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- 1. Definite Integrals
- 2. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define
- 3. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a b x=a
- 4. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a b x=a The upper limit of “integration” The lower limit of “integration”
- 5. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a b x=a The upper limit of “integration” The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”.
- 6. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a For example,(x2)' = 2x, b x=a The upper limit of “integration” The lower limit of “integration” to be the “definite integral of f(x) from x = a to x = b”.
- 7. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a b x=a 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 The upper limit of “integration” The lower limit of “integration” 0
- 8. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a b x=a 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 The upper limit of “integration” The lower limit of “integration” 0
- 9. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a b x=a 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 The upper limit of “integration” The lower limit of “integration” 0 Note that we would obtain the same answer regardless of what the integration constant k is, hence we may set k = 0.
- 10. Definite Integrals Let f(x) be a continuous function over the closed interval [a, b]. With F'(x) = f(x), we define b f(x) dx = F(x) | = F(b) – F(a) ∫ x=a b x=a 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 The upper limit of “integration” The lower limit of “integration” 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 f(x) gives the “area” bounded by f(x) and the x axis as we shall see.
- 11. Sums and Approximations
- 12. 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.
- 13. 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”.
- 14. 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.
- 15. 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
- 16. 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 1 hence we may conclude that 2 < π. Inscribed Square four right triangles each with area ½,
- 17. 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 1 hence we may conclude that 2 < π. Inscribed Square four right triangles each with area ½, Next we inscribe the circle with a hexagon for a better approximation.
- 18. Sums and Approximations 1 Inscribed Hexagon
- 19. Sums and Approximations The hexagon consists of six equilateral triangles with sides 1 1 Inscribed Hexagon
- 20. Sums and Approximations Inscribed Hexagon The hexagon consists of six equilateral triangles with sides 1. Each triangle has area = (√3) /4. 1 √3 2 1
- 21. Sums and Approximations Inscribed Hexagon The hexagon consists of six equilateral triangles with sides 1. Each triangle has area = (√3) /4. 1 √3 2 1 So the area of the hexagon is 6(√3)/4 = 3(√3)/2 and we conclude that 3(√3)/2 ≈ 2.598.. < π.
- 22. Sums and Approximations Inscribed Hexagon The hexagon consists of six equilateral triangles with sides 1. Each triangle has area = (√3) /4. 1 √3 2 1 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,
- 23. Sums and Approximations Inscribed Hexagon The hexagon consists of six equilateral triangles with sides 1. Each triangle has area = (√3) /4. 1 √3 2 1 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, then π = lim (area of the inscribed n sided reg–polygon) http://en.wikipedia.org/wiki/Pi n ∞
- 24. Sums and Approximations Inscribed Hexagon The hexagon consists of six equilateral triangles with sides 1. Each triangle has area = (√3) /4. 1 √3 2 1 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, then π = lim (area of the inscribed n sided reg–polygon) http://en.wikipedia.org/wiki/Pi n ∞ The Riemann sums we will define shortly are also related to calculations of areas.
- 25. Sums and Approximations Inscribed Hexagon The hexagon consists of six equilateral triangles with sides 1. Each triangle has area = (√3) /4. 1 √3 2 1 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, then π = lim (area of the inscribed n sided reg–polygon) http://en.wikipedia.org/wiki/Pi 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..
- 26. On Areas
- 27. We are to find the area of a given enclosed region R. R On Areas
- 28. 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 On Areas x
- 29. 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 On Areas x a b
- 30. 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 On Areas Let L(x) = cross–sectional length at a generic x. x a b
- 31. 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 On Areas L(x) x a x b
- 32. 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. xi–1 xi R a b x L(x) x Δx
- 33. 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 [x, x] be a small interval and Δx = x– x= the width i–1ii i–1 of the interval. Let xbe an arbitrary point in the interval with i L(x* ) as the cross–sectional length taken at x* . iiR a x b i–1 xi x L(x) x * Δx
- 34. 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 xi. L(xi *) xi * xi–1 xi R * cross–sectional length at xi * a b x * * L(x) x Δx
- 35. 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 [x, x] be a small interval and Δx = x– x= the width i–1ii i–1 of the interval. Let xbe an arbitrary point in the interval with i L(x) as the cross–sectional length taken at x. iiThen L(x)Δx approximates the area in R that is spanned from ixto x. i–1 iΔx L(xi *) xi * xi–1 xi R * cross–sectional length at xi * a b x * * * L(x) x A = L(xi*)Δx Δx
- 36. Riemann Sums Let I = [a, b] be the interval a < x < b.
- 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.
- 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
- 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 to be the length of each subinterval.
- 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 to be the length of each subinterval. (Note saying that n ∞ is the same as saying Δx 0 since Δx = (b – a)/n.)
- 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 to 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. 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 to 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. 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 to 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: 1/3 2/3 Δx=1/3 0 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. Riemann Sums Let y = f(x) be a continuous function over [a, b].
- 45. Riemann Sums Let y = f(x) be a continuous function over [a, b]. y = f(x) a=x0 b=xn
- 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]. a=x0 y = f(x) b=xn
- 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]. a=x0 x i-1 x i x1 x2 y = f(x) b=xn x n-1
- 48. 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 in each subinterval [x i – 1, xi], a=x0 x i-1 x i x1 x2 y = f(x) b=xn x n-1 xi* Riemann Sums
- 49. 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 in each subinterval [x i – 1, xi], x1 * a=x0 x i-1 x i x1 x2 y = f(x) b=xn x n-1 xi* Riemann Sums
- 50. 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 in each subinterval [x i – 1, xi], x2 * a=xxxx x 0 1 2 i-1 i x1 * y = f(x) b=xn x n-1 xi* Riemann Sums
- 51. 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 in each subinterval [x i – 1, xi], x x2 1 * * * a=xxxx x 0 1 2 i-1 i xi y = f(x) b=xn x n-1 xi* Riemann Sums
- 52. 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 in each subinterval [x i – 1, xi], is called a Riemann sum. x2 * a=xxxx x 0 1 2 i-1 i xi * the sum f(x1)Δx+ x1 * * f(x2* )Δx+ … f(xn* )Δx = nΣ i=1 f(xi*)Δx y = f(x) b=xn x n-1 xi* Riemann Sums
- 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 in each subinterval [x i – 1, xi]. the sum f(x1)Δx+ is called a Riemann sum. * f(x2* )Δx+ … f(xn* )Δx = nΣ i=1 f(xi)Δx xi* If f(x) > 0, this sum represents a rectangular approximation of the area between f(x) and the x-axis. x x2 1 * * * a=xxxx x 0 1 2 i-1 i xi y = f(x) b=xn x n-1
- 54. 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 in each subinterval [x i – 1, xi]. nΣ f(xi f(x *)Δx 2* )Δx+ … f(xn* )Δx = 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. a=xx 0 i-1 b=xn xi* x i x2 * x1 x2 xi * y = f(x) the sum f(x1)Δx+ x1 * * i=1 x n-1 Δx f(x1) Riemann Sums
- 55. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I
- 56. The Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC) Part I Let f(x) be a continuous function over [a, b].
- 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. 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 Σ f(xi *)Δx i=1
- 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. lim n Σ f(xi)Δx i=1 n∞ * (or Δx0)
- 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. lim n Σ f(xi)Δx i=1 n∞ * b = ∫ x=a (or Δx0)
- 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. lim n Σ f(xi)Δx i=1 n∞ * b f(x) = ∫ x=a (or Δx0)
- 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. lim n Σ f(xi)Δx i=1 n∞ * b f(x) dx = ∫ x=a (or Δx0)
- 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. lim n Σ f(xi)Δx i=1 n∞ * b f(x) dx = ∫ x=a (or Δx0) (The theorem also holds for Riemann sums of partitions that are not regular as long as max Δx0.)
- 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. lim n Σ f(xi)Δx i=1 n∞ * b f(x) dx = ∫ x=a (or Δx0) (The theorem also holds for Riemann sums of partitions that are not regular as long as max Δx0.) The limits of The Riemann sums have concrete interpretations in many applied situations.
- 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. lim n Σ f(xi)Δx i=1 n∞ * b f(x) dx = ∫ x=a (or Δx0) (The theorem also holds for Riemann sums of partitions that are not regular as long as max Δx0.) The limits of The Riemann sums have concrete interpretations in many applied situations. FTC passes the calculation of the limits to integrals.
- 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 x-axis, from x = a to b.
- 67. When f(x) > 0, the limit of the Riemann sums is the area bounded between the curve y = f(x) and the x-axis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis.
- 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 x-axis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. Graph y = –x2 + 2x.
- 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 x-axis, 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. 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 x-axis, 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 0 2 x y y = -x2 + 2x
- 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 x-axis, 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 so a = 0 and b = 2. 0 2 x y y = -x2 + 2x
- 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 x-axis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. 0 2 Hence the bounded area is 2 –x2 + 2x dx ∫ x=0 x y Graph y = –x2 + 2x. y = -x2 + 2x Solve –x2 + 2x = 0, x = 0, 2 so a = 0 and b = 2.
- 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 x-axis, from x = a to b. Example A. Find the area bounded by y = –x2 + 2x and the x-axis. 0 2 Hence the bounded area is 2 –x2 + 2x dx ∫ x=0 = –x3/3 + x2 | 2 x=0 x y Graph y = –x2 + 2x. y = -x2 + 2x Solve –x2 + 2x = 0, x = 0, 2 so a = 0 and b = 2.
- 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 x-axis, 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 so a = 0 and b = 2. 0 2 Hence the bounded area is 2 –x2 + 2x dx ∫ x=0 = –x3/3 + x2 | 2 = (–8/3 + 4) – 0 = x=0 4 3 x y y = -x2 + 2x
- 75. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer.
- 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. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. Example B. Calculate y = 2x + x2 0 ∫ 2x + x2 dx y x -2 0 x= –2
- 78. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate y = 2x + x2 x -2 0 2x + x2 dx 0 0 ∫ x= –2 = x2 + x3/3 | x= –2
- 79. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate y = 2x + x2 x -2 0 2x + x2 dx 0 0 ∫ x= –2 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3)
- 80. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate y = 2x + x2 x -2 0 2x + x2 dx 0 0 ∫ x= –2 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3) = –4/3
- 81. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate y = 2x + x2 x -2 0 2x + x2 dx 0 0 ∫ x= –2 = x2 + x3/3 | x= –2 = 0 – (4 – 8/3) = –4/3 We call this the signed area of the region.
- 82. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y Example B. Calculate y = 2x + x2 x -2 0 2x + x2 dx 0 0 ∫ x= –2 = 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.
- 83. The Fundamental Theorem of Calculus If f(x) < 0, the integral produces a negative answer. y y = 2x + x2 x -2 0 Example B. Calculate 2x + x2 dx 0 0 ∫ x= –2 = 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.
- 84. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 a. x – 1dx
- 85. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 a. x – 1dx 0 2
- 86. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 2 a. x – 1dx = x2/2 – x| 0 0 2
- 87. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 x – 1dx = x2/2 – x| ∫ x=0 2 = 2 – 2 = 0 a. 0 0 2
- 88. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 x – 1dx = x2/2 – x| ∫ x=0 2 = 2 – 2 = 0 0 2 a. signed area = +1/2 signed area = -1/2 0
- 89. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 2 x – 1dx = x2/2 – x| = 2 – 2 = 0 0 2 a. signed area = +1/2 signed area = -1/2 3/2 b. x – 1dx ∫ x=0 0 0 3/2
- 90. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 2 x – 1dx = x2/2 – x| = 2 – 2 = 0 0 2 a. signed area = +1/2 signed area = -1/2 3/2 x – 1dx = x2/2 – x | ∫ x=0 3/2 0 b. 0 0 3/2
- 91. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 2 x – 1dx = x2/2 – x| = 2 – 2 = 0 0 2 a. 3/2 x – 1dx = x2/2 – x | 0 3/2 signed area = +1/2 signed area = -1/2 ∫ x=0 = 9/8 – 3/2 = –3/8 3/2 0 b. 0
- 92. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 2 x – 1dx = x2/2 – x| = 2 – 2 = 0 0 2 a. signed area = +1/2 signed area = -1/2 3/2 x – 1dx = x2/2 – x | ∫ x=0 = 9/8 – 3/2 = –3/8 3/2 0 b. 0 0 3/2 signed area = -1/2 signed area = +1/8
- 93. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 2 x – 1dx = x2/2 – x| = 2 – 2 = 0 0 2 a. signed area = +1/2 signed area = -1/2 3/2 x – 1dx = x2/2 – x | ∫ x=0 = 9/8 – 3/2 = –3/8 We use the term area for the traditional notion of area-measurement which is always positive. 3/2 0 b. 0 0 3/2 signed area = -1/2 signed area = +1/8
- 94. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. 2 ∫ x=0 2 x – 1dx = x2/2 – x| = 2 – 2 = 0 0 2 a. signed area = +1/2 signed area = -1/2 3/2 x – 1dx = x2/2 – x | ∫ x=0 = 9/8 – 3/2 = –3/8 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. 3/2 0 b. 0 0 3/2 signed area = -1/2 signed area = +1/8
- 95. The Fundamental Theorem of Calculus Example C. Calculate and interpret the answer. ∫ x=0 2 x – 1dx = x2/2 – x| = 2 – 2 = 0 signed area = +1/2 3/2 x – 1dx = x2/2 – x | 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 f(x) dx ∫a 2 0 2 a. signed area = -1/2 ∫ x=0 = 9/8 – 3/2 = –3/8 3/2 0 b. 0 0 3/2 signed area = -1/2 signed area = +1/8 gives the signed area.
- 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. 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. 1 0 2 x y 3 y = –x2 + 2x
- 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. The signed area is 3 ∫ –x2 + 2x dx x=0 1 0 2 x y 3 y = –x2 + 2x
- 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. The signed area is 3 ∫ –x2 + 2x dx x=0 3 = –x3/3 + x2 | x=0 1 0 2 x y 3 y = –x2 + 2x
- 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. The signed area is 3 ∫ –x2 + 2x dx x=0 3 = –x3/3 + x2 | x=0 = (–9 + 9) – 0 = 0 1 0 2 x y 3 y = –x2 + 2x
- 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. The signed area is 3 ∫ –x2 + 2x dx x=0 3 = –x3/3 + x2 | x=0 = (–9 + 9) – 0 = 0 1 0 2 x y 3 y = –x2 + 2x b. Find the (real) area bounded by y = –x2 + 2x and the x-axis from x = 0 to x = 3.
- 102. Fundamental Theorem of Calculus 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. The signed area is 3 ∫ –x2 + 2x dx x=0 3 = –x3/3 + x2 | x=0 = (–9 + 9) – 0 = 0 1 0 2 x y 3 y = –x2 + 2x 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.
- 103. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2.
- 104. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0
- 105. 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 | x=0
- 106. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = (–8/3 + 4) = 4 = –x3/3 + x2 | x=0 3
- 107. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = (–8/3 + 4) = 4 = –x3/3 + x2 | x=0 3 The negative area spans from x = 2 to x = 3. 3
- 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 = (–8/3 + 4) = 4 = –x3/3 + x2 | x=0 3 The negative area spans from x = 2 to x = 3. It’s 3 –x2 + 2x dx ∫ x=2
- 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 = (–8/3 + 4) = 4 = –x3/3 + x2 | x=0 3 The negative area spans from x = 2 to x = 3. It’s 3 –x2 + 2x dx ∫ x=2 3 = –x3/3 + x2 | x=2
- 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 = (–8/3 + 4) = 4 = –x3/3 + x2 | x=0 3 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)
- 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 = (–8/3 + 4) = 4 = –x3/3 + x2 | x=0 3 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. The Fundamental Theorem of Calculus The positive area spans from x = 0 to x = 2. It’s 2 ∫ –x2 + 2x dx x=0 2 = (–8/3 + 4) = 4 = –x3/3 + x2 | x=0 3 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 3 + | 8 3 – 4 3 | =

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