Section 5.1. Areas and Distances. (1) The Area Problem (2) The Distance Problem (3) Summation Notation

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1 Section 5.1 Areas and Distances (1) The Area Problem (2) The Distance Problem (3) Summation Notation MATH 125 (Section 5.1) Areas and Distances The University of Kansas 1 / 19

2 Area Area is a measure of the size of 2-dimensional shapes. Area is preserved under cutting, gluing, sliding, and rotating. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 2 / 19

3 Area Area is a measure of the size of 2-dimensional shapes. Area is preserved under cutting, gluing, sliding, and rotating. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 2 / 19

4 Area Area is a measure of the size of 2-dimensional shapes. Area is preserved under cutting, gluing, sliding, and rotating. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 2 / 19

5 Area Area is a measure of the size of 2-dimensional shapes. Area is preserved under cutting, gluing, sliding, and rotating. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 2 / 19

6 Area Area is a measure of the size of 2-dimensional shapes. Area is preserved under cutting, gluing, sliding, and rotating. There are standard formulas for the areas of common shapes: Rectangle: A = bh Triangle: A = 1 2 bh Circle: A = πr 2 But what about more complicated shapes? MATH 125 (Section 5.1) Areas and Distances The University of Kansas 2 / 19

7 The Area Problem The motivation for this chapter is the problem of calculating the area of more general regions, such as the area under the graph of a function y = f (x). MATH 125 (Section 5.1) Areas and Distances The University of Kansas 3 / 19

8 The Area Problem The motivation for this chapter is the problem of calculating the area of more general regions, such as the area under the graph of a function y = f (x). When we studied tangent lines, we soon discovered that we needed to use limits to calculate them in a mathematically rigorous way. This led to the concept of a derivative. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 3 / 19

9 The Area Problem The motivation for this chapter is the problem of calculating the area of more general regions, such as the area under the graph of a function y = f (x). When we studied tangent lines, we soon discovered that we needed to use limits to calculate them in a mathematically rigorous way. This led to the concept of a derivative. Similarly, calculating area in a rigorous way will also require limits and will lead us to a new mathematical concept: the integral. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 3 / 19

10 The Area Problem Let f (x) be continuous and positive on a closed interval [a, b]. What is the area of the region bounded by the graph of f (x), the vertical lines x = a and x = b, and the x-axis? y = f(x) a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 4 / 19

11 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. y = f(x) a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

12 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. y = f(x) a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

13 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. y = f(x) a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

14 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. y = f(x) a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

15 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. y = f(x) a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

16 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. y = f(x) a x x x x x x b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

17 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. Form a rectangle of height f (x i ) on each segment. y = f(x) a x x x x x x b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

18 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. Form a rectangle of height f (x i ) on each segment. y = f(x) a x x x x x x b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

19 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. Form a rectangle of height f (x i ) on each segment. y = f(x) a x x x x x x b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

20 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. Form a rectangle of height f (x i ) on each segment. Then A f (x 1 ) x + f (x 2 ) x f (x n ) x. y = f(x) a x x x x x x b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

21 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. Form a rectangle of height f (x i ) on each segment. Then A f (x 1 ) x + f (x 2 ) x f (x n ) x. y = f(x) a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

22 The Area Problem The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles. Divide the domain [a, b] into n segments of length x = b a n. Inside each segment, choose a value x i. Form a rectangle of height f (x i ) on each segment. Then A f (x 1 ) x + f (x 2 ) x f (x n ) x. y = f(x) x f(x ) i a x i b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 5 / 19

23 Example 1: Approximate the area under y = x 2 on [1, 4] using 6 segments. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 6 / 19

24 Example 1: Approximate the area under y = x 2 on [1, 4] using 6 segments. Here a = 1, b = 4, n = 6, and x = b a n = 1 2. The 6 segments of the domain are [1, 1.5] [1.5, 2] [2, 2.5] [2.5, 3] [3, 3.5] [3.5, 4] MATH 125 (Section 5.1) Areas and Distances The University of Kansas 6 / 19

25 Example 1: Approximate the area under y = x 2 on [1, 4] using 6 segments. Here a = 1, b = 4, n = 6, and x = b a n = 1 2. The 6 segments of the domain are [1, 1.5] [1.5, 2] [2, 2.5] [2.5, 3] [3, 3.5] [3.5, 4] x i = right endpoint of i th interval: i x i f (x i ) f (x i ) x Area MATH 125 (Section 5.1) Areas and Distances The University of Kansas 6 / 19

26 Example 1: Approximate the area under y = x 2 on [1, 4] using 6 segments. Here a = 1, b = 4, n = 6, and x = b a n = 1 2. The 6 segments of the domain are [1, 1.5] [1.5, 2] [2, 2.5] [2.5, 3] [3, 3.5] [3.5, 4] x i = left endpoint of i th interval: i x i f (x i ) f (x i ) x Area MATH 125 (Section 5.1) Areas and Distances The University of Kansas 6 / 19

27 Example 1: Approximate the area under y = x 2 on [1, 4] using 6 segments. Here a = 1, b = 4, n = 6, and x = b a n = 1 2. The 6 segments of the domain are [1, 1.5] [1.5, 2] [2, 2.5] [2.5, 3] [3, 3.5] [3.5, 4] x i = midpoint of i th interval: i x i f (x i ) f (x i ) x Area MATH 125 (Section 5.1) Areas and Distances The University of Kansas 6 / 19

28 Example 1: Approximate the area under y = x 2 on [1, 4] using 6 segments. Here a = 1, b = 4, n = 6, and x = b a n = 1 2. The 6 segments of the domain are [1, 1.5] [1.5, 2] [2, 2.5] [2.5, 3] [3, 3.5] [3.5, 4] Choice of x i Right endpoints Left endpoints Midpoints Estimate of area (too high) (too low) (closest) MATH 125 (Section 5.1) Areas and Distances The University of Kansas 6 / 19

29 iclicker Question 1 Approximate the area under f (x) = x 1 on the interval [2, 4] by dividing it into n = 4 segments and choosing x 1,..., x 4 to be the midpoints of the segments. (A) (B) (C) 1 (D) (E) MATH 125 (Section 5.1) Areas and Distances The University of Kansas 7 / 19

30 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

31 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. n = 5 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

32 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. n = 10 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

33 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. n = 15 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

34 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. n = 20 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

35 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. n = 25 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

36 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. n = 30 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

37 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. The exact area is given by a limit. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

38 Area Expressed as a Limit The area A under the graph of f between x = a and x = b can be approximated as the total area of n rectangles: A R n {}}{ f (x 1 ) x + f (x 2 ) x f (x n ) x As n gets larger and larger, the approximation R n gets better and better. The exact area is given by a limit. The area A under the graph of a continuous function f between x = a and x = b equals the limit of the sum of the areas of approximating rectangles: A = lim R ( n = lim f (x1 ) x + f (x 2 ) x f (x n ) x ) n n MATH 125 (Section 5.1) Areas and Distances The University of Kansas 8 / 19

39 Calculating Distance Let v(t) be the velocity of an object at time t. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 9 / 19

40 Calculating Distance Let v(t) be the velocity of an object at time t. The area under the graph of v(t) on a time interval [a, b] measures the net distance traveled, or displacement, between times a and b. The units make sense: units of area under the graph of v(t) = units of t units of v(t) = time distance time = distance. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 9 / 19

41 Calculating Distance Let v(t) be the velocity of an object at time t. The area under the graph of v(t) on a time interval [a, b] measures the net distance traveled, or displacement, between times a and b. Example 2a: If v = v 0 on [a, b], then the region under the graph is a rectangle with area v 0 (b a). v(t) = v 0 a b MATH 125 (Section 5.1) Areas and Distances The University of Kansas 9 / 19

42 Calculating Distance Let v(t) be the velocity of an object at time t. The area under the graph of v(t) on a time interval [a, b] measures the net distance traveled, or displacement, between times a and b. Example 2b: An object starts at rest and accelerates at a constant rate of 1 m/s 2 for 10 seconds. Then v(t) = t m/s. Displacement = area under the curve = 1 2 (10 s)(10 m/s) = 50 m. v(t) = t m / s 0 10 seconds MATH 125 (Section 5.1) Areas and Distances The University of Kansas 9 / 19

43 Example 3: You are driving across Missouri. In order to stay awake, you estimate how far you have traveled from your speedometer readings: 2:00 PM 70 mph (the speed limit) 2:15 PM 65 mph (up a small hill) 2:30 PM 75 mph (down the hill) 2:45 PM 55 mph (careful, is that a speed trap?) 3:00 PM 80 mph (vroom!) You can now estimate 1 the maximum and minimum possible distance you have traveled during this hour: 1 Assuming that in each 15-minute interval, your max and min speeds occur at the endpoints. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 10 / 19

44 Example 3: You are driving across Missouri. In order to stay awake, you estimate how far you have traveled from your speedometer readings: 2:00 PM 70 mph (the speed limit) 2:15 PM 65 mph (up a small hill) 2:30 PM 75 mph (down the hill) 2:45 PM 55 mph (careful, is that a speed trap?) 3:00 PM 80 mph (vroom!) You can now estimate 1 the maximum and minimum possible distance you have traveled during this hour: Max: 1 4 (70) (75) (75) (80) = 75 miles 1 Assuming that in each 15-minute interval, your max and min speeds occur at the endpoints. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 10 / 19

45 Example 3: You are driving across Missouri. In order to stay awake, you estimate how far you have traveled from your speedometer readings: 2:00 PM 70 mph (the speed limit) 2:15 PM 65 mph (up a small hill) 2:30 PM 75 mph (down the hill) 2:45 PM 55 mph (careful, is that a speed trap?) 3:00 PM 80 mph (vroom!) You can now estimate 1 the maximum and minimum possible distance you have traveled during this hour: Max: 1 4 (70) (75) (75) (80) = 75 miles Min: 1 4 (65) (65) (55) (55) = 60 miles 1 Assuming that in each 15-minute interval, your max and min speeds occur at the endpoints. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 10 / 19

46 Example 3: You are driving across Missouri. In order to stay awake, you estimate how far you have traveled from your speedometer readings: 2:00 PM 70 mph (the speed limit) 2:15 PM 65 mph (up a small hill) 2:30 PM 75 mph (down the hill) 2:45 PM 55 mph (careful, is that a speed trap?) 3:00 PM 80 mph (vroom!) You can now estimate 1 the maximum and minimum possible distance you have traveled during this hour: Max: 1 4 (70) (75) (75) (80) = 75 miles Min: 1 4 (65) (65) (55) (55) = 60 miles The actual distance traveled is somewhere between these two estimates. 1 Assuming that in each 15-minute interval, your max and min speeds occur at the endpoints. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 10 / 19

47 iclicker Question 2 The table below gives a car s velocity as a function of time. Time (minutes) Speed (mph) Town A is 60 miles, town B is 70 miles, town C is 73 miles, and town D is 80 miles from the starting point. By estimating the minimum and maximum distance traveled, you can predict that the car is... (A) between towns A and B. (B) between towns B and C. (C) between towns C and D. (D) between towns A and D, but you can t be any more specific. (E) past town D. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 11 / 19

48 Summation Notation We have encountered expressions like f (x 1 ) x + f (x 2 ) x f (x n ) x that are sums of many similar-looking terms. We need a notation for writing sums in a compact form. Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. j=m MATH 125 (Section 5.1) Areas and Distances The University of Kansas 12 / 19

49 Summation Notation We have encountered expressions like f (x 1 ) x + f (x 2 ) x f (x n ) x that are sums of many similar-looking terms. We need a notation for writing sums in a compact form. Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. j=m is the Greek letter Sigma (mnemonic for sum. ) MATH 125 (Section 5.1) Areas and Distances The University of Kansas 12 / 19

50 Summation Notation We have encountered expressions like f (x 1 ) x + f (x 2 ) x f (x n ) x that are sums of many similar-looking terms. We need a notation for writing sums in a compact form. Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. j=m is the Greek letter Sigma (mnemonic for sum. ) The notation n tells us to start at j = m and to end at j = n. j=m MATH 125 (Section 5.1) Areas and Distances The University of Kansas 12 / 19

51 Summation Notation We have encountered expressions like f (x 1 ) x + f (x 2 ) x f (x n ) x that are sums of many similar-looking terms. We need a notation for writing sums in a compact form. Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. j=m is the Greek letter Sigma (mnemonic for sum. ) The notation n tells us to start at j = m and to end at j = n. j=m a j is called the general term and j is the summation index. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 12 / 19

52 Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. Examples: j=m 100 j=1 j = MATH 125 (Section 5.1) Areas and Distances The University of Kansas 13 / 19

53 Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. Examples: j=m 100 j=1 785 j=4 j = j 2 = MATH 125 (Section 5.1) Areas and Distances The University of Kansas 13 / 19

54 Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. Examples: j=m 6 (j 3 j 1) j=4 100 j=1 785 j=4 j = j 2 = MATH 125 (Section 5.1) Areas and Distances The University of Kansas 13 / 19

55 Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. Examples: j=m 100 j=1 785 j=4 j = j 2 = (j 3 j 1) = ( ) + ( ) + ( ) j=4 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 13 / 19

56 iclicker Question 3 Calculate the sum (A) 28 (B) 0 (C) 12 (D) 16 (E) 14 5 j=2 ( ( π )) j + 2 sin 2 j. MATH 125 (Section 5.1) Areas and Distances The University of Kansas 14 / 19

57 Summation Notation and Area Summation Notation n The notation a j means a m + a m+1 + a m a n 1 + a n. j=m Therefore, our estimate for the area under the graph of a continuous, positive function f (x) on an interval [a, b] is n R n = f (x 1 ) x f (x n ) x = f (x j ) x and the exact area is j=1 A = lim n n = lim 1) x f (x n ) x) n n = lim n f (x j ) x. j=1 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 15 / 19

58 Properties of Summations If you understand addition, you understand summation! n n n (a j ± b j ) = ± j=m n (ca j ) = c j=m j=m a j j=m b j n a j (for any constant c) j=m n c = c(n m + 1) j=m MATH 125 (Section 5.1) Areas and Distances The University of Kansas 16 / 19

59 Properties of Summations For example: (3j 2 5j + 3) = 3j 2 5j + 3 j=1 j= = 3 j 2 j=1 j=1 j=1 j= j Fortunately, there are nice formulas for the sum of the first n numbers, squares, cubes, fourth powers,... MATH 125 (Section 5.1) Areas and Distances The University of Kansas 17 / 19

60 Summation Formulas n j = (n 1) + n = j=1 n(n + 1) 2 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 18 / 19

61 Summation Formulas n j = (n 1) + n = j=1 n(n + 1) 2 n j 2 = (n 1) 2 + n 2 = j=1 n(n + 1)(2n + 1) 6 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 18 / 19

62 Summation Formulas n j = (n 1) + n = j=1 n(n + 1) 2 n j 2 = (n 1) 2 + n 2 = j=1 n(n + 1)(2n + 1) 6 n j 3 = (n 1) 3 + n 3 = n2 (n + 1) 2 j=1 4 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 18 / 19

63 Summation Formulas n j = (n 1) + n = j=1 n(n + 1) 2 n j 2 = (n 1) 2 + n 2 = j=1 n(n + 1)(2n + 1) 6 n j 3 = (n 1) 3 + n 3 = n2 (n + 1) 2 j=1 You don t have to memorize these formulas, but the first one has a very elegant explanation! 4 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 18 / 19

64 The Sum of the First n Integers N N MATH 125 (Section 5.1) Areas and Distances The University of Kansas 19 / 19

65 The Sum of the First n Integers N 2 x ( N) MATH 125 (Section 5.1) Areas and Distances The University of Kansas 19 / 19

66 The Sum of the First n Integers N+1 N N 2 x ( N) N x (N+1) MATH 125 (Section 5.1) Areas and Distances The University of Kansas 19 / 19

67 The Sum of the First n Integers N+1 N N 2 x ( N) N x (N+1) N = N(N+1) / 2 MATH 125 (Section 5.1) Areas and Distances The University of Kansas 19 / 19

4.1 Areas and Distances. The Area Problem: GOAL: Find the area of the region S that lies under the curve y = f(x) from a to b.

4.1 Areas and Distances. The Area Problem: GOAL: Find the area of the region S that lies under the curve y = f(x) from a to b. 4.1 Areas and Distances The Area Problem: GOAL: Find the area of the region S that lies under the curve y = f(x) from a to b. Easier Problems: Find the area of a rectangle with length l and width w. Find