Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 78 pot

23.2 Characteristics of the Area Function 751 f t 1 1 –1–2 2 2 3456 Figure 23.9 SOLUTION Let’s begin with −2 A f (x) because none of the domain is to the left of the anchor point. −2 A f (x) is increasing on [−2, 2] and decreasing on [2, 6]. −2 A f (x) is concave down on [−2, 6] because f is decreasing. graph of –2 A f sign of f –2 2 –+ 6 Finding a Formula for −2 A f (x). First consider −2 A f (x) for x on the interval [−2, 2]. −2 A f (x) is the area of the trapezoid bounded by f(t)=2−t, the t-axis, t =−2, and t = x. (See Figure 23.10.) Its area is 1 2 (h 1 + h 2 ) · b = 1 2 [4 + (2 − x)] · (x − (−2)) = 1 2 (6 − x) · (x + 2) = 1 2 (12 − x 2 + 4x) =− x 2 2 + 2x + 6. −2 A f (x) =− x 2 2 + 2x + 6on[−2, 2]. f t x 1–1–22 2 34 4 56 2 + x length = 2–x f t x 1–1–22 2 1 34 4 56 2 + x length = 2–x (where x is negative) x is negative Figure 23.10 For x on (2, 6] to find −2 A f (x) we must subtract from the area of the triangle above the t-axis the area of the appropriate triangle below. (See Figure 23.11.) 752 CHAPTER 23 The Area Function and Its Characteristics f t –22 2 4 4 length = 0 – (2–x) = x – 2 f(x) = 2–x x – 2 x Figure 23.11 (area of triangle above the t -axis) − (area of triangle below the t-axis) =  1 2 · 4 · 4  −  1 2 · (x − 2) · (x − 2)  = 8 − 1 2 (x 2 − 4x + 4) = 8 − x 2 2 + 2x − 2 =− x 2 2 + 2x + 6 So −2 A f (x) =− x 2 2 + 2x + 6 on the entire domain given. Once we have found −2 A f (x), finding 0 A f (x) and 1 A f (x) is not difficult because the functions differ only by additive constants. Finding a Formula for 0 A f (x). 0 A f (x) =  x 0 f(t)dt. To relate 0 A f (x) to −2 A f (x) we use the splitting interval property,  b a f(t)dt =  c a f(t)dt +  b c f(t)dt.  x −2 f(t)dt =  0 −2 f(t)dt +  x 0 f(t)dt −2 A f (x) =  0 −2 f(t)dt + 0 A f (x), where  0 −2 f(t)dt is the area of the trapezoid shaded in Figure 23.12.  0 −2 f(t)dt = 1 2 · (4 + 2) · 2 = 6 −2 A f (x) = 6 + 0 A f (x) So 0 A f (x) = −2 A f (x) − 6. −2 A f (x) =− x 2 2 + 2x + 6; therefore, 0 A f (x) =− x 2 2 + 2x. 23.2 Characteristics of the Area Function 753 f t –2 2 –2 A f (0) 4 Figure 23.12 Finding a Formula for 1 A f (x). Similarly,  x −2 f(t)dt =  1 −2 f(t)dt +  x 1 f(t)dt. −2 A f (x) =  1 −2 f(t)dt + 1 A f (x), where  1 −2 f(t)dt is the area shaded in Figure 23.13. t f –2 –11 2 –2 A f (0) Figure 23.13 −2 A f (1) = 1 2 (4 + 1) · 3 = 15 2 −2 A f (x) = 15 2 + 1 A f (x) So 1 A f (x) = −2 A f (x) − 15 2 . −2 A f (x) =− x 2 2 + 2x + 6; therefore, 1 A f (x) =− x 2 2 + 2x − 1.5. x –2 A f 1 A f 0 A f A f –1 –1 123456 6 –2 Figure 23.14 Alternatively, we could have found 1 A f (x) by starting with 1 A f (x) =− x 2 2 + 2x + K. Knowing that 1 A f (1) = 0 we can solve for K. 754 CHAPTER 23 The Area Function and Its Characteristics 0 =− 1 2 + 2 + K K =− 3 2 Therefore, 1 A f (x) =− x 2 2 + 2x − 1.5. Notice that A f (x) =− x 2 2 + 2x + K, where the value of K depends upon the anchor point. The area function graphs are vertical translates of one another. In fact, d dx A f (x) = d dx  − x 2 2 + 2x + K  =−x+2=f(x). ◆ Interesting Howmuch of this can we generalize? We begin by generalizing the observation that area functions of f with different anchor points differ only by an additive constant. This can be shown using the splitting interval property of integrals. Let a and c be constants in the domain of f . Then  x a f(t)dt =  c a f(t)dt +  x c f(t)dt,so a A f (x) =  c a f(t)dt + c A f (x), where  c a f(t)dt,the area under f from a to c, is a constant. 3 How are A f and f related? Let’s gather our observations from the previous examples and see what we can make of them. Observations Area functions for f are vertical translates of one another. Where f is positive, A f is increasing; where f is negative, A f is decreasing. Where f is increasing, A f is concave up; where f is decreasing, A f is concave down. These statements are true regardless of the anchor point (so we’ve omitted it from the notation). In addition, from the examples in which we were able to arrive at algebraic expressions for the area functions (Examples 23.1 and 23.4), we made the following intriguing observation. 3  c a f(t)dt can also be written as a A f (c). 23.2 Characteristics of the Area Function 755 In two examples we find the derivative of the area function A f is f itself. For the time being this is only an observation of a phenomenon that presented itself in two examples. It does not have the status of a fact. This last observation ties together the first three. Where the derivative of a function is positive, the function is increasing; where the derivative is negative, the function is decreasing. Where the derivative of a function is increasing, the function is concave up; where the derivative is decreasing, the function is concave down. If two functions have the same derivative, then they differ by a constant. 4 We have a conjecture in the making. Conjecture The derivative of the area function A f is f itself. We will prove this conjecture in the next section and find that it has beautiful and far-reaching consequences. PROBLEMS FOR SECTION 23.2 1. Let A f (x) be the area function given by A f (x) =  x 0 f(t)dt,where 0 ≤ x ≤ 11. The graph of f is given below. f t 1 2 34567891011 (a) On what interval(s) is the function A f (x) increasing? (b) On what interval(s) is the function A f (x) decreasing? (c) What is A f (0)? (d) Is A f (x) ever negative on the interval [0, 11]? (e) On [0, 11], where is A f (x) maximum? Minimum? 4 This is the Equal Derivatives Theorem. For the proof, refer to Appendix C. 756 CHAPTER 23 The Area Function and Its Characteristics 2. Below is the graph of f(t).f is an odd function. f t (a) Which of the graphs below could be the graph of 0 A f (x) =  x 0 f(t)dt?Explain your reasoning. (b) Which of the graphs below could be the graph of 2 A f (x) =  x 2 f(t)dt?Explain. x x x x –222 AB CD 3. Below is the graph of f(t). f t 1–1–223456 (a) If F(x)=  x 0 f(t)dt,which of the graphs given below could be a graph of F(x)? Explain your criterion. (b) If G(x) =  x −2 f(t)dt,which of the graphs given below could be a graph of G(x)? A t 1–1–223456 B t 1–1–223456 C t 1–1–223456 23.3 The Fundamental Theorem of Calculus 757 23.3 THE FUNDAMENTAL THEOREM OF CALCULUS We are now on the threshold of obtaining a truly wonderful and important result. At the end of the previous section, we conjectured that the derivative of the area function A f is f . In this section, we give a geometric justification of this conjecture. Then we’ll see what an amazing and extremely useful result it is. Proof that if f is a Continuous Function, then d dx A f (x) = f(x) Let f be a continuous bounded function and c a constant in the domain of f . Consider the function A f (x) =  x c f(t)dt, the signed area between the graph of f and the t-axis between t = c and t = x. To calculate d dx A f (x) let’s begin by looking at A f (x + x) − A f (x), the numerator of the difference quotient. We will show that lim x→0 A f (x + x) − A f (x) x = f(x) by using the “Squeeze Theorem” or “Sandwich Theorem.” For simplicity’s sake, let’s begin by assuming that f is positive and increasing on the interval [x, x + x], where x is positive. Then we can represent A f (x + x) − A f (x) by the region shaded in Figure 23.15(b). (Later we will drop the requirements that f is positive and increasing; our argument can be adapted easily to the general case.) f t f t A f (x) (a)(b) A f (x + ∆ x) A f (x + ∆ x) – A f (x) x + ∆ xx + ∆ xxxcc Figure 23.15 We can write the following inequalities. (Refer to Figure 23.15b.) area of smaller ≤ area under f(t) ≤ area of larger rectangle on [x, x + x] rectangle f (x)x ≤ A f (x + x) − A f (x) ≤ f(x+x)x It’s legal to divide by x because x is positive. f(x) ≤ A f (x+x)−A f (x) x ≤ f(x+x) Take the limit as x tends toward zero. lim x→0 f(x) ≤ lim x→0 A f (x+x)−A f (x) x ≤ lim x→0 f(x+x) The middle expression is the definition of derivative. lim x→0 f(x) ≤ d dx A f (x) ≤ lim x→0 f(x+x) Evaluate the limits. f(x) ≤ d dx A f (x) ≤ f(x) 758 CHAPTER 23 The Area Function and Its Characteristics d dx A f (x) is being “squeezed” on both sides by f(x);itmust be equal to f(x). This argument can be adapted to deal with functions that are not necessarily positive and increasing on [x, x + x]. The fact that [x, x + x] is a closed, bounded interval and f is continuous means we can always find M and m, the absolute maximum and minimum values of f on [x, x + x], respectively. We can set up the inequality  minimum value of f on [x, x + x]  x ≤ A f (x + x) − A f (x) ≤  maximum value of f on [x, x + x]  x and proceed as above. 5 We obtain lim x→0  minimum value of f on [x, x + x]  ≤ d dx A f (x) ≤ lim x→0  maximum value of f on [x, x + x]  . f is continuous, so as x → 0 the minimum and maximum values of f on [x + x] both approach f(x). The conclusion is that for any continuous function f , d dx A f (x) = f(x).Inwords, this means that the rate of change of the area function at any point is the height of the function f at that point. Think about this geometrically for a minute; it makes sense. The Fundamental Theorem of Calculus, version I If f is continuous on [a, b] and c ∈ [a, b], then the function A f (x) =  x c f(t)dt x∈[a,b] is differentiable on (a, b) and d dx A f (x) = f(x). Wecan rewrite the Fundamental Theorem of Calculus without explicit reference to the area function: If f is continuous on [a, b], then  x a f(t)dt is differentiable on (a, b) and d dx  x a f(t)dt =d(x). In the next chapter we will see that this result is amazingly useful. 5 This inequality assumes x > 0. If x < 0, simply switch the inequalities to ≥ and switch again after dividing by x. 23.3 The Fundamental Theorem of Calculus 759 PROBLEMS FOR SECTION 23.3 1. The graph of f(t)is given below. Let F(x)=  x −6 f(t)dt and G(x) =  x 1 f(t)dt. f t –6 –2–4 21 3.5 46 (a) Where on [−6, 6] is F(x)increasing? Decreasing? Concave up? Concave down? (b) Where on [−6, 6] is G(x) increasing? Decreasing? (c) Where on [−6, 6] does F have a local minimum? (d) Explain why the local extrema of F and G occur at the same values of x. (e) How are the graphs of F and G related? 2. Let C(q) be the cost of producing q widgets. The graph below gives dC dq , the mar- ginal cost of producing widgets. Economists interpret the term “marginal cost” as the additional cost of producing an additional item. dC dq q Assume that fixed costs, the cost when no widgets are being produced, is F . Graph C(x) = F +  x 0 dC dq dq. 3. Below is a graph of r(t), the rate of flow of liquid in and out of a tank. t = 0 corresponds to noon. r(t) in liters/hr. t –2 –14563217 760 CHAPTER 23 The Area Function and Its Characteristics Suppose you know the following.  7 −2 r(t) dt = 4  6 2 r(t) dt = 0  7 2 r(t) dt = 1    (a) What is the net change in the amount of liquid in the tank between 10:00 a.m. and 6:00 p.m.? (b) If there are 50 liters in the tank at 2:00 p.m., how many liters are in the tank at 7:00 p.m.? 4. A jug of cold lemonade is loaded into a cooler to be brought on a summer picnic. The lemonade is 40 degrees at the start of the trip. During the hour and a half drive to the park the lemonade gets steadily warmer at a rate of 1 degree every 15 minutes. Upon arriving at the park it is carried (in the cooler) to the river via a 40-minute hike. During the hike the lemonade gains 1 degree every 10 minutes. Once it reaches the river it is poured into cups. Now that it is out of the cooler the lemonade warms at a rate proportional to the difference between the temperature of the air and that of the liquid. Half an hour later it has warmed 12 degrees. Let L(t) be the temperature of the lemonade at time t, where t is the number of minutes into the trip. (a) Sketch dL dt versus time. Label important points. (b) Sketch L(t) versus time. Label important points. Be sure your pictures of L and dL/dt are consistent. . interval and f is continuous means we can always find M and m, the absolute maximum and minimum values of f on [x, x + x], respectively. We can set up the inequality  minimum value of f on [x,. bounded function and c a constant in the domain of f . Consider the function A f (x) =  x c f(t)dt, the signed area between the graph of f and the t-axis between t = c and t = x. To calculate d dx A f (x). to c, is a constant. 3 How are A f and f related? Let’s gather our observations from the previous examples and see what we can make of them. Observations Area functions for f are vertical translates