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# For what values of $m$ do the line $y = mx$ and the curve $y = \frac{x}{(x^2 + 1)}$ enclose a region? Find the area of the region.

## $m-1-\ln (n)$

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Applications of Integration

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this problem is a really difficult one. And it wants to find the area of the region between Why equals X over X squared, plus one? And why equals MX, where, um is greater than zero, but less than one. And the best thing we could do here is to graft this problem. Oh, that's a very nice tree line. We grab this problem. Let's do why equals X over experience one first and we keep graphing calculator, do it or keep it for ourselves. It would be something like this. And of course, all across the world are not there but across right there. And then we know that our Y equals MX because it's between zero and one. The line is always going to create this region with again this house across the origin. It's always going to create this region with y equals X over X squared, plus one. If the ah slope with any bigger, it would come outside and it wouldn't create the same region of like, this propeller shape. So that's basically what it means when it says M is between zero and one. So what we're going to do now is we're going to integrate. We're going to integrate from this negative point to the sponsor point. Or we could agree from zero to the positive point. Were they in a second? Just walk away to what? I'm gonna do that. So we have zero to whatever this boundary point is with top function minds of modern function the top function being X over X squared plus one minus M ax D X. And to find this point where they intersect we just set these two equal to each other. So if we set X over X squared plus one equal to M X but we can do is we can divide out a extra both sides because we have it. We divide this, we have one here on the right is one way of that. And then we can ah divide both sides by end and then multiply both sides by excrement everyone. So I just would get X squared. Well, sworn equals one over M X squared equals worn over. And this one So we'd have X equals plus or minus one was going to be positive here, square of one over and minus one, and that would be our bound report we learned over, um, minus one. And so this problem's really ugly. But lets integrate the integral of X over X squared plus one. We would have to you something that I don't think most familiar with is you substitution so quickly. I'm going to do that when I get you equals X squared. This one to you equals two X. So to integrate this, I would have to do 1/2 Ellen of X squared plus warm. And this is something unfamiliar concept to most of us. But that's what happens when you integrate it and then minus m x squared over two. I'm gonna just move this and cross it out so that we don't care about that right now. And then we would use the end points that we want one over M minus one, and we were right. This is getting really ugly. Gonna have 1/2 ln x squared plus one. Really checking in minus MX or most MX squared over two from square root of one over M minus 10 in an eternal bliss. But two after. So if we play this and we get 1/2 Ellen of this squared of the square root just makes it What's inside? So one over M minus one plus one. So one over M minus again squared. So it just squares it sometimes. This is, um and I wanna remember cancels us just one. So one minus m one Linus over two and then minus all the other plenty and zero. If we plug ins, you here, we don't get anything. But if you plug into your here we get Ellen of one and Ellen of one who is still zero. So my and zero so altogether. If you want of us, this all by two as well. We get Ellen of one over M minus one minus m. And that gives us a mme. Ah, We get, um, minus one plus Ellen of one over M. And to clear that up even more, we get M minus one minus ln of aa Lot of concept in this question, we haven't discussed yet, but it's a tricky question. Is the last one in this section. So yeah, definitely a harder one. And really, it's just a good question to look at to expose yourself to new our topics.

Cornell University

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Applications of Integration

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