22.4 Properties of the Definite Integral 741 c a f(t)dt = b a f(t)dt + c b f(t)dt. abc t f cba t f acb t f cab t f cba t f cba t f Figure 22.33 6. a −a f(t)dt =0iff is odd; a −a f(t)dt =2 a 0 f(t)dt if f is even. f odd t f even t Figure 22.34 PROBLEMS FOR SECTION 22.4 For Problems 1 and 2, evaluate the following. 1. (a) 0 2 xdx (b) −1 4 (x + 1)dx 2. (a) 1 0 √ 1 − x 2 dx (b) −1 1 √ 1 − x 2 dx (Hint: Think about circles.) 3. Choose the correct answer and explain your reasoning. (a) a −a 1 1+x 2 dx = (i) 0 (ii) a 0 1 1+x 2 dx (iii) 2 a 0 1 1+x 2 dx (iv) − a 0 1 1+x 2 dx 742 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral (b) a −a x 1+x 2 dx = (i) 0 (ii) a 0 x 1+x 2 dx (iii) 2 a 0 x 1+x 2 dx (iv) − a 0 x 1+x 2 dx 4. (a) Explain why 3 0.5 1−lnx x 2 +1 dx > 4 0.5 1−ln x x 2 +1 dx. (Hint: Look at the sign of the integrand.) (b) Put in ascending order: 0, 1 1/e 1 − ln x x 2 + 1 dx, 2 1/e 1 − ln x x 2 + 1 dx, e 1/e 1 − ln x x 2 + 1 dx, 4 1/e 1 − ln x x 2 + 1 dx. 5. (a) Assume that a<b.Insert a “≥” or “≤” sign between the expressions below as appropriate. Explain your reasoning clearly. b a |f(t)|dt and b a f(t)dt (b) Under what circumstances will the two expressions be equal? 6. For each of the following, sketch a graph of the indicated region and write a definite integral (or, if you prefer, the sum and/or differences of definite integrals) that gives the area of the region. (a) The area between the horizontal line y = 4 and the parabola y = x 2 (b) The area between the line y = x + 1 and the parabola y = x 2 − 1 7. Put the following four integrals in ascending order (from smallest to largest). Explain, using graphs, how you can be sure that the order you gave is correct. π 0 sin tdt, π 0 2sin tdt, π 0 sin(2t) dt, 0 π sin tdt 8. Suppose 5 0 f(t)dt =10. Evaluate four out of the five expressions that follow. One of them you do not have enough information to evaluate. (a) 5 0 7f(t)dt (b) 5 0 (f (t) + 7)dt (c) 5 0 f(t)dt +7 (d) 5 0 7f(t +7)dt (e) −2 −7 7f(t +7)dt 9. If 5 0 f(t)dt =10 and 5 0 g(t) dt = 3 does it follow that (a) f (t)>g(t)for all t between 0 and 5? Explain. Mathematicians might write this statement in mathematical symbols as follows: f (t)>g(t)∀tin [0, 5]. The symbol “∀” is read “for all.” (b) f (t)>g(t)for some t between 0 and 5? Explain. Mathematicians might write this statement in mathematical symbols as follows: ∃ t in [0, 5] such that f (t)>g(t).The symbol “∃” is read “there exists.” 23 CHAPTER The Area Function and Its Characteristics The Big Picture Our two fundamental interpretations of the definite integral b a f(x)dx are: the signed area between the graph of f and the horizontal axis on the interval [a, b]; the net change in amount A(x) between x = a and x = b if f(x)is the rate of change of A(x). Note the parallel in construct between our interpretations of the definite integral and our interpretations of the derivative evaluated at a point. We have a geometric interpretation and a physical interpretation. (For the derivative at a point these were (i) the slope of a curve at a point and (ii) instantaneous rate of change of a quantity.) Not only did we study the derivative evaluated at a point, and the significance of this number, but we studied the derivative function, and interpreted it as the slope function or instantaneous rate of change function. The definite integral b a f(x)dx, where a and b are constants, is a number— whether interpreted as signed area or net gain (or loss) in amount. In this chapter we look at the area (or amount) function. We’ll introduce this function via some examples. 23.1 AN INTRODUCTION TO THE AREA FUNCTION x a f(t)dt ◆ EXAMPLE 23.1 Consider x −1 f(t)dt, where f is the constant function f(t)= 4. Is x −1 f(t)dt itself a function? If so, can it be expressed by an algebraic formula? 743 744 CHAPTER 23 The Area Function and Its Characteristics SOLUTION f t 4 4 –112 3 f(t) = 4 f t 4 4 –13 4 f(t) = 4 f t 4 4 –1 π π + 1 f t x 4 4 –1 x + 1 4dt –1 x ∫ 4dt –1 π ∫ 4dt –1 3 ∫ 4dt –1 2 ∫ Figure 23.1 Indeed, x −1 4 dt is a function. It is the function that gives the signed area under f(t)=4 between t =−1and t = x. This area depends upon x and is uniquely determined by x.In fact, because the area is a rectangle, we can compute it by multiplying the height, 4, by the base, x + 1, to obtain the function 4 · (x + 1) = 4x + 4. We could name this area function A(x), but that wouldn’t tell us that it’s the area function for f with anchor point −1, so we dress it up a little more and write −1 A f (x). 1 The formula −1 A f (x) = 4x + 4 holds for x ≤−1aswell as x>1. For x =−1wehave −1 A f (−1)= −1 −1 4dt = 0. Any area function will be zero at its anchor point. For x<−1weuse the endpoint reversal property of definite integrals to write x −1 4 dt =− −1 x 4dt =− 4(−1−x) = 4 + 4x. x 4 3 2 1 –1 –1 A f (x) Figure 23.2 For example, we evaluate −1 A f (−2) as follows: −1 A f (−2) = −2 −1 4dt =− −1 −2 4dt =−(4)(1) =−4. ◆ Computing Distances: General Principles In the previous example, and in many applications of integration to come, you will be computing both vertical distances and horizontal distances. 1 This chapter was inspired by Arnold Ostebee and Paul Zorn’s treatment of the area function in their calculus text. Calculus From Graphical, Numerical, and Symbolic Points of View, Saunders College Publishing, 1997. The notation A f is theirs. 23.1 An Introduction to the Area Function x a f(t)dt 745 vertical distance = (high y-value) − (low y-value) horizontal distance = (right x-value) − (left x-value) It is as simple as that. The location of the x- and y-axes is completely irrelevant. Be sure this makes sense to you as you look at Figure 23.3. 2 –1 y x 2 – (–1) = 3 –7 –3 –3 y y x x –3 – (–7) = –3 + 7 = 4 –1 – (–3) = –1 + 3 = 2 f(x) = y f(x 1 ) – g(x 1 ) y 1 q(y 1 ) – r( y 1 ) x = q(y) x = r(y) y x y x x 1 g(x) = y –1–2 1 – (–2) = 1 + 2 = 3 y x 1 Figure 23.3 Let’s return to the area function from Example 23.1 but change the anchor point from −1. Below we look at the area functions −2 A f (x), 0 A f (x), and 1 A f (x), where f(t)= 4 and the anchor points are −2, 0, and 1, respectively. f x x + 2 x – 1 t –21 4 f x x t 4 f x x t 4 8 –4 A f –2 A f –2 A f (x) = 4dt = 4 • (x + 2) 1 A f 0 A f The graph of A f is a straight line with slope of 4. –2 x ∫ 1 A f (x) = 4dt = 4 • (x – 1) 1 x ∫ 0 A f (x) = 4dt = 4x 0 x ∫ Figure 23.4 Notice that changing the anchor point only changes the area function by an additive constant. EXERCISE 23.1 Argue that the formulas given for the functions −2 A f (x), 0 A f (x), and 1 A f (x) above are valid for x to the left of the anchor point. Definition For a continuous function f and a constant “a” in the domain of f we define a A f (x) to be x a f(t)dt. a A f (x) gives the signed area between the graph of f and the t-axis from t = a to t = x. Therefore, we will refer to it as the area function. 746 CHAPTER 23 The Area Function and Its Characteristics Amount Added, Accumulation, Accruement, In this chapter we will look at the characteristics of the function a A f (x) = x a f(t)dt.While our predominant interpretation of a A f (x) will be as the area function, we could give it other interpretations as well. For example, suppose t = 0 corresponds to noon, t is measured in hours, and f(t)gives the rate, in gallons per hour, at which water is leaving a storage tank. Then −1 A f (x) = x −1 f(t)dtis the amount of water that leaves the tank between 11:00 a.m. and time x. For instance, −1 A f (3) = 3 −1 f(t)dtis the amount of water that has left the tank between 11:00 a.m. and 3:00 p.m. When f gives a rate, A f (x) represents the net change in amount. So, we can think of the A as standing for Area (signed area), or as standing for Amount where by “amount” we mean “amount added.” The “A” can stand for accretion (an increase by addition), accumulation (a collection over time), or accruement (an amount added). Certainly “A” is a convenient letter for this function. PROBLEMS FOR SECTION 23.1 1. Let f(t)=7. We’ll define three area functions. The difference between their definitions is the anchor point. We’ll denote them as follows. 0 A f (x) = x 0 f(t)dt 2 A f (x) = x 2 f(t)dt 3 A f = x 3 f(t)dt (a) Find a formula (not involving an integral) for 0 A f (x), where x ≥ 0. What is 0 A f (−3)? Does your formula work in general for negative x? (b) Find a formula (not involving an integral) for 2 A f (x), where x ≥ 2. Does your formula work for x<2aswell? (c) Find a formula (not involving an integral) for 3 A f (x), where x ≥ 3. Does your formula work for x<3aswell? (d) Graph the functions 0 A f (x), 2 A f (x), and 3 A f (x). (e) Differentiate 0 A f (x), 2 A f (x), and 3 A f (x). 2. Let f(t)=2t.We’ll define three area functions. The difference between their defini- tions is the anchor point. We’ll denote them as follows: 0 A f (x) = x 0 f(t)dt 2 A f (x) = x 2 f(t)dt 3 A f = x 3 f(t)dt (a) Find a formula (not involving an integral) for 0 A f (x), where x ≥ 0. What is 0 A f (−3)? Does your formula work in general for negative x? (b) Find a formula (not involving an integral) for 2 A f (x), where x ≥ 2. Does your formula work for x<2aswell? (c) Find a formula (not involving an integral) for 3 A f (x), where x ≥ 3. Does your formula work for x<3aswell? (d) Graph the functions 0 A f (x), 2 A f (x), and 3 A f (x). (e) Differentiate 0 A f (x), 2 A f (x), and 3 A f (x). 3. Let f(t)=sin t. Let A f (x) = x 0 sin tdt,where x ≥ 0. 23.2 Characteristics of the Area Function 747 (a) Put the following in ascending order, with “<” or “=” signs between them. A f (0), A f π 2 , A f (π), A f 3π 2 , A f (2π) (b) For what values of x is A f (x) = 0? (c) For what values of x is A f (x) negative? (d) Fow what values of x is A f (x) maximum? 4. The rate that water is entering a tank is given by f(t)=40 − 2t gallons/minute where t is measured in minutes past noon. (a) Interpret 0 A f (x) in words, for x ≥ 0. (b) For what values of x is 0 A f(x) increasing? (c) At what time is the water level in the tank the same as it was at noon? At this time, what is the value of 0 A f (x)? 23.2 CHARACTERISTICS OF THE AREA FUNCTION ◆ EXAMPLE 23.2 Let 1 A f (x) = x 1 f(t)dt, where the graph of f is drawn below and the domain of A f is restricted to 1 ≤ x ≤ 12. f t 1–1–2–3 2 3 4 5 6 7 8 9 10 11 12 Figure 23.5 (a) On what interval(s) is the function 1 A f (x) increasing? What characteristic of f ensures that 1 A f (x) is increasing? (b) On what interval(s) is the function 1 A f (x) decreasing? What characteristic of f ensures that 1 A f (x) is decreasing? (c) Where on [1, 12] does 1 A f (x) have its local maxima and minima? Where on [1, 12] does 1 A f (x) attain its absolute maximum and minimum values? (d) On what interval(s) is the function 1 A f (x) concave up? Concave down? What characteristic of f ensures that 1 A f (x) is concave up and what characteristic ensures that 1 A f (x) is concave down? SOLUTIONS (a) 1 A f (x) is increasing on [1, 6] and [10, 12]. Wherever f is positive, the area function increases with x, because a positive contribution to area is being made. (b) 1 A f (x) is decreasing on [6, 10]. Wherever f is negative, the accumulated area de- creases with x, because a negative contribution is being made. 748 CHAPTER 23 The Area Function and Its Characteristics (c) On the number line below we indicate where 1 A f (x) is increasing and where it is decreasing. graph of 1 A f sign of f 1 6++–10 12 The local maximum occurs at x = 6; the local minimum at x = 10. To identify where 1 A f (x) takes on its absolute maximum and minimum values, we must also consider the endpoints of the domain, x = 1 and x = 12. The absolute minimum occurs either at x = 1orx=10. We must compare the value of the function at these two points. 1 A f (1) = 1 1 f(t)dt =0. For 1 <x≤12, 1 A f (x) is always positive, since between t = 1 and any value greater than 1 there is always more positive accumulated area than there is negative accumulated area. Therefore the absolute minimum value of 1 A f on [1, 12] is 0 and it is attained at x = 1. To find out where 1 A f attains its absolute maximum value, we must compare the values at x = 6 and x = 12. The area between 6 and 10 gives a negative contribution and is larger in magnitude than the positive contribution between 10 and 12, so 1 A f (12) is smaller than 1 A f (6). The absolute maximum is attained at x = 6. f t 1–1–2–3 2 3 4 5 6 7 8 9 10 11 12 Figure 23.6 (d) We’ll break this discussion into cases. We begin by looking at 1 A f on regions where f is positive. Where f is constant, area is accumulating at a constant rate; thus, 1 A f is increasing at a constant rate and is neither concave up nor concave down. Where f is positive and increasing, the area function is increasing at an increasing rate; 1 A f is concave up. Where f is positive and decreasing, the area function is increasing at a decreasing rate; 1 A f is concave down. Now let’s look at 1 A f on regions where f is negative. Where f is constant, the area function is decreasing at a constant rate and is neither concave up nor concave down. Where f is negative and increasing (i.e., f is getting closer to zero), 1 A f is decreasing by smaller and smaller amounts, so 1 A f is concave up. Where f is negative and decreasing, 1 A f is decreasing more and more steeply, so 1 A f is concave down. 23.2 Characteristics of the Area Function 749 We conclude that where f is increasing, 1 A f is concave up; where f is decreasing, 1 A f is concave down. 1 A f is concave up on [1, 2), (4, 5), (8, 12] and concave down on (2, 4), (5, 8). ◆ ◆ EXAMPLE 23.3 Let 1 A f be the area function given by x 1 f(t)as in Example 23.2, but this time let’s enlarge the domain to [−3, 12]. We already know the behavior of 1 A f on [1, 12], so we’ll concentrate on the interval [−3, 1]. We’re interested in determining whether the conclusions we drew above when considering 1 A f (x) for x to the right of the anchor point x = 1 also hold to the left of this anchor point. In particular, we’d like to answer the following questions. (a) Where f is positive, is 1 A f increasing? Where f is negative, is 1 A f decreasing? (b) Where f is increasing, is 1 A f concave up? Where f is decreasing, is 1 A f concave down? The graph of f is reproduced below. f t 1–1–2–3 2 3 4 5 6 7 8 9 10 11 12 Figure 23.7 SOLUTION Direct analysis of the area function to the left of the anchor point can sometimes make your head spin. 2 Instead, it will be simpler just to move the anchor point so that there are no values of the domain to the left of it. We’ll look at the function −3 A f and then show that 1 A f and −3 A f differ only by an additive constant; −3 A f and 1 A f are vertical translates. −3 A f is simple to analyze using the same logic as in Example 23.2. Where f is positive, −3 A f is increasing; where f is negative, −3 A f is decreasing. Where f is increasing, −3 A f is concave up; where f is decreasing, −3 A f is concave down. How are the functions −3 A f and 1 A f related? How can we apply our knowledge of the former to the latter? −3 A f (x) = x −3 f(t)while 1 A f (x) = x 1 f(t)dt. From the splitting interval property we know that b a f(t)dt = c a f(t)dt + b c f(t)dt. Therefore, 2 You need to use the fact that 1 A f (x) = x 1 f(t)dt =− 1 x f(t)dtin combination with the idea that if a sequence of numbers is increasing (e.g., 1, 2, 3), then when negated the sequence is decreasing (−1, −2, −3). 750 CHAPTER 23 The Area Function and Its Characteristics x −3 f(t)dt = 1 −3 f(t)dt + x 1 f(t)dt −3 A f (x) = 1 −3 f(t)dt + 1 A f (x) −3 A f (x) − 1 −3 f(t)dt = 1 A f (x). f t 1–1–2–3 2 3 4 5 6 7 8 9 10 11 12 –3 A f (1) Figure 23.8 1 −3 f(t)dt is a constant,so 1 A f (x) = −3 A f (x) + K for some constant K. 1 A f (x) and −3 A f (x) differ by a constant, so their graphs are vertical translates of one another. Therefore, where −3 A f (x) is increasing, 1 A f (x) is also increasing. Where −3 A f (x) is decreasing, 1 A f (x) is also decreasing. Similarly, where −3 A f (x) is concave up, 1 A f (x) is also concave up, and likewise for concave down. We summarize below. Where f is positive, 1 A f (x) is increasing; where f is negative, 1 A f (x) is decreasing. Where f is increasing, 1 A f (x) is concave up; where f is decreasing, 1 A f (x) is concave down. In fact, any anchor point “a” in the interval [−3, 12] could be chosen and a A f (x) would simply be a vertical translate of −3 A f (x). For any constant “a” we know that x −3 f(t)dt = a −3 f(t)dt + x a f(t)dt x −3 f(t)dt =K + x a f(t)dt. ◆ Let’s look at a case that’s more complicated than the constant function of Example 23.1 but more concrete and computational than Example 23.2. ◆ EXAMPLE 23.4 Let f(t)=2− t. Find formulas for −2 A f (x), 0 A f (x), and 1 A f (x), the area functions anchored at −2, 0, and 1, respectively. How are these area functions related? Restrict the domains of these functions to the interval [−2, 6]. . net change in amount. So, we can think of the A as standing for Area (signed area), or as standing for Amount where by “amount” we mean “amount added.” The “A” can stand for accretion (an increase. geometric interpretation and a physical interpretation. (For the derivative at a point these were (i) the slope of a curve at a point and (ii) instantaneous rate of change of a quantity.) Not only did. just to move the anchor point so that there are no values of the domain to the left of it. We’ll look at the function −3 A f and then show that 1 A f and −3 A f differ only by an additive constant; −3 A f and 1 A f are