And now what you're thinking is Oh, no, I'm gonna have to put this back into the outer Integral on. I have no idea how to integrate a absolute value bar and a squared function within a another function. And I'm here to tell you that you gotta you gotta You gotta simplify a little bit first before you tackle the problem. And it it helps to do so because we're thinking of it from a different perspective, or Teoh deal with each part. So let's deal with the absolute value bars and for this problem, actually were able to drop the's observation bars on. The reason is because we have inside function being X squared. So the reason we have the House of value bars is to allow Ellen to be defined. Right, because you can't have negative numbers plugged into a log function because basically, we have a negative number. What we're saying here is what what power would I have to raise E to in order to get negative one, And it's just not possible. So the reason we have these houses, body bars is for that reason. But because we have X square on the inside, any real number that we plug in should always result. Like when it's squared, it should always be positive. So it doesn't matter that we have these absolute value bars here. The inside number is always gonna be positive. So we can rewrite this as Ln of just x squared, okay. And then, in order to deal with this squared part, we're just gonna use the property of logs, and we're gonna rewrite this as to l n of X, and that makes it a lot simpler. Okay, And now we want to put this back into our African girl and our our Inter goal is from 1 to 2 of to l n of x dx. Um, this part could be a little tricky to, because then you're like, What the heck I can I don't know how to integrate Ellen of X. Like what? What is that? Impossible? Great. And again by my big brain, I'm gonna tell you that it is possible. And there is a, um, little thing you just have to memorize. Basically, you're just too. Take the entry anti derivative of L Innovex. It's actually just Ellen X Ln X minus X plus C Um, and this is a enough natural. And that's why your plus city. But for this definitely, girl, we don't need to worry about the plus C. So basically, what I'm and they do is to tires ex felon, X minus X evaluate from one to, and then we just plug in, like, usual, and that's that's it. That's the little trick he's got. Memorize this. I know, I know. It's very, uh very, uh, you know, niche. But it's just good to memorize this for no reason. Um, so plugging into, we get two Ln of to minus two on the play and one we get Ah, Ln of one which is just zero minus one. Yes, we have two times to Ellen to minus two minus one. This should be, uh Did you? That's right. Wait. Oh, wait. This should be a addition here. So negative zero and then plus one. So we have two times to Elena, too, Uh, minus one. Then we just smoked by the two hours we have four Ellen of to minus two, and we're gonna just reverse the trick we did above by putting this coefficient afford into the log. So it's to Ellen of Tuesday, the fourth minus to which is just equal to Ellen of 16 minus two. That is our final answer.

Rutgers, The State University of New Jersey