Here again we are writing poor results. x, up from A to B.. Y is hard to C to D values. If you receive this order in which to y ago after After or before by x to y with respect to x based on. Please note here the integral of the limits to y There are limits to y because it is inside. Before the first integral is with respect to x to x There are limits based. They obviously need to pay attention to. But limits on topics of interest each time will not be a rectangle. This non-rectangular regions as well functions with limits fixed here y = y are given in the above G1X y = G2x If g1x't integral to the G2x. Here again we see the constant value x. How in the rectangular region where the first x fixed, the limit fixed in the y values were receiving. Here, as if they were fixed value of y are on duty, but this time connected to x. Here comes the x, such that x is inside x for coming. In the subsequent second step, these We calculate the integral. Before or its equivalent with respect to x Taking derivatives wherein h1y'y this time in terms of function x We have to account. Because xi accept variable y We choose a fitting. Here you have to take limits accordingly. Be made clear that if in reverse, because here we take the integral with respect to x, y hard. Y with respect to x is here also limits in terms has been working as hard. We will see that already in practice. In a univariate function again We knew in the following integral, if f and g x as the collecting function two function the integral thereof If we take them one by one f's and g's is the sum of the integral. Or from A to B, from B to C in If we're going to go from A to C of this total is equal to the integral. These are functions of one variable from the We know. Quite readily apparent from this collection why this would happen. Wherein the values of f and g is totals totals for the gun before getting on to combine totals take on the same It is clear that the results given. From A to B where A, up to a point go there to get the total mean integral, then a b c from c when the sum to get from A to C, or total up to take the total is the same thing. As a result of a total of says will not change. The same thing in two variable function the sum of f and g f and g are individually integrally is the sum of the integral. Similarly, a region on A1 integral See as herein, a a b, b a c from those who like to go as an integral a1 on the + a2 on the overall integration of the integral a2 with A1 by integrals on compound also show that. Easier to see on the rectangle but this is true in any area. Make the process more difficult here but could be significant is a collection of the integral Remembering you want to collect on a1 a2 on the structure before after do, both merge or combine with A1 to A2 Make collecting on it. This theorem says that give the same result. As I also have a theory. In many cases a function of x and y The function of only x with g (x), and only as a function of y a h (y) of two functions, two one valued function is the product. If the rectangular region that x fixed of values, fixed values y function of this product is between two one The integration of functions of one variable As we can see the product. See supremely easy to account, before Let's write what we know. Let's take for example the integral on the first year. y based on the integral of g (x) a is constant, so affected. Thus h (y) is the integral. h (y) is the integral of the function g (x) that affected. After computing these values then receive integral on bringing gun. As you can see they can be separated. This ease of calculation in many cases provides. So far, all with two variables We tried. Although you again three variables x, y and z directions picking up small range, a range of each Select the point that points obtained with prismatic volume Collect hit. It then limits the receipt of the three story collection into triple integrals income. This is also the region where the function is defined. Now this just an example of what we do concretely. Let f be such a function. X = 2 in the area with 0, and y = 1-up of 2 Whether the rectangle defined. As you can see here between x = 2 with 0, Y is 2 to 1's of the resulting this rectangular region. Function something like that. So we know what it must , but the z-direction has such a surface. This is under the surface, and this rectangle find the volume of the receiving base body we want. Just before our discussions x fix the integral y on Before we can get or we fix y x We can get on the integral. Then the remaining integral. See also the two third way functions function of the total. Accordingly, the first function of the integral plus account of the second function as integral We can. Further wherein x and y in terms of the function can be separated. Wherein x and y terms can be separated. Therefore, separation of variables method and it can be calculated. Thus, three different calculation methods occurs. Now let's see them. Before we take the integral over y. y are writing the limits of 1 and 2. We are writing the limits of x 0 and 2. y on x fixed, while the integral role is doing. So where x fixed stands. 2 y squared divided by the integral of y, here y 2 divided by the square are writing. Where x squared, x squared in the second term doing the hard tasks. y the y integral of the cube of four divided by four. X had not been here that you know, just wherein the single-ply If we were to account the above integral limits were put in, 2 and 1. Here are putting these limits also bring. When the value of this limit account only a function of x will remain backward. Let us briefly these values. See where we put y = y 2 is the square Four divided by two. So the two. If you put y = 1, y 1/2, 2 1/2 I remove 3.2. But x, x, of course, remains. When it comes to the latter term is equal to 2 years When we put y 4, 16. When we put y instead of 1 year 4, 1. That means 16 - 1 = 15, but the x squared and 1.4 remains as it is. So, as x square and 4.1 remains. Here we found the 15. So. Place the remainder of the year here x function as you can see here. Now this single variable functions Take the integral of three two had split, x squared x squared divided by two, x squared two means divided by three divided by four is going In the denominator multiplied by two more. Here there were fifteen divided by four. the integral of x squared x cube divided by three. So the denominator is a three coming on a more two going on, there is a de x cube. Wherein X is zero to two We're going. When we put the two of x squared four four in the denominator to simplify. Here we have two bays instead of x when eight. We are writing here. If you make this simplification quadruples takes three stays from the first term. The second term of from eight to four times two If he thought three of four in the denominator is turning twelve. When you think of the trilogy following five fifteen divided by three to five finds. But I had stayed here a couple of the eight, Five times two, ten. So come on in here, and the result as we find thirteen. Straightforward accounts of sophistication. But the main idea is the essence:Each In step one variable function we calculate the integral. Where x are temporarily freezes. Only a year in terms of integrals 're doing. You are account where x is staying back. Here in the real sense of a single storey remains integral. We calculate it, it's that way to us gives the result. So every step of the single-layer integral 're getting. How in the derivatives at each step one If we take the derivative of variable functions, at every step here a univariate We're taking the integral of the function. For that reason, once you know the variables two Variable pass supremely simple step. We keep in mind one basic principle. At each step, there are also integral derivative in Whether a single variable freeze We relegates variable functions. Now see here, we take a different number of We bought integral y on here before. where d x d y had got inside. Border year was about the limit where x on that. If we see this integration also accounts where y is as hard this time I'll keep. the integral of x squared divided by x two here again As hard will keep y y three remains. the integral of x squared the x-cube divided by three. As limits of two and zero. Subject to the following limits will not give anything, When the x, all will fall to zero. And when you see two bays instead of x from two to four Division two will stay here. As in years standing. When you give, where x eight three two As stands, eight divided by three. As in years standing. As you can see the cube one year in real terms We went down to the integral story. The reality integral single storey When we say the terms of the account Change x to which we y constant gene because only acts We are an integral single storey account. Here the immediate integral when he calculated y integral y squared divided by two and two is The dilemma remains for simplification y squared. four years from the cube remains divided in four years, the denominator than three times four is twelve. Here also if we make the simplifying you'll see this time limit, such as one and two. When we put four minus two y Then following a three stays from the border. The second term here eight divided by ten there are two. When we put on two to six y y fourth power minus to force a fourth one is Fifteen remain. They still work if you simplify this simplification again overtakes them again in slightly different order giving. Divide by three to five stays here, three Divided by four remain. The quartet also refine by eight two remain. So twice five to ten again thirteen as We reach the same conclusion. In the account of a third way By forming the calculation of variables the two functions in the first step The total integral of the integral We are writing as a total. Because of its own, where this total variables can not be parsed. But this is where you can see the type of As xa The integral according to y integrally will occur. Single-storey, single storey completely true sense single-ply a single storey integral integral x on them a x of y on them here at work the x squared integral year divided by the square of the integral over two y We find divided by two. Limits a little different. x equals from zero to two, zero Up to two, here than if The upper limit of four split up into two two remain two. Where y is three minus one in four divided by two remains. sadeleşip we see that three twos. Again in the second integral in terms of x only story y in terms of an integral single storey integration. This account is also completely true when you have made sense where the product of single-storey integral on giving. These two have collected the previous page in two different This we do as an integral thirteen the value of giving. Now here we were subjected to simple examples 've done. The next main section than derivatives mixed partial derivatives of the function Methods to calculate We will develop. Still more complex integrators In order to calculate We will develop methods and they also We will see applications. Now it before the following theoretical We need not answer the question. But before it you also a In order to provide an opportunity to review a We're taking a break. Opinion until the next session goodbye. Please review these topics also eyeing the difficulty does not occur.