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# 1 areas, riemann sums, and the fundamental theorem of calaculus

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### 1 areas, riemann sums, and the fundamental theorem of calaculus

1. 1. Sums and Approximations http://www.lahc.edu/math/frankma.htm
2. 2. Sums and Approximations Differentiation is the process of dividing smaller and smaller differences of the outputs versus the differences in the inputs to get better and better rates.
3. 3. Sums and Approximations Differentiation is the process of dividing smaller and smaller differences of the outputs versus the differences in the inputs to get better and better rates. Integration refers to summing more and more smaller and smaller terms to obtain better and better “totals”.
4. 4. Sums and Approximations Differentiation is the process of dividing smaller and smaller differences of the outputs versus the differences in the inputs to get better and better rates. Integration refers to summing more and more smaller and smaller terms to obtain better and better “totals”. The calculation for the value of π, which is the area of the circle with radius 1, is an example of integration.
5. 5. Sums and Approximations Differentiation is the process of dividing smaller and smaller differences of the outputs versus the differences in the inputs to get better and better rates. Integration refers to summing more and more smaller and smaller terms to obtain better and better “totals”. The calculation for the value of π, which is the area of the circle with radius 1, is an example of integration. We find a lower bound by inscribing a square in the circle of radius 1. 1 Inscribed Square
6. 6. Sums and Approximations Differentiation is the process of dividing smaller and smaller differences of the outputs versus the differences in the inputs to get better and better rates. Integration refers to summing more and more smaller and smaller terms to obtain better and better “totals”. The calculation for the value of π, which is the area of the circle with radius 1, is an example of integration. We find a lower bound by inscribing a square in the circle of radius 1. The area of the square consists of four right triangles each with area ½, 1 hence we may conclude that 2 < π. Inscribed Square
7. 7. Sums and Approximations Differentiation is the process of dividing smaller and smaller differences of the outputs versus the differences in the inputs to get better and better rates. Integration refers to summing more and more smaller and smaller terms to obtain better and better “totals”. The calculation for the value of π, which is the area of the circle with radius 1, is an example of integration. We find a lower bound by inscribing a square in the circle of radius 1. The area of the square consists of four right triangles each with area ½, 1 hence we may conclude that 2 < π. Next we inscribe the circle with a hexagon for a better approximation. Inscribed Square
8. 8. Sums and Approximations 1 Inscribed Hexagon
9. 9. Sums and Approximations 1 Inscribed Hexagon The hexagon consists of six equilateral triangles with sides 1
10. 10. Sums and Approximations 1 √3 2 Inscribed Hexagon 1 The hexagon consists of six equilateral triangles with sides 1. Each triangle has area = (√3) /4.
11. 11. Sums and Approximations The hexagon consists of √3 six equilateral triangles 2 with sides 1. Each triangle 1 has area = (√3) /4. Inscribed Hexagon So the area of the hexagon is 6(√3)/4 = 3(√3)/2 and we conclude that 3(√3)/2 ≈ 2.598.. < π. 1
12. 12. Sums and Approximations The hexagon consists of √3 six equilateral triangles 2 with sides 1. Each triangle 1 has area = (√3) /4. Inscribed Hexagon So the area of the hexagon is 6(√3)/4 = 3(√3)/2 and we conclude that 3(√3)/2 ≈ 2.598.. < π. Continuing in this fashion, if n = number of sides of the regular polygons, 1
13. 13. Sums and Approximations The hexagon consists of √3 six equilateral triangles 2 with sides 1. Each triangle 1 has area = (√3) /4. Inscribed Hexagon So the area of the hexagon is 6(√3)/4 = 3(√3)/2 and we conclude that 3(√3)/2 ≈ 2.598.. < π. Continuing in this fashion, if n = number of sides of http://en.wikipedia.org/wiki/Pi the regular polygons, then π = lim (area of the inscribed n sided reg–polygon) 1 n∞
14. 14. Sums and Approximations The hexagon consists of √3 six equilateral triangles 2 with sides 1. Each triangle 1 has area = (√3) /4. Inscribed Hexagon So the area of the hexagon is 6(√3)/4 = 3(√3)/2 and we conclude that 3(√3)/2 ≈ 2.598.. < π. Continuing in this fashion, if n = number of sides of http://en.wikipedia.org/wiki/Pi the regular polygons, then π = lim (area of the inscribed n sided reg–polygon) 1 n∞ The Riemann sums we will define shortly are also related to calculations of areas.
15. 15. Sums and Approximations The hexagon consists of √3 six equilateral triangles 2 with sides 1. Each triangle 1 has area = (√3) /4. Inscribed Hexagon So the area of the hexagon is 6(√3)/4 = 3(√3)/2 and we conclude that 3(√3)/2 ≈ 2.598.. < π. Continuing in this fashion, if n = number of sides of http://en.wikipedia.org/wiki/Pi the regular polygons, then π = lim (area of the inscribed n sided reg–polygon) 1 n∞ The Riemann sums we will define shortly are also related to calculations of areas. However in applications, “area” may represent concrete measurements such as Work, Revenue etc..
16. 16. On Areas
17. 17. On Areas We are to find the area of a given enclosed region R. R
18. 18. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, R x
19. 19. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. R x a b
20. 20. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and and assume that R spans x = a to a to x end,assume that R spans from from x =x = b. = b. Let L(x) = cross–sectional length at a generic x. R x a b
21. 21. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. R L(x) x a x b
22. 22. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the [xi-1, xi] be a small interval and Δx = xi – xi-1 = the width of the interval. R L(x) x a x xi–1 xi Δx b
23. 23. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the [xi-1, xi] be a small interval and Δx = xi – xi-1 = the width of the interval. Let xi* be an arbitrary point in the interval with L(xi) as the cross–sectional length taken at x*. i R L(x) x a x xi–1 xi Δx b
24. 24. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the [xi-1, xi] be a small interval and Δx = xi – xi-1 = the width of the interval. Let xi* be an arbitrary point in the interval with L(xi) as the cross–sectional length taken at x*. i * cross–sectional length at x* i R * L(xi) L(x) a * xi x xi–1 xi Δx x b
25. 25. On Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the [xi-1, xi] be a small interval and Δx = xi – xi-1 = the width of the interval. Let xi* be an arbitrary point in the interval with L(xi) as the cross–sectional length taken at x*. i Then L(x*i)Δx approximates the area in R that is spanned from cross–sectional xi–1 to xi. Δx length at x* i R * L(xi) L(x) A = L(xi* )Δx a * xi x xi–1 xi Δx x b
26. 26. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define
27. 27. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a
28. 28. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b The upper limit of “integration” b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a The lower limit of “integration”
29. 29. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b The upper limit of “integration” b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”.
30. 30. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b The upper limit of “integration” b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”. For example,(x2)' = 2x,
31. 31. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b The upper limit of “integration” b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”. 1 1 For example,(x2)' = 2x, so ∫ 2x dx = x2 | x=0 0
32. 32. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b The upper limit of “integration” b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”. 1 1 For example,(x2)' = 2x, so ∫ 2x dx = x2 | = (1)2 – (0)2 = 1 x=0 0
33. 33. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b The upper limit of “integration” b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”. 1 1 For example,(x2)' = 2x, so ∫ 2x dx = x2 | = (1)2 – (0)2 = 1 x=0 0 Note that we would obtain the same answer regardless of what the integration constant k is, hence we may set k = 0.
34. 34. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b] and if that F'(x) = f(x), we define b The upper limit of “integration” b |= ∫ f(x) dx = F(x) x=a F(b) – F(a) x=a The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”. 1 1 For example,(x2)' = 2x, so ∫ 2x dx = x2 | = (1)2 – (0)2 = 1 x=0 0 Note that we would obtain the same answer regardless of what the integration constant k is, hence we may set k = 0. The definite integral of a function f(x) gives the “area” bounded by f(x) and the x axis.
35. 35. Riemann Sums Let I = [a, b] be the interval a < x < b.
36. 36. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals.
37. 37. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals. We will use n to denote the number of the subintervals
38. 38. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals. We will use n to denote the number of the subintervals and Δx be the length of each subinterval.
39. 39. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals. We will use n to denote the number of the subintervals and Δx be the length of each subinterval. (Note saying that n ∞ is the same as saying Δx  0)
40. 40. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals. We will use n to denote the number of the subintervals and Δx be the length of each subinterval. (Note saying that n ∞ is the same as saying Δx  0 since Δx = (b – a)/n.)
41. 41. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals. We will use n to denote the number of the subintervals and Δx be the length of each subinterval. (Note saying that n ∞ is the same as saying Δx  0 since Δx = (b – a)/n.) For example, {0, 1/3, 2/3, 1} gives a regular partition of [0, 1] with n = 3:
42. 42. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals. We will use n to denote the number of the subintervals and Δx be the length of each subinterval. (Note saying that n ∞ is the same as saying Δx  0 since Δx = (b – a)/n.) For example, {0, 1/3, 2/3, 1} gives a regular partition of [0, 1] with n = 3: Δx=1/3 0 1/3 2/3 1
43. 43. Riemann Sums Let I = [a, b] be the interval a < x < b. A regular-partition of I is a division of I into equal size subintervals. We will use n to denote the number of the subintervals and Δx be the length of each subinterval. (Note saying that n ∞ is the same as saying Δx  0 since Δx = (b – a)/n.) For example, {0, 1/3, 2/3, 1} gives a regular partition of [0, 1] with n = 3: Δx=1/3 0 2/3 1/3 1 {0, ¼, ½, ¾, 1} gives the regular partition of [0, 1] with n = 4: Δx=1/4 0 ¼ ½ ¾ 1
44. 44. Riemann Sums Let y = f(x) be a continuous function over [a, b].
45. 45. Riemann Sums Let y = f(x) be a continuous function over [a, b]. y = f(x) a=x0 b=xn
46. 46. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. y = f(x) a=x0 b=xn
47. 47. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. y = f(x) a=x0 x1 x2 x i-1 x i x n-1 b=xn
48. 48. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. By selecting an arbitrary point xi* in each subinterval [x i – 1, xi], y = f(x) a=x0 x1 x2 x i-1 x i x n-1 b=xn
49. 49. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. By selecting an arbitrary point xi* in each subinterval [x i – 1, xi], y = f(x) * x1 a=x0 x1 x2 x i-1 x i x n-1 b=xn
50. 50. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. By selecting an arbitrary point xi* in each subinterval [x i – 1, xi], y = f(x) * x1 a=x0 * x2 x1 x2 x i-1 x i x n-1 b=xn
51. 51. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. By selecting an arbitrary point xi* in each subinterval [x i – 1, xi], y = f(x) * x1 a=x0 * x2 x1 xi* x2 x i-1 x i x n-1 b=xn
52. 52. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. By selecting an arbitrary point xi* in each subinterval [x i – 1, xi], n * * * the sum f(x1)Δx+ f(x2)Δx+ … f(xn)Δx = ∑ f(x*i)Δx i=1 is called a Riemann sum. y = f(x) * x1 a=x0 * x2 x1 xi* x2 x i-1 x i x n-1 b=xn
53. 53. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. By selecting an arbitrary point xi* in each subinterval [x i – 1, xi]. n * * * the sum f(x1)Δx+ f(x2)Δx+ … f(xn)Δx = ∑ f(xi)Δx i=1 is called a Riemann sum. If f(x) > 0, this sum represents a rectangular approximation of the area between f(x) and the x-axis. y = f(x) * x1 a=x0 * x2 x1 xi* x2 x i-1 x i x n-1 b=xn
54. 54. Riemann Sums Let y = f(x) be a continuous function over [a, b]. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]. By selecting an arbitrary point xi* in each subinterval [x i – 1, xi]. n * * * the sum f(x1)Δx+ f(x2)Δx+ … f(xn)Δx = ∑ f(x*i)Δx i=1 is called a Riemann sum. If f(x) > 0, this sum represents a rectangular approximation of the area between f(x) and the x-axis. y = f(x) Δx f(x1) * x1 a=x0 * x2 x1 xi* x2 x i-1 x i x n-1 b=xn
55. 55. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I
56. 56. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b].
57. 57. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b]
58. 58. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n ∑ f(x*i)Δx i=1
59. 59. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n lim ∑ f(x*i)Δx n∞ i=1 (or Δx0)
60. 60. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n lim ∑ f(x*i)Δx n∞ i=1 b = ∫ x=a (or Δx0)
61. 61. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n lim ∑ f(x*i)Δx n∞ i=1 b = ∫ f(x) x=a (or Δx0)
62. 62. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n lim ∑ f(x*i)Δx n∞ i=1 b = ∫ f(x) dx x=a (or Δx0)
63. 63. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n lim ∑ f(x*i)Δx n∞ i=1 (or Δx0) b = ∫ f(x) dx x=a (The theorem also holds for Riemann sums of partitions that are not regular as long as max Δx0.)
64. 64. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n lim ∑ f(x*i)Δx n∞ i=1 (or Δx0) b = ∫ f(x) dx x=a (The theorem also holds for Riemann sums of partitions that are not regular as long as max Δx0.) The limits of The Riemann sums have concrete interpretations in many applied situations.
65. 65. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b]. The limit of the Riemann sums of f(x) as n  ∞ (Δx  0) is the definite integral over [a, b] i.e. n lim ∑ f(x*i)Δx n∞ i=1 (or Δx0) b = ∫ f(x) dx x=a (The theorem also holds for Riemann sums of partitions that are not regular as long as max Δx0.) The limits of The Riemann sums have concrete interpretations in many applied situations. FTC passes the calculation of the limits to integrals.
66. 66. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b.
67. 67. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis.
68. 68. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. Graph y = –x2 + 2x.
69. 69. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. Graph y = –x2 + 2x. Solve –x2 + 2x = 0, x = 0, 2
70. 70. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. y y = -x + 2x Graph y = –x2 + 2x. Solve –x2 + 2x = 0, x = 0, 2 2 0 2 x
71. 71. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. y y = -x + 2x Graph y = –x2 + 2x. Solve –x2 + 2x = 0, x = 0, 2 so a = 0 and b = 2. 2 0 2 x
72. 72. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. y y = -x + 2x Graph y = –x2 + 2x. Solve –x2 + 2x = 0, x = 0, 2 so a = 0 and b = 2. Hence the bounded area is 2 0 2 ∫ x=0 –x2 + 2x dx 2 x
73. 73. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. y y = -x + 2x Graph y = –x2 + 2x. Solve –x2 + 2x = 0, x = 0, 2 so a = 0 and b = 2. Hence the bounded area is 2 0 2 ∫ –x2 + 2x dx x=0 = –x3/3 + x2 | 2 x=0 2 x
74. 74. The Fundamental Theorem of Calculus When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the xaxis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. y y = -x + 2x Graph y = –x2 + 2x. Solve –x2 + 2x = 0, x = 0, 2 so a = 0 and b = 2. Hence the bounded area is 2 0 2 ∫ –x2 + 2x dx x=0 = –x3/3 2 + x2 | x=0 4 = (–8/3 + 4) – 0 = 3 2 x
75. 75. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer.
76. 76. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. Example B. Calculate 0 ∫ 2x + x2 dx x= –2
77. 77. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate 0 ∫ 2x + x= –2 x2 dx -2 y = 2x + x2 0 x
78. 78. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate 0 ∫ 2x + x2 dx x= –2 0 = x2 + x3/3 | x= –2 -2 y = 2x + x2 0 x
79. 79. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate 0 ∫ 2x + x2 dx x= –2 0 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3) -2 y = 2x + x2 0 x
80. 80. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate 0 ∫ 2x + x2 dx x= –2 0 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3) = –4/3 -2 y = 2x + x2 0 x
81. 81. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate 0 ∫ 2x + x2 dx -2 x= –2 0 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3) = –4/3 We call this the signed area of the region. y = 2x + x2 0 x
82. 82. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate 0 ∫ 2x + x2 dx -2 y = 2x + x2 0 x= –2 0 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3) = –4/3 We call this the signed area of the region. The negative sign indicates that part or all of the region is below the x-axis. x
83. 83. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate 0 ∫ 2x + x2 dx -2 y = 2x + x2 0 x= –2 0 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3) = –4/3 We call this the signed area of the region. The negative sign indicates that part or all of the region is below the x-axis. The definite integral gives the signed area which is the sum of the positive and negative areas over the interval. x
84. 84. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ a. x – 1dx x=0
85. 85. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ a. x – 1dx x=0 0 2
86. 86. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ a. x – 1dx = x2/2 x=0 0 2 2 – x|0
87. 87. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ a. x – 1dx = x=0 x2/2 2 – x|0 =2–2=0 0 2
88. 88. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ a. x – 1dx = x=0 x2/2 2 – x|0 =2–2=0 0 2 signed area = +1/2 signed area = -1/2
89. 89. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 3/2 ∫ a. x – 1dx = x=0 x2/2 2 – x|0 =2–2=0 0 2 signed area = +1/2 signed area = -1/2 b. ∫ x – 1dx x=0 0 3/2
90. 90. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 3/2 2 ∫ a. x – 1dx = x2/2 – x| 0 x=0 =2–2=0 0 2 signed area = +1/2 signed area = -1/2 b. ∫ x – 1dx = x2/2 x=0 0 3/2 3/2 –x| 0
91. 91. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 3/2 2 ∫ a. x – 1dx = x2/2 – x| 0 x=0 =2–2=0 0 2 signed area = +1/2 signed area = -1/2 b. ∫ x – 1dx = x=0 x2/2 –x| = 9/8 – 3/2 = –3/8 0 3/2 3/2 0
92. 92. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 3/2 2 ∫ a. x – 1dx = x2/2 – x| 0 x=0 =2–2=0 0 2 b. ∫ x – 1dx = x=0 3/2 –x| 0 = 9/8 – 3/2 = –3/8 0 signed area = +1/2 signed area = -1/2 x2/2 3/2 signed area = +1/8 signed area = -1/2
93. 93. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 3/2 2 ∫ a. x – 1dx = x2/2 – x| 0 x=0 =2–2=0 0 2 b. ∫ x – 1dx = x=0 3/2 –x| 0 = 9/8 – 3/2 = –3/8 0 signed area = +1/2 signed area = -1/2 x2/2 3/2 signed area = +1/8 signed area = -1/2 We use the term area for the traditional notion of area-measurement which is always positive.
94. 94. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 3/2 2 ∫ a. x – 1dx = x2/2 – x| 0 x=0 =2–2=0 0 2 b. ∫ x – 1dx = x=0 3/2 –x| 0 = 9/8 – 3/2 = –3/8 0 signed area = +1/2 signed area = -1/2 x2/2 3/2 signed area = +1/8 signed area = -1/2 We use the term area for the traditional notion of area-measurement which is always positive. The area in example a is 1, its signed area is 0.
95. 95. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 3/2 2 ∫ a. x – 1dx = x2/2 – x| 0 x=0 =2–2=0 0 2 b. ∫ x – 1dx = x=0 3/2 –x| 0 = 9/8 – 3/2 = –3/8 0 signed area = +1/2 signed area = -1/2 x2/2 3/2 signed area = +1/8 signed area = -1/2 We use the term area for the traditional notion of area-measurement which is always positive. The area in example a is 1, its signed area is 0. b ∫a f(x) dx gives the signed area.
96. 96. The Fundamental Theorem of Calculus Example D. a. Find the signed area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3.
97. 97. The Fundamental Theorem of Calculus Example D. a. Find the signed area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. y = –x2 + 2x y 0 1 2 3 x
98. 98. The Fundamental Theorem of Calculus Example D. a. Find the signed area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. y = –x + 2x y The signed area is 2 3 ∫ –x2 0 + 2x dx x=0 1 2 3 x
99. 99. The Fundamental Theorem of Calculus Example D. a. Find the signed area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. y = –x + 2x y The signed area is 2 3 ∫ –x2 = –x3/3 0 + 2x dx x=0 + x2 3 | x=0 1 2 3 x
100. 100. The Fundamental Theorem of Calculus Example D. a. Find the signed area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. y = –x + 2x y The signed area is 2 3 ∫ –x2 0 + 2x dx x=0 –x3/3 x2 3 = + | x=0 = (–9 + 9) – 0 = 0 1 2 3 x
101. 101. The Fundamental Theorem of Calculus Example D. a. Find the signed area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. y = –x + 2x y The signed area is 2 3 ∫ –x2 0 + 2x dx x=0 –x3/3 x2 2 3 3 = + | x=0 = (–9 + 9) – 0 = 0 1 b. Find the (real) area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. x
102. 102. The Fundamental Theorem of Calculus Example D. a. Find the signed area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. y = –x + 2x y The signed area is 2 3 ∫ –x2 0 + 2x dx x=0 –x3/3 x2 2 3 3 = + | x=0 = (–9 + 9) – 0 = 0 1 b. Find the (real) area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3. To find the (real) area, we find the positive area and the negative area separately then add their absolute values. x
103. 103. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2.
104. 104. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ x=0 –x2 + 2x dx
105. 105. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ x=0 –x2 + 2x dx 2 = –x3/3 + x2 | x=0
106. 106. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ x=0 –x2 + 2x dx 2 = –x3/3 + x2 | = (–8/3 + 4) = 4 3 x=0
107. 107. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ x=0 –x2 + 2x dx 2 = –x3/3 + x2 | = (–8/3 + 4) = 4 3 x=0 The negative area spans from x = 2 to x = 3.
108. 108. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = –x3/3 + x2 | = (–8/3 + 4) = 4 3 x=0 The negative area spans from x = 2 to x = 3. It’s ∫ 3 x=2 –x2 + 2x dx
109. 109. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = –x3/3 + x2 | = (–8/3 + 4) = 4 3 x=0 The negative area spans from x = 2 to x = 3. It’s ∫ 3 x=2 –x2 + 2x dx 3 = –x3/3 + x2 | x=2
110. 110. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = –x3/3 + x2 | = (–8/3 + 4) = 4 3 x=0 The negative area spans from x = 2 to x = 3. It’s ∫ 3 x=2 –x2 + 2x dx 3 = –x3/3 + x2 | x=2 = (–9 + 9) – (–8/3 + 4)
111. 111. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = –x3/3 + x2 | = (–8/3 + 4) = 4 3 x=0 The negative area spans from x = 2 to x = 3. It’s ∫ 3 –x2 + 2x dx x=2 3 = –x3/3 + x2 | x=2 = (–9 + 9) – (–8/3 + 4) =– 4 3
112. 112. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = –x3/3 + x2 | = (–8/3 + 4) = 4 3 x=0 The negative area spans from x = 2 to x = 3. It’s ∫ 3 –x2 + 2x dx x=2 3 = –x3/3 + x2 | x=2 = (–9 + 9) – (–8/3 + 4) =– 4 3 Hence the real area is 4 + | – 4 | = 8 3 3 3