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22.4 Properties of the Denite Integral 741 . f f f a b c t a c b t c a b t f f f c b a t b a c t b c a t 6. 0 if is odd; Figure 22.33 2 0 if is even. f odd f even t t Figure 22.34 P R O B L E M S F O R S E C T I O N 2 2 . 4 For Problems 1 and 2, evaluate the following. 1. (a) 0 (b) 1 1 2. (a) 0 1 2 (b) 1 1 2 (Hint: Think about circles.) 3. Choose the correct answer and explain your reasoning. (a) 12 (i) 0 (ii) 0 12 (iii) 2 0 12 (iv) 12 742 CHAPTER 22 Net Change in Amount and Area: Introducing the Denite Integral (b) 12 (i) 0 (ii) 0 12 (iii) 2 12 (iv) 12 4. (a) Explain why 0.5 1ln 0.5 1ln . (Hint: Look at the sign of the integrand.) (b) Put in ascending order: 0, 1 1 ln , 1 2 1 ln , 1 1 ln , 1 4 1 ln . 1 5. (a) Assume that . Insert a “” or “” sign between the expressions below as appropriate. Explain your reasoning clearly. and (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 denite integral (or, if you prefer, the sum and/or differences of denite integrals) that gives the area of the region. (a) The area between the horizontal line 4 and the parabola 2 (b) The area between the line 1 and the parabola 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. sin , 0 2 sin , 0 0 sin2, sin 0 8. Suppose 5 10. Evaluate four out of the ve expressions that follow. One of them you do not have enough information to evaluate. (a) 5 7 (b) 5 7 (c) 5 7 (d) 0 7 7 (e) 7 7 7 9. If 5 10 and 5 3 does it follow that (a) for all between 0 and 5? Explain. Mathematicians might write this statement in mathematical symbols as follows: in [0, 5]. The symbol “” is read “for all.” (b) for some between 0 and 5? Explain. Mathematicians might write this statement in mathematical symbols as follows: in [0, 5] such that . The symbol “” is read “there exists.” C H A P T E R The Area Function and Its Characteristics The Big Picture Our two fundamental interpretations of the denite integral are: the signed area between the graph of and the horizontal axis on the interval [, ]; the net change in amount between and if is the rate of change of . Note the parallel in construct between our interpretations of the denite 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 signicance of this number, but we studied the derivative function, and interpreted it as the slope function or instantaneous rate of change function. The denite integral , where and 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 EXAMPLE 23.1 Consider 1 , where is the constant function 4. Is 1 function? If so, can it be expressed by an algebraic formula? itself a 743 744 CHAPTER 23 The Area Function and Its Characteristics SOLUTION f f f f 4 f(t) = 4 4 f(t) = 4 4 4 4 4 4 4 –1 1 2 t –1 3 t 3 4 2 3 4dt 4dt –1 –1 –1 t –1 x t + 1 x + 1 x 4dt 4dt –1 –1 Figure 23.1 Indeed, 4 isafunction.Itisthefunctionthatgivesthesignedareaunder 4 between 1 and . This area depends upon and is uniquely determined by . In fact, because the area is a rectangle, we can compute it by multiplying the height, 4, by the base, 1, to obtain the function 4 14 4. We could name this area function , but that wouldn’t tell us that it’s the area function for with anchor point 1, so we dress it up a little more and write 1 .1 The formula 1 4 4 holds for 1 as well as 1. For 1 we have 1 1 1 4 0. Any area function will be zero at its anchor point. For 1 we use the endpoint reversal property of denite integrals to write 1 4 4 41 1 4 4. –1Af(x) 4 3 2 1 –1 x Figure 23.2 For example, we evaluate 1 2 as follows: 2 1 1 2 4 4 414. 1 2 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 is theirs. 23.1 An Introduction to the Area Function 745 vertical distance (high -value) (low -value) horizontal distance (right -value) (left -value) It is as simple as that. The location of the - and -axes is completely irrelevant. Be sure this makes sense to you as you look at Figure 23.3. y y y f(x) = y 2 2 – (–1) = 3 –1 x –3 –7 –3 – (–7) = –3 + 7 = 4 x1 x g(x) = y f(x1) – g(x1) y y x = r(y) x = q(y) y q(y1) – r(y1) 1 – (–2) = 1 + 2 = 3 –1 – (–3) = –1 + 3 = 2 y1 –2 1 x –3 –1 x x 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 , 0 , and 1 , where 4 and the anchor points are 2, 0, and 1, respectively. Af 0Af f f f –2Af 8 1Af 4 –2 x x + 2 x –2Af (x) = 4dt = 4  (x + 2) –2 x 4 t x x 0Af (x) = 4dt = 4x 0 1 x 4 t x – 1 x 1Af (x) = 4dt = 4  (x – 1) 1 x –4 The graph of Af is a straight line with slope of 4. Figure 23.4 Noticethatchangingtheanchorpointonlychangestheareafunctionbyanadditiveconstant. EXERCISE 23.1 Argue that the formulas given for the functions 2 , 0 , and 1 above are valid for to the left of the anchor point. D e f i n i t i o n For a continuous function and a constant “a” in the domain of we dene to be . givesthesignedareabetweenthegraphof andthe-axisfrom to . Therefore, we will refer to it as the area function. ... - tailieumienphi.vn