Now that we started to get a feel of what this definite integral is, is accumulation under the graph, the net area under a graph. Let's talk about some of its algebraic properties. The first one I want to talk about is what happens if you switch the bounds? Consider a definite integral from a to b of some function. Again, in your head, this is a number, I want us to find, but it's usually defined. If it's defined at some number, it's the net area underneath the graph. The question is, what happens if I switch the endpoints of the function? Does it matter? Does it do nothing? If you think about the limit definition of what's happening behind this, where a and b play a role is in the formulation of the delta x, the width of the base of the rectangle. We normally define delta x to be b minus a over n, the right end point minus the left point over n. If you switch the numbers around, If you had five minus three and all of a sudden to three minus five, if you switch the numbers around, then all of a sudden you get a minus sign pop out. You factor out the minus, you get good old a minus b. If you switch it around, a minus sign pops out. I'll can put it in red so you see it. By switching the bounds produces a minus sign that comes immediately from the formula by looking at the delta x term. Here's another one that's both algebraic and hopefully little geometric intuitive at this point. What if I have the same bound? Some crazy integral, f of x, dx and the bounds are the same; the lower and upper bounds are the same. In terms of delta x, in this case, the right endpoint is equal to the left endpoint. You get like a minus a over n, which is zero. The width of this rectangle is zero. No, not approaching zero, it is actually zero and then in that case, if you start computing areas of rectangles with a base zero, you get a lot of zeros. You can also think of it as like what is the area under a curve, under a single line? There's no area here, there's just the distance. This is zero. When the bounds are the same, you get zero. Other things that we can use to figure out. What happens if I have a definite integral of a constant function? Now we did one of these, but it's good to have it down. Let us take a function, let's say we're going from a to b of a constant function y equals some number seven Pi, you get back a rectangle in that case because the graph is a horizontal line. In that case, we have the formula for the area of a rectangle and this could be positive, it could be negative depending on if the graph's above or below, but the c, the constant will capture that. In this case, you can do the whole limit definition. You can actually check this, but it works out to just be, not surprisingly, the area of the rectangle, it's the base times the height, constant b minus a. Other things that we're going to use, so there's some geometric intuition there. Remember these things are limits. These things are at the end of the day limits, like most of everything important in calculus, remember derivatives were limits too, so I need some calculus, the study of limits. If I have a different integral over a sum or a difference, because limits distribute over sum and differences, so do definite integrals. You can take the definite integral of either piece. Just don't change the bounds as you distribute. You can say like integral distributes over addition and for all the same reasons, if I have a definite integral times a constant times a function, then the constant will actually just pop right out. This should feel like derivatives. If you had a constant when doing derivative, the constant pop right out. Same thing with doing definite integrals, the constant pops right out as well. Another property that we can use is if I have a large interval from a to c, now c is another constant, some number of a function, and I have some number in the middle, then you can break this up from a to b of f of x, dx plus b to c of f of x, dx. Let me draw a picture just to motivate this one along. If I have some endpoints a to c and I want to put like b in the middle somewhere, it could be anywhere in the middle, doesn't have to be the exact piece, then what this property says is that the area of the entire accumulation, the area under a curve, that area from a to c can be thought of. There's no reason we can't break it up as two pieces, as the area from a to b added to the area from b to c. If you put these together, you get the total area. You can break up an interval as much as you want. Then there's one more property that comes from limit laws. See if I can squeeze it in here. If you remember, this was like when you did squeeze theorem in Calc 1, if you have a limit, or if you have functions that have an inequality, if you take the limit of functions on an equality, the equality doesn't change. If you have f of x is greater than or less than, doesn't matter, but if you have some inequality, you can integrate both sides and not have to worry about something going wrong. Inequality, remember, you got to be just a little careful sometimes if you multiply by a negative. You got to flip the sign. Integrals are limits. Limits don't change. Limits don't change inequality so you can integrate an inequality. There's some nice properties of the definite integral. We can do an example here to play around with these things. Notice, if I have a lower bound, let's do an example. Let's say I have a function that is trapped between zero and three. So me crazy function, I don't know what it's doing, but I know its lower bound is zero and its upper bound is three. Y is 0 less than or equal to 1,4, f of x, dx less than or equal to 9. I can bound the area. If I know something about the function, maybe I can take something about the area. If I don't know things that are exactly I can at least bound it or approximate it. The idea with this is like why isn't true? Why would that possibly be true? Taking the inequality and integrate all three pieces. The integral, and they're giving us from 1 to 4, so we'll use the same thing of 1 to 4 of 0 dx, would be less than or equal to the integral of 1 to 4 of f of x, dx, which is less than or equal to the integral of 1 to 4 of 3, dx. Now we have three integrals, but I can evaluate to the outside too, because they're just constants. Remember the integral of a constant is the constant times the difference of the bounds. I can evaluate the two outside piece and this becomes 3 times 4 minus 1, and then you get exactly what you would expect, the answers predicting. We're using how integrals behave over inequalities. You get the result that you wanted. That's just an example of using inequalities. We will use more as we go and do lots of these things, but our goal in this class is to take lots and lots of definite integrals. Let's do another one. I'm going to give you some definite integrals to get used to seeing these things. Let's go from 0 to 2pi of sine of x, dx. Definite integral. I really would not like to do the limit definition and work this whole thing out with sine. That's going to get gross. Before I do that, I'm going to remember that the definite integral captures the area under a curve, the net area under a curve and sine of x is one of these nice functions that I'm supposed to know. We're going from 0 to 2pi, so one period of this graph, and it goes from 1 to minus 1. Remember what the definite integral captures, it captures the net area. It takes all the area above; the top area and it subtracts all the area below. You say, Wait a minute, I see what's going on here. The area above is equal to the area below. There's symmetry going on here, I have the same one period, I have the same area above as I do below, brilliant, zero. This integral's zero. Not zero because there's no area, but because the area above is equal to the area below and some things net out and they cancel. Use the graph. Remember this is geometric when you can, a lot of times the function, it's just too complicated and you're going to need to resort to other things. But if you do another graph, then you can work it out. Keep that in mind. Let's do another example. From minus 1 to 2 of the absolute value of x, dx. I gave a way here, but absolute values is another graph that we should know. It's the one that looks like a V, goes right through the origin. Y equals absolute value of x, and it goes from minus 1 to 2. There's two and there's one. Remember what this does, it captures the area under a graph. You got to be careful, am I adding to the total area or am I subtracting because my areas below? If you draw the picture and shade in the two shapes you want, you get two triangles. You say triangles, I can do those. Area of this triangle of course, is one half, area of the left triangle, the smaller one, base times height. Since we are going to negative one, the base is one and then the height is also one, that's just one half. The left triangle gives you area one half. The other side of that triangle is again one half, base times height. The base in this case, because I'm going to two, remember that was given from one to two. So one half is 2, and then the height we're going all the way up to two as well. Clean all that, you just get two. The final answer here, you add the two. Now they're both about the x axis, they're both positive, they both contribute positive amounts, we add them together. one half plus 2, you can write as 2.5 or you can write it as five over two however you want, but that's the total area. Keep these in mind as you go through. Use your geometric understanding of what the definite integral captures. If you see something that's zero, remember it doesn't mean the function is zero, just means that the net area is zero. We'll get lots of practice as we go through and our goals is to get better at these as we go along. See you in the next video.