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# Sketch the region enclosed by the given curves and find its area.$y = \frac{1}{x}$ , $y = x$ , $y = \frac{1}{4}x$ , $x > 0$

## $\ln 2$

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Applications of Integration

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### Video Transcript

we want to find the area of the enclosed region, uh, closed by the functions y equals one of her ex ah y equals X and y equals 1/4 times X. So doing a quick sketch shows us that the region is this triangle over here. So we do have to figure out what are these three points so that we know what will be our limits of integration. So if we label these points 12 and three, let's take a look at one first. So this point is very easy. It's easy to see it's the origin because this is the intersection of the points of the lines X and 1/4 X solving. For this toeses X equals zero. Let's take a look at the second point. This is the intersection of the line X with the function one over X solving for this gives us X squared equals one which gives us exes either plus one or miners one. But since we're only considering X greater than zero, this tells us that X is equal to one is the second point. And now if we take a look at the third point, this is going to be the intersection of the line 1/4 X, uh, with the function one over X. So solving for this we get X squared equals four. And then again, since excess positive X is equal to two so we can label these points over here with, um zero X equals zero X equals one and X equals two. So now we want to integrate to figure out the area. So let's open up. A new page area is equal to first. We're integrating from 0 to 1, so it's going to be the top function, minus the bottom function. The top here is X, and the bottom is 1/4 x so x minus 1/4 X on all of this D X plus the next part We're integrating from 1 to 2 and we look at the top function here is one over X, and the bottom function is 1/4 X. So we have one over X minus 1/4 X d x. So this is going to give us the area so we just integrate this. We can do this in one piece, because X minus 1/4 X is just 3/4 x. So here we have the into the anti derivative of 3/4 X is three over eight x squared. This is going from 0 to 1 and then plus, here we take the anti derivatives of these two pieces. The 1st 1 is the anti derivative of one over X Is the natural longer them or lawn of X on. Then we have minus one overeat X squared and this is going from 1 to 2. So after we plug in everything So let's write this out. This is three over eight minus zero. Plus, here we have a lot of two minus So two squared is 44 overeat is 1/2 and then we minus. Now we plug in one loan of one is zero and then minus one over eight. So we see that things cancel because here we have a positive one over eight, and here we have a three overeat. So that works out to for over eight, which is 1/2 which cancels with this negative half. And then we're left with the answer, which is Lung of two

University of Toronto

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Applications of Integration

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