00:01
We want to sketch these curves and then find the area of the enclosed region.
00:06
So first of all, let's take a look at the first equation.
00:10
Let's label that 1.
00:12
So 1.
00:13
So we see that there are horizontal asymptotes because as x goes to positive infinity, this becomes like 1 over 1.
00:22
And as x goes to negative infinity, this, which equals 1.
00:28
And as x goes to negative infinity, this looks like negative 1 over 1.
00:32
Negative 1.
00:33
So we see that we have these horizontal asymptotes.
00:37
So we can sketch that.
00:38
Let's just erase this quickly.
00:43
So let's sketch our horizontal asymptotes.
00:48
This is x equals 1.
00:51
And then here we have 1 at x equals negative 1.
00:56
And we see that the denominator is always larger than the numerator, because the square of x squared plus 1 is always greater than the square of x squared, which is just x.
01:06
So the numerator is always greater than the denominator, which means that our function always lies within our asymptotes.
01:13
So let's draw a quick sketch of what they look like.
01:16
Of course we know it has to cross the origin.
01:22
Okay, so that's our first function.
01:25
Our second function, let's take a look at it.
01:31
So we want to look at vertical asymptotes since the denominator can be zero.
01:36
So we see that the denominator is zero at the values x equals 3 and x equals minus 3.
01:41
So we're going to get these vertical asymptotes.
01:43
So let's sketch those in quickly.
01:51
So this is x, y, sorry, these were y equals 1 and y equals negative 1.
02:01
This is going to be x equals negative 3 and x equals 3.
02:10
And now if you do a quick sketch of this, of this function, we have our behavior near the asymptotes.
02:20
And we know it also crosses the origin.
02:23
So it has this kind of behavior.
02:26
And of course we are only considering x greater than 0.
02:30
So that means that this region over here is the area we want to determine.
02:35
So we need to determine what are our points, our limits of integration.
02:39
We know one point here will be the origin, but we also need to solve for this point.
02:43
So we solve for these by equating our two functions.
02:47
So let's open up a new page and do that.
02:49
Our first function is x over square root 1 plus x squared.
02:58
The second one is x over square root of 9 minus x squared.
03:07
So let's just square both sides and then multiply by the denominators.
03:12
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, and we're going to have our 9 minus x squared.
03:24
Since we squared, we don't have the squared anymore.
03:27
Here we also have a x squared.
03:28
And then we also have this 1 plus x squared coming from the denominator on the left.
03:34
Now we just expand everything and then we are going to solve for x.
03:41
So let's do that.
03:42
9x squared minus x to the 4 equals x squared plus x to the 4.
03:48
Let's bring everything to one side.
03:50
We have 2x to the 4s plus no, that's a minus.
03:56
Let's erase that.
04:00
We have minus 8 x squared.
04:05
0 is equal to, we have 2x squared multiplied by x squared minus 4, which is equal to 2x squared times x plus 2, x minus 2, because it's a difference of squares.
04:27
So what we have are solutions, x equals 0, so we expected that.
04:31
X equals negative 2 and x equals 2.
04:37
So we're actually interested in integrating from 0 to 2.
04:42
Of course, minus 2 comes from the fact that they do cross again over here on the left, but we're not interested in that point.
04:49
So we're just going to integrate from 0 to 2...