22.3 The Definite Integral: Qualitative Analysis and Signed Area 731 (a) The lead the chicken has when the goat begins to move (b) The distance between the goat and the chicken at time t = 1.5 (c) The distance between the goat and the chicken at time t = 3 (d) The time at which the goat is farthest ahead of the chicken (e) Approximate the time(s) at which the chicken and the goat pass one another on the path. (f) Approximate the time at which the distance between the goat and the chicken is increasing most rapidly. At this time, what is the relationship between v g and v c ? 6. Suppose we want to approximate 2 0 f(x)dx by partitioning the interval [0, 2] into n equal pieces and constructing left- and right-hand sums. Let f(x)=x 3 . (a) Put the following expressions in ascending order. 2 0 f(x)dx, L 4 , R 4 , L 20 , R 20 , L 100 , R 100 (b) Find |R 4 − L 4 |. (c) Find |R 100 − L 100 |. (d) How large must n be to assure that |R n − L n | < 0.05? (e) Write out R 4 , once using summation notation, once without. Evaluate R 4 . 7. Turn back to Problem 1 of Problems for Section 22.1 on page 723. Express the following quantities using a definite integral or the sum or difference of definite integrals. (a) The length of the line at 11:00 a.m. (b) The length of the line at its longest (c) The number of people who came to the clinic for flu shots (d) The number of people actually served by the clinic (e) The length of the line at 3:00 p.m. (f) The amount of time a person arriving at noon has to wait in line (g) The amount of time by which the clinic must extend its hours in order to serve everyone who is in line before 4:00 p.m. 8. Repeat parts (a) through (e) of Problem 6, letting f(x)= 1 x+1 . 22.3 THE DEFINITE INTEGRAL: QUALITATIVE ANALYSIS AND SIGNED AREA In Section 22.1 we established that the signed area under a rate graph is of particular interest to us; this area corresponds to the net change in amount. In Section 22.2 we established that if f is a reasonably well-behaved function, 10 we can, in theory, find the signed area under the graph of f(t)on the interval [a, b] as follows. Partition the interval [a, b] into n equal pieces, each of length t = b−a n , and label as shown on the following page. 10 A continuous function on a closed interval is reasonably well-behaved. 732 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral t 0 t n t n–1 t 1 t 2 ab The signed area is defined to be lim n→∞ n i=1 f(t i )t, provided this limit exists, 11 and is denoted by b a f(t)dt.Inthis section we interpret b a f(t)dt as signed area (positive for f>0and negative for f<0)and use our knowledge of functions and their graphs, along with a bit of basic geometry, to approximate some definite integrals and evaluate others. 12 REMARK: The definite integrals b a f(t)dt, b a f(x)dx, b a f(w)dw, and b a f(s)ds are all equivalent. Each gives the signed area between the graph of f and the horizontal axis from a to b. The variables t, x, w, and s are all dummy variables. ◆ EXAMPLE 22.7 The function r(t) gives the rate at which water is flowing into (or out of) a backyard swimming pool. The graph of r(t) is given below. At time t = 0 there are 200 gallons of water in the pool. 120 –150 59 14 20 (5, 120) t (hrs) r (t) (gal/hr) Figure 22.20 (If w(t) is the number of gallons of water in the pool at time t, where t is given in hours, then r(t) = dw dt .) Express your answers to the questions below using definite integrals wherever appro- priate and evaluate these integrals. (a) When is the amount of water in the pool greatest? At that time, how many gallons of water are in the pool? (b) What is the net flow in or out of the pool between t = 4 and t = 12? (c) At what time is the pool’s water level back at 200 gallons? (d) Sketch a graph of the amount of water in the pool at time t. SOLUTIONS (a) The amount of water is maximum at t = 9 because water is entering the pool from t = 0 to t = 9 and leaving for t greater than 9. 11 If f is continuous we can guarantee that this limit exists. 12 We will rely on the following bits of geometry: The area of a disk is πr 2 . The area of a trapezoid is 1 2 (h 1 + h 2 ) · b, where 1 2 (h 1 + h 2 ) is the average of the heights. We can think of a triangle as a trapezoid with one “height” of length zero and a rectangle as a trapezoid with both heights of equal length. 22.3 The Definite Integral: Qualitative Analysis and Signed Area 733 The amount of water entering the pool on the interval [0, 9] is 9 0 r(t)dt, the area under the graph from 0 to 9. 9 0 r(t)dt = the area of the rectangle + the area of the triangle = (5)(120) + (1/2)(4)(120) = 600 + 240 = 840 59 t r (t) Figure 22.21 840 gallons of water have entered the pool in those 9 hours. Adding this to the original 200 gallons gives a total of 1040 gallons of water in the pool. (b) The net flow in or out of the pool between t = 4 and t = 12 is given by 12 4 r(t)dt. Compute this by summing the signed area under r(t) from t = 4tot=5, from 5 to 9, and from 9 to 12. In other words, compute 12 4 r(t)dt by expressing it as follows. 12 4 r(t)dt = 5 4 r(t)dt + 9 5 r(t)dt + 12 9 r(t)dt = area of rectangle + area of triangle + (− area of triangle) = 120(1) + (1/2)(120)(4) − (1/2)(3)(height of triangle) 13 = 120 + 240 − (1/2)(3)(90) = 360 − 135 = 225 The pool has a net gain of 225 gallons of water. 5 9 12 90 4 t r (t) Figure 22.22 13 The slope of the line forming the triangles is (−120/4) =−30, so between t = 9 and t = 12 the value of r(t) drops by 90, making the height of this triangle 90. 734 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral (c) There are many ways to approach this question. From part (a), we know that by time t = 9 the pool has gained 840 gallons of water. After t = 9 the pool loses water. Between t = 9 and t = 14 the pool has lost (1/2)(5)(150) = 375 gallons of water, so it now has a net gain of 840 gallons − 375 gallons = 465 gallons. Subsequently water is flowing out at a rate of 150 gallons per hour. In t more hours the pool will have lost an additional 150t gallons of water. For what t will 150t = 465? When t = 465/150 = 3.1. Therefore, at time t = 14 + 3.1 = 17.1, (17 hours and 6 minutes) the pool will be back at its original level of 200 gallons. Using integral notation, we have solved the following equation for T . T 0 r(t)dt = 0 Or, equivalently, 9 0 r(t)dt =− T 9 r(t)dt. 5 9 t r (t) areas must be equal Figure 22.23 (d) Here is a graph of w(t), where w(t) is the amount of water in the pool at time t. w(t) t (hrs) (5, 800) (9, 1040) (14, 665) (17.1, 200) 5 9 14 17.1 1000 800 600 400 200 Figure 22.24 ◆ ◆ EXAMPLE 22.8 Evaluate the following definite integrals by interpreting each as a signed area. (a) 2π 0 cos xdx (b) a −a sin xdx (c) 7 −7 x x 4 +x 2 +1 dx (d) 3 −2 |x|−2dx SOLUTIONS (a) 2π 0 cos xdx=0.The “positive” area and the “negative” area cancel. 22.3 The Definite Integral: Qualitative Analysis and Signed Area 735 cos x x 1 –1 ππ 2 3π 2π 2 Figure 22.25 (b) a −a sin xdx=0.Sine is an odd function (sin(−x) =−sin x), so again the negative and positive areas cancel regardless of the value of a. x sin x Figure 22.26 (c) The integrand is an odd function: (−x) (−x) 4 +(−x) 2 +1 =− x x 4 +x 2 +1 . Thus, the negative and positive areas cancel and we conclude that 7 −7 x x 4 +x 2 +1 dx = 0. y x x x 4 + x 2 + 1 –77 –.4 .4 y = Figure 22.27 (d) Sketch the graph of |x|−2. The area of the triangle lying below the x-axis is (1/2)(4)(2) = 4. Therefore, 2 −2 (|x|−2)dx=−4. The area of the triangle lying above the x-axis between x = 2 and x = 3is(1/2)(1)(1) = 1/2. We conclude that 3 −2 (|x|−2)dx= 2 −2 (|x|−2)dx+ 3 2 (|x|−2)dx =−4+0.5 =−3.5. 736 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral x 2–23 f(x) = |x| –2 Figure 22.28 ◆ ◆ EXAMPLE 22.9 Put the following expressions in descending order. A = −π −2π e −0.1x sin xdx, B= 0 −π e −0.1x sin xdx, C= 2π 0 e −0.1x sin xdx, D= 3π 0 e −0.1x sin xdx, E= 2π π e −0.1x sin xdx, F= π π e −0.1x sin xdx SOLUTION Sketch the graph of f(x)= e −0.1x sin x. f(x) x –2π 2π 3π–ππ Figure 22.29 We begin by grouping the expressions above by sign: F = π π e −0.1x sin xdx=0: a a f(x)dx is always zero; there is no area contained between x = a and x = a. The following expressions are positive: A = −π −2π e −0.1x sin xdx, C= 2π 0 e −0.1x sin xdx, D= 3π 0 e −0.1x sin xdx. The following expressions are negative: B = 0 −π e −0.1x sin xdx, E= 2π π e −0.1x sin xdx. Convince yourself, by considering the graphical interpretation of each of these definite integrals, that the integrals should be ordered as follows. −π −2π e −0.1x sin xdx> 3π 0 e −0.1x sin xdx> 2π 0 e −0.1x sin xdx> π π e −0.1x sin xdx> 2π π e −0.1x sin xdx> 0 −π e −0.1x sin xdx A>D>C>F >E>B ◆ 22.3 The Definite Integral: Qualitative Analysis and Signed Area 737 PROBLEMS FOR SECTION 22.3 1. By interpreting the definite integral as signed area, calculate the following. (a) 5 0 xdx (b) 2 −2 xdx (c) 2 −2 |x| dx (d) 3 −1 3 dx (e) 2π 0 sin tdt (f) π −π cos zdz (g) 2 −2 (x + 1)dx (h) 2 −2 |x + 1| dx 2. Let f be the function whose graph is given below. f x (3, 2) 36–6–3 3 −6 f(x)dx is closest to which of the following? Explain your reasoning. (a) −18 (b) −9 (c) −3 (d) 1.5 (e) 3 (f) 9 (g) 18 3. Put the following integrals in ascending order, placing “<” or “=” signs between them as appropriate. (Strategy: First determine which integrals are positive, which are negative, and which are zero.) (a) π 0 sin(t) dt (b) 2π −π sin(t) dt (c) 2π −π cos(t) dt (d) π/2 0 cos(t) dt (e) π 0 cos(2t) dt (f) 3π/2 0 | sin(t)| dt 4. Put the following in ascending order, placing “<” or “=” signs between them as appropriate. (a) π/2 0 sin(t) dt (b) π/2 0 tdt (c) 0 −π/2 tdt (d) 0 −π/2 sin(t) dt (e) π/2 0 1 dt (f) π/2 Below is the graph of a function f(x). Problems 5 and 6 refer to this function. The curve from (0, 0) to (6, 0) is a semicircle. f x (3, –3) (–2, 4) 42–2 –2 –4 –4–6–8 2 4 6 Use this graph to evaluate the following: 738 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral 5. (a) −2 −8 f(x)dx (b) 0 −8 f(x)dx (c) 6 0 f(x)dx (d) 6 3 f(x)dx 6. (a) 3 −2 f(x)dx (b) 6 −8 f(x)dx (c) 6 0 |f(x)|dx (d) 6 −8 |f(x)|dx 7. (a) What is the equation of a circle of radius 2 centered at the origin? (b) Write a function, complete with domain, that gives the equation of the top half of a circle of radius 2 centered at the origin. (c) Let f(x)= √ 4 − x 2 for −2 ≤x ≤ 0, 2x for x>0. Evaluate 2 −2 f(x)dx. 22.4 PROPERTIES OF THE DEFINITE INTEGRAL From the definition of the definite integral and its interpretation as signed area, we obtain the following properties. 14 1. Constant Factor Property For any constant k, b a kf (t) dt = k b a f(t)dt. Aconstant factor can be “pulled out” ofadefinite integral. 2. Dominance Property If f ≤ g on the interval [a, b], then b a f(t)dt ≤ b a g(t) dt. 3. Endpoint Reversal Property a b f(t)=− b a f(t)dt 4. Additive Integrand Property b a [f(t)+g(t)] dt = b a f(t)dt + b a g(t) dt 5. Splitting Interval Property c a f(t)dt = b a f(t)dt + c b f(t)dt (This is true regardless of the relative positions of a, b, and c.) 6. Symmetry Property a −a f(t)dt =0iff is odd; a −a f(t)dt =2 a 0 f(t)dt if f is even. These properties are illustrated below. Properties Illustrated 1. For any constant k, b a kf (t) dt = k b a f(t)dt. We can pull a constant multiple out of a definite integral. This is a property we use repeatedly in our work. (a) Interpretation as the integral of a rate function Suppose f(t) is the rate that water is entering a pool. Doubling the rate at which water enters the pool over a certain time interval will double the amount of water 14 Assume f is bounded on the intervals of integration and all integrals exist. 22.4 Properties of the Definite Integral 739 added to the pool in that time interval. Similarly, halving the rate will halve the net change in water. (b) Interpretation as area under f from t = a to t = b If the height of f is doubled, the area it bounds doubles. (If this doesn’t make sense intuitively, consider partitioning the region and approximating its area with rectangles; the height of each rectangle will double, so the area of each rectangle will double.) 2f f t ba f(x) f(x) Figure 22.30 2. If f ≤ g on the interval [a, b], then b a f(t)dt ≤ b a g(t) dt. (a) Interpretation as the integral of a rate function Suppose f(t)and g(t) give the velocities of two cars. If the velocity of the second is greater than or equal to that of the first, then the second car will have the same or greater net change in position. In other words, if the rate of change of position of the second car is greater, the net change in position will also be greater. (b) Interpretation as area under f from t = a to t = b a g(t) dt a b b g f ab g f f(t) dt ≤ a b g(t) dt a b f(t) dt ≤ a b more negative less negative ∫ ∫ ∫ ∫ Figure 22.31 3. a b f(t)dt =− b a f(t)dt. (a) Interpretation as the integral of a rate function Suppose f(t)is the rate at which water is entering a pool and t = a corresponds to 10:00 a.m., while t = b corresponds to noon. b a f(t)dt is the net change in water in the pool between 10:00 a.m. and noon. Let’s say this is 1000 gallons. Then the net change in water between noon and 10:00 a.m. (2 hours earlier) is −1000 because there were 1000 gallons less water at 10:00 a.m. than at noon. 10 12 f(t)dt =− 12 10 f(t)dt.When you see a b f(t)dt,where a<b,you can think of running a movie backward. 740 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral (b) Interpretation as area under f from t = a to t = b Suppose a<b.Think back to the approximating rectangles; each rectangle has base t, where t = b−a n is positive. If we switch the roles of a and b, then our measure of t becomes negative, so the value of the integral reverses sign. Advice: Whenever you come across c b f(t)dt, where b>c, switch the endpoints and negate immediately. For instance, when dealing with −2 0 f(t)dt, write it as − 0 −2 f(t)dt. 4. b a [f(t)+g(t)] dt = b a f(t)dt + b a g(t) dt (a) Interpretation as the integral of a rate function A pool is being filled with two hoses. Suppose f(t)is the rate at which water is entering through one hose and g(t) is the rate that water is entering via another hose. Then the total amount of water that enters the pool is the water that enters through one hose plus the water that enters through the other. (b) Interpretation as area under f from t = a to t = b Think about the approximating rectangles. In the bottom graph of Figure 22.32 the rectangles are positioned on top of one another; their positioning relative to one another does not affect their sum. f f t g y g g g g t f+g f+g f f f g t t } } } } } Figure 22.32 5. c a f(t)dt = b a f(t)dt + c b f(t)dt (a) Interpretation as the integral of a rate function Suppose f(t)is the rate at which water is entering a pool. The total amount of water that enters the pool from 9:00 a.m. to 5:00 p.m. is the amount that enters from 9:00 a.m. to noon plus the amount that enters from noon to 5:00 p.m. (b) Interpretation as area under f from t = a to t = c Using the endpoint reversal property, (3), you can convince yourself that the split- ting interval property holds regardless of whether or not b lies between a and c. The figures on the following page illustrate the statement for different relative positions of a, b, and c. Areas shaded with gray indicate a positive weighing in; areas shaded with hash marks indicate a negative weighting due to endpoint reversal. Work through each scenario slowly to assure yourself that in each case . (positive for f> 0and negative for f<0 )and use our knowledge of functions and their graphs, along with a bit of basic geometry, to approximate some definite integrals and evaluate others. 12 REMARK:. relationship between v g and v c ? 6. Suppose we want to approximate 2 0 f(x)dx by partitioning the interval [0, 2] into n equal pieces and constructing left- and right-hand sums. Let f(x)=x 3 . (a). from t = a to t = b Think about the approximating rectangles. In the bottom graph of Figure 22.32 the rectangles are positioned on top of one another; their positioning relative to one another does