Hello. In the previous session, a simple two- We saw storey integral account. Furthermore, the region of the rectangular structure x or are fixed in this case, y also be constant values with variable limits for the I have seen can be given. We have seen very briefly on each of these Birinie but also that you need to follow a different order 've seen. Now here we will see significant examples. Let's talk about the type of problem once and There are two general types of steps followed. One of them is given in the region. This order is issued a decision on the region supposed to give. Before going to do the integral with respect to x to y? Once you've decided it's integral will determine the boundaries. For example, x of y secure borders done; will be variable. I might. And will account for one-storey integral it then again an integral single storey account You will. The third step of this integral calculation. In a second type of problems not given, but the limits of the integral may be given. The first thing to understand here in To determine the region. Yet here then in the integral in which they made to understand that. And then the integral calculation. I want to say this little two-note. The order does not change the end of the integration. General theorems about it already the concept of a collection of integrally If you think you want before After collecting going on x y on this result, total does not change. However, in terms of ease of calculation careful There are behaving benefits. Some derivatives, such as the importance of may not but many of them are in the order of importance. They now hold in our remembrance of a Let's start with examples. Now here again given region, No geometry, such a function In the example just in a previous session we do the same function in the sample. x is zero, y is one and A line as determined by x to y is equal to region. Now how is something this area it first thing we need to understand. You'll recall a previously fixed to x we were giving a second x we're having a hard're having a hard y y would give a second hard. As the situation here is different. been one constant value for x. a fixed unit, one for the variable y given. Now we need to think of this region provide them. Y means x is equal to zero not too difficult axis means. parallel to the x axis y represents a mean is true. we know y equals x bisecting angle It is actually. Now let's see this region. Now we say in the region x is equal to zero. x is equal to zero along the y axis in this line. Y represents a y is the last one horizontal line. this line bisecting angle y is equal to x. When these three lines intersect at As you can see in this triangle off occurs. Now here before we fix y on x integral and get it, we can calculate or x Before we fix y variable that takes x We can work on by integrals. As you can see here from the beginning If you want to be in this order I get you want in a single species in this order border can identify with. Here variables constant going up and like everywhere in all the x values. Here, too, the value of y is constant in all the variable goes. The structure of the integral limits a bit different but the same result will be achieved. The first job of course this intersection find points. It's very easy. First, reset the X, y is equal to x is equal to Get one corner we find this hard value. Then y is equal to x correct the other two the truth intersect in the corner she and a We find the point. In more complicated cases a little more thinking may require this intersection before, but the process the point is to find. Now before this integral over x To calculate integrals of. Now here given the general integral of this Now let's write limits. x on we will calculate integrals means that the variable y by y If we recall the previous page xi We constants. y x, y will change to a will go. As you can see one of x to y going. After the calculation of this integral of x A reset of the borders. This is the one we were given a zero horizontal of this line with the line bisecting angle We found ourselves at the intersection. Yet, as we did in the rectangular region We're forgetting the second integral. Before the integration of a single storey are calculated. Therefore temporary variables x, where y doing a hard task. We calculate this integral a previous as such. But here the variable y is equal to the limits x is a fixed upper limit. We calculate an upper limit of y is equal to We give. We give minus y is equal to x. Here are four upper limit minus x squared ago lower limit minus minus this year We put y in terms of a cube. We put x to x. X cube consists of a minus. If we arrange them see here There are constants a four-Less This 12 minus 11 divided by three divided by three is going. There are four x minus X, here are the terms of this is coming. There are x where x is the square squared term From the second term No, but one cube of x squared term in x here are a plus x terms have a cube one The term also has triple x cube that splits it giving. As you can see now only terms X, remained. Just that when he calculated this integral we know the integral single storey 11 to 11 divided by the integral of the integral three of X, who is possessed of. Four x is the integral of x squared divided by two for Coming two x squared minus twos sadeleşip This x cube the x-four divided by four is integral, but In the meantime, a more simple quadruples x squared x cube divided by three is going there. When we put them in place One upper limit value of x and x is the number obtained, the lower limit of zero We are. Results divided by three to five turns. Now the same integral different this time Let's do that again here in general we wrote integral, integral two-storey x on first integrals are doing. Wherein Y is a hard task, thus I will see him combine fours. Back to the variable x squared remains. that y is equal to the limits of x by x x is the inverse function of y equals. X is equal to y, where x is equal to zero We're coming. Given x, where x is equal to zero because x is equal to y come. Yet we forget the other integral. When the integral of X, where it accounts simple algebraic After working for four years minus four divided by three y cube is staying. When this integration is now also accounts we know the integral single storey simple. And y squared divided by two is happening. Go back a couple of twos of Four remains. The cube is divided by four years. Quadruples are more simple. Remains a divided by three. Limits zero and one. First, we calculate the upper limit. The first term gives the two. The second term gives a negative divided by three. The lower value of y is equal to zero if the course When these terms are falling. As you can see we make this account again six less than one third divided by three to five turns. We found that a previous result is the same. Which it's supposed to be. Here you need to pay attention to two points. You need to look at the first integral outer constants of integration. In the second account of both the outermost integral Although the year again, time constants one. Can not be otherwise, because the outer the integral is calculated for subsequently limits should be fixed on the end of a fixed we can find the number. Because we are looking at the end of a number. If you put it in reverse, this time as a You can find the function. So it does not mean anything. Do not make sense. Although here in function of x and y can be separated in the first integral four in the second x square to square in the third year can separate the For variables that limits method of separation of variables n / A. And these two integral single layer of ply We can become the integral of the product. Now we want to make a second example. Now see here the other to the limits are very similar. In a preceding x is zero, y is y is equal to one and one has had x line. Here y is equal to zero on the contrary, x represents a. Thus as little difference Despite appearing in different It is obvious that. Now how will these regions, however, Let's think. y is zero, along the x axis a line. x equals a, parallel to the y-axis, x a straight line is equal to the previous one. y is equal to x bisector line again. See Applying them y is zero x axis. x is equal to one, this vertical line. This bisector of x and y are equal. As you can see here in the triangle again but in the region following a previous upper triangular came out. The lower triangle in this region involved. So this is the message that you want to I need to pay attention to the border. Similar numerals even very different regions might be. Here comes the result is the same, but this x in our function and our fully is symmetric with respect to y and from bisector of use comes from our region as the boundary. Same as the previous one that it goes symmetric and function of selected regions from being symmetrical. Although these two results we collected a where the integral of the previous integrals, then this rectangle calculations on and we have made indeed made on rectangular integration took ten divided by three. The sum of these two gives ten divided by three. I leave it to you as homework. Again, we have an instance like this. However, the x and y values are simple constants. But this time y equals x squared. So again, y is zero, x axis. x is equal to one, a vertical line. a parabola y equals x squared. Now using them in more Let the concrete. y is equal to the zero line, the first frontier. x is equal to a second boundary line. And this parabola y equals x squared. Again one side of a triangle but this time we obtain a triangular curve. I want to calculate this integral. Two types, see sabitleyip ago x According to y we can calculate the contrary, we fix y According to x can be calculated. Both as a result of integrals will give. To do this in two different I'm waiting. This account is quite detailed as before we saw no significant difference from the account. Y equals x squared only going to scratch is here. X to y is equal to zero at the other was going. Wherein the x, y fixed whereby these time x, y one will go to the square root. This puts limits on the account after a would not be difficult. That followed a very similar problem again with the boundaries of this time x is zero, Y represents a given. There are still a parabola. Y is zero in the previous example, x a was equal. However, if parabola. Now we can see it immediately. Again, we have the parabola, our main curve. But x is equal to zero, vertical axis, y equals y is also a horizontal line passing one. In the previous example the following curvilinear We found the triangle. In this example, little difference in spite of its integral curvilinear triangle involved. Where the result will be different, of course. One hundred and fourteen divide the face in the previous example while five Five hundred thirty-six divided by two hundred here involved. These two have collected them again divided in three is a rectangle, rather here square is going on. On this square value of the integral We'll find. It takes a hash value. Both positions us to better understand show. Because it upright, square to the integral over means any of the lower part with integral frame that complements other have collected an integral part of The total area will be on the integral means. Again, we give an example. Again, this function is always the same hold here because mixed if I put something in your hand that is important it is not. The important thing right side boundaries to write. Then the integrals in the right order to calculate. Now see here again the simple limits is there. Here's a hard one, this time two There are variable limits. Two fixed by the previous examples, a There was one border of the variable used. Let's see how this will effect. Now we can of variation in this region. y is zero, x axis. y is equal to x, bisector. the slope of y equals two minus x minus one a right again. See, when you have made them equal to y x y equals minus two equals zero, the right of x and y. As you can see in these three right from the intersection area of the triangle again involved. Now, in this example, only a very serious There are differences in others. In others, you keep constant at x, y you keep a constant In the same difficulty as the first integral or was published in the same ease of integration. Because with only one kind of zone boundaries You could describe. Here, first, we fix y, x on integral As you can see we calculate this directly from this correctly, that this right to direct a The only region in this general We can cover. Held constant, whereas x to y integral hesaplasak ago, see Up here in the region, the first in the region up to the intersection y is equal to x y is zero Are you going but the second region y is equal to zero, we now also two Are you going to minus x. Consequently only one of these regions by integrals Your kapsayamazs. For him, this election, the first figure The choice is appropriate. One setpoint calculation by integrals can be. But here but the account of two integrals You can. Accordingly, as our account. First, if we take the integral over x, then y, If that is secure before x on y If we take, as you can see that our borders This comes directly from the right. In the outer integral limits are fixed. That's when we do these calculations, but similar to the integral than ever before precisely where the integral calculation the same size, but limits In the interim because it is variable We arrive to a different value. When we calculate it here when you put two minus x for y We are here to give these values. Minus one divided by three, x cubic again Set limits by the following values of the second kind 're getting. Here we can compile them a bit. Collection of both of you have a problem, a serious end one does not matter. But we have seen in the last line follows by Derley such as constant, linear, quadratic terms and diced occurs. But ultimately we know the hard limited, a simple integral comes. limits on the integral constant in y, hence a constant figure at the end of a number We'll find. X on x does not do the same process before As we do we fix would be y on. See in this first region y yx actually goes up to scratch. But this is a point x is equal to one Or After arriving boundary changes. Here again, from scratch, two minus x to the y income. x the limits of revenues in pieces. So the first in the region x, becomes a zero. X in the second zone, one to two goes. In this way, in all these We can cover. But this time, two integral, where which was written These two independently calculated integral must be collected. This account is made, an appropriate choice of Although it is not the same conclusions. I'm giving it to you as homework. This is calculated by integrating the first two integrals is the same exit, theorem is provided for you to see. Another example, always I'm getting the same function, because a simple better to work with the function. Because it is important for complex integration calculate is not the main issue here, our borders the right to write. See this time is equal to x a, y x. X is equal to zero in the previous did. The other two borders are the same. Wherein y is zero X represents a with the change. Here again it will be two lines y is equal to x, this is still bisector. This function is a little different than the other. At the other, where there were two minus x two x is there. And x represents a perpendicular line that at even We're cutting. These two lines are already at zero each other They're cutting. So here is made in this triangle. Now, before which we thought appropriate x If we fix the See the top of this sub-function y is going to function. However, if we do the reverse order, ie, before If y constants of this sub- in the region than under the AC line Rather, Or equivalently y is equal to one, that y is equal to two from the line actually come to y is equal to x, where limits of such a kind occurs. Here, the boundaries are changing. Although the first boundary y still equals two a second boundary, even if x is equal to x in Ends. This account can be done in two ways again. As you can see here a longer account because you will need Even if have two integral You will need to calculate and collect. Here, the interests of single integrals. It would be useful to you when you do your homework I'm giving results. And that both give the same result Calculate your vision of the theorem to ensure that you get Now most of them in similar ways can take. Refer now fixed, where x equals y It is equally hard not given. Y is equal to x and y is equal to a line x square that the parabola defining borders we see. Calculate the integral within these limits we say. Now this is not difficult. y is equal to x, we know bisector. Y equals x squared Öbürkü also parabola. We easily see right here. y is equal to x is y equals x squared rather parabola, indeed they intersect at two points and occurs in a closed area. The first job of course this is the point of intersection finding. This said, we find the points of intersection of course; We say x is equal to x squared. As you can see here, x is equal to zero interests. y when x is zero zero. This O points. One of the x is equal to x squared x is equal to a involved. That this point. After you identify these integrals Want to fix the y on xi ago x or y by integration by fixing can be calculated based on the integral. This is the inverse function of y equals x x are inherently equal y. y is equal to the inverse function of the square x x y is equal to the square root. Using this identity, this integral are written in both in and out of the way just as easily are integral. Because I can easily say the same in both a single integral is sufficient. Now here I give a little more homework. They are a little bit faster now We can now. The whole purpose of this curve given limits the right to write. Or here until the end of the integration To calculate it does not matter so much. Even a very complex integration you Accounts can be obtained on you computer You can put accounts. What is important to limit spelled correctly. As you can see in the first region xi better to go on because we fix y This can come directly to this parabola. However, if you make the opposite, y the If you fixate on the two different regions to consist of two different integrally account You have to. Therefore, this would be the first choice of appropriate. But if you do it the same result as Theorem guarantees. Now here's an example of a slightly different We'll see. See two had fixed. y is zero, x axis. x is equal to pi divided by two, a vertical line. And the bisector y is equal to x. So, we encountered more prior Not unlike the triangle area. Here we draw the triangle region is composed here. y is zero, x is equal to two and divided by pi y is equal to x. The intersection of these three lines as you can see This brings us to the triangle area. First, fix the x on y integral What if we can be calculated by a single-integrals, area can be covered. Again with vertical strips hesaplasak, ie, x y sabitleyip ago if we changed again, this kind of all regions We can cover with strips, we can close. But here is something different will encounter. Let's start in the following order. This means that in order to begin by fixing y x on integral, then do y on the Integral to do. reset the line of y divided by two pi , the x's borders x is equal to y x is equal to two pi divided outgoing limits. But here the issues to be seen immediately sinus x divided by the integral of x do not know. To calculate such an integral open impossible to calculate. Thus, although in here both in terms If there is a difference between a in this region terslikl've encountered. Because we do not go beyond the first step. Because the calculation of this integral We do not know. We can not do that simply comes to mind for the way, Does a convenience wonder in reverse? Now when we come to it in reverse order time fix the scratch x to y x We're taking. Then the zero x pi divided into two going up. So it's a two-storey integral outer limits this time of x, whereas in the previous example external borders of the year was. Constants out. When we came here sinus x divided by x, to y Accordingly, because we fix the x in this region We're just changing y. Serves as a constant. So you divide the integral sine x x remaining constant, is y. If there are limits to zero in y and x. When we put the upper limit of y is equal to x here comes the upper limit. y is equal to zero when you put it in the falls term. See the following occurs only pleasant thing. We had sinus x divided by x, from the top this year an x and x are coming in at the border cancel out. And so all that's left sinus x remains. This is integral to our we can calculate size. Get it by minus cosine integral We know that. Because of the negative sine cosine derivatives would give. We put upper and lower limits. The cosine of pi divided by the upper limit of two, ninety- cosine of zero degrees, the drops. While the bottom one is the cosine of zero degrees. Therefore, zero, minus one but outside the to a negative value because it is a movement has taken place is. We can see from this example, therefore, each how regions in what we do in terms of a not important to account for the integral we want our progress in one of the function structure does not allow. However, in the other as a simplification able to reach conclusions. This was another example of a species was. I've also got to sample the two types of regions. One of them is given in the area, boundaries We want to find. In the second kind given integral boundaries, at one given function f. It does not matter what kind of function f here is, integrally, but two of the integral given as the sum. Now how do we find this area? Now this kind of problem before moving on to a I want to break. Because I've seen a lot of species. These regions so simple. Some of the variables, but borders still simple. But you need to think which in some as is appropriate? In some cases, integration of two or even three The sum of integrals may be required. In some cases result in a single integrals can have. Therefore, learn how to make this choice need. They can think a little, samples If you pass out, To be able to take a break assignments promise. In an opinion until the next session of Please goodbye.
