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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.

$m-1-\ln (n)$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

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Yeah. In this problem we are really using the definition of an integral which is finding the area under a curve or between two curves. So in this case we are given a function Y equals X over x squared plus one. And we need to find the values of M. Such that we enclose the region of the area. So something that we can do pretty easily is just find the derivative of this function and we can use the quotient rule to do that. We would get why prime equals one minus X squared over X squared plus one squared. Now the reason why you could do that is you could graph why and why prime. And to visualize this dysfunction we could have some idea of what area were in closing but in this case we don't need the derivative because we are integrating. So we are also given the equation of a general line, Y equals mx. And so we can set mx equal to r Y function M X will be equal to X over X squared plus one. And now from here on out, it's basically just some algebra. We can multiply to get X times the quantity M x squared plus m equal to X. Now we can do some factoring, we can have x times M x squared plus m minus one. All equal to zero. And why did we do this factoring? Well we can have actually put a zero as one solution now since we know x zero. We can take the art in er function if you will or our inner quadratic and set it equal to zero. Mx squared plus m minus one equal to zero. Now if we do that we can get X squared by itself by manipulating our ems. So we would have x squared equals one minus em all over em and now if we take the square root of each side we can get X. So we would get X equals plus or minus the square root of one minus M over M. But in this case it doesn't really make a lot of sense to have a negative area. So we'll take the positive route in this case but remember when we're finding the area under a curve or in between a region we use an integral and in this case we're going to say let U equal to r X. So we're going to say that U equals the square root of one minus em. All over. Em So then using the integral definition of area, we could see our area a equals two times the integral evaluated from zero to you. Of our function X over X squared plus one minus mx with respect to X. And now, if we just integrate we get two times one half times the natural log of X squared plus one minus one half M x squared. Again evaluated from zero to you. So everywhere would we see an X would essentially plug into you because if we subtract zero, that doesn't change our function. So we'll get a equals the natural log of U squared plus one minus M U squared. But remember we already know what you is. Well, set U equal to the square root of one minus M over M. So we can plug that in and we'll get a equals the natural log of one minus M over M plus one minus M times one minus M over M. So all we did here was plug in what we said you was. But remember this is squared. So essentially our square root cancels. Now if we simplify this a little bit more using the properties of logs, we get A equals M minus the natural log times n minus one. So I hope that this made sense. Again, this is just using our definitions of area and integration. If you have any questions, please feel free to comment.

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