In today's video, we go through two important and quite subtle calculations, both of which were known to the ancient Greeks. We use the perimeter to find a formula for the area of a circle, and calculate precisely a certain area bounded by a parabola in the x, y plane. We begin with two very familiar areas. The area of a rectangle, of course is the base times the height, and in fact this is how one defines multiplication of two positive real numbers in the first place. The area of the triangle, I expect you know, is half the base times the height. Already something interesting is happening as this enables one in principle to find areas of any geometric figure bounded by straight lines, since such figures can be split up into triangular pieces. To see why this formula holds, one has to relate a typical triangle in some way to rectangles. In this particular figure, one can form altitude of height h from the apex of the triangle to make the horizontal base of length b, and then use it to form two rectangles that enclose the triangle. One of width b1, and the other one width b2 say. Notice that the area of the triangle, colored beige, splits up into two pieces. The first of which is exactly half the area of the rectangle with width b1, and the second of which is exactly half the area of the rectangle with width b2. So, putting together the areas of these rectangles, and using the fact that b1 plus b2 equals b, we get to the beige area is 1.5 of b times h. This isn't the most general diagram for a triangle, but you can easily adapt this argument to all triangles. What about areas bounded by curves that are not straight lines? The most classic example would be a circle say of radius r. I expect you know the formula for the area of a circle, but do you know why it works? We discussed the perimeter in a very early video which we can call p, and note the fact that p equals Pi times diameter or 2 Pi times radius. We are going to engage in a thought experiment used by the ancient Greeks. Imagine dividing a circle into N pieces called sectors, which are like pieces of pizza. Here for example, N equals four. Dividing all the pieces in half we can get N equals eight, and in half again we can get N equals 16. We can imagine going on dividing the sectors in half indefinitely and they get thinner and thinner. Suppose we stop halving the segments at some stage, and join up the points on the circle using successive small line segments where all the radii emanating from the centre meet the circle, to form this grain polygonal shape that appears to be hugging the perimeter. Suppose we do this so that all of the sectors have exactly the same shape, and here's a picture of one enlarged, including a small green line segment joining up the two points on the perimeter. The drawing is not too thin to make it easier to follow the mathematics associated with the diagram. We have the two radii of length R emanating from the center, which form a triangle together with the green line segment of length b say. Denote the height or altitude of the triangle by h say. Notice that h is approximately equal to the radius r, because the altitude emanating from the center of the circle almost reaches the perimeter. If you imagine making more and more thinner sectors, by allowing the number N of sectors to become arbitrarily large, then you can see h getting closer and closer to the radius. We can express this precisely using limit notation. The limit as N goes to infinity of h is R. What about the area of the sector? It falls naturally into two pieces, the area of the triangle colored green bounded by the two radii and the line segments, and a little bit left over between the line segment and the perimeter colored beige. The area of the sector is approximately the area of the triangle, which is half the base b times the height h. Here's our circle again, split up into N sectors with the same size and shape, each with area half the base b times the height h, and the limit as N goes to infinity of h is the radius R. Observe that if we circumnavigate the circle using the grain line segments of length b, we get N lots of b approximating the perimeter P. The more we subdivide the circle into thinner and thinner sectors, the closer N times b will be to the actual perimeter P. In fact the limit as N goes to infinity of N times b is P. How does all of this information relate to the area of the circle? Here, all the triangles approximating the sectors have been colored in green. There is only a tiny bit of area of the circle not included, which you can hardly say between the green line segments and the perimeter. So, the area of the circle is approximately the sum of the areas of the triangles, which is N times a half bh, which we can rewrite as a half of Nb in brackets times h. The approximation gets better and better as N gets larger, and in fact the area is precisely the limit of this expression as N tends to infinity. But we have enough information to evaluate this limit. The constant half comes out the front and the limit of a product is the product of the limits. The limit of Nb is just the perimeter P, and the limit of h is the radius R. But the perimeter is 2Pi times R. So, the expression a half PR becomes half of 2Pi R times R. Which simplifies quickly to Pi R squared. Thus, we answer our earlier question the formula for the area of a circle is A equals Pi R squared. Look carefully at this in light of our knowledge of the derivative. The area A is Pi R squared, and the perimeter P is 2 Pi R. The derivative da dr is also 2 Pi R. This isn't a coincidence. We call the area an antiderivative of the perimeter, and soon you'll see this fact in a wider context. Functions associated with areas bounded by curves, are intrinsically related to the functions describing the curves by going backwards from differentiation. The Greeks also thought carefully about areas associated with another classical curve the parabola. However, a parabola goes on forever and doesn't close up like a circle. Nevertheless we can position a parabola in the x, y plane and consider the area between the curve and the x-axis over some particular interval. Let's simplify everything as much as possible and just ask for the area under the parabola y equals x squared for x between zero and one, shaded here in green. If you use graph paper as in this diagram, you can get a very good estimate just by adding up tiny sub squares and fragments of sub squares. If you do this, you can probably make an educated guess as to what the answer might be. But we'll try to emulate the Greeks and find out the exact answer without guessing. Remember in the previous video, we estimated displacement from the velocity curve by using rectangular approximations. We do something similar here. First of all, divide the interval from zero to one on the x-axis into n equal sub-intervals of width one on n where n is a positive integer. You can think of n as large, though in this particular diagram n equals six. Then move up to the curve and build rectangles that sit above the curve. It's enough for us to consider upper rectangles only though you can perform a similar calculation using lower rectangles if you wish. Then we shade in the areas inside the rectangles which includes a green area under the curve and extra areas colored in beige which sit above the curve but still inside the rectangles. Note that all the rectangles have a width of one on n units, adding up the areas of these upper rectangles, we get one on n times one over n squared for the first rectangle since the height is the value X squared when x equals one on n plus one on n times two on n squared for the second rectangle. Plus one on n times three on n squared for the third rectangle and so on all the way up to one on n times n on n squared for the final rectangle. Notice how we get one on n cubed times one squared plus two squared plus three squared all the way up to n squared. Denote the green area under the curve by A. Then A is approximated by this expression for the areas of the rectangles. As n gets larger and larger, the rectangles get thinner and thinner and the extra beige area becomes vanishingly small so that A becomes the limit of this expression as n goes to infinity. We have a chance of evaluating this limit if we can understand or control the sum of all those squares inside the brackets. A very succinct way of writing a sum like that is to use Sigma notation, using the Greek letter capital sigma as an abbreviation for repeated addition. Here we're adding up i-squared as i ranges from one up to n. If you're not sure about Sigma notation, I suggest you pause the video and read about it in the accompanying notes. We're going to use it a lot in the rest of the video. I should also warn you that what follows is quite advanced material and we move very quickly. Don't worry if you don't follow every step and pause the video frequently if you want to check each of the manipulations slowly and carefully. Now, we do something that looks totally bizarre and unexpected, known as the telescoping sum method. Though we want to sum squares, we look at n cubed and open it up like a telescope by successively adding together a cubed takeaway, the cube of one less, climbing all the way down to one cubed take away zero cubed. Now, this looks absurdly complicated, but we're just playing around with zero. You might remember how we used expansions of zero in the trick of completing the square and also in proving the product rule for differentiation. Here, we're using the same trick repeated many times to produce this huge telescope. You can see how terms cancel out producing zeros all the way to the end where we get to zero cubed which is also zero. So, this huge long telescope really is just n cubed expressed in a complicated way for the purpose of discovering something quite remarkable and surprising. Now, we use Sigma notation to express all of this very concisely. Again, if you're not used to Sigma notation, please pause the video and read about it in the accompanying notes. Here, we can rewrite n cubed as a sum of i cubed minus i minus one cubed as i ranges from one to n. Let's play with this expression that uses Sigma notation. What follows is quite intricate and you should pause the video to check the steps. If I go too fast, there are also more detailed explanations in the notes. We first expand the difference of the two cubes and see that it simplifies to three i squared minus 3i plus one because of the cancellation of i cubed with minus i cubed. But the sigma notation means a gigantic sum so we can split this into three separate pieces. The first piece becoming three times the sum of i squared. The second piece becoming minus three times the sum of i, and the third piece, the sum of one n times which is just n. So, now we get this even more complicated way of expressing n cubed. Where is all this leading? There's another surprise often referred to as the trick of Gauss. There's a story about Gauss when he was a child. The teacher asked the class to add up all the numbers from one to a 100. Almost immediately, the pupil Gauss answered 5,050 astonishing everyone. How could a child add up a 100 numbers so quickly inhis head? Well, let's think more generally about adding up all the numbers from one to a very large number n and call the total sum S. Gauss's trick is to write the same sum S but backwards and then add it to itself. Notice how the numbers line up and you get n copies of n plus one on the left and twice S on the right. So, the two times S is n times n plus one. So, dividing by two, the sum S becomes this nice formula n times n plus one over two. So, that's how Gauss did it. He multiplied 100 by a 101 and divided by two which is 5,050. What's the relevance is this? You can see in our complicated expression for n cubed, we have the sum of i as i ranges from one to n. Which is just as sum S that we evaluated using the trick of Gauss. So, let's sub that expression in and see what happens. We get yet another complicated expression for n cubed. But before throwing up our hands in despair, notice that we've made remarkable progress. The thing we're really interested in finding related to our area problem for the parabola is the sum of squares and you can see it trying to hide inside this equation. But that's not a problem for us because we just rearrange the furniture using algebraic manipulation and isolate the sum of the squares and then just carefully simplify the expression which involves powers of n. You can pause the video perhaps and check each step if you like. The upshot is, that we get this very elegant formula for the sum of squares. Now, we can complete our calculation of the area under the parabola. Remember, we call the area A and saw before that A is the limit, as n goes to infinity of one over n cubed times the sum of the squares, one squared up to n squared. So, now, using our expression for the sum of squares, A becomes a limit of one over n cubed times this expression. We can cancel one of the n's and just use our tricks for evaluating such limits dividing through the top and the bottom by n squared. Seeing that the answer quickly simplifies to one third, thus finally, we get the answer. In fact, discovered by the Greeks thousands of years ago that the area under the parabola between zero and one is one third exactly. No approximations. You can say that this was a really difficult calculation and much more difficult than finding the area of a circle. Later in this module, we'll learn a method that exploits the antiderivative of x squared and the answer of one third will drop out in a couple of steps. However, I wanted you to see this extended bare handed calculation as it involves so many different ideas and techniques that you can apply in all sorts of other mathematical settings. In today's video, we discussed two examples of finding areas bounded by curves. Firstly, to use the perimeter to find a formula for the area of a circle, and secondly to calculate the area under the parabola Y equals X squared in the XY plane over the interval from zero to one which turns out to be exactly one third. It's remarkable that by using fairly abstract ideas involving approximations and taking limits, we get exact answers in both cases. In the second example, we use sigma notation which is an abbreviation for adding lots of things together. Telescoping sums, which involves elaborate ways of playing around with zero. The Gauss trick of adding successive numbers together and lots of algebraic manipulation and our techniques of evaluating limits. Please read the notes and when you're ready, please attempt the exercises. Thank you very much for watching. I look forward to seeing you again soon.