 Hello again and welcome to another screencast where we're going to be thinking about computing the exact values of a definite integral. This time we're going to be focusing on how to use definite integral properties to help us calculate definite integral values. We're going to start with this function h of x here and as you can see this goes from zero to six on the x-axis and move this over here so you can see the y-axis and it's composed entirely out of straight line segments. So I want to give you a little review question to start with. As a review I would like for you to pause the video and try to evaluate the integral from zero to six of h of x, dx. Okay so this might take a few minutes because there's several things to do. If you've forgotten how to do this go back to the previous screencast and just do a quick review but once you've done the calculations come on back, unpause the video, think about what your answer is and let's check our work. So the answer here is going to be that the integral from zero to six of h of x is exactly six and the reason that is is because this is a very nicely geometrically behaved function here pretty clearly. It's been made entirely up out of line segments and so I can just block off chunks of area that are easy to calculate. Like for example pretty clearly I might want to go from zero to two. Okay the area there that's just a rectangle and so its area is equal to six. It's got a base of two and a height of three. So the integral just from zero to two would be six. Now the integral from two to four, that's just a simple triangle with a base of two and a height of three. So its area is three and then this triangle here from four to six is a triangle. So its area is one half base times height. Its base is equal to two and its height is technically negative three. Now remember we always treat area that is below the horizontal axis as being negative and so this is an integral value of negative three and so this total area comes out to six plus three plus negative three and then ends up equaling six. So now we're going to use this fact that the integral from zero to six of h of x is equal to six to calculate three other integrals that are related to this. Okay and we're going to be using the properties of integration that are given to you in your textbook to do this and importantly we want to see what the limitations of those properties are. Like what do the properties not say in addition to what they do say. So let's begin with the first integral here in the blue. The integral from zero to six of two times h of x dx. Well the property that pertains to the situation says it's a constant multiple rule which says just as with derivatives if I have a constant multiple times a function I can simply remove that constant multiple out in front of the of the integral just like I can do it in front of a derivative. So this is going to be two times the integral from zero to six of h of x dx and now the integral from zero to six of h of x dx is known to be equal to six. So this is two times six which is twelve in the end. And we used an existing integral right up here to calculate a new integral and I just realized that I forgot to put in a dx right here. Don't make that mistake. That should be the integral from zero to six of h of x times dx to get the notation exactly right. Now under the second integral here in the blue it's the integral from three to three of h of x dx. Now you might think at first glance that we have to go back to the picture and look at the area from three to well to three and if you do that that's not wrong but just realize what area would that be? That would be this quote unquote area. Well that's just a straight line segment that has no area and that's the just behind the rule in your textbook that says that the integral of any function from a point to itself is automatically equal to zero. Now finally let's look at this last one here. This is kind of an interesting one. What is the integral from zero to six of x times h of x dx? The answer here is that we don't know yet. We don't have a rule that tells us how to handle this situation because this is a variable and not a constant. Okay very importantly we have no rule for integration that tells me that I can pull a variable out in front of an integral. It's also not going to be the case that this is equal to the integral from zero to six of x times dx times the integral from zero to six times h of x dx. That is not true because we have no rule for it and it's going to turn out not to be true for better reasons than that but we don't have anything that warrants splitting apart an integral by a product like this so we do not know how to answer that particular question yet. Now on to some questions that we do know how to answer. Let's throw in another function into the mix here. We're going to continue using our h of x function from the from the slide from the plot and now we're going to suppose that f of x is another function and such that it's integral from zero to six and here again I left off my dx. Can I erase that and start over? The integral from zero to six of f of x dx is equal to ten. So just take my word forward that this integral is equal to ten. We don't know what f looks like but we do know the value of its integral so can we compute the rest of these and I think we can. So what's the integral from zero to six of f of x plus h of x? Well we have a rule for integrals that says I can split this up into one integral f of x dx plus the other integral. Unlike products we can split up integrals along sums. Again this is a lot like a derivative rule and the fact those two are very closely related. Okay so we know the value of the first integral. We said that was equal to ten and the value of the second integral we know is equal to six. So the value of the sum f of x plus h of x its integral from zero to six is sixteen. In the next one we're going to combine some of these rules together. What's the integral from zero to six of three times f of x minus four times h of x? Well we can do a couple of steps here. I can split this up into two integrals zero to six three f of x dx minus here's the second integral zero to six four h of x dx. Okay just so I can split up by a sum I can also split up by a difference. And now I have constant multiples so I can pull the three out in front and the four out in front here. So this is going to be three times the integral of f of x dx and that is equal to ten minus four times the integral of h of x from zero to six and that is equal to six. So that comes out to be thirty minus twenty four and that is equal to six. Finally let's get down here at the bottom. That's what's the integral from six to zero. This looks a little funny right here from six to zero of h of x minus two f of x. Well let's first of all note that I'm going from the lower limit of integration is larger than the upper limit of integration. So one of the rules that we have says that I can swap those limits of integration out as long as I introduce a negative sign. Swapping the limits of integration puts a negative sign on the value of the integral and this is the integral of h of x minus two f of x dx. And now much of this can now proceed as we normally think of it. This is going to be negative. The integral of h of x was six. The two times the integral of f of x is going to be two times ten and this works out to be positive fourteen. Okay so there's some examples of if you know the value of certain definite integrals then you can compute the value of related definite integrals in some cases but not all cases as we remember using the rules that are developed in the section. Thanks for watching.