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# Sketch the region enclosed by the given curves and find its area.$y = \frac{\ln x}{x}$ , $y = \frac{(\ln x)^2}{x}$

## $\frac{1}{6}$

#### Topics

Applications of Integration

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### Video Transcript

We want to do a quick sketch of these two functions and then find the area of the enclosed region. So let's take a look at the 1st 1 Let's draw it in green, so it's long necks over X. We know that lawn expert grows very slowly so that as X goes to infinity, um, it's, uh, the function will go to zero s o the end behavior here will boat a zero. Um, but for smaller X did, um, looks like lawn of X. So that's what lan of X over X looks like. Now let's draw the 2nd 1 in blue a lot of X squared. So here we're alone. X is going to be positive. So we're actually going from appear and it goes like that. So we see that we have this enclosed region and we want to figure out what our limits of integration. So it's a sulfur that we just equate the two equations lawn X over X equals lan X squared over X. So we see that there's two possibilities. Ah, In one case, we have lawn X equals zero, which gives us X is equal to one. So that's one solution in the other case, Lawn of X is not equal to zero so that we can divide by both a bylaw next on both sides. So in this case, we can cancel one lawn X. So this case will be in red. This allows us to cancel one lung X. So then what we have is one over X equals lawn X over X. We can cancel these exes so we just have long X equals one or raising both sides by e So we have e to the one is equal to X. So that's our second point. So we want we're going to be integrating from this left most point here, X equals one to this point. Over here, X equals E, which is about 2.7. So let's carry the integral to figure out the area area is equal to integral. We're going from one to e of art top function. Here. The top function was the green one. So that's the lawn X over Rex. And the bottom function is lan X squared over X. So we're going to subtract one x over X minus lawn X squared over X DX to sulfur. This, um we're going to do a U substitution. Let's actually write these over the same. Uh, okay, this is her a minus here, right? Because we can just combine the fractions. So there we go. Now we can do a U Substitution of U equals Lawn X, and this gives us d'you equals one over x d x. It's already we see here this d x over X, that's are do you and we can just plug in our you Fairlawn X. So of course we're going to need our new limits of integration. So we see that when x are so looking over here when X is one, we know that you was zero. And when ex's e we have u equals lot of e, which is just one. Okay, so now we have lawn X, which is this you minus lan X squared. So you minus you squared and in D X over exes are do you so we can sell for this? But, uh, quite simply, the anti derivative of you is half you squared and two driven off you squared is one over three, you cubed. We're going from 0 to 1. So we just put in our numbers. We have half times one minus 1/3 times one minus zero minus zero, which is equal to half minus 1/3 which is 1/6 so the area of the enclosed region is one over six.

University of Toronto

#### Topics

Applications of Integration

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