1. Copyright © Cengage Learning. All rights reserved. 6 The Integral

2. Copyright © Cengage Learning. All rights reserved. 6.3 The Definite Integral: Numerical and Graphical Viewpoints

3. 3 Example 1 – Oil Spill Your deep ocean oil rig has suffered a catastrophic failure, and oil is leaking from the ocean floor wellhead at a rate of dollars per hour. thousand barrels per day where t is time in days since the failure. Use a numerical calculation to estimate the total volume of oil released during the first 20 days.

4. 4 Example 1 – Solution The graph of v(t) is shown in Figure 3. Let’s start with a very crude estimate of the total volume of oil released, using the graph as a guide. The rate of change of this total volume at the beginning of the time period is v(0) = 60 thousand barrels per day. Figure 3

5. 5 Example 1 – Solution If this rate were to remain constant for the entire 20-day period, the total volume of oil released would be Total volume = Volume per day  Number of days = 60  20 = 1,200 thousand barrels. Figure 4 shows how we can represent this calculation on the graph of v(t). cont’d Figure 4

6. 6 Example 1 – Solution The volume per day based on v(0) = 60 is represented by the y-coordinate of the graph at its left edge, while the number of days is represented by the width of the interval [0, 20] on the x-axis. Therefore, computing the area of the shaded rectangle in the figure gives the same calculation: Area of rectangle = Volume per day  Number of days = 60  20 = 1,200 = Total volume. cont’d

7. 7 Example 1 – Solution But, as we see in the graph, the flow rate does not remain constant, but goes down quite significantly over the course of the 20-day interval. We can obtain a somewhat more accurate estimate of the total volume by re-estimating the volume using 10-day periods—that is, by dividing the interval [0, 20] into two equal intervals, or subdivisions. We estimate the volume over each 10-day period using the flow rate at the beginning of that period. cont’d

8. 8 Example 1 – Solution Cost of first period = Volume per day × Number of days = v(0)  10 = 60  10 = 600 thousand barrels Cost of second period = Volume per day × Number of days = v(10)  10 = 28  10 = 280 thousand barrels Adding these volumes gives us the more accurate estimate v(0)  10 + v(10)  10 = 880 thousand barrels. cont’d Calculation using 2 subdivisions

9. 9 Example 1 – Solution In Figure 5 we see that we are computing the combined area of two rectangles, each of whose heights is determined by the height of the graph at its left edge: cont’d The areas of the rectangles are estimates of the volumes for successive 10-day periods. Figure 5

10. 10 Example 1 – Solution Area of first rectangle = Volume per day  Number of days = v(0)  10 = 60  10 = 600 = Volume for first 10 days Area of second rectangle = Volume per day  Number of days = v(10)  10 = 28  10 = 280 = Volume for second 10 days. cont’d

11. 11 Example 1 – Solution We can get an even better estimate of the volume by using four divisions of [0, 20] instead of two: v(0)  5 + v(5)  5 + v(10)  5 + v(15)  5 = 300 + 210 + 140 + 90 = 740 thousand barrels. cont’d Calculation using 4 subdivisions

12. 12 Example 1 – Solution As we see in Figure 6, we have now computed the combined area of four rectangles, each of whose heights is again determined by the height of the graph at its left edge. Notice how the volume seems to be decreasing as we use more subdivisions. More importantly, total volume seems to be getting closer to the area under the graph. cont’d Estimated Volume Using 4 Subdivisions The areas of the rectangles are estimates of the volumes for successive 5-day periods. Figure 6

13. 13 Example 1 – Solution Figure 7 illustrates the calculation for 8 equal subdivisions. The approximate total volume using 8 subdivisions is the total area of the shaded region in Figure 7: v(0)  2.5 + v(2.5)  2.5 + v(5)  2.5 + · · · + v(17.5)  2.5 = 675 thousand barrels. cont’d Calculation using 8 subdivisions Estimated Volume Using 8 Subdivisions The areas of the rectangles are estimates of the volumes for successive 2.5-day periods. Figure 7

14. 14 Example 1 – Solution Looking at Figure 7, we still gets the impression that we are overestimating the volume. If we want to be really accurate in our estimation of the volume, we should really be calculating the volume continuously every few hours or, better yet, minute by minute, as illustrated in Figure 8. cont’d Every Six Hours (80 subdivisions) Volume 619.35 thousand barrels Every Minute (28,800 subdivisions) Volume 613.35 thousand barrels Figure 8

15. 15 Example 1 – Solution Figure 8 strongly suggests that the more accurately we calculate the total volume, the closer the answer gets to the exact area under the portion of the graph of v(t) with 0  t  20, and leads us to the conclusion that the exact total volume is the exact area under the rate of change of volume curve for 0  t  20. In other words, we have made the following remarkable discovery: Total volume is the area under the rate of change of volume curve! cont’d

16. 16 The Definite Integral: Numerical and Graphical Approaches In general, we have a function f (such as the function v in the example), and we consider an interval [a, b] of possible values of the independent variable x. We subdivide the interval [a, b] into some number of segments of equal length. Write n for the number of segments, or subdivisions. Next, we label the endpoints of these subdivisions x 0 for a, x 1 for the end of the first subdivision, x 2 for the end of the second subdivision, and so on until we get to x n, the end of the nth subdivision, so that x n = b.

17. 17 The Definite Integral: Numerical and Graphical Approaches Thus, a = x 0 < x 1 < · · · < x n = b. The first subdivision is the interval [x 0, x 1 ], the second subdivision is [x 1, x 2 ], and so on until we get to the last subdivision, which is [x n – 1, x n ]. We are dividing the interval [a, b] into n subdivisions of equal length, so each segment has length (b – a)/n.

18. 18 The Definite Integral: Numerical and Graphical Approaches We write  x for (b – a)/n (Figure 9). Having established this notation, we can write the calculation that we want to do as follows: For each subdivision [x k – 1, x k ], compute f (x k – 1 ), the value of the function f at the left endpoint. Figure 9

19. 19 The Definite Integral: Numerical and Graphical Approaches Multiply this value by the length of the interval, which is  x. Then add together all n of these products to get the number f (x 0 )  x + f (x 1 )  x +· · ·+ f (x n – 1 )  x. This sum is called a (left) Riemann sum for f. In Example 1 we computed several different Riemann sums. Here is the computation for n = 4 we used in the oil spill example (see Figure 10): Figure 10

20. 20 The Definite Integral: Numerical and Graphical Approaches Left Riemann sum = f (x 0 )  x + f (x 1 )  x +· · ·+ f (x n – 1 )  x = f (0)(5) + f (5)(5) + f (10)(5) + f (15)(5) = 60(5) + 42(5) + 28(5) + 18(5) = 740 Because sums are often used in mathematics, mathematicians have developed a shorthand notation for them. We write f (x 0 )  x + f (x 1 )  x +· · ·+ f (x n – 1 )  x as f (x k )  x.

21. 21 The Definite Integral: Numerical and Graphical Approaches The symbol  is the Greek letter sigma and stands for summation. The letter k here is called the index of summation, and we can think of it as counting off the segments. We read the notation as “the sum from k = 0 to n – 1 of the quantities f (x k )  x.” Think of it as a set of instructions: Set k = 0, and calculate f (x 0 )  x. Set k = 1, and calculate f (x 1 )  x. … Set k = n – 1, and calculate f (x n – 1 )  x. f (0)(5) in the above calculation f (5)(5) in the above calculation f (15)(5) in the above calculation

22. 22 The Definite Integral: Numerical and Graphical Approaches Then sum all the quantities so calculated. Riemann Sum If f is a continuous function, the left Riemann sum with n equal subdivisions for f over the interval [a, b] is defined to be Left Riemann sum = f (x k )  x = f (x 0 )  x + f (x 1 )  x +· · ·+ f (x n – 1 )  x = [f (x 0 ) + f (x 1 ) +· · ·+ f (x n – 1 )]  x where a = x 0 < x 1 < · · · < x n = b are the subdivisions, and  x = (b – a)/n.

23. 23 The Definite Integral: Numerical and Graphical Approaches Interpretation of the Riemann Sum If f is the rate of change of a quantity F (that is, f = F ), then the Riemann sum of f approximates the total change of F from x = a to x = b. The approximation improves as the number of subdivisions increases toward infinity. Quick Example If f (t) is the rate of change in the number of bats in a belfry and [a, b] = [2, 3], then the Riemann sum approximates the total change in the number of bats in the belfry from time t = 2 to time t = 3.

24. 24 The Definite Integral: Numerical and Graphical Approaches Visualizing a Left Riemann Sum (Non-negative Function) Graphically, we can represent a left Riemann sum of a non-negative function as an approximation of the area under a curve: Riemann sum = Shaded area = Area of first rectangle +Area of second rectangle + · · · + Area of nth rectangle = f (x 0 )  x + f (x 1 )  x + f (x 2 )  x +· · · + f (x n – 1 )  x

25. 25 The Definite Integral: Numerical and Graphical Approaches Quick Example In Example 1 we computed several Riemann sums, including these: n = 1: Riemann sum = v(0)  t = 60 × 20 = 1,200 n = 2: Riemann sum = [v(t 0 ) + v(t 1 )]  t = [v(0) + v(10)]  (10) = 880

26. 26 The Definite Integral: Numerical and Graphical Approaches n = 4: Riemann sum = [v(t 0 ) + v(t 1 ) + v(t 2 ) + v(t 3 )]  t = [v(0) + v(5) + v(10) + v(15)]  (5) = 740 n = 8: Riemann sum = [v(t 0 ) + v(t 1 ) +· · ·+ v(t 7 )]  t = [v(0) + v(2.5) +· · ·+ v(17.5)]  (2.5) = 675

27. 27 The Definite Integral: Numerical and Graphical Approaches The Definite Integral If f is a continuous function, the definite integral of f from a to b is defined to be the limit of the Riemann sums as the number of subdivisions approaches infinity: In Words: The integral, from a to b, of f (x) dx equals the limit, as n→, of the Riemann Sum with a partition of n subdivisions.

28. 28 The Definite Integral: Numerical and Graphical Approaches The function f is called the integrand, the numbers a and b are the limits of integration, and the variable x is the variable of integration. A Riemann sum with a large number of subdivisions may be used to approximate the definite integral. Interpretation of the Definite Integral If f is the rate of change of a quantity F (that is, f = F), then is the (exact) total change of F from x = a to x = b.

29. 29 The Definite Integral: Numerical and Graphical Approaches Quick Example If f (t) is the rate of change in the number of bats in a belfry and [a, b] = [2, 3], then is the total change in the number of bats in the belfry from time t = 2 to time t = 3. Visualizing the Definite Integral Non-negative Functions: If f (x)  0 for all x in [a, b], then is the area under the graph of f over the interval [a, b], as shaded in the figure.

30. 30 The Definite Integral: Numerical and Graphical Approaches General Functions: is the area between x = a and x = b that is above the x-axis and below the graph of f, minus the area that is below the x-axis and above the graph of f: = Area above x-axis – Area below x-axis

31. 31 The Definite Integral: Numerical and Graphical Approaches Quick Example

32. 32 Computing Definite Integrals

33. 33 Example 4 – Estimating a Definite Integral From a Graph Figure 16 shows the graph of the (approximate) rate f (t), at which the United States consuming gasoline from 2000 through 2008. (t is time in years since 2000.) Use the graph to estimate the total U.S. consumption of gasoline over the period shown. f'(t) Gasoline consumption (100 million gals per year) Figure 16

34. 34 Example 4 – Solution The derivative f '(t) represents the rate of change of the total U.S. consumption of gasoline, and so the total U.S. consumption of gasoline over the given period [0, 8] is given by the definite integral: Total U.S. consumption of gasoline = Total change in f (t) = and is given by the area under the graph (Figure 17). Figure 17

35. 35 Example 4 – Solution One way to determine the area is to count the number of filled rectangles as defined by the grid. Each rectangle has an area of 1  0.5 = 0.5 units (and the half-rectangles determined by diagonal portions of the graph have half that area). Counting rectangles, we find a total of 41.5 complete rectangles, so Total area = 20.75. cont’d

36. 36 Example 4 – Solution Because f '(t) is in 100 million gallons per year, we conclude that the total U.S. consumption of gasoline over the given period was about 2,075 million gallons, or 2.075 billion gallons. While counting rectangles might seem easy, it becomes awkward in cases involving large numbers of rectangles or partial rectangles whose area is not easy to determine. cont’d

37. 37 In a case like this, in which the graph consists of straight lines, rather than counting rectangles, we can get the area by averaging the left and right Riemann sums whose subdivisions are determined by the grid: Left sum = (2.5 + 2.5 + 3 + 3 + 2.5 + 2.5 + 2.5 + 2.5)(1) = 21 Right sum = (2.5 + 3 + 3 + 2.5 + 2.5 + 2.5 + 2.5 + 2)(1) = 20.5 Average = = 20.75. Example 4 – Solution cont’d

38. 38 Example 4 – Solution To see why this works, look at the single interval [1, 2]. The left sum contributes 2.5  1 = 2.5 and the right sum contributes 3  1 = 3. The exact area is their average, 2.75 (Figure 18). cont’d Figure 18

39. 39 Example 4 – Solution The average of the left and right Riemann sums is frequently a better estimate of the definite integral than either is alone. cont’d