00:01
So we have two functions, f of x, which is x plus 2.
00:06
And we've got another one, g of x, which is x squared.
00:10
So let's go ahead and draw those out.
00:11
So x squared is the parabola 0, 0, 0, 1, 1, 1, and then 2, 4, and then negative 2 pounds of 4.
00:20
So it's going to come down like this, something like that.
00:27
And then our function here is y intercept of 2, and then slope of 1.
00:35
So you can see that's going to intersect here and here.
00:40
So the points to 4 and negative 1, 1 for those points of intersection.
00:49
And the region we're concerned with is right here, this region.
00:54
And then what we're going to do is talk about this region as far as a type 1 or type 2 region.
01:01
So a type 1 region, type 1 is when you're, you can define with vertical lines meaning like you can find the region bound by two vertical lines so here we've got one we've got x is negative 1 to x is 2 so our region is going to be negative 1 x is between negative 1 and 2 and then it's bounded above and below by the functions the upper bound is function f of x so this is y y is bounded above above by the functions x squared below and then x plus 2 above.
02:00
Alright, so here we're going, so x is left right, but if you're looking vertically, it's going like this.
02:07
The lower bound is this parabola going up to the linear function.
02:13
We can also view it in terms of a horizontal function, horizontal lines, that's what type 2 is.
02:22
So this is we're going to look at a horizontal line.
02:25
So its horizontal bounds would be y is 0 for this line, the x -axis, and then y is 4.
02:36
However, if you're looking at this, if you're defining the x values, because the y goes from 0 to 4, looking at the x then left to right, it's right here we have two different functions, right? we've got this little division line here where this bottom portion is bounded on the left and the right.
02:53
By this is a root by this portion and this portion but then it changes here the lower bound to be here the upper bound is here so we need to define these functions in terms of y so what i mean by that is here so we can think about f of x is y is x plus 2 and same thing as y is x squared but in terms of x is we need to solve these each for x so this would be y or x equals y minus 2 and this would be x equals plus minus root y and so it's bounded by the left by root x by negative root y and this would positive root y so we have two functions for type 2 we've got the first one will go from y is 0 up to y is 1 which is this first region right in here and x is bounded below by this function or the the left on this function, to the right on this function, so negative root y, lower bound and the upper bound is positive root y.
04:03
They have a second region going from 1, y is 1 to 4.
04:08
And here the lower bound to be this linear function, which is y minus 2, up to this upper function which is root y.
04:20
Great.
04:20
Now what we're going to do is we're going to define the integral, and this is the double integral.
04:27
And calculate the double integral, x, y, d, x, y, d, x, and then we're going to do the same thing, but we're going to change the order of integration to be x, y, d, x, d, y, where we integrate with respect to x first.
04:43
And these numbers, these should get us the same value.
04:46
So let's go ahead and do that.
04:47
So we need to define the bounds here.
04:49
So this, d, y, x means your inner integral will be with y equals and your outer is x.
04:56
Likewise, when you're integrating with respect to x first, your inner integral will be with respect to x and then your outer will be y equals.
05:06
So if you're looking at this one, with respect to y first, this one we're going to look at a type 2, this type, excuse me, type 1 region.
05:13
So it's going to be the integral from x going from negative 1.
05:17
This is x is negative 1 to x is 2.
05:21
The inner integral is going to go from x, y is x squared up to y, y is x plus 2 and then we have x, y, d, d, y, d, d x.
05:33
Great.
05:35
Now this one, we're going to actually have two of them because we have two type 2 regions.
05:39
So we've got two integrals...