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# Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves.$y = \frac{x}{(x^2 + 1)^2}$ , $y = x^5 - x$ , $x \ge 0$

## $\approx 0.5901$

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Applications of Integration

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We want to find the area of the region enclosed by these two functions as well as exterior than or equal dessert. The first function y equals X over X plus X squared plus one squared is in green and the second function y equals X to the five minus X is in blue. Since we're interested in executed in zero, the region that we're interested in is this region over here. So we used a graphing calculator to approximate the intersection points. We see that the first intersection point is that X equals zero and the 2nd 1 is approximately X is 1.52 So we want to find the area. Were which means that we're going to have to take an integral area equals integral. We're going from our left most point, which is X equals zero to our right, most points, which is X equals 1.52 We're going to do the top function minus the bottom function. Here we see that the top function is the green one, and the bottom function is the blue one. So we're just going to subtract them. Okay, so if we solve this integral will have our approximation of the area. Remember, it's an approximation because this upper bound here is an approximation giving to us by the graphing calculator so we can break this up into two separate into girls. 1.52 x squared plus one squared under X d X minus into grow 1.52 of X to the five minus x d x. So the second integral is easy because we can take the anti derivative of this. It's just a polynomial, but for the 1st 1 it's a little trickier. We wanted to do a U substitution of u equals X squared plus one. Ah, so that do you equals two x d x. We have this x d x appearing in the Inter Grant, but we want to to exit DX, so we're gonna have to multiply by two end and divide by two. And now we can carry out this first integral as well as the second Integral. So we read that out equals half integral. We, um we have to change our limits of integration. So, looking over here, we see that when you is when x zero, you is one. So are lower limit knows one. And we see that when x is 1.52 you is, um, 1.52 squared plus one. So remember, that's just a number. Um, so it looks a little strange, but of course, it's just the numbers so we can write it like that. And then we have this two x d. X, that's our d'you. And then we have one over you squared in the gnome. We have a use credit in the denominator, So we have a you squared and the denominator on a D u up there and this integral stays the same. Okay, so now let's just carry out these anti derivatives. We have 1/2 and then we want to do the anti derivative of one over you squared, which is negative one over you. This is going from 1 to 1 plus 1.52 squared. We have a minus and the anti derivative of X to the five minus X is 1/6 x to the six minus half X squared. This one is going from 0 to 1.52 Okay, so we just plug in our numbers and we'll have your approximation for the area. Have. First we have negative one plus 1.5 two squared under one minus minus, which becomes a plus. Ah, one over one, which is just the one. So we have this money is here. First we plug in 1.52 1.5 two squared and then we have this minus. But then we have a minus again. So it becomes plus and then we plug in zero, which is just zero. And this gives the approximation of the area as 0.5901 That's the approximate area enclosed by the two functions.

University of Toronto

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Applications of Integration

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