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# Sketch the region enclosed by the given curves and find its area.$y = \frac{x}{1 + x^2}$ , $y = \frac{x^2}{1 + x^3}$

## $\frac{1}{6} \ln 2$

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Applications of Integration

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##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

we want to do a quick sketch of these two equations, and then we want to determine the area of Dean closed Region s. So let's take a look at our 1st 1 So we know that as X goes to infinity and also X goes to negative infinity, we know that the symbiotic behavior is that it goes to zero. So for this one, we know that we're going to zero both on the right and the left. Okay? And we also know that it's going to cross the origin. What else do we know? We know that the denominator one plus X squared is always greater than the numerator, uh, for all acts. So we're never going to pass the values between minus one and one. So this is what the the equation looks like. Now, let's take a look at the 2nd 1 Um, for the 2nd 1 we also see the same behavior is X goes to plus and minus infinity. We see that it's going to zero because excuses in the denominator, it's a good in the next squared. Ah, but we also have this one vertical aspirin tote. Since the denominator can be zero at X equals negative one since negative one cubed is minus one. So we should draw in that ass in tow. This is that X equals minus one. Let's draw this one in blue. So to the left, we see the X is going to minus infinity and it goes to zero. But on the right, we also see that this function must touch the origin. And it exhibits this kind of behavior. It kind of goes up, and then it goes back down. Eso we do see that there's this enclosed region right here, and that's the area that we want to determine. So we need to figure out our limits of integration. We no one will be the origin, and we need to figure out the other. So to solve for these, we equate that you equations. So let's do that. So our 1st 1 was X over one plus X squared. The 2nd 1 was X squared over one plus x cubed. So we're just going to multiply the denominators. So here we have X times one plus execute equals X squared times one plus x squared Good. Okay, X plus X to the fore equals X squared plus X to the fore. Okay, so we see that these x to the force cancel. We bring everything to one side. X squared minus X equals X times X minus one. So we see that our values, our X equals zero, which we expected and X equals one. So we see that this point over here is X equals one, and this one is X equals zero. So we're just going to integrate from X equals zero X equals one area equals into rose 0 to 1 of our top function minus or bottom function. Our top function here was the green one. So this was our 1st 1 And then the bottom function is our blue one. That's our second function. So we're just going to subtract those. The 1st 1 was X over one plus X squared. The 2nd 1 was X squared over one plus execute DX. Okay, let's break this up into 200 girls and salvage separately. X over one plus X squared DX minus into grows 0 to 1 of one plus X cube under X squared D x. For this 1st 1 we're gonna do a U. Substitution of U equals one plus x squared so that d u equals two x d x And over here we're going to do also u substitution Yukos one plus X cubed and d'you equals three x squared DX. So we see that we don't have a three up here, So we're going to multiply by three and then divide by three. And so now we can just carry out these into girls. Oh, uh, also, over here, we don't have a to X in the top, so we're gonna multiply by two and divide by two. And now we can carry out these into girls. Okay, Equals here we had 1/2 integral. We want to switch to the U variable now. So, looking over here, when x zero you is one and when X is one us too. So here we're integrating from 1 to 2. Now we have this two x d. X Over here, that's a d u. And we have a you in the bottom. So this is just d'you over you. We're adding there was a 1 30 year into girl. Oh, it's a subtraction, not adding, uh, minus. Here we have our three x squared DX. That's our d'you and then we have a u N the denominator. So this is d'you in the numerator. You in the denominator. Oh, and we need our limits of integration. So, looking over here, we see that when x zero you is one and when x is one used to So again, we're going from 1 to 2. Okay, so actually receive that these two into girls are the same. So oh, um, what we can do is just combine them into one piece. So we have 1/2 minus 1/3 which works out to be will have his 36 and whether it is to six. So we have 1/6 integral from 1 to 2 of do you over you. Okay, so let's just calculate this one over six. The anti derivative of one over you is the natural longer them or lawn of you. And we're going from 1 to 2. So we have won over six multiplied by lan of two minus long off. One loan of one is just zero. And so we have. Our final solution equals one over six, one of two. That's the area of the enclosed region.

University of Toronto

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Applications of Integration

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