This is the final video for this module on integral calculus and indeed for the entire course. There's no new material and there are no exercises tied to this video. So, afterwards you can move directly to the main quiz for this final module. Nevertheless, I think it's beneficial to have a video that creates unity by bringing ideas together. You'll notice many themes from the entire course permeate the narrative. Even just to let the story wash of you should be beneficial. If you have time to pause the video and check details here and there, you should find this video helpful in terms of general revision and developing your fluency with the techniques. Today I want to do something very special and come full circle to the discovery of the integral calculus in the 17th century by Isaac Newton, motivated by his remarkable estimate of the escape velocity of a rocket. This must have seemed like ridiculous science fiction at a time when space travel would have been unthinkable and absurd. In fact, Newton was thinking in terms of cannonballs, not rockets. There are apocryphal stories of him sitting under a tree, gazing serendipitously at the moon, stirred into action by an apple falling on his head. Who knows exactly what happened? In any case, we're the great beneficiaries of Newton's incredibly inventive imagination and vision and the rapid evolution of mathematics in the intervening centuries, supported by the great minds and genius of people such as Leibniz, Euler, Gauss, Agnesi, Riemann, and many others. What follows is an extended example that takes us back to the birth of the integral calculus in the 17th century. Recall the witch of Maria Agnesi in its simplest form with rule y equals 1 over x squared plus 1, with global maximum 1 where it crosses the y axis. What happens if we replace 1 in the denominator with a half? The curve gets stretched in the vertical direction and the y-intercept moves up to 2. What happens if we replace a half with a third? The curve stretches some more and the y-intercept moves up to 3. We can replace a third by a quarter and the y-intercept moves up to 4. We quickly see that if the fraction added to x squared in the denominator is 1 on k, then the y-intercept becomes k. What happens then if k heads off towards infinity? Well, the y-intercept of the witch must also head off towards infinity. But infinity isn't a number, so the curve can't cross the y-axis anymore. There must be some kind of explosion. And when the dust settles, the denominator becomes just x squared. So, we get the curve y equals 1 on x squared, which can be rewritten as x to the minus 2. And it's good revision to check the shape of the curve against the sign diagrams for the first and second derivatives. The first derivative is negative 2 x to the minus 3, that is, negative two over x cubed, with sign diagram indicating that the curve is increasing from the left and decreasing to the right with undefined derivative at x equals 0. The second derivative is 6 x to the minus 4, that is, 6 over x to the fourth, with sign diagram indicating that the curve is concave up, both from the left and to the right with undefined second derivative at x equals 0. So finally we do get a traditional witch's hat with a sharp point somewhere off in the infinite distance. The simple rule, y equals 1 on x squared, is especially important in mathematics and science and in particular features in the inverse square law used to model gravitation. Suppose we have two celestial bodies such as planets or stars separated by a distance of x units. I've drawn them as perfect spheres but they can be irregular, it doesn't matter. In any case, they have centers of mass somewhere and the distance x is measured between them. The body on the left has mass capital M and the body to the right has mass little m. Denote the force of gravitational attraction by F, which is a function of x. It makes sense that F should be proportional to both of the masses. The greater the mass, the greater the gravitational attraction. It also makes sense that F should be inversely proportional to the square of the distance. Remember, the surface area of a sphere is 4 Pi times the square of the radius. So you can imagine the gravitational force radiating uniformly outwards from a spherical body, distributed evenly over 4 Pi times the square of the distance from the center. The gravitational force on any unit of area is therefore proportional to the inverse of the total surface area of an expanding sphere, which in turn is proportional to the inverse of the square of the distance to the center of the sphere. The overall constant of proportionality is denoted by capital G, also known as the universal gravitational constant. This was all known to Newton in the 17th century. There's some controversy about who stated this fact first, but certainly Newton was the first person to explore the mathematical consequences. Armed with this information about gravitational forces, let's consider a rocket moving away from the Earth. Denote the radius of the Earth by capital R, and the distance from the center of the Earth to the position of the rocket, wherever it happens to be, by capital D. We can ask how much energy was expended in getting the rocket from the surface of the earth to its present position. The total energy expended can be described by the following definite integral, where the distance x ranges from R, when the rocket was being launched on the surface of the earth to to its present position D. The integrand is F of x, the gravitational force of attraction with the earth against which the rocket is propelling itself. In physics, if a force acts on an object that moves a certain distance, then the work done or energy expended is the product of the force with the distance. So multiplying the gravitational force F of x by some tiny change in the distance, represented by the differential dx, gives a tiny bit of energy required to move that tiny bit of distance. Adding up all these bits of energy, which we think of as being done continuously, is captured by the definite integral. Just think of the curly S integral symbol as summing up all of the energies captured by the products F of x times dx as x varies from R to D. But F of x is given by the formula G times capital M times little m over x squared. The little m is the mass of the rocket and capital M is the mass of the earth. We have all of the tools to evaluate this integral. The constants come out the front. We can rewrite the integrand as x to the minus 2 and apply the fundamental theorem of calculus by evaluating the anti-derivative minus x to the minus 1 between R and D, which becomes in a few steps the expression G capital M times little m times 1 over R minus 1 over D. Thus we find an expression for the amount of energy expended getting the rocket to its present position. We can ask what happens to this expression for the energy required as a rocket travels further out into space towards the distant stars as D approaches infinity and the Earth vanishes to a single point. As D goes to infinity, one over D goes to zero. Thus, to reach the stars, in principle, the energy required becomes G capital M times little m times one over R. We can rewrite this as the fraction G capital M times little m over R. At the time, Newton was in fact thinking about cannonballs rather than rockets making use of a continuous supply of fuel. You might recall in a very early video, we considered the trajectory of a cannonball shot directly upwards from the Earth's surface, and the graph of its displacement function turned out to be approximately a parabola. This all works out fine assuming we're close to the Earth's surface where the gravitational force may be regarded as constant. Up to this point, that was basically Newton's experience of cannonballs also, but he wanted to think about moving far away from the earth where the gravitational force can vary. For Newton, the question was how much energy should you transfer to a cannonball at the moment it's fired that it manages to escape the earth's gravitational field? The question then becomes what should be the initial velocity given to a cannonball such that it has enough energy to escape? The energy required is given by our formula, but now little m denotes the mass of the cannonball rather than the mass of the rocket. We just need to link the velocity v that we're seeking with the formula for the energy required. There's one more fact from physics: the energy of the cannonball is a half m v squared. So we have a half m v squared equals G times capital M times little m over R. So, to find the velocity, we just have to unravel this equation. Observe that the mass little m cancels from both sides so the velocity will, in fact, be independent of the mass of the cannonball. Rearranging and taking the positive square root yields a formula v equals the square root of two times G times M over R. But we want an actual number for the escape velocity. Getting back to the surface of the Earth, we have the universal formula for the force F, which we can rewrite as little m times capital G times capital M over R squared expressed as mass times acceleration. And Newton had observed that, near the surface of the Earth, the acceleration is about 9.8 metres per second squared. So, we have, at least, G M over R squared approximately equal to 9.8. The last part of the puzzle is having an estimate of the radius R of the Earth, but happily this was provided by the Greek scholar and mathematician Eratosthenes in about 200 BC, using clever measurements and some geometry, and was known to Newton and is about 6.37 times 10 to the sixth metres in scientific notation. Putting this all together gives v approximately equal to the square root of two times the acceleration 9.8 times the radius of the Earth, which comes to about 11,000 metres per second. Thus, to escape the earth, the projectile has to be given an initial velocity of about 11 kilometres per second, and the mathematics also shows this to be independent of the mass of the projectile. This is the final video in this course, An Introduction to Calculus, and I wanted you to get a glimpse of how it all began and what motivated Isaac Newton, who, in parallel with Gottfried Leibniz, invented calculus in the 17th century. We've spoken at length in earlier videos about Leibniz notation, and it appears implicitly in the exposition above leading to Newton's estimate of the escape velocity. Newton didn't have access to Leibniz notation nor did he have access to the smooth and elegant framework for formulating the mathematics, which evolved subsequently over the 18th and 19th centuries in which we, in modern times, too easily take for granted. This makes Newton's achievements and profound insights all the more remarkable. The first two modules of this course introduced ideas and techniques from pre-calculus, preparing for the third and fourth modules which introduced the differential calculus, which is really all about the study of slopes of tangent lines to curves. This fifth and final module introduced and discussed integral calculus, which is really all about the study of areas under curves. Remarkably, slopes of tangent lines and areas under curves turn out to be intrinsically linked, leading to the Fundamental Theorem of Calculus. We began this final module by illustrating how one can use areas under velocity curves to estimate displacement, using averages of lower and upper rectangular approximations, and replicated thought experiments, originally due to the ancient Greeks, involving limits of approximations, to discover the formula for the area of a circle and the area under a parabola. We then formalized the method of Riemann Sums using rectangular approximations to areas under curves over a given interval, and the definite integral, which is defined to be the limit of the Riemann sums, as the widths of subintervals, in partitions of the interval, go to zero. The definite integral captures precisely areas under a curve and has a symbolic description using an integral symbol like a stylized S which we can think of as a continuous sum, an integrand representing the rule of a function, a differential indicating the variable with respect to which we're finding the area, and terminals which tell us the endpoints of the relevant interval. We can calculate the definite integral exactly under certain conditions using the Fundamental Theorem of Calculus. This provides a simple and elegant formula involving an antiderivative of the integrand, that is, a function whose derivative is the integrand. Antiderivatives are captured symbolically using indefinite integrals, the result of removing terminals from the definite integral, and provide convenient descriptions of integration formulae. These typically involve a constant of integration, due to the fact that all antiderivatives of a given function differ by a constant. We introduced and illustrated the method of integration by substitution, closely related to the chain rule for differentiation. We discussed odd and even functions, related to rotational and reflectional symmetry, and noted, in particular, that the area under the curve of an odd function evaluates automatically to zero over a symmetric interval. We discussed logistic functions which are solutions of the logistic equation, which modifies the usual exponential growth model, by introducing an inhibition factor, which provides a maximal limiting size to the population. The graph of any logistic function is a sigmoid curve with a 180 degree rotational symmetry about its point of inflection. The purpose of this last video was to provide you with some context and insight concerning interacting ideas and motivation that led to the development of calculus from the 17th century. The mathematical treatment was only the briefest sketch and included important ideas from physics. You'll see a more thorough and rigorous treatment if you undertake further courses in advanced calculus. There are some notes accompanying this video which you can read out of interest and to support your mathematical development. But there are no new techniques required and no specific exercises. Thank you very much for watching, and especially for your sustained interest and participation over many weeks. I wish you well in your studies and look forward very much to seeing you again soon, exploring new horizons perhaps in future courses in mathematics.
