[MUSIC] As a first example, let us consider a shape which is formed by a polynomial expression for this side. So let's say y = alpha x to the n power, so it's an nth order polynomial. And we will try to calculate the area at the center of the shape enclosed by this polynomial, the x axis and the vertical axis passing through point B. Let's assume that this height here is h. Now, based on what we said earlier, the area can be calculated as the integral of little a dA whereas little dA, we can consider in this case an infinitesimal strip with width dx and with height y y(x) given by the expression above. So this becomes integral of y(x) alpha x to the n power times the x, integral from zero to b. This becomes alpha xn plus 1 over n plus 1. So alpha b to the n + 1 over n + 1. Since the expression for y(x) is given here, it is clear that h is obtained when we apply x = b. So it is alpha b to the n power. So alpha b to the n power is equal to h. So this can be rewritten as bh over n+1. So what we found is that the area of such a polynomial shape is given by base times height over the order of the polynomial plus 1. This actually can be easily confirmed, that it agrees with results. We know, basic results we know. That's the case of a rectangle which is the case when the order of the polynomial is equal to zero. So when the order of polynomial is equal to zero, when n = 0, then this becomes a constant so we end up with a rectangular shape. And what we found, assuming the base is b and the height is h, we find that the area is b times h over 0 plus 1, over 1. In the case of a linear shape, we obtain a triangle again with base b and height h. This is the case of first order polynomial and in that case, the area based on the expression we calculated is b times h over 1 + 1 over 2. Next, in the case of a parabolic shape, You see that when n = 2, the area becomes the bh over 2 plus 1, 3, etc. And we can see that as we move to higher orders of polynomials, this shape, this scale becomes more and more pushed to the right. And the limit of the area, when n goes to infinity, Is 0. That means this area becomes smaller and smaller. Now next, let us calculate the central distance, xc. The centroid, c, so the central distance from here to here xc. Xc as we learned is the weighted average of xs, so it's x times dA divided by the integral of dA which is the area. The numerator here also is referred to as first moment, Of a with respect to the y-axis. So first moment, that means the moment of A, first moment, because x is to the first power, okay? Integral of x dA is referred as the first moment of A. Let us calculate, for this particular case, what is the integral of x dA, the first moment? Integral of x dA becomes x dA, we said is alpha x to the n power times dx, integral from 0 to b. This becomes integral of alpha x to the n plus 1. The result is alpha x to the n plus 2 over n plus 2 from zero to b. So we end up with alpha b to the n + 2 over n + 2. And given that alpha b to the n is equal to 8, this can be written as b square, 8 over n +2. Therefore using this result, the central distance is the ratio of the first moment, which we just calculated, Divided by the area, which we calculated before, bh over n + 1. bh cancels out and we obtain n + 1 over n + 2 times b. Let's look at the numbers that this expression gives for this particular cases. When n equal to zero, this centroid, this case, xc central distance, xc, is one half of b. In the case of a triangle where n equal to one, we get that xc is two thirds of b. In the case where n equal to 2, xc = three quarters of b. And clearly As we consider So c moves from one-half to two-thirds to three-quarters. And the limit of xc as n goes to infinity is clearly equal to b. That means the centroid approaches the right boundary of this set.