22.1 Finding Net Change in Amount: Physical and Graphical Interplay 721 Let’s return for a moment to the problem of calculating the cheetah’s net change in position and approach it with a different mindset. Let s(t) = the cheetah’s position at time t. Then v(t) = s (t). In other words, ds dt = 2t + 5; s(t) is a function whose derivative is 2t + 5. Because s (t) is linear, we might suspect that s(t) is quadratic. If s(t)= t 2 + 5t, then s (t) = 2t + 5. In fact, if two functions have the same derivative, then the functions must differ by an additive constant; if f (x) = g (x), then f(x)= g(x) + C for some constant C. Therefore, s(t) = t 2 + 5t + C for some constant C. Then the change in position from t = 1tot=4isgiven by s(4) − s(1) = 4 2 + 5(4) + C − [1 2 + 5(1) + C] = 16 + 20 + C − 6 − C = 30. ◆ In Chapter 24 we will explore the relationship between the two mindsets presented in Example 22.5b and arrive at a wonderful theorem that unifies them. For the time being, there is a lot to be learned from the first mindset; we will stick with it for a while. ◆ EXAMPLE 22.6 A gazelle’s velocity is given by the graph below. v(t) is increasing on [0, 3] and decreasing on [3, 6]. How can we find the net change in the gazelle’s position over the interval [0, 5]? v(t) v(t) is increasing on [0, 3] and decreasing on [3, 6]. t 12345 6 Figure 22.13 SOLUTION Because v is increasing on the interval [0, 3], we can find lower bounds for the gazelle’s net change in position by using left-hand sums (inscribed rectangles) and upper bounds by using right-hand sums (circumscribed rectangles). v(t) t 53 v(t) t 53 Figure 22.14 722 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral Because v is decreasing on the interval [3, 5], we can find lower bounds for the net change in position by using right-hand sums (inscribed rectangles) and upper bounds by using left-hand sums (circumscribed rectangles). See Figure 22.14. We can obtain lower and upper bounds for the gazelle’s net change in position on [0, 5] by treating the intervals [0, 3] and [3, 5] independently. For instance, L = (L n on [0, 3]) + (R m on [3, 5]) gives a lower bound, while U = (R n on [0, 3]) + (L m on [3, 5]) gives an upper bound. If we compute the limit as n and m increase without bound, L and U will both approach the area under the velocity curve from t = 0tot=5. This area corresponds to the gazelle’s net change in position on [0, 5]. If we did not treat the intervals [0, 3] and [3, 5] independently, we could still look at left- and right-hand sums to approximate the gazelle’s net change in position. We could not, however, label them as under- or overestimates. v(t) t v(t) t Figure 22.15 Nevertheless, lim n→∞ (R n − L n ) = 0; in fact, lim n→∞ R n and lim n→∞ L n both correspond to the area under the velocity curve. If we were to partition [5, 6] into n equal pieces and compute L n and R n , they would give us upper and lower bounds, respectively, for the displacement. See Figure 22.16. Both L n and R n will be negative for all n>1. lim n→∞ L n = lim n→∞ R n = the signed area under the curve. 56 56 56 R n gives a lower bound (more negative than the actual displacement). L n gives a upper bound (less negative than the actual displacement). Figure 22.16 ◆ Examples 22.5 and 22.6 illustrate the interplay between the graphical and physical problems posed at the beginning of the section. The net change of a quantity can be represented as the signed area under the graph of the rate of change function. The question of how to find the area under the graph of a function is an important one, and is of interest on its own merits. It will be the focus of the next section. 22.1 Finding Net Change in Amount: Physical and Graphical Interplay 723 PROBLEMS FOR SECTION 22.1 Do Problem 1; it’s a key problem and worthy of discussion. 1. It’s flu season and a health clinic has set up a flu shot program for its patients. The clinic is open on Saturday from 8:00 a.m. to 4:00 p.m. (16:00) giving flu shots on a first-come, first-serve basis. The clinic has the capacity to serve 30 patients per hour. The function r(t), whose graph is given below, gives the rate at which people are arriving at the clinic for shots. r(t) (people/hr) t(time) 8910 15 30 45 11 12 13 14 15 16 17 Explain your answers to the questions below by relating them to points, lengths, or areas on the graph. 5 (a) At what time does a line start forming? (b) At approximately what time is the length of the line increasing most rapidly? (c) At approximately what time is the line the longest? (The answers to (b) and (c) are different. Explain why.) (d) When the line is longest, approximately how many people are in line? (e) Approximate the longest amount of time a person could wait for a shot. (Assume that doors close to new arrivals at 4:00 p.m. but everyone who has arrived by 4:00 p.m. is served.) (f) Approximately how long is the line at 3:00 p.m.? (g) Approximate the number of people who came to the clinic for a flu shot this day. 2. Maple syrup is being poured at a decreasing rate out of a tank. By taking readings from the valve on the tank, we have the following information on the rate at which the syrup is leaving the tank. t (seconds) 0 2468 rate (in cm 3 /sec) 10 9742 (a) Find a good upper bound for the amount of maple syrup that has been poured out between time t = 0 and t = 8. (b) Find a good lower bound for this same amount. 3. An industrial chemist is making a mixture in a large container. A certain chemical, B, is being introduced into the mixture at an ever-increasing rate. Some of the rates have 5 This problem, like Problem 8 in Section 22.2, was inspired by Peter Taylor’s wicket problem, from Calculus: The Analysis of Functions by Peter Taylor, Wall and Emerson, Inc., 1992, p. 393. 724 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral been registered below. Time t = 0 marks the first introduction of this chemical into the mixture. time (in minutes) 037101315 rate (in grams/min.) 10 12 20 23 25 29 Determine reasonable upper and lower bounds for the number of grams of chemical B in the mixture at time t = 13. 4. (a) By partitioning the interval [0, π/2] into four equal pieces and using the areas of inscribed and circumscribed rectangles as appropriate, find upper and lower bounds for the area between the graph of sin x and the x-axis for x in the interval [0, π/2]. Draw a picture illustrating what you have done. (Note: You will have to use your calculator to get some of the values of sin x and to get a numerical answer.) (b) Using the work you did in part (a), find upper and lower bounds for the area under the graph of sin x between x = 0 and x = π. Explain what you have done using a picture. (c) Using the work you did in part (b), give upper and lower bounds for the area under the graph of cos x between x =−π/2 and π/2. 5. Suppose velocity (in miles per hour) is given by v(t) = 3t, where t is measured in hours. We are interested in the distance traveled from t = 0tot=k,where k is a constant. (a) By solving the differential equation ds/dt = 3t and using the initial condition s(0) = s 0 , find the distance function s(t). Using s(t), find i. s(0). ii. s(k). iii. the distance traveled between t = 0 and t = k. (b) Find the area under the graph of v(t) from t = 0tot=k.Verify that your answers to part (a) iii and (b) are the same. 6. Suppose velocity (in miles per hour) is given by v(t) = mt + c, where m and c are positive constants. (a) Using your knowledge of the area of a trapezoid, find the area under the graph of v(t) on the interval [a, b], where a and b are positive constants. (b) By solving the differential equation ds/dt = mt + c and using the initial condition s(0) = s 0 , find the distance function s(t). Using s(t) find i. s(a). ii. s(b). iii. the distance traveled between t = a and t = b. Verify that your answers to parts (a) and (b) iii are the same. 7. Below is the graph of the velocity of a bee traveling in a straight line from a clover to a hive. Find the following. (a) the distance traveled between t = 1 and t = 3. 22.2 The Definite Integral 725 (b) the distance traveled between t = 0 and t = 8. (c) the distance traveled between t = 3 and t = 5. v(t) t 12 2 34 4 56 6 78 8 (1, 4) (5, 8) (8, 2) 8. (a) The velocity of an object at time t is given by v(t) = t 2 ft/sec. Partition the time interval [0, 3] into 3 equal pieces each of length 1 second. Find upper and lower bounds for the distance the object traveled between time t = 0 and t = 3. (b) Illustrate your work in part (a) by graphing v(t) and using areas of inscribed and circumscribed rectangles. Draw two pictures, one illustrating the upper bound and the other the lower bound. (c) Repeat part (a), but this time partition the interval into 6 equal pieces, each of length 1/2. Make a sketch indicating the areas you have found. (d) What is the difference between R n and L n if the interval is partitioned into 50 equal pieces? 100 equal pieces? (e) Into how many equal pieces must we partition [0, 3] to be sure that the difference between the right- and left-hand sums is less than or equal to 0.01? 9. Suppose v(t) gives the velocity of a trekker on the time interval [0, 3] and suppose that v(t) is positive and decreasing over this interval. If we use a left-hand sum to approximate the distance she has covered over this time interval, will the approximation give a lower bound or an upper bound? 22.2 THE DEFINITE INTEGRAL Suppose we want to find the signed area under the graph of a continuous function f on the interval [a, b]. We’ll use the method of successive approximations and then apply a limit process. We divide the interval [a, b] into n subintervals of equal width, each subinterval being of width x = b−a n . We label as shown where x k = a + kx for k = 0, 1, 2, ,n. The subintervals are [x 0 , x 1 ], [x 1 , x 2 ], ,[x n−1 ,x n ] 6 . 6 Notice that x n = a + nx = a + n b−a n = a + b − a = b. 726 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral ∆ x ∆ x x 0 x 1 x n–1 x n x 2 x 3 = a = b . . . x 1 x 0 x n–1 x n x 2 x 3 = a = b . . . Notice that x n = a + n(∆x) Figure 22.17 On each subinterval, we approximate the function f(x)by a constant function and approx- imate the (signed) area under f by the (signed) area of a rectangle. For a left-hand sum, we approximate f(x)on each subinterval by a constant function whose height is the value of f at the left endpoint of that subinterval. For example, on the second subinterval, [x 1 , x 2 ], the height of the rectangle is f(x 1 ) because x 1 is the x- coordinate of the left endpoint of that subinterval. The width of every rectangle is x. Accordingly, we write the left-hand sum using n subintervals as follows. L n = f(x 0 )x + f(x 1 )x +···+f(x n−1 )x = n−1 i=0 f(x i )x We write the right-hand sum using n subintervals in a similar way, but this time beginning with x 1 and ending with x n . R n = f(x 1 )x + f(x 2 )x +···+f(x n )x = n i=1 f(x i )x x 1 x 0 x n–1 x n x 2 . . . left-hand sum x 1 x 0 x n–1 x n x 2 . . . right-hand sum Figure 22.18 22.2 The Definite Integral 727 Notice that we are not making any statement about these approximating sums being upper or lower bounds for the exact area. We are taking f to be a generic continuous bounded function, so we don’t know where f is increasing and where f is decreasing. The left- and right-hand sums are special cases of what is called a Riemann sum, after the mathematician Bernhard Riemann. We will focus our attention on left- and right-hand sums, partitioning [a, b] into equal pieces, but Riemann sums are approximating sums that allow for more freedom of construction than R n and L n . To construct a Riemann sum for a bounded function f on [a, b], chop the interval [a, b], into n subintervals (not necessarily equal in width). Label the chops consecutively a = x 0 <x 1 <x 2 <···x n =b.From each of the subintervals choose an x-value in that interval. Call these x-values x ∗ 1 , x ∗ 2 , ,x ∗ n where the subscript k indicates an x-value from the kth interval. Let x k be the width of the kth subinterval. x k = x k − x k−1 for k = 1, 2, ,n.Anysum of the form n k=1 f(x ∗ k )x k is referred to as a Riemann sum. It is more compact to use summation notation than to write out the individual terms, and this way of expressing the sums is suggestive of the notation we will soon introduce. Finding the Exact Area To get better approximations to the area, we increase n, the number of subdivisions. To get an exact value for the area, we look at the limit as n increases without bound. When we did so in Example 22.5b, L n and R n both approached the same value, this being the exact area under the curve. In the general case the left- and right-hand sums will approach the same limiting value provided that f is continuous. 7 The size of their difference is given by |R n − L n |=|(f (x 1 )x + f(x 2 )x +···+f(x n )x) − (f (x 0 )x + f(x 1 )x +···+f(x n−1 )x)| =|f(x n )x − f(x 0 )x| =|f(b)−f(a)|· b−a n . f(b),f(a),b,and a are all fixed quantities; they do not change as n grows without bound. Thus, the difference between the left- and right-hand sums approaches zero as n grows without bound. lim n→∞ R n = lim n→∞ L n . We define the signed area under f from a to b to be lim n→∞ R n (or lim n→∞ L n ) provided the limit exists. The signed area under f on the interval [a, b] is called the definite integral of f from a to b. Signed area means that we consider the area between f and the horizontal axis to be positive where f is positive (where f lies above the horizontal axis) and negative where f is negative (where f lies below the horizontal axis). 7 Being continuous is sufficient to guarantee that f is integrable, i.e., that the limit of the Riemann sums exists. This is proven in more advanced texts. 728 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral f x 2 1 –1 –112345–2–3 The shaded areas have a negative sign. Figure 22.19 The definition given above agrees with our intuitive notion of signed areas in those instances in which we have such a notion. For f a reasonably well-behaved function, the interval [a, b] can be partitioned into regions on which either f is increasing or f is decreasing. 8 By doing so, we can guarantee that within each region left- and right-hand sums will provide upper and lower bounds for the signed area within the region. lim n→∞ R n equals lim n→∞ L n , so we can conclude that these quantities must also equal the exact value of the signed area. If f is not reasonably well-behaved, then the signed area under f on [a, b]isdefined to be lim n→∞ R n . Definition The definite integral of f(x) from x = a to x = b is written b a f(x)dx and read as “the integral from a to b of f(x)”.Wedefine it as follows. Subdivide the interval [a, b] into n equal subintervals of width x = b−a n and label these subintervals [x 0 , x 1 ], [x 1 , x 2 ], ,[x n−1 ,x n ], where x i = a + ix, for i = 0, 1, 2, n. b a f(x)dx = lim n→∞ n i=1 f(x i )x = lim n→∞ n−1 i=0 f(x i )x, provided the limits exist. If f is continuous on [a, b], then this limit is guaranteed to exist. 9 The symbol b a f(x)dx is laden with meaning. is a script S, reminding us that the definite integral is the limit of a sum. It recalls the Greek letter for summation in the Riemann sums approximating the area. We approximate the area with areas of rectangles; f(x) represents all the heights f(x i ), and dx represents the widths, the x, in the Riemann sums. You can think of the limiting process as an agent of metamorphoses from n i=1 f(x i )x to b a f(x)dx. 8 More precisely, by “increasing” we really mean nondecreasing, and by “decreasing” we mean nonincreasing. Intervals on which f is constant can be included in one or the other. 9 The definite integral b a f(x)dx can be defined more generally as the limit of a general Riemann sum. When the limit is computed |x k | must approach zero for all x k . If the limit of the Riemann sum as the width of the largest x k goes to zero exists, it is equal to b a f(x)dx.Again, if f is continuous on [a, b], then the limit is guaranteed to exist. 22.2 The Definite Integral 729 Notice that the numbers a and b, called the endpoints of integration, do not appear explicitly in the Riemann sums n i=1 f(x i )x and n−1 i=0 f(x i )x.Atfirst this may surprise you, but on more careful examination you can see that x 0 = a and x n = b.As nincreases without bound, x 1 gets arbitrarily close to a and x n−1 gets arbitrarily close to b. Terminology The function f(x)being integrated in b a f(x)dx is called the integrand. The values a and b are called the endpoints of integration or the lower and upper limits of integration. Interpretations of the Definite Integral As we saw in the example of computing distance traveled, the definite integral, b a f(x)dx, can be interpreted as the signed area under the graph of f(x)between x = a and x = b,or the net change in the amount A(x) between x = a and x = b if f(x)is the rate of change of A(x). EXERCISE 22.1 Let f be the function graphed in Figure 22.19. Its graph is composed of three semicircles. Evaluate the following definite integrals by interpreting the definite integral as signed area. (a) 3 −1 f(x)dx (b) 5 3 f(x)dx (c) 5 −1 f(x)dx Answers (a) 2π (b) −π 2 (c) 2π − π 2 = 3π 2 EXERCISE 22.2 Suppose that f is continuous on the interval [−1, 3]. We partition [−1, 3] into n equal subintervals, each of length x, and form left- and right-hand Riemann sums, L n and R n , respectively. Explain and illustrate the following. If f > 0, then L n < 3 −1 f(x)dx <R n ;and if f < 0, then R n < 3 −1 f(x)dx <L n . These inequalities hold regardless of the sign of f . PROBLEMS FOR SECTION 22.2 1. On the following page are the graphs of the velocities of three runners on the interval [0, 5]. Express the distance each runner has traveled in this interval in terms of a definite integral. Who has traveled the greatest distance in this time interval? Who has traveled the smallest distance in the time interval? 730 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral v(t) in ft/min t (in minutes) 5 A: velocity = f(t) C: velocity = h(t) B: velocity = g(t) 2. Summation notation review (a) Write the following in summation notation. i. 3 − 4 + 5 − 6 + 7 −···−300 ii. 2 + 4 + 6 +···+1000 iii. 1 + 3 + 5 +···+999 iv. 2 3 − 2 9 + 2 27 − 2 81 +···+ 2 3 15 v. x + x 2 + x 3 + x 4 +···+x 40 vi. 1 2 + 2 2 + 3 2 +···+100 2 vii. a 0 + a 1 x + a 2 x 2 + a 3 x 3 +···+a n x n (b) Write out the following sums. i. 5 i=2 i 2 ii. 4 k=0 2 k iii. 3 j=0 a j x j 3. Find the following and express your answer as simply as possible. (a) 10 k=1 k 5 2 − 9 k=0 k 5 2 (b) n k=1 k n 2 − n−1 k=0 k n 2 4. Find upper and lower bounds for each of the following definite integrals by calculating left- and right-hand Riemann sums with the number of subdivisions indicated. (a) 2 0 x 3 dx (n = 4) (b) 3 1 1 t dt (n = 6) 5. Below are the velocity graphs for a chicken and a goat. Assume that at time t = 0 the goat and the chicken start out side by side and they both travel along the same straight dirt path. Answer the questions below. When the quantity corresponds to an area, describe this area on the graph and then give it using the appropriate definite integral or sums and differences of definite integrals. The graphs intersect at t = 1.5 and t = 6.5. The graph of v g (t) is maximum at t = 4.5. velocity t 1 2 3 4 5 6 7 8 v g (t): velocity of goat v c (t): velocity of chicken . quantity can be represented as the signed area under the graph of the rate of change function. The question of how to find the area under the graph of a function is an important one, and is of. lim n→∞ R n and lim n→∞ L n both correspond to the area under the velocity curve. If we were to partition [5, 6] into n equal pieces and compute L n and R n , they would give us upper and lower. graph of sin x and the x-axis for x in the interval [0, π/2]. Draw a picture illustrating what you have done. (Note: You will have to use your calculator to get some of the values of sin x and to