0:00
All right, hello.
00:01
In this question we're told that we have an area bounded by these two functions, f of x and g of x, and we want to figure out what the area of that is.
00:08
So what i'm going to do is i'm going to go ahead and graph these functions on graphing software and then post that image just so we can walk through exactly what's happening and we're going to have a visual aid with us.
00:19
So if i graph them, this is what it's going to look like, and it's not necessary to graph this, but i just want to explain very clearly what's happening.
00:25
So you see as i graph this, i have two points where they intersect, and on this graph there is a region that is bounded, the region of area between the two.
00:34
So this is the region that we're looking for, r.
00:37
Now in order to figure out how we're going to find this area, we're going to use the fact that the integral of a function is the area under that function, but it's the integral between that function and the x axis.
00:50
If we have two integrals then and subtract one from the other, we're going to be able to find the area left here.
00:56
But first we need to figure out what are these two points of intersection.
01:00
So that's where f of x is going to equal g of x.
01:05
So we're going to have four times the square root of x equals four thirds x plus eight thirds, and we want to go ahead and solve this.
01:12
Now since i have a square root of x over here, the best way i think i can go about this is going to be to just square both sides.
01:19
So i'm going to get 16 x on that side, and then i'm going to have to foil this out.
01:24
So i'm going to have to multiply this by itself and foil that out.
01:28
If i do that i should get something that looks like this.
01:31
I then want to put this in quadratic form, so with the x squared out front, and then i'm going to combine these two terms and just have it set equal to zero.
01:40
And look, it's something that looks like this.
01:41
I can go ahead and get rid of all of those nines, and i note that all of these are actually divisible by 16.
01:48
So i can just have x squared out front minus 5x plus 4, so that's 16 nines cancels out of everything.
01:57
That's going to equal zero.
01:59
And then i can de -factor this pretty easily.
02:01
I'm going to have x minus 4 and x minus 1 is going to equal zero.
02:06
So x is going to be 1 or 4.
02:08
And sure enough, if we look at our graph here, which actually has the points labeled, we see that x equals 1 and 4 are going to intersect.
02:16
So we have x equals 1 over here and x equals 4 here...