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Antiderivatives and the Area Problem Chapter 7 antiderivatives and the area problem Let me begin by defining the terms in the title: 1. an antiderivative of f is another function F such that F 0 = f. 2. the area problem is: ”find the area of a shape in the plane" This chapter is concerned with understanding the area problem and then solving it through the fundamental theorem of calculus(FTC). We begin by discussing antiderivatives. At first glance it is not at all obvious this has to do with the area problem. However, antiderivatives do solve a number of interesting physical problems so we ought to consider them if only for that reason. The beginning of the chapter is devoted to understanding the type of question which an antiderivative solves as well as how to perform a number of basic indefinite integrals. Once all of this is accomplished we then turn to the area problem. To understand the area problem carefully we'll need to think some about the concepts of finite sums, se- quences and limits of sequences. These concepts are quite natural and we will see that the theory for these is easily transferred from some of our earlier work. Once the limit of a sequence and a number of its basic prop- erties are established we then define area and the definite integral. Finally, the remainder of the chapter is devoted to understanding the fundamental theorem of calculus and how it is applied to solve definite integrals. I have attempted to be rigorous in this chapter, however, you should understand that there are superior treatments of integration(Riemann-Stieltjes, Lesbeque etc..) which cover a greater variety of functions in a more logically complete fashion. The treatment here is more or less typical of elementary calculus texts. 245 246 CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM 7.1 indefinite integration Don't worry, the title of this section will make sense later. 7.1.1 why antidifferentiate? The antiderivative is the opposite of the derivative in the following sense: Definition 7.1.1. antiderivative. If f and F are functions such that F 0 = f then we say that F is an antiderivative of f. dF Example 7.1.2. Suppose f(x) = x then an antiderivative of f is a function F such that dx = x. We could 2 d 2 try x but then dx (x ) = 2x has an unwanted factor of 2. What to do? Just adjust our guess a little: try 1 2 d 1 2 1 d 2 1 F (x) = 2 x . Note that dx ( 2 x ) = 2 dx (x ) = 2 (2x) = x: kt 1 kt Example 7.1.3. Let k be a constant. Suppose g(t) = e then we guess G(t) = k e and note it works; d 1 kt kt kt 1 kt dt ( k e ) = e therefore g(t) = e has antiderivative G(t) = k e . d Example 7.1.4. Suppose h(θ) = cos(θ). Guess H(θ) = sin(θ) and note it works; dθ (sin(θ)) = cos(θ). Obviously these guesses are not random. In fact, these are educated guesses. We simply have to think about how we differentiated before and just try to think backwards. Simple enough for now. However, we 1 2 should stop to notice that the antiderivative is far from unique. You can easily check that F (x) = 2 x + c1, 1 kt G(t) = k e + c2 and H(θ) = sin(θ) + c3 are also antiderivatives for any constants c1; c2; c3 2 R. Proposition 7.1.5. antiderivatives differ by at most a constant. If f has antiderivatives F1 and F2 then there exists c 2 R such that F1(x) = F2(x) + c. dF1 dF2 dF1 dF2 Proof: We are given that dx = f(x) and dx = f(x) therefore dx = dx . Hence, by Proposition 6.1.10 we find F1(x) = F2(x) + c. To understand the significance of this constant we should consider a physical question. Example 7.1.6. Suppose that the velocity of a particle at position x is measured to be constant. In particular, dx dx suppose that v(t) = dt and v(t) = 1. The condition v(t) = dt means that x should be an antiderivative of v. For v(t) = 1 the form of all antiderivatives is easy enough to guess: x(t) = t + c. The value for c cannot be determined unless we are given additional information about this particle. For example, if we also knew that at time zero the particle was at x = 3 then we could fit this initial data to pick a value for c: x(0) = 0 + c = 3 ) c = 3 ) x(t) = t + 3 : For a given velocity function each antiderivative gives a possible position function. To determine the precise position function we need to know both the velocity and some initial position. Often we are presented with a problem for which we do not know the initial condition so we'd like to have a mathematical device to leave open all possible initial conditions. 7.1. INDEFINITE INTEGRATION 247 Definition 7.1.7. indefinite integral. If f has an antiderivative F then the indefinite integral of f is given by: Z 0 f(x)dx = fG(x) j G (x) = f(x)g = fF (x) + cjc 2 Rg: However, we will customarily drop the set-notation and simply write Z f(x)dx = F (x) + c where F 0(x) = f(x): The indefinite integral includes all possible antiderivatives for the given function. Technically the indefinite integral is not a function. Instead, it is a family of functions each of which is an antiderivative of f. Example 7.1.8. Consider the constant acceleration problem1; we are given that a = −g where g = 9:8m=s2 dv and a = dt . We can take the indefinite integral of the equation: dv Z = −g ) v(t) = −g dt = −gt + c : dt 1 dy Furthermore, if v = dt then dy Z 1 = −gt + c ) y(t) = −gt + c dt = − gt2 + c t + c : dt 1 1 2 1 2 Therefore, we find the velocity and position are given by formulas 1 v(t) = c − gt y(t) = c + c t − gt2: 1 2 1 2 If we know the initial velocity is vo and the initial position is yo then v(0) = vo = c1 − 0 ) v(t) = vo − gt 1 y(0) = y = c − 0 − 0 ) y(t) = y + v t − gt2 o 2 o o 2 These formulas were derived by Galileo without the benefit of calculus. Instead, he used experiment and a healthy skepticism of the philosophical nonsense of Aristotle. The ancient Greek's theory of motion said that if something was twice as heavy then it falls twice as fast. This is only true when the objects compared have air friction clouding the dynamics. The equations above say the objects' motion is independent of the mass. 1here F = ma is −mg = ma so a = −g but that's physics, I supply the equation of motion in calculus. You just have to do the math. 248 CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM Remark 7.1.9. redundant comment (again). The indefinite integral is a family of antiderivatives: R f(x) = F (x) + c where F 0(x) = f(x). The following equation shows how indefinite integration is undone by differentiation: d Z f(x) dx = f(x) dx the function f is called the integrand and the variable of indefinite integration is x. Notice the constant is obliterated by the derivative in the equation above. Leibniz' notation intentionally makes you think of cancelling the dx's as if they were tiny quantities. Newton called them fluxions. In fact calculus was sometimes called the theory of fluxions in the early 19-th century. Newton had in mind that dx was the change in x over a tiny time, it was a fluctuation with respect to a time implicit. We no longer think of calculus in this way because there are easier ways to think about foundations of calculus. That said, it is still an intuitive notation and if you are careful not to overextend intuition df df du it is a powerful mnemonic. For example, the chain rule dx = du dx . Is the chain rule just from multiplying by one? No. But, it is a nice way to remember the rule. A differential equation is an equation which involves derivatives. We have solved a number of differential equations in this section via the process of indefinite integration. The example that follows doesn't quite fit the same pattern. However, I will again solve it by educated guessing2 . Example 7.1.10. A simple model of population growth is that the rate of population growth should be directly proportional to the size of the population P . This means there exists k 2 R such that dP = kP: dt Fortunately, we just did Example 7.1.3 where we observed that Z 1 ekt dt = ekt + c k 1 kt So we know that one solution is given by P (t) = k e . Change variables by substituting u = ln(P ) so du 1 dP dP du du du dt = P dt thus dt = P dt . Hence we can solve P dt = kP or dt = k instead. This we can antidifferentiate kt+c1 c1 kt to find u(t) = kt + c1. Thus, ln(P ) = kt + c1 hence P (t) = e = e e . If the initial population is given c1 kt to be Po then we find P (0) = Po = e thus P (t) = Poe : The same mathematics govern simple radioactive decay, continuously compounded interest, current or voltage in an LR or RC circuit and a host of other simplistic models in the natural sciences.
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