 Hi there, and let's do an example where we're going to calculate a definite integral exactly without any estimation using the graph of the function. So what you see here in front of you is the graph of a function we're going to call g of x. And we're going to calculate the integral from 0 to 8 of g of x dx. Now, just to worry about g of x here, as you can see, it's pretty geometrically nice. Every part of g of x is either a line segment or what I intend for this little piece down here to be, this is a semicircle. So this graph is completely composed of line segments and the semicircle here. And we're going to calculate this definite integral not estimated with a Riemann sum, but calculate its exact value. Now, how do we do such a thing? Well, go back and remember what the definite integral actually tells you. The definite integral from 0 to 8 of g of x dx is going to be the exact value of the area that is between the graph of g, the x-axis. And x equals 0 over here on the left and x equals 8 over here on the right. So this total area that I'm shading in right now, even the stuff that's below the x-axis, that total area is going to be the exact value of the definite integral. So since we have a very geometrically nice graph here, I think we can calculate that total area right on the nose without estimation just by using geometry. So we're going to split this up into four distinct regions here. One going from 0 to 2, one going from 2 to 3, another going from 3 to 6, and then another region going from 6 to 8. The reason we're choosing those intervals is because something changes in the geometry of g at those points, at 2, and 3, and 6. So let's take a look at each of these four little blocks of area here. Now from 0 to 2, we're looking at this area here in this little trapezoid region here. You can think of this as a triangle that sits on top of a rectangle. Now what's the area combined of those two regions here? Well, the rectangle here on the bottom is pretty clearly two units by three units. So that's six units of area inside there. This little triangle up here, just recall that the area of a triangle is simply one half base times height, one half base times height. The base of that triangle is two units wide and its height is also two units high. So the total area of that triangle is simply two goes in there. And so the combined area of those two figures there is eight. In integral language, we would say that the definite integral from 0 just to 2 of g of x dx would equal 6 plus 2 or 8. Now let's move on to this region right here. This is obviously a triangle, kind of a long skinny triangle. Its area is also one half base times height. The base of this triangle is one and the height, you can read off the graph, goes all the way up here to five. So that area is five over two. And again, in the language of an integral, we would say that the integral just from two to three of g of x dx would equal that exact area, which is five halves. Now let's move on to the third region, which is this triangle here that sits below the x axis. Let's calculate its area, but remember with a definite integral, this is a really important thing about definite integrals. Any area that appears below the horizontal axis, like this area does here, is going to be considered negative area. In other words, we're defining area here to have sort of a direction. Area that sits on top of the x axis is going to be considered positive, like usual. Area that sits below the x axis is going to be considered negative area. So let's calculate the area here and remember to make sure it's negative at the end. Well, this is just another triangle, this long triangle here. So its area is one half base times height again. And the base goes from three to six, so that's three units long. And the height is actually negative two, okay? So it's going to have a negative height value, and that comes out to be negative three. So, and again, in the language of an integral, this integral just from three to six of g of x dx would equal the signed value of the area, which is going to be negative three. Very importantly, remember to make sure that the area is negative if it lives below the x axis. Lastly, we're going to deal with the semi-circle here. This is half of a circle, and the radius of it, as you can see, is one unit. So its area is one half of the area of a circle, which is pi r squared. The radius is one, so this just comes out to one half pi, okay? And once again, just in the language of the integral, we would say that the integral from six to eight of g of x dx would be one half pi, which we also write as pi over two. So now on the next slide, we're just going to put all these areas together to get us the integral from zero to eight. So here's the setup for our integral. We have an integral property from an earlier section that says, if I am integrating from one point to another on the x axis, I can pick any point in between and split the integral into two, and then add the results. That's what we're going to do here, except we're going to split multiple times. And all we're doing is saying that the integral from zero to eight can be broken up into several pieces, four blocks, in this case, of area. And we're just going to calculate each of those blocks separately. I think you can see why, because each of those four blocks we have is like a separate geometry problem that unto itself is pretty easy. Now, we've already calculated all these, so I'm just going to put these back into this formula here. The integral from zero to two, that was the area of that small triangle on top of a rectangle, which was eight. The area from the integral, which is the same thing as the area from two to three, was that tall skinny triangle, and it had an area of five over two. The integral from three to six was the triangle that was underneath the x-axis. So this area was minus three, very importantly, to have that negative sign right there. And then finally, the integral from six to eight was the area of the semi-circle, and that was pi over two. So we're adding all these things up. I'm going to have, I'll skip the arithmetic here, but that will come out to be 15 plus pi divided by two. And that is the exact value of the integral. Let us not an estimate, we calculated the exact value because the graph we started with was very geometrically well behaved. That's all, thanks for watching.