Problem

Sketch the region enclosed by the given curves an…

07:15

Problem 30 Hard Difficulty

Sketch the region enclosed by the given curves and find its area.

$ y = \frac{x}{\sqrt{1 + x^2}} $ , $ y = \frac{x}{\sqrt{9 - x^2}} $ , $ x \ge 0 $

Answer

$\approx 0.47$

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 6

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Video Transcript

we want to sketch these curves and then find the area of the enclosed region. So first of all, let's take a look at the first equation. Let's label that one. So one. So we see that there are horizontal Assen totes because his ex ghost to positive infinity. Uh, this becomes like one over one. And as X goes to negative infinity, this which OK equals one. And as X goes to negative infinity, this looks like negative one over one, So negative one. So we see that we have these horizontal assam totes so we can sketch that. Let's just erase this quickly. Okay, So let's sketch are horizontal as in coats, this is X equals one. And here we have one at X equals negative one. And we see that in the nominators, always larger than the numerator, because discreet of X squared plus one is always greater than discredit of X squared, which is just X. So the numerator is always greater than in the nominator, which means that our function always lies within our s and goats. So let's draw a quick sketch of what they look like. Of course, we know has across the origin. OK, so that's our first function. Our second function. Let's take a look at it. So we we want to look at vertical awesome totes since the denominator can be zero. So we see that the denominator zero at the values X equals three and X equals minus three. So we're going to get these vertical Assen dotes, So it's a sketch dozing quickly. So this is why three these were Wyck was one and y equals negative one. This is going to be X equals negative three and X equals three. And now, if you do a quick sketch of this, uh, of this function, uh, we have our behavior near the Assam coats and we know it also crosses the origin. So it has this kind of behavior and of course, we are only considering X a greater than zero. So that means that this region over here is the area we want to determine. So we need to determine what are our points are limits of integration. We know one point here will be the origin, but we also need to solve for this point. So we saw for these by acquitting are two functions so Let's open up a new page and do that. Our first function is X over a screw one plus X squared. The 2nd 1 is X over, screwed of nine minus X squared. So let's just square both science and then multiply by their didn't buy the denominators. So here we're going on the left side. We're going to have an X squared by multiplying the top by squaring the left side. Uh, and we're going to have our nine minus X squared since we scared we don't have the squared anymore here. We also have this X squared. And then we also have this one plus x squared coming from the denominator on the left. Now we just expand everything and then we are going to solve for X. So let's do that. Nine X squared minus X to the floor equals X squared plus X to the fore. Let's bring everything to one side. We have two x to the force, plus no, that's a minus. It's erase that we have minus eight x squared. Uh, zero is equal to We have two x squared, multiplied by X squared minus four, which is equal to two x squared Times X plus two x minus two Because it's a difference of squares. So what? We have our solutions. X equals zero. So we expected that, um, X equals negative too. Um, and X equals two. So we're actually interested in integrating from 0 to 2. Of course, minus two comes from the fact that they do cross again over here on the left. Ah, but we're not interested in that point. So we're just going to integrate from 0 to 2. Let's open up. A new page area is equal to integral 0 to 2. It's going to be the top function, minus the bottom function. Here are top function is our first function. Uh, this one and the bottom function is our second function. So we're just going to subtract those and integrate. So the 1st 1 we said was X over the screwed of one plus X squared, and we're subtracting X over the square root of nine minus X squared DX. Okay, so once we calculate, this will have our area. So let's break this up into two separate into girls and solve them separately. So 0 to 2 of X over screwed one plus x squared DX. Let's put the DX bit higher the ex and we're going to subtract the integral of go from 0 to 2 nine minus X squared under in the denominator x in the numerator d x Okay, let's hold them separately for this one. We're going to want to do a U substitution. Let's say you is equal to one plus x squared one plus X squared so that d U equals two x d x Uh, we see it. We don't have a two appears so we're gonna mult multiplied by two and divide by two. Um, and let's take a look at this one. It's also gonna be a U substitution you is nine minus x squared and then we have d'you equals negative two x d x again we needed to. So we're gonna multiply by two and divide by two. Not that we're going to. All right, bring this minus inside so that we see this negative two x d x and we can calculate each of these intervals separately. So let's continue. Okay, Maybe we could fit in. Okay, let's open up a new page. Okay, So the first integral becomes we had 1/2. Let's check that. Yes, we have 1/2 integral. So our limits of integration change. We see that by looking at this equation we want to plug in when x zero. We see that U equals one. So are lower limit is one. And we see that when X is too. We have u equals five. So now we're interviewing from 1 to 5. This two x d x here, this is gonna be a d u. And we have a square roots of you in the denominator. So we have. Do you over screwed you, and we have a plus. There was also 1/2 here. We're integrating from 0 to 2 in the X variable. So we have to find out. What are you? Limits are so again looking at this equation when x zero you is nine. So that's our lower limit. And when X is too, we have y minus two squared. So y minus four is five. So our upper limit this time is five. And what do we have here? We have this negative two x d x. That's r D u. And then we have a square root of you in the denominator. So do you. Over Square, You Okay? So we can, uh we can calculate each of these. So let's carry on the anti derivative of one over. Screw you. So that's you to the negative, too. Is we have this negative too? Uh, you two, the half. Let's double check that. If we bring this half down, the tubes cancel, and Oh, there's no negative. Okay, we're going from 1 to 5, and we have this plus half again. It's to you to the half. And this one's going from 9 to 5. Okay, So this half castles with this too. Same thing here. And then let's just plug in our numbers. So this becomes screwed. Five minus screw. One plus screw. Five minus screwed. Nine. Of course, we see that this is just the three, and that's just the one. So we have here. Let's he's black. Equals we have two squared fives and then we have a minus three and minus one. So that's minus four, which is equal to weaken. Factor out the two. Route five minus two. And in decimals, this works out to be approximately, uh, zero point for seven

Kenneth K.

University of Toronto